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Prepared by Michael McGoodwin, Autumn 2008
Abell 2218, a cluster of mostly spiral and elliptical galaxies at 2 billion light years located in the constellation Draco.
Hubble/NASA Wide Field and Planetary Camera 2 image made January 2000.
The many arcs seen are images of even more distant galaxies (5 - 10 times further away)
that are distorted by gravitational lensing caused by the interposition of the Abell 2218 galaxy cluster.
|Introduction and Acknowledgements||Units, Constants, and Quantities||Laws of Kepler and Newton|
|Fundamental Principles Of Relativity||Inertial Frames Of Reference Properties||Timing, Synchronization, and Spatially Locating Spacetime Events|
|Spacetime Events and Intervals||Lorentz Transformations||Spacetime Maps, Worldlines,
Types of Intervals, and
Proper Times & Proper Lengths
|Special Effects of Special Relativity||Relativistic Mechanics: Mass, Energy, Momentum, Nuclear & Particle Reactions||Additional References|
In normal human experience, speeds are low, gravitational fields are weak, and distances are short, and as a result properties of space and time are adequately described by the classical laws of Newtonian physics (i.e., to an extremely high level of accuracy). Motions and other phenomena can still be relative for humans, however, at least in the Galilean sense discussed here and here, but when the unqualified term Relativity is otherwise used in this webpage, it refers to Einsteinian Relativity.
Einstein's Theory of Relativity deals largely with phenomena outside this everyday human-scale realm, and generally becomes a significant consideration only at very high speeds (typically speeds greater than 1/10 the speed of light) or in very strong gravitational fields, and often in astronomical-scale or atomic-scale settings. Newton's laws of motion and his views of space and time preceded Einstein's Theory of Relativity, and serve as the limiting case for systems having velocities much less than the speed of light and modest gravitation.
The concepts and quantities of relativity (time, mass, momentum, energy, etc.) can be subtle, elusive, and hard to pin down. Although I majored in physics as an undergraduate in the 60s, I have had to approach studying relativity beginning with the most basic of subjects. In learning or relearning some of the fascinating aspects of relativity, I found myself wondering often, "what is solid and real that I can really count on?" As a result, I have devoted considerable electronic ink to definitions, tedious and wordy clarifications, and repetitive and even redundant descriptions that attempt to make the same point or that provide a detail from slightly varying points of view or phraseology. The concept of the invariant quantity (Lorentz interval, momenergy, invariant mass) is especially important in relativity when everything is in motion and often relative.
Even though it has been a topic of latent interest my entire intellectual life, I remain a beginner in the study of Einstein's Theory of Relativity. I have prepared this summary primarily to improve my understanding of the concepts discussed in the course, textbooks, and resources listed below. The focus for now is on Special Relativity (SR)—the conditions of which are defined here—but some General Relativity (GR) topics are also mentioned. Regrettably, the course I took on relativity had insufficient time to take up GR in any detail. I must therefore defer to a later date learning about the mathematics of GR, including the marvels of tensor calculus, Riemannian geometry, the curvature of space, black holes, Hawking radiation, the precession of the perihelion of Mercury, gravitational waves, cosmological implications, etc.
I hope this presentation might prove useful to others interested in learning more about Relativity, but obviously it is not comprehensive. The examples presented in fact barely scratch the surface for the many topics of potentially great interest. In order to keep my time expended and the size of this single page within reasonable bounds, I have not attempted to provide many images, graphs, or elegantly formatted equations, but I have tried to partially offset this lack by providing many links to webpages that do present these aids to learning.
Greek letters (αβγδΔπζεηνθκλμρστφχψω etc.) and unusual mathematical symbols (°≈≠≡∂∏∑√•∫±♦¶⇒∠∝∇∗) that are not all available in older versions of HTML (prior to 4.0) and may not viewable in older browsers are utilized herein, so be sure your browser is up-to-date and able to view characters specified in the HTML 4.0 standard. Some mathematical symbols used that have multiple meanings or might be potentially confusing are defined as follows:
≈ "is approximately equal to"
≡ "is defined as" or "is equal to by definition" or "is identically equal, therefore equal for all values of the variables contained"
Δ "the change in"
⇒ "implies" or "the expression on the left leads to the expression on the right"
⇔ "left side is true if and only if the right side is true", also known as "iff"
| Q | "the scalar magnitude of vector Q" or "the absolute value of Q ignoring sign". (4-vector magnitude for 4-vectors)
This is a complex project, likely containing errors and/or misconceptions, and I would be pleased to be notified of any errors or other constructive suggestions. Please send to MCM at McGoodwin period NET [converting this address to standard format first].
For more information about the life and scientific writings of Albert Einstein relating to relativity, see Abraham Pais's Subtle Is The Lord and Walter Isaacson's Einstein: His Life and Universe.
I wrote this primer as I audited the University of Washington course Physics 311 "Relativity" in Autumn 2008, taught by Professor Aurel Bulgac. I thank Professor Bulgac for his willingness to allow me to audit his course. He is aware of this Web page but has not reviewed it in any detail and does not vouch for its accuracy. I have recently discovered the syllabus and PDFs for the same class taught previously by Prof. Boris Blinov, which is keyed to the same textbook.
This primer draws especially on the following textbooks:
(1) Edwin F. Taylor and John A. Wheeler. Spacetime Physics: Introduction to Special Relativity 2nd Edition. W. H. Freeman & Co. 1992. (Hereafter termed STP2, this is the course textbook selected by Prof. Bulgac.)
(2) Ray d'Inverno. Introducing Einstein's Relativity. Oxford 1996. (Hereafter termed IER, this text is technically difficult but concise and includes GR); and
(3) Leo Sartori. Understanding Relativity. Univ. of Calif. Press. 1996. (Hereafter termed UR, this textbook is also available to U of WA students as a Web based e-book.)
Numerous Web-based resources were also consulted and many of these are acknowledged through hyperlinks below. Authoritativeness of these linked sources, including many from Wikipedia, is not assumed or implied. Links to Web resources were current as of Autumn 2008.
The second (symbol s), under the International System of Units, is currently defined as the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the 133Cesium atom ([Xe]6s1: F=3 mF=0 to/from F=4 mF=0) at rest and at 0 K.
The meter (symbol m) was redefined by the International Bureau of Weights and Measures in 1983 as the distance traveled by light in absolute vacuum in 1 ⁄ 299,792,458 of a second.
This speed applies to all photons, including not just those having wavelengths, frequencies, and energy hν characteristic of visible light, but also photons from other parts of the electromagnetic radiation spectrum. (Correspondingly, throughout this Web page, statements about "light" may usually be considered to apply to all photons regardless of wavelengths.)
Light in space travels 1 astronomical unit (1 AU = 1.49598 x 1011 m or 92.956 x 106 miles) in 499 s ≈ 8.3 minutes.
Relativistic vs. Traditional Velocity Units: In relativity calculations used for instance in STP2, velocity magnitudes (speeds) are often represented as "relativistic units" defined as vrel ≡ V/c, where V and c are expressed in traditional units such as m/s, and the resulting ratio vrel is dimensionless. To convert equations in which velocities vrel appear as such dimensionless ratios, multiply vrel by c to obtain traditional units m/s. See here for more on velocity in relativity.
Reflecting the unity of spacetime, it is often convenient to express distances in relativity problems in units of light travel time (ltt) and/or times in units of distances traveled by light (dtl) in these times.
Distances (expressed as light travel times ltt) include:
- light-year of distance
- 9.4607304725808 x 1015 m (as defined exactly by the International Astronomical Union)
9.4607304725808 x 1012 km (same definition)
63241 astronomical units (approximate value)
- light-day of distance
- 2.59021 x 1013 m
- light-minute of distance
- 1.79875 x 1010 m
- light-second of distance
- 2.99792 x 108 m = 299,792,458 m ≈ 3x108 m
- light-nanosecond of distance
- 2.99792 x 10-1 m = 0.299792 m ≈ 0.3 m ≈ 1 foot
Time increments (expressed as distances traveled by light dtl) include:
- meter of time
- 1/299,792,458 s = 3.33564 x 10-9 s
- kilometer of time
- 3.33564 x 10-6 s
- mile of time
- 5.3682 x 10-6 s (note that because the inch is defined as exactly 2.54 cm, 1 mile is exactly 1,609.344 m)
First Law: It is possible to select a set of reference frames, called inertial reference frames, in which a particle moves without any change in velocity if no net force acts on it. In simpler terms: A particle will stay at rest or continue at a constant velocity unless acted upon by an external unbalanced net force.
Second law: Observed from an inertial reference frame, momentum p is the product of mass and velocity, thus p = mv.
In such an an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum: F = d (mv) / dt = dp/dt. Force and momentum are vector quantities and the resultant force is found from all the forces present by vector addition. In simpler terms:
♦ F = ma
(i.e., the net force on an object is equal to the mass of the object multiplied by its acceleration).
Third Law: Whenever a particle A exerts a force on another particle B, B simultaneously exerts a force on A with the same magnitude in the opposite direction. The strong form of the law further postulates that these two forces act along the same line. This law is often simplified as "Every action has an equal and opposite reaction."
♦ F = GM1 M2 / r2
F is the magnitude of the vector gravitational mutual force of attraction between the two point masses (in Newtons N)
G is the gravitational constant,
M1 is the mass of the first point mass (kg),
M2 is the mass of the second point mass (kg),
r is the distance between the two point masses (m)
Eötvös-type experiments including those of Dicke in the 1960s, which essentially measure the correlation between inertial mass (m=F/a) and gravitational mass (satisfying the law of Universal Gravitation), have shown that the force of gravity varies for different density objects (aluminum vs. gold, etc.) by no greater than |Δg|/g = 3 x 10-11.
First Law: The orbit of every planet is an ellipse with the Sun/star at one of the two foci. [More accurately, it is the common center of gravity or barycenter of the planet and Sun/star that is located at one of the two foci, and for Jupiter this lies outside the Sun's surface.]
Second law: A line joining a planet and the Sun/star sweeps out equal areas during equal intervals of time.
Third Law: The squares of the orbital periods of planets are directly proportional to the cubes of the axes of the orbits. If the orbital period is represented by P and the semi-major axis of the orbit by a, then P2 = ka3, k being a constant. Specifically, (P/2π)2 = a3 / G( Ms + mp ) where G is the gravitational constant, Ms the mass of the Sun/star and mp the mass of the planet.
Related Orbital Formulas:
The magnitude of the centripetal force Fc on a orbiting mass M is given by Fc = Mv2 /r where r is the radius of orbit and v is the tangential velocity of the mass, or Fc = 2π•r/T (where T is the period of orbit). The magnitude of the centripetal acceleration is v2 /r.
The mass MS of a star having a planet of much smaller mass and in circular orbit about it is given approximately by MS = (4π2 /G)•(r3 /T2), where r is the radius between the star's center and the planet's center, and T is the period of orbit (which can be directly observed).
For extrasolar planets having mass MP and orbiting about a star with mass MS, one can derive Mp (sin i) = MS K/VP, where K is the maximal observed radial (line of sight) velocity of the star with respect to the observer (measured using Doppler shift of spectral lines) and VP is the maximal radial velocity of the planet. The orbit has inclination i with respect to the viewer (an orbit seen side-on has i = 0 degrees), so this formula gives the lower limit for the planet's mass.
Galileo Galilei first stated that a person observing various animal behaviors in the stateroom of a ship moving at uniform speed would not detect any difference compared to their behaviors when the ship was at rest in port. (Dialogue Concerning the Two Chief World Systems, 1639) In other words, the physics inside the stateroom appears to be the same in these two states of the ship, moving or at rest. (A detailed discussion of Galilean Relativity, including topics such as coordinate transformations, stellar aberration, covariance of physical laws, and conservation of momentum, may be found in the first chapter of UR.)
I. Einstein Principle of Relativity: "All the laws of physics are the same in every free-float (inertial) reference frame." (STP2 p. 55). The actual Einstein 1905 principle is quoted in translation as follows: "The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of co-ordinates in uniform translatory motion." Rephrased in the negative, "No test of the laws of physics provides any way whatsoever to distinguish one free-float frame from another."
Although the physical laws are the same in different inertial free-float frames (e.g., the law of acceleration F = ma), there are many physical quantities that do differ among the various frames: spatial position of events, time between events, velocity, acceleration, force, electric and magnetic fields, kinetic energy, etc. These measured quantities may vary (e.g., m and a), but their interactions through laws of physics (e.g., F = ma) remain valid. Fundamental constants such as the speed of light c, the electron charge, the rest (invariant) mass of the electron, and the order of the elements in the periodic table also remain constant, as otherwise a violation of the relativity principle could result.
II. Einstein Principle of Invariance Of Speed Of Light In Vacuum (c or c0): The actual Einstein 1905 principle is quoted in translation as follows: "We will raise this conjecture (the purport of which will hereafter be called the 'Principle of Relativity') to the status of a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely, that light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body. T"
The Michelson-Morley experiment of 1887 put to rest the theory that light as a wave required a conducting medium called the "luminiferous ether", finding instead that there was no compass direction resulting from Earth's orbital motion which affected the measured speed at which light travels. Search here for a link to their 1887 paper, and detailed discussion of this experiment and of other efforts to salvage the ether, including the positing of Fitzgerald-Lorentz Contraction.
For a number of reasons, the speed of matter having non-zero rest /invariant mass, or of information transfer using such matter, cannot equal or exceed the speed of light in vacuum c. Light signals, of course, do travel the speed of light and can carry information. Some phenomena can travel faster than light, but these superluminal phenomena cannot transfer mass or information at superluminal speeds except in science fiction. See additional comments below.
III. Equivalence Principle: This fundamental principle of General Relativity is not a part of Special Relativity, other than its role in helping decide what constitutes an inertial frame. It may be expressed in various ways:
"The outcome of any local gravitational experiment in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime." or
"The outcome of any local experiment, whether gravitational or not, in a laboratory moving in an inertial frame of reference is independent of the velocity of the laboratory, or its location in spacetime."
Note that this equivalence principle is not the same as the equivalence of mass and energy (E = mc2).
Events in spacetime are the real stuff of nature, unique and independent of any particular frame, whereas the frames are arbitrary human constructs that have no fundamental existence. There is no preferred or fundamental frame in Nature.
The rules and laws of Special Relativity apply specifically to inertial frames of reference (hereafter, frames) for which relative motion, if any, between the frames is at constant velocity, and in which there is no acceleration (gravitational or otherwise) or rotation. Such frames ultimately can only be found in deep outer space, far from gravitationally attracting stars and planets. However, it is possible to use a "free-float" frame that is freely falling in a gravitational field to dispense with the effects of gravity and simulate an inertial frame within a delimited space. For actual experiments testing special relativity, a local free-floating frame serves as a valid substitute for an inertial frame only within a limited spatial extent and duration. The greater the required precision of the experiment, the more constricted the allowable spatial and time separations become. In fact, Taylor and Wheeler (STP2 p. 31) define a reference frame to be an inertial reference frame (Lorentz reference frame) specifically by the requirement that test particles of negligible mass released at rest or in motion remain at rest or unchanged in their initial motion (speed and direction within the frame) within the limits of precision and detectability for the test equipment employed. (The size of the frame and duration of the test will affect the degree to which the definition is satisfied.)
In setting up Special Relativity problems, two inertial frames (say, S and S') are usually considered that are in motion at constant unaccelerated non-rotating velocity with respect to each other. One of the frames S is often the "Earth" frame which is considered stationary for an Earth-based observer, such as in a laboratory. S' is said to be moving with respect to S, and is often termed the "rocket" frame, but it is also equally the case that S is moving with respect to S'. No frame can therefore be considered absolutely at rest, though an Earth-based observer can be said to be at rest in the Earth frame. The axes are chosen so that that the constant relative motion between the frames is along the x-axis only, and specifically the motion of S' is usually in the positive x-direction (S' is less precisely but often referred to as moving "away from" S). Therefore, Δx between two events changes in one of the frames S' but Δy and Δz are fixed in value and do not change. This choice of axes and direction of motion is sometimes called the standard configuration of inertial frames.
(1) Motion when initial separation of test objects is transverse to the direction of fall: If two [frictionless] ball bearings of negligible mass are separated by 20 meters within a local free-floating frame (say, a falling railway coach car), they fall toward the center of Earth. Because these are slightly different directions of acceleration, the distance separating the balls gradually decreases as they converge toward Earth's center. Such a frame is an inertial frame if the incremental distance Δx over which the balls move closer together is below the measurement precision of the test apparatus. If the balls and the frame fall 315 meters to the Earth's surface, and the radius of Earth is 6.371 x 106m, the movement Δx between the balls may be computed by similar triangles as follows:
Δx/2 / 315 m = (10m / [6.371 x 106m + 315 m]) so
Δx ≈ 0.001 m = 1 mm
Thus the balls come closer together by 1 mm of motion transverse to the fall direction in a fall of 315 meters.
Note that if the balls have mass of 100 g, the amount of displacement toward each other due to mutual gravitational attraction during the 8 second fall is calculated by Newton's law of gravitation as 1.07 x 10-13 m, clearly negligible on the scale of this experiment.
(2) Motion when initial separation of test objects is in the direction of fall: If two ball bearings are separated by 20 meters vertically within a local free-floating frame (say, a falling railway coach car), they fall toward the center of Earth along the same line extending to Earth's center. The ball closer to the center of Earth will always feel a slightly larger tug of gravity and will accelerate faster than the ball further away. Therefore the separation between the two balls will increase with time. A frame is an inertial frame if the incremental distance Δx over which the balls move further apart is below the measurement precision of the test apparatus. If the balls and the frame fall 315 meters to the Earth's surface over 8 seconds, and the radius of Earth is 6.371 x 106m, the movement Δx between the balls may be computed as follows:
The gravitational acceleration g in m s-2 is given by F/m where F is the force on each ball and m is its mass. By Newton's law of universal gravitation,
g = F/m = GM/r2 = (GM/r0 2)(r0 2/r2) = g0r02/r2
Here, G is the gravitational constant given above, and g0 is the acceleration of gravity at Earth's surface, also given above.
The incremental change of g, namely Δg, as a function of altitude r of the balls, may be approximated by the first derivative:
dg/dr = (-2) g0 r0 2/r3
Δg = -2g0 r0 -1Δr
The difference in distance fallen Δy may be expressed as
Δy = 1/2Δgt2
Δy = 1/2 *(-2g0 r0 -1Δr)t2 = 0.002 m = 2 mm
Thus the balls move 2 mm apart in vertical distance during the fall of 315 meters.
Professor Bulgac notes that problems related to Earth gravity like this may be approached also using Newton's Shell theorem, which states that for a spherically symmetrical body:
(1) External objects are attracted gravitationally to the body as though all of the body's mass were concentrated at a point at the body's center;
(2) If the body is a a hollow ball, no gravitational force is exerted by the body's shell on any object inside, regardless of the object's location within the shell; and
(3) Inside a solid sphere of constant density, the gravitational force varies linearly with distance from the center, becoming zero at the center of mass. (However, most large celestial bodies will have increasing density closer to the center.) Perturbation theory—in which an approximate solution to a problem which cannot readily be solved exactly is arrived at by starting from the exact solution of a related problem—also may be used to facilitate an approximate calculation.
In general, particles that are separated transverse to the direction of fall come closer together in free fall, whereas particles that are separated vertically with respect to the direction of fall move apart in free fall. An initially circular pattern of test balls that are allowed to fall freely will distort into an ellipse with shorter axis oriented transversely and longer axis oriented in the direction of the fall (STP2 p. 33).
Forces that act in different directions or with different magnitudes on particles separated spatially in a non-uniform gravitational field are termed tidal forces in STP2, a broadening of the usage pertaining to the generation of tides. (The "tidal" distortion experienced by the human body is minor in Earth's gravitational field, but would be so severe as to tear a person apart on approaching the event horizon or Schwarzschild radius of a black hole.)
Spacetime Event Positions and Displacements: Spacetime events are located in spacetime at positions defined by their spatial and time coordinates (positions are always given in some specified coordinate system and frame of reference). For instance, events E1 and E2 may be said to have positions defined by coordinates (x1 , y1 , z1 , t1) and (x2 , y2 , z2 , t2). Sometimes the t coordinate is given before the x coordinates, but the order of listing of these components is arbitrary. Events that have the same spacetime coordinates coincide in spacetime, and may be said to be the same event (or they may represent a collision or decay involving 2 or more particles simultaneously, etc.) The spacetime separation or displacement between two spacetime events is called the Lorentz or spacetime interval. (The uses and properties of Four-vectors, including the displacement 4-vector, are discussed in greater detail here and here.)
Synchronizing Clocks: This is a critical concept in Relativity. Here is one way to synchronize clocks in a 3-D grid in an inertial frame (according to STP2): Place slave clocks at regularly spaced distance intervals (say, 1 m of light travel time or ltt) in a 3-dimensional grid in an inertial or free-float frame. Let one clock be the reference clock. Set it to t = 0 at precisely the time when it emits a flash of light. Set the other (slave) clocks according to their distance measured from the reference clock. For instance, a slave clock known to be located at spatial positions x=6 m ltt, y=8 m ltt, z=0 m ltt with respect to the reference clock is 10 m of light travel time away from the reference clock (measuring along the 3-D diagonal). Set this slave clock to 10 m of dtl (that is, to 10/c seconds) and start it running just as the light flash passes it. The slave clocks will all correctly reflect the light travel time with respect to the reference clock and if one allows for the delay in light propagation of a light signal, they can be said to read the same time. (Of course, if they were separated by astronomical distances, it could take a long time to confirm their synchronization.) This method is termed Einstein synchronization.
The observer is merely a "shorthand way of speaking about the whole collection of recording clocks associated with a free-float frame"—it is "a person who goes around reading out the memories of all recording clocks" after an event has occurred and the various clocks have recorded when the event occurred. The observer does not peer into the distance to look at light arriving in the distance from remote events. (STP2 p. 39)
Sartori (p. 61 in UR) suggest that synchronization could be confirmed somewhat differently using the following technique: Send out a light pulse from the reference clock A, say at 7:00. When it arrives at a clock station B, a TV camera at B makes an image of B's clock, which it immediately beams back to A. Upon arrival of the image at A, if A receives the image at 9:00 and the image shows a clock set to 8:00, A is confident that B's clock is properly synchronized. This technique has the advantage that no lengths are actually measured, and that A is able to be certain that synchronization has occurred. (However, some way of initially synchronizing the clocks must still be found that does make use of the distances between the clocks.)
Note that it is invalid to perform the synchronization by carrying a mobile clock that is synched to the reference clock to each slave clock, setting these slave clocks to the time displayed on the mobile clock. The problem with this approach, aside from its inconvenience at great distances, is that it will introduce time dilation errors because the clock must be accelerated to reach the slave positions. Such a clock will no longer read correctly upon returning to the reference clock.
As Sartori concludes (UR p. 66): "If a set of clocks, synchronized in the frame of reference [S] in which they are at rest, is examined by observers in any other frame [that is moving relative to S], the clocks will be found to be out of synchronization..." Even if they can somehow start out synchronized, they will not remain synchronized.
Relativity of Simultaneity: As a result, simultaneity becomes relative for observers in inertial frames that are in relative motion. Two events which appear simultaneous in one frame S will not generally appear simultaneous in a different frame S' that is moving with respect to S. This relativity of simultaneity plays a critical role in producing many of the surprising effects of Relativity.
Length Contraction or Lorentz Contraction: The relativity of simultaneity affects how we measure lengths along the direction of motion (see UR p. 86). Measurements made in frame S (i.e., using a measuring stick at rest in S) of the length of a moving object (at rest in S') must always be made by measuring the start and end point events simultaneously in S (these events will not be simultaneous in the moving frame S'). If the object at rest in S' is measured with a measuring stick also at rest in S', this requirement is not needed. An object at rest in frame S' and in motion with respect to an observer's frame S will appear, when measured in S (with a measuring stick at rest in S that is not moving with S'), to be contracted along the direction of motion (see below or here for details).
In Newtonian 3-dimensional space, 3 dimensions may be used to express a location, and time is considered universal and often ignored. For example, a location might be expressed by orthogonal coordinates (x , y , z). It was the insight of Einstein—drawing on the ideas of the Dutch physicist Hendrik Antoon Lorentz 1853-1928 and Henri Poincaré (1854 - 1912), and formalized by Hermann Minkowski (1864-1909)—that 4 dimensions must be used to exactly specify the location of an event in what Minkowski ultimately termed spacetime. Spacetime is the inextricable union of space and time. Minkowski said, "Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality." An event is defined at a location in 4D spacetime. To be actually present at an event, an observer must be present not just at (or nearby to) its spatial location where it occurs but present precisely when the event occurs in time.
In 3-dimensional space, the distance Δs between 2 locations is given by
Δs2= Δx2 + Δy2 + Δz2 , or
Δs = (Δx2 + Δy2 + Δz2)1/2
Note that Δs2 is somewhat ambiguous shorthand for (Δs)2, but will be used in this way throughout this webpage. This distance Δs is invariant with respect to the particular coordinate frame chosen in which to measure the orthogonal distances Δx, Δy, and Δz. So for example, if one used a different set of orthogonal coordinates (x',y',z') and measured Δx', Δy', and Δz', these values would differ from Δx, Δy, and Δz, but the computed value Δs' equals Δs when measuring distance between the same 2 locations.
In contrast, the 4-dimensional spacetime interval defines the spacetime separation, displacement, or "interval" between the two spacetime coordinates, say (x1 , y1 , z1 , t1) and (x2 , y2 , z2 , t2), representing 2 events or positions. The spacetime interval is a 4-vector quantity also called the Lorentz Interval or displacement vector, and represented loosely as Δs or ds. The magnitude of this vector is given by (Δs2)1/2 = |Δs| where |Q| signifies magnitude of vector Q, and is often written loosely as Δs and referred to as the "Lorentz Interval". (I find this terminology involving "intervals" confusing and inconsistently used, and apologize that I too have used "Lorentz interval" to refer to the magnitude of this interval at times on this webpage.) The magnitude of Δs satisfies:
♦ Δs2 = c2Δt2 - Δx2 - Δy2 - Δz2 or
♦ Δs2 = c2Δt2 - Δr2 where Δr2 = Δx2 + Δy2 + Δz2
♦ Δs2 = c2Δt2 - Δx2 for standard configuration where Δy = Δz = 0
The spacetime interval Δs or s is sometimes expressed as Δσ or σ. The Δt is the time separation and the other three terms Δx, Δy, and Δz express the space separation components. This quantity Δs2 is invariant (unchanging) when computed in two different frames of reference for which the intervals are measured, provided the frames are moving at constant velocity with respect to the other. (For a simplified proof, see STP2 p. 67 or UR p. 106) This provision defines what are termed inertial frames. (Examples of non-inertial frames include those in which rotation or acceleration is present. These motions introduce fictitious forces, also called inertial forces, pseudo-forces and d'Alembert forces, including centrifugal force, the Coriolis force or effect, and the Euler force.)
As given by this formula, the magnitude of Δs obtained by taking the square root may be real or imaginary. See here regarding timelike intervals and spacelike intervals. In computing the Lorentz intervals, the choice of + and - signs on time and spatial quantities (which must be opposite) is subject to mild controversy (see sign convention and discussion here.)
The Lorentz interval formula reduces to
♦ Δs2 = Δt2 - Δr2
provided identical units are used, either
a. Δr is measured in distance units such as meters, and Δt is measured in equivalent units of light travel distance (ltd) such as meters
(where t [meters ltd] = c [meters/second] • t [seconds]) , or
b. Δt is measured in time units such as seconds, and Δr is measured in equivalent units light travel time (ltt) such as seconds
(where r [seconds] = r [meters] / c [meters/second])
For standard configuration where Δy = Δz = 0, this formula reduces to
♦ Δs2 = Δt2 - Δx2.
¶ One often-used example by which one might understand the spacetime interval Δs and why it is invariant is as follows (see UR p. 69 and STP2 p. 71): A very high speed rail car moves in the x-direction, in which a passenger flashes a light beam (event E1) that travels from the floor to reflect from a ceiling mirror positioned at height L = Δy = 3 meters straight up from the floor source. The light beam reflects at the mirror (event E2) and arrives back at the floor where it started from (event E3) Consider half the path of the light, between events E1 and E2, as viewed from inside the train car (frame S'), and also as viewed from alongside the track (frame S'). Note that for the train frame S', the emission and reception events are at the same place. Assume the train speed is such that the time of flight of the light beam to reach the ceiling as viewed from the Earth frame S is Δt = 5 meters of time, so that the round trip time is 10 m of time. Using the same units and the Pythagorean theorem to compute one side given the hypotenuse and the other side of a right triangle, the net component of distance the light beam travels in the direction of motion parallel to the x-axis in the Earth frame S to reach the ceiling must be Δx = 4 m of time, or 8 m for the round trip. The time the light travels in the moving train frame to reach the ceiling and return is Δt' = 2 * 3 = 6 m of time, and Δx' = 0 in the moving train frame (the light source and the ceiling mirror do not move with respect to each other). Thus, for the Earth frame S, Δs2 = Δt2 - Δx2 - Δy2 = 52 - 42 - 02 = 25 - 16 = 9 = 32. For the train frame of the passenger, Δs'2 = Δt'2 - Δx'2 - Δy'2 = 9 - 0 - 0 = 32. For any other standard configuration frame having a different relative velocity but viewing the same events, the Δt''2 and Δx''2 will be different, but the quantity Δt''2 - Δx''2 - Δy''2 will remain the same, namely 32. The spacetime interval Δs2 = Δs'2 = Δs''2 in this example thus expresses the unchanging distance of light travel time (3 m) that light travels one-way in the transverse (perpendicular) direction, and is a quantity which is invariant with respect to any frame in standard configuration.
The following are sample Special Relativity calculations using the invariance of Lorentz intervals in inertial frame examples (i.e., not employing the Lorentz Transformations discussed further below):
¶ PROBLEM: Determine the "time dilation" for a moving particle A proton at rest in an inertial frame S' zips by an observer in a laboratory frame S at 3/4 of the speed of light (0.75c as measured by the lab observer) and is detected at 2 events E1 and E2 spatially separated by 2 meters in the lab frame S. In the "(unpowered) rocket frame" S' moving along with the proton, the two events are observed at the same spatial location with respect to the proton (so Δx' is 0). Initial times and locations are synchronized in the 2 frames so t1 = t'1 and x1 = x'1 .
The elapsed time in the lab frame S is Δt and the spatial separation is given by Δx = 2 m. Δx' is 0 in the proton's frame. The elapsed time Δt' in the proton's rest frame S' is given by
c2Δt'2 - Δx'2 = c2Δt2 - Δx2
Δt'2 = (Δt2 - (1/c2)Δx2) because Δx' is 0
Δt'= Δt(1 - (1/c2)(Δx/Δt)2)1/2 where Δx/Δt = vrel between frames
Δt= Δt'/(1 - (1/c2)(Δx/Δt)2)1/2
Let γ = 1 / [1 - (vrel/c)2]1/2 where γ>=1. Then
♦ Δt = γΔt'
This is the formula for Time Dilation in which S' is the frame for which the events occur at the same place (they are colocal), and S is a frame moving with respect to S' along the x-direction. The frame S' will exhibit the shortest time interval Δt' for colocal events.
Here, γ = 1.511858
Δt = (2 m) / 0.75c = 8.895 ns
Then Δt' = Δt/γ = 5.8835 ns.
To rephrase and state the general properties of time dilation:
TIME DILATION: Measurement of the time interval Δt' in the frame S', for which the two events forming the interval are colocal (do not move) yields the shortest possible time interval, termed the proper time interval τ. For any other frame S that is moving at nonzero speed vrel with respect to frame S', the measured time interval using clocks in S for the same two events is Δt= γΔt' . This time interval is longer than τ because γ > 1. This effect is termed time dilation—time appears to be moving slower (is "dilated") and yielding longer time intervals when observing with clocks in a frame S events that are colocal in a relatively moving frame S'.
¶ PROBLEM: Time dilation for a rocket traveler: A rocket travels at 0.95c (as measured from Earth frame) to a star at 4.3 light years away. What is the space and time separation in the Earth frame between the 2 events (i.e., departure from the Earth and arrival at the star, ignoring accelerations)?
c2Δt'2 - Δx'2 = c2Δt2 - Δx2
Δx = 4.3 ly = 4.06811 x 1016 m
Δt = 4.3/0.95 = 4.5263 yr
Δt'2 = (Δt2 - (1/c2)Δx2) because Δx' is 0 (the rocket is at rest in its own moving frame and has colocal events)
Δt'2 = 20.48763 - 18.48998 = 1.99753 yr2
Δt' = 1.41334 yr of light travel time. Note that this is again much shorter than the time interval Δt for the Earth-based observer, 4.5263 yr.
¶ PROBLEM: Time dilation of muons generated in the upper atmosphere: This example is simplified from actual experimental results reported by Rossi and Hall in 1941 (see UR p. 71) and many others. A group of muons are generated by cosmic rays at 60 km above Earth and travel straight down (in this simplified idealized example). Their normal half-life in a rest frame is 1.5 x 10-6 s. How long will it take for them to reach the Earth, and what fraction will do so assuming initially that we should ignore relativity time dilation effects? Assume they are traveling at a speed "nearly that of light" c.
Δx = 60 km = 60 x 103 m (Earth frame separation between muon generation and arrival at the ground)
The time separation required to reach Earth in Earth's frame when traveling near speed c is:
Δt ≈ Δx / c = 60 x 103 m / 299792458 m/s = 2.00138 x 10-4 s
Now if they did not experience time dilation (i.e., if the decay times were the same in the Earth frame as for the frame of the muons at rest), they would go through
(2.00138 x 10-4 s / 1.5 x 10-6 s) half-lives = 133 half-lives, so virtually none would remain: (1/2)133 = 9x10-13
Assume however that the number of muons reaching Earth is actually 1/8 of the number created. What is their velocity expressed as a fraction of c in the Earth frame?
The fraction 1/8 of surviving muons indicates that the muons have experienced only 3 half-lives in their muon rest frame, so
Δt' = 3 x 1.5 x 10-6 s (time separation in muon frame)
Δx' = 0 m (space separation in muon frame—i.e., the muons by definition are not moving in their own rest frame)
Thus the spacetime interval s using muon space and time separations is given by:
s'2 = c2Δt'2 - Δx'2 = 1.82 x 106 m2
s' = 1.35 x 103 m.
In the Earth frame, velocity is defined:
VE = (Δx/Δt) so Δx = VEΔt
(again, the Earth frame velocity is expressed with Earth frame separations in distance and time)
The spacetime interval in Earth frame is given by
s2 = c2Δt2 - Δx2 = c2Δt2 - VE2Δt2
VE2/c2 = 1 - Δt'2/Δt2
VE/c = (1 - Δt'2/Δt2) 1/2
but using the binomial expansion approximation (1 ± ε) n ≈ 1 ± εn
(ignoring the much smaller higher order terms, a simplification that makes sense when |ε| << 1
Ve/c ≈ 1 - Δt'2/2Δt2
so for this specific problem:
Ve/c ≈ 1 - Δt'2/2Δt2 = 0.99975
¶ PROBLEM: A speedy space traveler ages less: Alice departs her twin brother on Earth on her 25th birthday and travels at constant speed to Proxima Centauri, 4.28 light years away (in Earth's frame). She arrives on her 28th birthday (in her frame), then immediately returns to Earth at the same speed. How old is Jeff (in Earth's frame) when she returns to Earth, and what speed did she travel at. (In this simplified example, ignore the acceleration and deceleration that Alice must undergo.)
What do we know:
Δt = unknown Δx = 4.28 light years (one way) Δt' = 3 years Δx = 0 (she does not move with respect to her rocket)
The Lorentz interval:
c2Δt'2 - Δx'2 = c2Δt2 - Δx2
but when time is expressed in years and distances also as years (of light travel time), this becomes:
Δt'2 - Δx'2 = Δt2 - Δx2
32- 0 = Δt - 4.282
Δt = 5.23 y (the time in Jeff's Earth frame that is taken in her one-way trip to the star)
2Δt = 10.46 y (the round-trip time in Jeff's Earth frame, thus he has aged to 35.46 y as she ages to 31 y.)
v = Δx / Δt = 4.28 ly / 5.23 y = 0.82c (her speed as seen from the Earth frame).
¶ PROBLEM: Decay of K+ mesons: K+ mesons have a speed of 0.868c and traverse 9m in an Earth/lab frame S. During this time of transit they decay from 1000 to 250 in numbers. What is their rest frame half-life.
Δx = 9 m Δt = v/Δx = 0.868/9 (also expressed in m dtl) Δt' = 2 * T1/2 (expressed in m dtl) Δx' = 0 m
Then T1/2 (rest frame) = 2.57 m dtl = 8.59 x 10-9 s
The Lorentz transformations (LT) relate the time and spatial coordinates of spacetime events in two different inertial frames. They describe spacetime events, not spacetime intervals. Often, prime superscripts are used to denote coordinates in a frame moving relative to unprimed coordinates in a frame considered stationary (typically from Earth perspective), but this is not always the case. One must always carefully consider and specify which direction one frame is moving with respect to the other.
Applications of the LT are found in several other sections of this Web page, including here.
In general, problems involving velocities in multiple frames cannot be easily computed without using the Lorentz Transformations (LT). They are useful in regularizing and "mechanizing" such computations. (Taylor and Wheeler appear to regard them as a mixed blessing, and emphasize that, while useful, they are not fundamental to Relativity. However, from my point of view, anything that helps place these computations on a structured footing is desirable.)
For standard configuration (in which there is uniform unaccelerated motion at speed vrel of the primed frame S' only and in the positive x-direction with respect to the unprimed frame S), the Lorentz Transformations are given by
♦ x = (x' + vrelt') / [1 - (vrel/c)2]1/2 or x = γ(x' + vrelt')
♦ t = (t' + vrelx'/c2) / [1 - (vrel/c)2]1/2 or t = γ(t' + vrelx'/c2)
♦ y = y' and z = z'
where γ = 1 / [1 - (vrel/c)2]1/2 (here γ is the Greek symbol gamma)
Note the positive or plus sign on vrelt'. Because S' is moving in the positive x direction, a fixed x' coordinate in S' will intuitively transform to a constantly increasing (toward the more positive) value of x in S. If S' were instead moving in the negative x direction, a fixed x' coordinate in S' will transform to a constantly decreasing (toward the more negative) value of x in S, and the sign on vrelt' would be minus, not plus.
Because [vrel/c] is always < 1 for material objects, [1 - (vrel/c)2] is always < 1, and therefore the "Lorentz factor" γ is always > 1. The Lorentz factor γ is also called the time stretch factor for apparent reasons (STP2 p. 155), because when x' does not change between events,
t = γt'
♦ γ = 1 / [1 - (vrel/c)2]1/2 = "time stretch factor" = "Lorentz factor" = dt/dτ
Of course γ is a function of vrel, but it is not affected by the direction of motion (that is, the sign of vrel). Some authors (IER p. 31) use the symbol β instead of γ, but most others reserve β for the ratio vrel/c.
As one would expect from the symmetry and the absence of primary inertial frames, the inverse Lorentz Transformation equations may be derived from these by simply reversing the sign of the relative velocity (but keeping the direction of motion of S' with respect to S unchanged), leading to
♦ x' = (x - vrelt) / [1 - (vrel/c)2]1/2 or x' = γ(x - vrelt)
♦ t' = (t - vrelx/c2) / [1 - (vrel/c)2]1/2 or t' = γ(t - vrelx/c2)
♦ y' = y and z' = z
where γ = 1 / [1 - (vrel/c)2]1/2
Note the negative or minus sign on vrelt. Because S' is moving in the positive x direction, a fixed x coordinate in S will intuitively transform to a constantly decreasing (toward more negative) value of x' in S'. If S' were instead moving in the negative x direction, a fixed x coordinate in S will transform to a constantly increasing (toward more positive) value of x' in S', and the sign on vrelt' would be plus, not minus.
We can check that these LT equations meet the required conditions as follows (per UR p. 101):
• Requirement #1: They are clearly linear.
• Requirement #2: The inverse transformation is symmetrical allowing for the change in sign of vrel t corresponding to the reversal of velocity direction.
• Requirement #3: They satisfy the constancy of light speed. Consider a photon emitted from x = 0 of frame S at t=0 and which travels along the positive direction of the x-axis. At time t, the photon has reached x = ct in S. The LT of this equation into S' frame coordinates is given by
x = ct
γ(x' + vrelt') = cγ(t' + vrelx'/c2) , so
x' + vrelt' = ct' + vrelx'/c) , or
x' (1 - vrel /c) = ct' (1 - vrel /c) , or
x' = ct'
Therefore, the LT formulas satisfy the Invariance Of Light Speed regardless of frame.
• Requirement #4: They reduce to the Galilean transformation in the limit of vrel approaching 0.
(1) The equations must be linear with respect to x and t. Why does this make sense? Because, we are transforming a linear motion in (t,x) into a linear motion in (t',x') coordinates.
(2) The spacetime interval (Lorentz interval) must have the same value when computed in the two frames.
The LT can be derived by a variety of techniques. Here is part of a derivation patterned on STP2 for the LT in the special case of standard configuration.
Consider the special case of standard configuration and for which the starting position at t = t' = 0 is x = x' = 0, and the relative velocity between the two frames is vrel.
The position coordinate x is given by x = vrelt (note that we started at position and time coordinates 0)
The difference in clock rates is derived from by the Lorentz intervals:
c2 t'2 - x'2 = c2t'2 = c2t2 - x2 = c2t2 - (vrelt)2 = t2(1-v2rel) , so
so t' = t (1 - (vrel/c)2)1/2
Then t = t' / (1 - (vrel/c)2)1/2 or t = γt'
Using formula x = vrel t, we get x = vrel γt'.
For the more general case in standard configuration where x' does not equal zero for the two events, start with the assumed linear relationships:
x = Ax' + Bct'
ct = Cx' + Dct'
where A B C and D are unknown coefficients to be determined.
By Lorentz intervals,
(ct)2 - x2 = (ct')2 - x'2
So x2 = (Ax' + Bct')2 = A2 x'2 + 2ABx'ct' + B2c2t'2
and C2x'2 + 2CDcx't' + D2c2t'2 - A2 x'2 - 2ABx'ct' - B2c2t'2 = c2t'2 - x'2
or (D2c2 - B2c2)t'2 - (A2-C2)x'2 +(2CDc - 2ABc) x't' = c2t'2 - x'2
In order for this equation to be satisfied for all x', t', the following must be true:
CD - AB = 0 (because the coefficient of the x't' term must be zero;
D2- B2 = 1 (because the coefficient of the c2t'2 term must be 1);
C2 - A2 = -1 (because the coefficient of the x'2 term must be 1)
These are 3 equations in 4 unknowns but subject to additional constraints. Using hyperbolic trig functions (sinh, cosh, and tanh, according to one approach), or algebraic/matrix manipulations (as seen here), one obtains the Lorentz Transformations for standard configuration.
The LT are often expressed in matrix form, which is convenient for representing the matrix multiplication (i.e., the ordinary matrix product) that is required. In standard configuration, the Lorentz transformation provides a "boost" in the x-direction. In the more general form (i.e., not in standard configuration), the LT provides a boost in an arbitrary direction (βx ,βy ,βz ) where β = v/c.
The invariance of spacetime intervals under LT can be proven (not shown here).
The addition of velocities formula is readily derived from the LT. (Elsewhere I have provided a derivation of the formula using a clever argument not employing the LT.) Consider this specific example (most intermediate steps of this derivation are omitted):
An observer in a rest Earth frame is represented by unprimed superscripts;
A rocket frame S' is moving at speed vrel with respect to the Earth observer frame S (and is represented by primed superscripts);
A bullet is fired from the front of the rocket. It exits the rifle with a known velocity v' in the rest frame of the rocket and the rifle.
We seek to know the velocity of the bullet in Earth frame, given by v = Δx/Δt.
The velocity of the bullet in the rocket frame is given by v' = Δx'/Δt'.
In the following derivation, calculate v = Δx/Δt where Δx = x2 - x1 and Δt = t2 - t1 using the LT, so that γ = 1 / [1 - (vrel/c)2]1/2 and x2 = γ(x'2 - vrel t'2), etc.
The velocity v (velocity of the bullet in Earth frame) is given by (proof omitted, see UR p. 110 which also provides the y and z components):
♦ v = Δx/Δt = [v' + vrel ] / [1 + v'vrel/c2 ]
The same formula may be restated in velocities relative to c as follows:
♦ v = [v' + vrel ] / [1 + v'vrel]
It is again apparent that if one or both of v' + vrel = 1 (the speed of light) the addition formula simply yields c or 1 (the speed of light).
Note that if the velocity to be "added" is in the opposite direction with respect to the direction of motion, the sign on vrel becomes negative. For example, if a rocket traveling at speed 0.9c with respect to Earth frame has a K0 particle at rocket rest which emits a π+ particle at speed 0.85c "backwards" (in the more negative direction), the resulting velocity of the π+ in the Earth frame is v = (-0.85 + 0.9) / (1 - (0.85)(0.9) ) = 0.2128c (which is still in the forward direction).
The addition of velocities was tested by the Fizeau experiment performed in moving water (see also UR p. 111).
The older Galilean transformations for coordinates in inertial frames are the limiting case of the LT for slow frame velocities vrel << c so that γ approaches 1 and vrel /c ≈ 0. The Galilean transformations are expressed as follows (for primed frame S' moving along x-axis away from (in positive x direction) with respect to unprimed frame S in which the observer is at rest):
x = x' + vconvt , and
t = t'
where t is (absolute) time expressed in seconds and vconv is the conventional velocity in m/s between the two frames. (See full discussion in UP p. 12).
Spacetime Diagrams (Spacetime Maps): Time has meaning only as it applies to individual histories of travel through spacetime from a particular starting point (STP2 p. 138). For any two events it is possible to have many different paths of travel between them. This fact can be most readily appreciated in viewing a planar graph of t versus x, thus displaying only one spatial dimension. Such a graph is a spacetime map or spacetime diagram, and pertains to a particular reference frame S, sometimes having a reference event O positioned at t = x = 0. The t axis represents the lattice-clock (frame) time in the particular frame S, and both dimensions are often measured in meters, years, or other appropriate common unit (or the t axis may actually be represented by ct to explicitly yield meters, etc.) Other events on the t axis or positioned on a vertical line parallel to it represent colocal events that occur at different times in S. Events falling on the x-axis or on a horizontal line parallel to it occur in different places but at the same frame time (i.e., they are simultaneous in S, or alternatively phrased, they lie on a line of simultaneity). A separate spacetime diagram can be used to represent the LT-transformed coordinates t' and x' with corresponding axes.
However, it can be useful though more confusing to represent the transformed time t' and x' coordinates on the same spacetime diagram. One variant is the symmetrical Loedel diagram described in UR p. 151, which is limited to 2 frames. However, this diagram does not appear to be in common use (for example, see here). A commonly used approach is a Minkowski diagram in which a reference frame is plotted with orthogonal axes, and the transformed coordinates are plotted in a non-rectangular parallelogram grid of lines representing fixed spatial locations (constant x') and lines of simultaneity (constant t') in S'. (See the succinct presentation in IER p. 35 and also the more detailed presentation here.)
An interesting animation of apparent paths traveled for oppositely emitted photons in moving cart frame (photon emitter positioned in cart frame S' moving at 0.8c) versus fixed frame (train track frame S) is given here. This displays simultaneously a graphical representation of the event coordinates in the 2 frames, and for that matter depicts the origins of SR's time dilation, length contraction, and breaking of simultaneity compared to Galilean transformations.
When combining the coordinates of the two frames in such a diagram so that any single event is plotted as the same diagram point for both frames, the length scales cannot be identical, due to Lorentzian geometry. To calibrate the scales, proceed as follows. Assume x = 0 and x' = 0 coincide at t = t' = 0, and the units of t and x are the same, S is at rest and plotted with orthogonal coordinates, and S' is in motion with respect to S.
Using the inverse LT for t', when t' = 0 we have
0 = t' = γ(t - vrel x) , so t = vrel x. This is the formula for plotting the x' axis (satisfying t' = 0). Note that in the limiting case where vrel= 0, t = 0 for all x and the x- and x'-axes coincide, and if vrel= 1, we have the light-line.
Similarly using the inverse LT for x', when x' = 0 we have
0 = x' = γ(x - vrel t) , so x = vrel t. This is the formula for plotting the t' axis (satisfying x'=0). Note that in the limiting case where vrel= 0, x=0 for all x, and t- and t'-axes coincide.
Now draw in the invariant hyperbolas given by x2 - t2 = x'2 - t'2= ±1, where 1 is a suitably chosen grid unit distance in each coordinate (the axes are expressed in the same units). (This calibration process was shown for example on page 3 of Prof. Robert Gardner's PDF for his course in Differential Geometry and Relativity. The hyperbola for the +1 value intersects the x-axis (when t = 0) at x = 1. The corresponding calibration point on the x' axis representing x' = 1 (when t' = 0, though t ≠ 0) is represented by the point where the same hyperbola crosses the x' axis (which can be seen in the diagram to be at a different distance away from the origin along the x'-axis). Similarly, the hyperbola for the -1 value intersects the t-axis (when x = 0) at t = 1. The corresponding calibration point on the t' axis representing t' = 1 (when x' = 0, though x ≠ 0) is represented by the point where the same hyperbola crosses the t' axis (which can be seen in the diagram to be at a different distance away from the origin along the t'-axis).
Such a S' coordinate system grid is stretched out obliquely in parallelogram fashion in relation to the orthogonal grid of S—increasingly so as γ increases. As expected, events simultaneous in S (lying parallel to the x-axis) are not necessarily simultaneous in S' (i.e., they do not lie parallel to the x'-axis). A unit of time in S is dilated in S', and a unit of time in S' is dilated in S by the same factor.
The Lorentz length contraction may also be visualized in such a diagram (though I struggle somewhat with the details.): Consider a rod lying on the x-axis with proper length x = 1 and with ends A at xA = 0 and B at xB = 1 (when t = 0 for both measurements). Then the transformed position in S' of x'B when x'A = 0, measured simultaneously in S', is found at the point where a vertical from xB = 1 on the x-axis intersects the x' axis. (The invariant hyperbola passing through x=1 intersects x'-axis at x'=1, further out.) At this intersection point, x' = 1/γ = (1 - vrel /c)1/2 < 1. Similarly, consider a unit rod lying on the x'-axis with the position in S' of end C at x'C = 0 and end D at x'D = 1 (when t' = 0 for both measurements). Its transformed position in S is found where a line passing through x' = 1 on the x-axis and parallel to the t'-axis intersects the x-axis. At this point, x = 1/γ = (1 - vrel /c)1/2 < 1.
Pole and barn paradox: This classic "paradox" (as given for instance in STP2 problem 5-4 and UR p. 166, though there are many variants) can be resolved with the aid of two 1-frame or one 2-frame spacetime diagram. (The presentation in UR is particularly clear, showing the effect with separate frame diagrams and also with a Loedel diagram p. 172.):
Consider a runner who carries a pole (frame S') with proper length say L = 10 m at high speed v toward a stationary barn (frame S) that is also L = 10 m in proper length. The speed of the runner is so fast that the pole appears in S to be contracted to a length of 5 m, therefore able to fit in the barn with ease. (This means that 1/γ = 1/2, or γ = 2, or and therefore v/c = [3/4]1/2 = 0.866.)
In barn frame S, the front end of the pole P first reaches the front of the barn F at event PF, then the back of the pole Q reaches the front of the barn F (event QF, at which point the pole is entirely inside the barn), then the front end of the pole P reaches the rear of the barn R at event PR, then the back of the pole Q reaches the rear of the barn R (event QR, at which point the pole has exited the barn).
But in runner/pole frame S', the runner sees the barn and the sequence differently. The barn appears contracted to the same degree as above, namely by a factor 1/γ = 1/2, so that the barn appears only 5 m in length to the runner in S' holding a pole of proper length 10 m.
Will the pole ever be fully contained by the barn in the S' frame? The diagrams make clear that resolution of this question ("paradox") comes from showing that the sequence of events is different in S' as a result of the relativity of simultaneity, as follows:
According to Sartori's analysis (which makes sense to me), the formulas for times in S and S' are given by velocity equations and LT. (Assume for discussion that we are setting t=0 in S and t'=0 in S' for event QF when the back of the pole Q reaches the front door F. Explicit calculations are added to convey a bit of reality to this remarkably unreal problem):
|Time t measured in S
|Time t' measured in S'
The computations as well as the diagrams show therefore that
(1) The events PF, QF, PR, and QR do not occur at the same times in S' as in S. In fact,
(2) In S the sequence of events given by the respective times t is
PF < QF < PR < QR,
and the pole is fully inside the barn between QF and PR.
But in S', the sequence of events given by the respective times t' is given by
PF < PR < QF < QR.
As a result, the forward end of the pole P exits the rear of the barn at PR before the back of the pole enters the barn at QF, so in S' the pole protrudes at both ends for part of the time (between t' = PR and QF). The paradox is resolved, the discrepancy being due to the relativity of simultaneity. The altered sequence of PR versus QF is succinctly displayed as a dual spacetime diagram on p. 6 of this Web document.
Another Web demo, which employs varying pole's colors in the S' frame with time, makes visually apparent that the barn observer S sees at one time in his frame parts of the pole that have given off light at different times in the runner's frame S'.
Other length paradox problems (the fast walker on a grid, etc.) can be similarly resolved with the aid of spacetime diagrams (e.g., UR p. 185 and as mentioned here.)
Proper Time: For 2 events that are found in some particular frame to occur at exactly the same place (are "colocal", or have the same spatial coordinates x, y, z), the spacetime interval measured between the events is called the proper time. (Clearly, proper time measured in this way would be in a frame at rest with respect to the two events.) Proper time is usually designated with the Greek letter tau—τ. Proper time is also sometimes called wristwatch time or, by Einstein, eigenzeit ("own time"). For events that are not colocal in a given frame, the proper time interval Δτ (if defined) is invariant and given by the formula
♦ Δτ2 = Δt2 - Δx2/c2 - Δy2/c2- Δz2/c2 (for 3 spatial dimensions)
♦ Δτ2 = Δt2 - Δx2/c2 (for 1 spatial dimension, as with standard configuration events for which y = z = 0)
♦ Δτ2 = Δt2- Δx2 (when units are chosen so that c = 1 or t and x are measured in the same units)
For colocal events (happening at the same place), Δx = 0 and the latter reduces to Δτ2 = Δt2. If the frame S is chosen so that a reference event is positioned at t = 0, x= 0, the formula for the second event becomes τ2 = t2 - x2 or t2 = x2 + τ2. (When Δt2- Δx2 < 0, the proper time Δτ becomes imaginary and undefined and we have a space-like separation, see below.)
Invariant Hyperbola: If the vertical axis of a spacetime map or diagram is taken to be t and the horizontal axis is x, and when t2 - x2 = Δτ2 > 0, then the two disconnected curves or arms of the graph of this relationship form a "North-South opening" invariant hyperbola with transverse axis (semi-major axis) being the t-axis (again assuming the reference event is positioned at x = t = 0). This hyperbola represents the possible values of (t, x) for a single 2nd event as observed in all possible inertial frames. The vertex of the hyperbola where it crosses its transverse axis (usually the t-axis) so that x has not changed relative to the reference event is the point defining the frame time t value termed proper time. (I am unsure as to the utility of the South-opening hyperbola defined also by this equation, as it appears to never be mentioned, though clearly it could represent events in the past referred to a later reference event at the origin.)
When proper time τ is imaginary and undefined because τ2 = t2 - x2 < 0 we use instead the conjugate hyperbola defined by the equation for proper length or proper distance (L or L0 ), namely L2 = x2 - t2. The two disconnected curves or arms of the graph of this relationship form an "East-West opening" invariant hyperbola with transverse (semi-major) axis being the x-axis rather than the t-axis. (This hyperbola is the conjugate hyperbola of τ2 = t2 - x2.) Again, assuming the reference event is positioned at (0,0), this hyperbola represents the possible values of (t, x) for a single 2nd event as observed in all possible inertial frames. The vertex of the hyperbola where it crosses the x-axis (so that t has not changed relative to the reference event and the events are therefore simultaneous in this frame) is the point defining the proper length or proper distance.
Worldlines: The worldline or world line (German weltlinie per H. Minkowski) for an object or particle moving through spacetime is the historical record of the sequential path of event spacetime locations corresponding to this motion (thus, not simply a physical path in space). The worldline is often depicted graphically as a worldline spacetime map, but the real worldline is the actual path of events, not the graphical representation of it.
As for any single event, worldlines representing multiple events exist independent of particular frames. Because time intervals between events are incorporated into a worldline, the true description of motion is given for the particle allowing for determination of its speed. Worldlines are straight in spacetime for particles moving at constant speed, but become curved in spacetime as a result of accelerations, even when only a single spatial dimension is being depicted. (The spacetime interval for a curving worldline is given by ∫ ds where ds is the infinitesimal 4-vector spacetime displacement).
The worldline for a light beam of photons rises at a ±45° angle with respect to a vertical line parallel to the t-axis, again assuming t and x have the same units. (These straight line loci may be considered to be degenerate hyperbolas.) Particles constrained as they must be to move at less than the speed of light exhibit worldlines having instantaneous slope angles falling between <45° and <-45° with respect to the vertical t-axis).
The worldlines that fill up spacetime intersect when particles and photons undergo collisions, absorptions, emissions, etc. The authors of STP2 (p. 160) note that a catalog listing the names and intersections defining such events, along with the proper times between them, define our reality, even without reference to a specific frame of reference.
Total Proper Time Along A Worldline: The total proper time ("TPT τ") along a worldline is the total wrist watch time or aging time experienced by a particle moving along this worldline.
(1) Straight vs. Curved: TPT τ varies with the shape of the worldline. For any two events, TPT τ is maximal for the uniquely straight worldline connecting these two events rather than any of the infinite number of curved connecting worldlines for the same events. (This fact, which is related to geodesy calculations used in GR, may be proved using the calculus of variations according to Prof. Bulgac, a technique I have not learned.) The straight worldline is that taken by a free particle (i.e., not subject to any accelerating forces), and this is the worldline of maximal aging. (In fact, this Principle of Maximal Aging is also true in GR with minor modifications for free particles moving in the vicinity of gravitational masses, per STP2 p. 150. Note that the principle of maximal aging for TPT τ is in contrast to spatial distance, which is minimal between two spatial points for a straight path compared to any curved spatial path. This difference arises from comparing Euclidean geometry for space versus Lorentzian geometry for spacetime.
(2) Invariance: For any particular fixed worldline, the total proper time TPT τ is invariant in all inertial frames. (STP2 p. 150)
(3) Computation of TPT τ : The length of a curved worldline (the TPT τ) is given by an integration process starting from
♦ TPT τ = ∫ 1/γ dt , or
TPT τ = ∫ [1 - v(t)2/c2]1/2 dt , or for one spatial dimension
TPT τ = ∫ [1 - (1/c2) (dx/dt)2 ]1/2 dt , or
TPT τ = ∫P [dt2 - dx2/c2 ]1/2 expressed as a path or curve integral, for which P is the path of the wristwatch along the worldline in spacetime.
This calculation process is further depicted here. For any given worldline, the TPT τ is invariant regardless of the frame in which it is observed.
For simpler problems in which the components of the worldline consist of various straight line segments at constant velocity, the computation consists of summing up the quantities [(Δt2- Δx2/c2)]1/2 for each segment of constant velocity, or
♦ TPT τ = ∑ [(Δt2- Δx2/c2)]1/2
This formula for example may be used in solving STP2 sample problem 5-1 and exercise 5-1. STP2 problem 5-2 deals with changing frames, and uses the relationships , t'B = γτ, and x'B = -vrel t'B as well as the previously mentioned formula.
(4) Light Beam (Photon): The TPT τ for a worldline traversed by a light beam is 0 (because Δt always = Δx and therefore Δτ2 = Δt2- Δx2 = 0.) (We may therefore envision that "time stands still" for photons.)
(5) Kinks: The TPT τ for a worldline having a sharp "kink"—at such a kink-associated event the velocity of the object rapidly changes due to an acceleration—is decreased compared to a worldline that does not have such a kink. This helps to explain the Paradox of Non-Identically Accelerated Twins, presented here.
Time Stretch Factor γ: The time stretch factor γ is the factor by which a proper time interval Δτ (measured between two colocal events in frame S' and therefore minimal compared to other frames) must be multiplied to obtain the frame time interval Δt between these same two events for a frame S in motion with respect to S at velocity vrel . γ is > 1 unless vrel = 0. The formula for γ is established by the Lorentz Transformations, or see STP2 p. 157. (The frame time interval Δt observed by an S frame observer appears longer—is dilated—compared to an observer in S', because in the frame S' the frame time Δt' = Δτ is minimal [Δt2 = Δx2 + Δτ2 = Δτ2 because Δx = 0], whereas at other points on the invariant hyperbola the frame time Δt is greater than Δτ because Δx is not zero and Δτ is unchanged. Note the potentially confusing contrast between what is minimal here and what is maximal above.) The time stretch factor γ is sometimes used to express the speed of a particle, for example:
γ = 100 is equivalent to a speed ratio V/c = (1-(1/γ2))1/2 = 0.9999499987, or V = 299777468 m/s. Other values of V/c versus γ include:
Time Stretch Factor γ
Time Stretch Factor γ
1 + .0000000050
2 + .2941573387
1 + .0000005000
3 + .2025630761
1 + .0000500038
7 + .0888120501
1 + .0050378153
22 + .366272042
1 + .1547005384
70 + .712445952
1 + .6666666667
223 + .60735677
Light-Like (Null) Intervals: When τ2 = t2 - x2 = 0 or equivalently t2 = x2, the interval is said to be a light-like interval (because the spatial separation matches the time separation based on propagation at the speed of light or photons). Thus, it is possible for there to be a cause and effect relationship between the two events, provided the causative agent moves at light speed. Photon-related events have light-like separation. (So do gravitons and neutrinos according to STP2 p. 152. However, though neutrino velocities have been found to be extremely close to c, they may be slightly less than c, and neutrino rest/invariant mass and therefore neutrino velocity is currently a somewhat unsettled question.)
Light Cone (Null Cone): Given a reference event ER, all those events EF which follow it at light-like intervals define the propagation into the future of one arm of a light cone, the future light cone of this event, for which frame time tF > tR, etc. (The term "cone" is most appropriate when referring to spacetime graphical depictions of 2 spatial dimensions along with a time dimension, but the concept extends to the third spatial dimension. The cones are the two arms of a degenerate hyperboloid of revolution revolved about an axis which coincides with or is parallel to the t-axis. Light cones are not so straightforward in GR.)
The following statements about the flow of time and the possibility of cause and effect apply to the comparison of a reference event ER to another event (EF, EP, or EX). The frame times tF , tP , tX of these compared events relative to the frame time of the reference event tR are described in context:
• Events EF on the future light cone arm (t2 - x2 = 0 with tF > tR) are in the absolute future of the reference event ER and might be caused by ER if ER emits a photon (or other luminal emission). These are separated by light-like intervals.
• Events EF contained "inside" the future light cone (t2 - x2 > 0 with tF > tR) are in the absolute future of ER and might be caused by ER. These are separated by time-like intervals.
• Events EP on the past light cone (t2 - x2 = 0 with tP < tR) are in the absolute past of ER and might have caused ER if EP emits a photon (or other luminal emission). These are separated by light-like intervals.
• Events EP "inside" the past light cone (t2 - x2 > 0 with tP < tR) are in the absolute past of ER and might have caused ER. These are separated by time-like intervals.
• Other events EX that are outside the two arms of the light cone (thus t2 - x2 < 0 with tX and tR otherwise irrelevant) and are considered "elsewhere", and EX can not have caused ER, nor can ER have caused EX. These are separated by space-like intervals. Such events are considered not to occur in each other's future or past, and neither can be the cause of the other. In fact, every such event can be made to look either earlier than or later than any other event in this region by a suitable choice of frame (STP2 p. 182).
Time-Like Intervals: When τ2 = t2 - x2 > 0 or equivalently t2 > x2, τ is defined and the interval is said to be a time-like interval (because they are separated in time), so that there may be be a cause-effect relationship between the two events. The frame with the shortest possible frame time separation is the one with colocal events yielding τ for the time separation. (See STP2 p. 172) Such an interval is said to have the time component dominate over the space component.
Space-Like Intervals: For some events, however, it is not possible to find a frame in which proper time can be expressed as a real number. Such events (again for standard configuration) have τ2 = t2 - x2 < 0, so that τ is imaginary and undefined (UR p. 71). These intervals may instead be expressed by σ2 = s2 = x2 - t2 > 0, and are said to be space-like (because the space separation dominates over the time separation). Their representation with the invariant conjugate hyperbola is discussed above, and as noted, the vertex of the hyperbola yields the proper distance or proper length.
Proper Length/Distance: The measurement of space-like separation is given by the proper distance or proper length, defined as the rod distance between space-like events in a frame at rest with respect to both events and for which the events are simultaneous in that frame. It is often represented by L0. In other inertial frames (in standard configuration), proper length is also an invariant quantity, and is given by the formula
♦ L02 = Δx2 - c2Δt2 or
♦ L02 = Δx2 - Δt2 for c = 1 or identical units
Note that when the measurement events are made in a frame S for which the events occur simultaneously, Δt = 0 and L0 is specified simply by the spatial separation measured, so L0 = Δx. Proper Length/Distance is also discussed here.
Lorentz Contraction (Length Contraction or Lorentz-Fitzgerald Contraction): For a frame S' in motion with respect to S at speed vrel and containing an object at rest, the formula for the measured length L of the object in the S frame becomes
♦ L = L0/γ where γ > 1 and L0 is the proper length of the object measured in the object's rest frame S'.
This formula is derived here and discussed here. A rod measuring length L0 in its rest frame S' appears shortened to length L = L0/γ when viewed from a frame S moving with respect to the rod's rest frame S'. This is a result of the relativity of simultaneity and of the time dilation effect. An alternate approach, based on measuring the time when each end passes by an observer in S (thus at different times), is given at STP2 p. 157.
Material objects (made of particles), photons, and transfer of energy or information cannot travel faster than the speed of light in vacuum c.
This assertion is not contradicted by "motion of effects" phenomena such as the apparent transverse motion of a distant illumination spot produced by the beam of light from a rapidly rotating spot lamp. Such a moving spot might indeed move faster than c, but each illuminated point receives its own photons of light moving at c, and any information carried by these photons travels from the lamp to the spot at speed c, rather than along the path of the spot moving at any higher speed.
Certain processes (such as phase velocities of electromagnetic waves, and group velocities of rapidly attenuating light beams) can move faster than c but do not carry information. Cosmological models of the early universe positing a period of rapid inflation are beyond my skill level to discuss, although the concept seems to be that space itself expands rapidly while the measurable speed of light embedded in that space remains constant.
The speed limit c is not a fundamental principle of Einstein's theory of special relativity, but it arises in various ways as a consequence. One consequence of superluminal travel speed, should that be possible, would be that there would always be a frame in which an event B that is known to follow an event A would be found instead to precede A, creating an impossible confusion of cause and effect. This is best illustrated with spacetime diagrams (STP2 p. 108 and Chapter 5).
Some galaxies or quasars (such as 3C273) emit jets traveling at relativistic speeds with respect to the Earth frame—these are thought to arise from the accretion disk of a central black hole. If a jet is emitted with relativistic velocity v (in the Earth frame) which has a large component in the direction of the Earth, then it is possible that the transverse component of motion of the jet may appear to exceed c (i.e., it appears to have superluminal velocity). However, this is an incorrect conclusion based on not allowing for the difference in the time light takes to reach us from separate points of emission. Consider 2 pulses of light emitted by a moving knot (plasma blob, as seen for instance by VLBI imaging) in the jet in time interval Δt in the Earth frame. If the angle made in Earth frame between the jet and the line of sight is termed θ, then the apparent elapsed time observed on Earth between these light pulses is Δtobs = Δt(1-(v/c)•cosθ). The apparent transverse motion of the knot is Δxobs = vΔt•sinθ. Therefore the apparent velocity in the transverse direction vx-obs is Δxobs/Δtobs = v•sinθ / (1-(v/c)•cosθ). This quantity is 0 for θ = 0 (quasar moving directly toward Earth) and v for θ = 90. However, for angles in between, vx-obs may exceed c. The maximum value of vx-obs occurs at θ = cos-1 (v/c) (see STP2 p. 91), yielding a vx-obs of v/(1-(v/c)2)1/2. When v = 0.99c, we obtain a maximal value of vx-obs = 7.1c. Observational data on the quasar 3C273 combined with the estimated Hubble distance of 2.6 x 109 ly yields a vx-obs of 9.6c.
According to the Galilean concept, for two speeds combined in the same direction, the sum of these two speeds is simply V1 + V2 , such as might be observed when throwing a baseball forward from a slowly moving ship. This simple formula remains valid for nonrelativistic speeds << c.
The following calculation is based on Problem 3-11 in STP2 and again makes use only of the Lorentz interval, not Lorentz transformations. (This law may be readily computed using the LT instead.) Alice is standing at the back of a very fast-moving train which has speed vrel with respect to the stationary Earth frame. She fires a high-speed rifle bullet toward the front of the train car in exactly the same direction as the direction of motion of the train (this therefore is a special case of the more general problem of adding velocities). The muzzle velocity of the bullet (as measured with the rifle at rest) is known to be v'bullet. John is observing in the stationary Earth frame. At the same time the rifle fires, a photon starts from the rifle's position forward to the front of the train car, and bounces back. What velocity does John observe the bullet to be moving at?
(In the following analysis, prime superscripts denote values in Alice's frame—that of the moving car—whereas unprimed quantities are as determined in John's stationary Earth frame.)
Let t1 be the time for the photon in John's Earth frame to travel forward to the mirror. Then the distance the photon travels in John's frame is ct1 = L1 + vrelt1 where L is the length of the rail car in John's frame. Then
(1) t1 = L/(c - vrel ).
Let x be the distance (in John's frame) from the front of the car that the photon travels in time t2 in the reverse direction after reflection at the front mirror until it encounters the bullet still moving forward. Then x = ct2 + vrelt2, again all in the Earth frame. Thus
(2) t2 = x / (c + vrel).
But when the bullet encounters the photon, they have equal displacement from the rear of the train. Therefore,
(t1 + t2) vbullet = c(t1 - t2),
Note that the times on the right hand are subtracted to determine the net resulting photon travel rather than the overall distance traveled roundtrip.
t1vbullet + t2 vbullet = ct1 - ct2, or
t2 (c + vbullet) = t1 (c - vbullet), or
(3) t2 = t1 (c - vbullet) / (c + vbullet)
Rearranging and substituting from the three equations (1), (2), and (3), we get
x = t2 (c + vrel) , and
L = t1 (c - vrel), so
x/L = t2 (c + vrel / t1 (c - vrel) , or
x/L = [(c - vbullet) / (c + vbullet) ] [ (c + vrel) / (c - vrel)]
Note that the same calculation can be performed in Alice's frame, assuming car length L', bullet distance traveled x', speeds v'rel = 0 and v'bullet , obtaining
x'/L' = [(c - v'bullet) / (c + v'bullet) ] [ (c + v'rel) / (c - v'rel)] = [(c - v'bullet) / (c + v'bullet) ] because v'rel = 0.
The fraction of L or L' traveled by the bullet must be the same in both frames, so x'/L' = x/L. Therefore,
[(c - vbullet) / (c + vbullet) ] [ (c + vrel) / (c - vrel)] = [(c - v'bullet) / (c + v'bullet) ]. Solving for vbullet, we get (after some seriously arduous work—multiply both sides by (c + v'bullet) and expand all the products, ending up with factors 2c2 top and bottom that cancel out...)
♦ vbullet = (v'bullet + vrel) / [1 + (v'bullet vrel / c2) ]
That is, the summed velocity of the bullet in the Earth frame is given here in terms of the muzzle velocity of the bullet v'bullet (as measured with the rifle at rest in the moving frame) and the velocity of the train as seen from the Earth frame. This specific derivation is discussed in Mermin (1983).
¶ Example: If v'bullet = (3/4)c and vrel = (3/4)c, then vbullet = 1.5 c / ( 1 + 9/16) = 0.96c
¶ Example: If v'bullet = c or vrel = c, then vbullet = c.
Note: The derivation of a more general formula for combining non-parallel velocities into x, y, and z components is found in UR p. 110.
The Doppler effect refers to a shift in frequency and wavelength of light or other wave resulting from relative motion of source and observer. (The alteration in the direction of light, termed aberration, resulting from motion is discussed below.) The term relativistic Doppler however refers specifically to change in frequency of photons including light caused by the relative motion of emitter and observer. The equations include the time dilation effect and do not apply to wave motion requiring a medium.
For Motion Along The X-axis (Radial Relativistic Doppler Effect)
This section considers the special case of propagation only in the "radial" direction
along the line of sight, as with receding galaxies, etc. The following calculation follows Problem 3-10 in STP2 and makes use simply of the Lorentz interval. Consider an object moving directly away from the observer (i.e., along the line of sight) at speed v in the vacuum of space. The object emits pulses of light every Δτ seconds (the proper time interval in the frame of the moving emitter). The distance for the first flash E1 is set to 0 in both frames, as are the clocks of the observer and emitter. After time Δτ in the emitters frame, the second flash E2 is emitted when the emitter has attained a distance from the observer of vΔt as measured in the observer's frame. Here, Δt is the time interval in the observer's frame between E1 and E2 (not including the additional time required for the light from E2 to reach the observer). The Lorentz intervals provide that c2Δt - v2Δt2 = c2Δτ2 (for the emitter there is no relative motion in its own frame). Therefore, Δt = γΔτ where γ ≡ 1/(1 - v2/c2)1/2. The total time between flashes arriving at the observer Δtrec is given as Δtrec = Δt + (v/c)Δt, because the light from E2 takes (v/c)Δt units of time to travel to the observer. Therefore, Δtrec = Δt + (v/c)Δt = γΔτ + (v/c)γΔτ = γΔτ(1 + v/c) = Δτ(1 + v/c) / (1 - v2 /c2)1/2 = Δτ(1 + v/c) / (1 - v /c )1/2(1 + v /c )1/2, so
Δtrec = Δτ(1 + v/c)1/2 /(1 - v/c)1/2.
The frequency of light fo as seen by the observer is 1/Δtrec and the frequency of light fe emitted by the emitter at rest in its own frame is 1/Δτ. Therefore, for emitter movement at speed v away from the observer,
♦ fo = fe (1 - (v/c))1/2 / (1 + (v/c))1/2 = fe [(1 - (v/c)) / (1 + (v/c) )]1/2 (redshifted for v positive)
(This equation is also given in STP2 ex. 8-18 as the formula for the LT of frequency arising from motion along the x-axis.) The factor (1 - v/c)1/2 /(1 + v/c)1/2 = [(1 - (v/c)) / (1 + (v/c) )]1/2 by which the emitter's frequency is multiplied to obtain the observer's frequency is termed the Doppler factor, and for distant quasars can be as high as 7.96 (see reference with z factor). Note that the object is moving away, so that the signs on the v's correctly lead to fo < fe. The relative speed of recession in Earth frame for a Doppler factor 7.96 quasar is given by 7.962 = (1 + v/c) / (1 - v/c) so v/c = (7.962 - 1) / 7.962 + 1) = 0.9689.
The relativistic radial redshift factor z (for movement away from the observer) is given by z + 1 = λo / λe or
♦ z = -1 + [(1 + v/c)1/2 /(1 - v/c)1/2].
Note: The classical non-relativistic radial Doppler formula, as would for example apply to sound emitted by an emitter having a relative velocity Ve and with speed of sound in a conducting medium Vs, is given for a stationary receiver as:
f' = f / (1 ± (Ve/Vs) ). Expressed as a redshift z, this would be given by z = (femit - fobs) / fobs. This non-relativistic formula is the limiting case for relativistic radial Doppler according to UR p. 120.
Police Radar: For radar reflection from an oncoming (rather than a receding) car moving at v << c (such as in STP2 problem 5-5), the sign of v would be negative, and the formula for a police radar gun frequency shift of the frequency shift of the incident beam on the car would be:
fo = fincident [(1 + (v/c) )/(1 - (v/c) )]1/2.
(The textbook STP2 p. 167 appears to erroneously list this formula without the square root exponent. Alternatively, the formula as printed is correct but only it pertains to femitted rather than fincident.)
Because the moving car is reflecting incident radar arriving from the radar gun (rather than emitting it on its own), the incident frequency is itself Doppler shifted by the motion , so that the total Doppler shift of the police radar beam by the moving car is given by multiplying the Doppler factor by itself (i.e., squaring it), so that for this specific circumstance:
♦ fo = fe [(1 + (v/c) )/(1 - (v/c) )]1/2 x [(1 + (v/c) )/(1 - (v/c) )]1/2 = fe [(1 + (v/c) )/(1 - (v/c) )] for a car coming toward the gun with velocity v.
This expression for v/c <<1 can be approximated by (1+v/c ) x (1-v/c)-1 = (1+v/c)(1+v/c + +smaller factors) ≈ (1+2v/c)
Then Δf/fe ≡ (fo - fe) / fe ≈ 2v/c
For example, a car doing 100 km/hr = 27.78 m/s, the ratio v/c = 9.266 x 10-8. The Δf/f for police radar (as given by the preceding formula, which is also listed in STP2 p. 167) is
Δf/f = 2v/c = 2*9.266 x 10-8 ≈ 1.85 x 10-7.
Therefore the change in frequency that must be detected for a police radar with carrier frequency fe = 10.525 x 109 Hz is:
Δf = fe * Δf/fe = 10.525 x 109 Hz * 1.85 x 10-7 = 1950 Hz. (Note: STP2 p. 301 says this answer should be 3136 Hz. However, another website also calculates the answer as 1949 Hz and the textbook appears to be in error.)
z factor: Astronomers usually express Doppler redshift using the z factor, defined as
♦ z = Δf/f = (fe - fo) / fo = Doppler factor - 1 = [ (1 + v/c) / (1 - v/c) ] - 1 , or equivalently
♦ z = Δλ/λ = (λo- λe) / λe
The highest Doppler redshift reported from spectroscopy is z = 6.96 for galaxy IOK-1 (as of 14 September 2006, according to the paper by Masanori Iye et. al.) This z value is equivalent to a Doppler factor of 7.96. Thus a galaxy with Doppler factor = 7.96 has a z factor = 6.96, and v/c = [(z + 1)2 - 1] / [(z + 1)2 + 1] or again, 0.9689. Values of z that are positive are referred to as "redshifted" whereas negative values are "blueshifted", so such a quasar is markedly redshifted (this terminology is used even for photons emitted or observed that have wavelength outside the visible light spectrum).
For Motion At Any Angle In X-Y plane
A more general case of the Relativistic Doppler Effect is as follows. Two similar and related problem setups and formulas are presented:
Variant I: The light emitting source at rest in frame S' is moving with velocity v along the x-axis (direction of relative motion) as seen in observer frame S and generally toward the observer, and light is emitted along a path forming a non-zero angle θ with respect to the direction of relative motion (x-axis) at the time of emission in the observer's frame. The relativistic Doppler formula for frequency shift is given (see UR p. 121) by
♦ fo = fe / [γ (1 - (v/c) • (cos θ) ] , using θ of the observer's frame, or
♦ fe = foγ [1 - (v/c) • (cos θ)] , also using θ of the observer's frame
where fe is the frequency of light or photons emitted by the moving source when measured at rest in its own frame, and fo is the frequency observed by the observer in observer frame.
• For example, if v/c = 0.2, fe = 876.283 Hz, and θ = pi/4 radians (45 degrees), then gamma = 1.0206, and fo = 1000 Hz (the observed photon is blueshifted because the emitter is moving generally toward the observer). If the inverse formula is applied, we recover the emitter frequency 876.283.
Variant II. In STP2 ex. 8-19, a similar formula can be derived from the LT on the momenergy relationships and E = hf and is shown on page 263. In this example's setup, the photon moves in lab frame along the x-y plane at an angle φ with respect to the x-axis. The rocket frame moves along the x-axis at positive speed vrel away from the emitter of the photon, and in this rocket frame, the photon forms an angle of φ' with respect to the x' axis. The frequency of the photon in lab frame f versus rocket frame f' is given by (STP2 p. 263):
♦ f ' = f γ [ 1 - (v/c) • (cos φ) ]
♦ f = f ' γ [ 1 + (v/c) • (cos φ') ]
• For example, if v/c = 0.2, fe = 1000 Hz, and φ = pi/4 radians (45 degrees), then f ' = 876.283 Hz (the photon is redshifted as expected, because the rocket frame is moving generally away from the emitter). If the inverse formula for f as a function of f ' is applied, we recover the original frequency 1000 Hz.
When θ = 0 and cos θ = 1 (pure "radial" motion along the line of sight), Variant I. reduces to the special case of radial relativistic Doppler discussed above, although here we are setting up the example so that the emitter is moving toward the observer:
fo = fe [(1 + (v/c)) / (1 - (v/c) )]1/2 (blueshifted for v positive)
For Transverse Motion (Transverse Relativistic Doppler Effect)
When θ = 90 degrees so that cos θ = 0 (motion of the emitter perpendicular to the line of sight), the general formula above reduces to
♦ fo = fe / γ
This is the Transverse Doppler Effect, which has no counterpart in classical theory. Note that because γ > 1, 1/γ is < 1 and fo is always lower than fe (that is, it is red-shifted) regardless which direction of transverse motion applies.
This is a special case of the relativistic aberration discussed below. It is not the same as (1) light deflection (bending) by gravity, nor is it the same as (2) stellar parallax (in which nearby stars appear to shift slightly in position relative to distant stars due to the variation in viewing angle as Earth orbits about the Sun), nor is it the same as (3) aberration arising from Earth's axial rotation or movement of the solar system in the galaxy, nor does it arise from (4) the transverse component of motion of the star ("proper motion", which must also be "corrected out" in evaluation stellar aberration).
As the Earth orbits about the Sun, a star appears to be shifted in position by an angle ψ due to the relative velocity of the Earth with respect to the sun. When viewed from the Sun's frame, the Earth is seen to move and the star light is unshifted. However, when viewed from the frame of the Earth, the star appears to an Earth observer to be shifted forward in the direction of the Earth's orbital motion by a small angle ψ. (This is similar to the impression that vertically falling raindrops appear to an automobile driver to be coming in at an angle from in front of the moving auto.) When the Earth has moved around 180 degrees in its solar orbit, the annual stellar aberration is reversed in direction. VE is the orbital velocity of the Earth, which is approximately 30 km/s. Using the Lorentz interval, the relativistic formula for stellar aberration is sin ψ = VE / c. (According to UR p. 115, the relativistic formula for stellar aberration should be tan ψ = γVE / c. I am not sure what accounts for this discrepancy.) Using this formula, ψ = 5.7 x 10-3 degrees or 20.6 arcsec or 1 x 10-4 radian, the maximum observable change in either right ascension or declination. The precise reported value of this constant of aberration κ = 20.49552 arcsec (at J2000).The classical (non-relativistic) annual aberration angle is given by tan ψ = VE / c, which for the small angles involved is nearly equal to sin ψ—in fact, the ratio tan ψ / sin ψ has not yet been experimentally distinguished from 1 for celestial sources.
The term aberration is generally used to refer to the alteration or displacement in the apparent direction of light resulting from various motions. Light undergoing relativistic aberration is also redshifted or blueshifted according to the Doppler formulas discussed above.
Regarding astronomical stellar aberration of light, the Earth frame is considered to be moving at velocity vrel relative to the motionless frame of the celestial sphere. In general aberration problems, velocities and angles in the x-y plane are considered, assuming the the z-component of velocity does not change, and the x-axis and x'-axis are aligned.) Unlike for particles with non-zero rest/invariant mass, photons are assumed to have the same velocity in both frames.
The formula for relativistic aberration of light was given here [link has gone dead], assuming the following specifications, as follows:
• The frame SE contains the reference light emission and the frame SO of the observer is moving at relative velocity v along the x-axis in the negative x direction at the time of emission, and
• Light is emitted by the source E at an angle θE in SE with respect to the x-axis, and
• θO is the angle of light emission with respect to the x-axis seen by the observer in SO (this is not the same angle as is usually employed with stellar aberration)
Then using formulas for the transformed xO and yO components of velocity
tan θO = u sin θE / γ[u cos θE + vrel] where u is the velocity for generic emitted particles (not necessarily photons) moving at speed u in S .
But for photons, u = c in all frames, so
(1) ♦ tan θO = sin θE / γ(cos θE + vrel/c)
For instance, for emission angle 30 degrees wrt [measured with respect to] the x-axis, v/c = 0.99, gamma = 7.0888, the resulting observer angle has shrunk to a mere 2.176 degrees wrt the x-axis.
Sartori in UR p. 116 gives a similar formula, but in his formula the angle
φ is expressed relative to the y-axis, which is of course perpendicular to the x-axis, and in addition the observer frame is moving in the positive direction with respect to the emitter frame. Expression of the angle φ in this manner is more consistent with the custom for expressing astronomical stellar aberration. His formula (adjusted for emission at the speed of light) is
(2) ♦ tan φO = γ(sin φE - vrel/c) / cos φE
This formula is equivalent to (1) when allowing for the different direction of frame motion and the different axis against which the angle is measured. For instance, for angle 60 degrees wrt y-axis, v/c = -0.99, gamma = 7.0888, the resulting observer angle is 87.82 degrees wrt the y-axis or again 2.176 degrees wrt the x-axis.
(3) A formula for light aberration is given by STP2 p. 115) for a flash of light emitted at angle θE from a rocket frame traveling at vrel in the positive direction with respect to observer Earth frame S:
♦ cos θO = [ cos θE + (vrel/c) ] / [1 + (vrel/c) cos θE ].
This formula gives the same result as (1). For instance, for emission angle 30 degrees wrt x-axis, v/c = 0.99, gamma = 7.0888, the resulting observer angle is 2.176 degrees wrt the x-axis. (In STP2 p. 263 ex. 8-19, this formula is derived from the LT on momenergy components.)
(4) Nearly the same formula as (3) is given as the following here:
cos θO = [ cos θE - (vrel/c) ] / [1 - (vrel/c) cos θE ]
but note the opposite vrel sign. The authors of (4) set up the problem so that the emitter is moving away from (in the positive x direction) the observer at speed vrel , so it is equivalent to (3) allowing for the differing setup. For instance, for emission angle 30 degrees wrt x-axis, v/c = -0.99, gamma = 7.0888, the resulting observer angle is 2.176 degrees wrt the x-axis
Additional sample calculations may be seen here.
As vrel /c increases from 0 for an observer, the observed angle θO changes from θO = θE (i.e., no aberration) to values closer and closer to θO = 0 degrees, indicating concentration of light in the direction of motion (beaming).
Note that the relativistic formula for stellar aberration given as STP2 problem 3-9 and discussed here (sin ψ = VE / c) can be rewritten as cos θO = VE / c where now θO is the angle formed between the incoming light and the direction of Earth motion. By comparison, the formula on STP2 p. 115 [equation (3) just above] reduces to the same equation, cos θO = VE / c, for light emitted perpendicular to the direction of Earth motion (for which cos θE = cos π/2 = 0). Therefore these two expressions are equivalent for this special case perpendicular to Earth's motion about the Sun.
In general, rays of light emitted by a moving object appear to a relatively nonmoving observer as if they have arisen concentrated into a forward angle more in the direction of the object's motion. Thus light emitted by a moving object appears to be compressed or concentrated into a cone about its direction of motion, an effect called relativistic beaming, and this can intensify the brightness of the beam. (This concentration of light in the object's forward direction is also referred to as the headlight effect).
Although it is unlikely we will actually have human observers moving at relativistic speeds anytime soon, the opposite already is probably being observed, in the form of high speed jets emitted by active galactic nuclei. One author, Wim Schuur, states (2007, link no longer available) "The effect of relativistic aberration may look a bit far fetched for having any practical purpose. But the reverse, namely that the object which emitted the light beam is traveling with a relativistic speed relative to the observer, has a profound significance. This is the case when we observe the jets that emerge from certain types of active galactic nuclei, such as quasars. In many cases one jet is clearly visible, while the other is not or only barely. It is known that jet material moves with relativistic speeds... If we assume that the approaching jet emits its radiation isotropically, than we observe the radiation from its facing radiating hemisphere as being highly concentrated to us. This is completely explained by the relativistic aberration..." The author includes a plot of θO vs θE for vrel /c =0.9 which shows that photons emitted at angles of up to 40 degrees from the radial axis (line of sight) are concentrated into an angle of 10 degrees around the line of sight (they are "beamed"), and therefore substantially more likely to be observed, certainly when compared to the jet being emitted in the opposite direction. The observed flux density will increase as γ2. "This is the reason why the jet which is directed toward us is observed being so much brighter than the other jet of the pair. In fact, if we take the relativistic Doppler effect into account, the luminosity [observed luminosity or better, the apparent brightness] of the jet will be boosted [over its intrinsic luminosity] by at least another factor of γ. So the total effect, together known as relativistic boosting, is an increase of luminosity in excess of a factor γ3 . Similar relativistic effects are responsible for the way we observe long-duration gamma ray bursts (GRB). The huge energy of the observed radiation is explained by applying relativistic beaming." Note that not only is the brightness of the forward jet intensified compared to its intrinsic luminosity, the observed brightness of the jet emitted in the opposite direction (away from the observer) is diminished compared to what would otherwise be expected given the same intrinsic luminosity as the forward jet.
Light received by a moving object from its nonmoving surroundings—e.g., the view from a very fast spacecraft of the fixed stars—also appears compressed into a cone projecting forward about the object's direction of motion. (See demo here.) This result is perhaps expected in the following intuitive sense. For externally emitted light to reach a very high speed rocket, it must be radiated in a direction close to the direction of motion, otherwise the rocket will have passed on by the time the light crosses the rocket's path. This aberration can distort the appearance of objects—they may appear bent, curved or twisted; long rods may appear as parabolic arcs bending away; cubes can present curved edges or non-parallel surfaces, etc. In some circumstances, the warped path taken by the light emitted can allow one to see behind normally opaque objects. A box that would be seen face-on if motionless appears rotated when moving at high speed (STP2 Ex. 3-17). This distortion is not due simply to length contraction.
Particles traveling in a medium m at speed VPm may have speed exceeding the speed of light VLm in that medium (where VLm < c, the speed of light in vacuum). Such superluminal particles generate a kind of shock wave in the form of light, termed Čerenkov radiation. By simple geometry of right triangles, the wavefront of the Čerenkov radiation forms an angle φ with respect to the direction of particle velocity, for which
♦ cos φ = VLm / VPm.
This angle may be experimentally measured and used to determine the speed of the particles. For a medium such as Lucite, VLm = 0.66c, whereas water has VLm = 0.75c. (The refractive index n for light in a medium is defined as the ratio of the phase velocity of light in a reference medium—usually vacuum—to the phase velocity VLm in the medium.)
Čerenkov radiation accounts for the blue glow seen in water surrounding a nuclear reactor emitting high speed particles. Experiments making use of Čerenkov radiation include the aborted Deep Underwater Muon Neutrino Detector (DUMAND) project, as well as the Baikal, AMANDA, ANTARES, and NESTOR projects for detection of cosmic neutrinos. These attempt to detect muons generated by energetic neutrinos that have passed through the Earth and that interact deep underwater. The muons are superluminal in sea water and therefore give off Čerenkov radiation.
Verification of this Einstein prediction of Special and General Relativity was historically one of the first and most celebrated tests of the theory. It was ostensibly first accomplished by Arthur S. Eddington et al in 1919, using starlight bending during a solar eclipse. (The validity of this confirmation, made from the island of Príncipe near Africa, is somewhat controversial.) Starlight grazing the surface of the Sun should be deflected according to the Equivalence Principle toward the center—otherwise, an observer in an elevator free falling at the surface of the Sun and through which the starlight passes would not see it move in a straight line.
Working from Special Relativity considerations (i.e., ignoring the effects of General Relativity subsequently described by Einstein of space curvature), the deflection may be estimated from the calculated time—4.67 seconds—that light takes in passing along the distance of the solar diameter 1.4 x 109 m at speed c. A free-falling inertial frame starting at rest will attain a final centripetal velocity V of 1284 m/s in such a field (using V = ast, where as = 275 m s-2 (the acceleration of gravity at the Sun's surface). The estimated angle of deflection (considering only pre-General-Relativity Newtonian light deflection and Euclidean space) is given as tan-1 (1284/c) = 4.3 x 10-6 radians = 0.898 arcsec. (Einstein initially predicted 0.83 arcsec, later revised this to 0.85 arcsec).
Subsequently, applying General Relativity principles, Einstein found that the deflection should be twice this, or approximately 1.75 arcsec. Using Very Long Baseline Interferometry and extragalactic radio sources, DE Lebach et al found in 1995 that the ratio between experimentally measured and predicted values for the General Relativity effect on light bending is extremely close to 1—specifically, the gravitational deflection is 0.9998 ± 0.0008 times that predicted by general relativity.
The photo at the top of this webpage illustrates astronomical gravitational light bending (lensing) of light from stars and galaxies into "Einstein rings" and other arc-like deformations.
Length Contraction (Lorentz Contraction)
Consider a rod of proper length L0 lying along the x'-axis of inertial frame S' in which the rod is at rest. (See UP p. 105 and IER p. 32) We measure the end positions of the rod in frame S' as at x'1 and x'2 and therefore the spatial separation of the ends in this frame is given by Δx' = x'2 - x'1 measured at simultaneous time t'. Assume S' is moving at velocity vrel in standard configuration with respect to an inertial frame S. We wish to determine Δx, the apparent length of the rod as measured in the S frame (the two ends must be measured at the same time t in S). Using the LT formula x' = γ(x - vrelt), we have for the two measurements in S':
x'1 = γ(x1 - vrelt1) and
x'2 = γ(x2 - vrelt2)
Then Δx' = x'2 - x'1 = γ(x2 - x1) - vrel (t2 - t1)
Let Δx = x2 - x1
Δx' = γΔx - vrel (t2 - t1)
But for length measurement in S, the two measurements in S must occur at the same time t, so Δt = t2 - t1 = 0. Therefore, the formula for Lorentz contraction is:
Δx' = γΔx , or
♦ Δx = Δx' / γ
Summarizing, because γ is greater than one, the rod length L = Δx measured in frame S (in which the rod is moving) is smaller (contracted) when compared to the proper length L0 as measured in the rod's rest frame S'. (The proper length L0 is always the maximal possible measured length when comparing length measurements made in various inertial frames, always measured at the same time within the frame.)
Note: The textbook STP2 (and UP p. 184) discusses in problem L-14 (p. 119) the fact that in real life, a rod is never perfectly rigid, and that a rod that accelerates due to a force exerted at one end undergoes transient compression distortion and propagation of the deformation along the rod at a speed that cannot exceed c but is usually far less.
Transformation of Angles
A rocket frame S' is moving at speed vrel with respect to the Earth frame S. (See STP2 problem L-6.) A meter stick lies at rest at an angle φ' in the rocket frame with respect to the x'-axis defining the direction of motion. (Note that here the proper length L0 = 1 m of the meter stick is said to be measured in the rocket frame, not the Earth frame.) The angle the stick makes with the x'-axis in the rocket frame may be expressed by φ' = arctan (Δy' / Δx'). The same rod is seen to be at an angle φ in Earth frame, for which φ = arctan (Δy / Δx). But in standard configuration, Δy = Δy' = y'2 - y'1. By the formula for Lorentz length contraction (STP2 p. 157), Δx = (1/γ)(x'2 - x'1) = Δx'/γ. Note that the spatial positions of the two ends are measured at the same time t in Earth frame. The component of the stick parallel with the x-axis is Lorentz contracted, because γ > 1. Then
tan φ = γΔy'/ Δx' = γ (Δy'/Δx') = γ tan φ', or
♦ φ = arctan (γ tan φ')
Therefore, because γ > 1, the angle φ measured in Earth frame is larger than that measured in the rocket frame φ', and for acute angles φ' the rod appears more tilted up at the forward end in the Earth frame compared to the rocket frame.
Transformation of an X-velocity
A rocket frame is moving at speed vrel with respect to the Earth frame along the x' and x axis. A particle is at rest in the rocket frame and moves along the x-axis. Note that y = y' = 0 at all times. The components of displacement and velocity of the particle in the Earth frame are given by:
vbullet = (v'bullet + vrel) / [1 + (v'bullet vrel / c2) ]
♦ vx = Δx/Δt = [v' + vrel ] / [1 + (v'vrel/c2)] = 0 + vrel / 1 = vrel
(using the formula for addition of velocities and taking the added velocity v' = 0,
because the particle is at rest in the rocket frame); and
♦ vy = Δy/Δt = 0
because y and y' are always 0 in both frames
The inverse velocity transforms are
♦ v'x = -vrel
♦ v'y = 0
Transformation of a velocity of arbitrary oblique direction in the x-y plane
(Adapted from textbook Understanding Relativity By Leo Sartori, p. 109, Univ. of Calif. Press, 1996 and the Wim Schuur essay [link no longer available].) A rocket frame S' is moving at speed vrel along the x-axis with respect to and "away from" or in the positive direction with respect to the Earth frame S. (Thus we are in standard configuration. In all formulas that follow, if instead the frame S' is moving "toward" or in the negative x direction with respect to S, change the sign of vrel.) A particle moves at speed v in the Earth's S frame x-y plane with components vx = Δx / Δt and vy = Δy / Δt. Assume that at t = t' = 0, x =x'=y=y'=0. The components of velocity of the particle in the rocket frame S' are given by:
♦ v'x = (vx - vrel) / (1 - vxvrel/c2)
using the sum of velocities formula.
For the y-component of transformed velocity, we have
y' = y = vy t = vyγ(t' + vrelx'/c2) by LT, so dividing by t'
y'/t' = v'y = vyγ(1 + vrel v'x /c2)
Note that v'x has appeared in the formula for the v'y component, which does not occur in Galilean transformations.
Substituting the value above for v'x, we obtain the final result:
♦ v'y = vy / γ [1 - (vrelvx/c2) ]
The inverse velocity transforms (when frame S' is moving in the positive direction with respect to S) are
♦ vx = (v'x + vrel) / (1 + v'xvrel/c2)
♦ vy = v'y / γ [1 + (vrelv'x/c2) ]
Transformation of a y-velocity
(STP2 problem L-7, see also UP p. 110) A rocket frame S' is moving at speed vrel with respect to the Earth frame S in standard configuration. A particle rises at speed measured in S' of v'y = Δy' /Δ t' in a direction parallel with the y'-axis, thus perpendicular to the x'-axis. Assume starting conditions as follows: at t = t' = 0, that x = x' = y = y' = 0. The components of velocity in the Earth frame S (using the formula for vx from the general oblique velocity example above modified for the current problem) are given by:
♦ vx = (v'x + vrel) / (1 + v'xvrel/c2) = vrel
because v'x = 0
vy = v'y / γ [1 - (vrelv'x/c2) ] , but because v'x = 0:
♦ vy = v'y / γ
The inverse velocity transforms (when frame S' is moving in the positive direction with respect to S) are
♦ v'x = - vrel
♦ v'y = vy / γ
Tilted Meter Stick
(STP2 problem L-10) Assume a meter stick initially lies along the x-axis in frame S centered on x=0 at t=0 (so that the angle formed with the x-axis φ = 0). It is not moving in the x-axis direction in S, but is rising in the y-axis direction in S at velocity vy while remaining parallel to the x-axis, so the two ends therefore remain at x = -0.5m and x = +0.5m, respectively. At time t = 0, the y-position in S of each end and the center is taken to be 0, and the y-position at time t is therefore given by vy t. The frame S' is moving at speed vrel with respect to the frame S in standard configuration along the x-axis away from S. How does the meter stick appear in S'?
The following is taken from the derivation of Prof. Gerhard Müller shown here. Consider these three spacetimes events:
At Event 1, the right end crosses the x'-axis, and this event has coordinates x1, y1, t1, and x'1, y'1, t'1, respectively. Using the LT:
x'1 = γ(x1 - vrel t1) = γ(0.5) , because x1 = 0.5 and t1 = 0
y'1 = y1 = 0 because y1 = 0 at time 0 and y'1 = y1
t'1 = γ(t1 - vrelx1/c2) = -γ(0.5m)(vrel/c2) because t1 =0
At Event 2, the center of the stick crosses the x'-axis, for which the coordinates are x2, y2, t2, and x'2, y'2, t'2, respectively. Using the LT:
x'2 = γ(x2 - vrel t2) = 0 because x2 = 0, t2 = 0
y'2 = y2 = 0
t'2 = γ(t2 - vrelx2/c2) = 0 because x2 = 0, t2 = 0
At Event 3, the right end of the stick is observed as the center crosses the x'-axis, for which the coordinates of the right end are x3, y3, t3, and x'3, y'3, t'3, respectively. Then:
x'3 = x'1 + v'x(t'3 - t'1) expressing values entirely in S', using the starting value of x' (x'1), the velocity in x' (v'x), and the elapsed time (t'3 - t'1)
y'3 = y'1 + v'y(t'3 - t'1)
t'3 = γ(t3 - vrelx3/c2) = 0 using LT, because x3 = 0, t3 = 0
The velocity components of the stick in S' are given by adapting the y-velocity transforms above:
v'x = −vrel
v'y = vy / γ
Therefore, x'3 = x'1 + v'x(t'3 - t'1) = γ(0.5) + (−vrel) (0 - (-0.5)(vrel/c2) γ) = γ(0.5) - (0.5)(vrel/c2)γ
x'3 = (0.5)(1 - vrel/c2) / (1 - vrel/c2)1/2 = (0.5)(1 - vrel/c2)1/2
y'3 = y'1 + v'y(t'3 - t'1) = 0 + vy (1 - vrel/c2)1/2 (0 + γ(0.5vrel/c2)
y'3 = vy (1 - vrel/c2)1/2 γ(0.5vrel/c2) / (1 - vrel/c2)1/2
y'3 = 0.5vy vrel/c2
Then if φ' is the angle in S' that the meter stick makes with the x'-axis,
♦ tan φ' = y'3 / x'3 = 0.5vy (vrel/c2) / (0.5)(1 - vrel/c2)1/2 = γvy vrel/c2 or
♦ φ' = arctan ( γvy vrel/c2)
which is also that result given by Shaw 1962.
From the perspective of S', the meter stick appears tilted up at the right end—the end lying in the forward direction that S' is moving with respect to S—so that it forms an angle φ' with respect to the x'-axis. This arises from the relativity of simultaneity, as the left and right meter stick ends cross the x'-axis in S' at different times. As this problem is configured, the stick is also seen to be moving in S' to the left and upward.
Shaw also shows that a stick moving relative to a rising table with a hole in it—similar to the problem of the meter stick and rising one meter wide manhole in STP2 problem L-11—can pass through the hole due to the angle formed between the stick and the hole. (Similar length "paradoxes", including the Pole and Barn Paradox and the Paradox of the Fast Walker, are discussed here (and in UR p. 167 - 192 and STP2 problem 5-4).
Transformation of an Acceleration
Consider two frames S and S' in standard configuration, with S' moving along the x-axis at speed vrel with respect to S. For the special case of constant acceleration A of a particle along the x-axis in frame S, the LT of this acceleration is given by (UR p. 117):
♦ A' = A / γ3 [1 - vrel v/c2 ]3
where v is the instantaneous velocity in the x-axis direction at the time the transformation is made. It is apparent that transformed acceleration A' is dependent on instantaneous velocity (a non-Newtonian outcome, but to be expected in order to keep the velocity attained subluminal), and is also not constant even though A is constant.
The more general case of transforming an acceleration in arbitrary direction is discussed in IER p. 36 and here (although I have not studied these derivations in detail).
Time dilation affects any clock traveling in a moving frame S', including biological clocks associated with aging. Therefore, if you can travel arbitrarily fast relative to Earth frame S (limited only by the speed of light c), you can slow your aging rate down to as small a rate as you like (as judged in the Earth frame S), even though you will experience a normal aging rate in your frame S' in which you are at rest. However, you will return to find that your Earth world and relatives in S have dramatically aged. (This is discussed in STP2 Chapter 4.) If one twin Earl stays on Earth, and the other Cathy travels at high speed out to star Canopus, turns around, and then returns to Earth, Cathy will have aged less (in Earth frame) than Earl who remained on Earth. This is the well-known original Twin Paradox (though Einstein did not regard this as paradoxical, only peculiar). Its apparent lack of symmetry of outcome is caused by the fact that the twin Cathy who travels to Canopus undergoes an acceleration to get up to speed for the outbound leg, is decelerated on turning around, and undergoes a further acceleration to get up to speed for the inbound leg—while Earl experiences no accelerations. The voyager Cathy can never return to find Earl at her same age—such "time travel" is one-way. A complete resolution requires General Relativity (unless one envisions doing the acceleration by stepping between an infinite succession of faster and faster rockets... see STP2 p. 132). The problem is considered in in STP2 p. 169 (as well as UR p. 192), which where it is demonstrated that there are differing "lines of simultaneity" in Earth frame for the traveler arriving at versus departing from Canopus, specifically how these are different lines of simultaneity correspond to different inertial frames.
EXPERIMENTAL VERIFICATION OF TIME DILATION: There have been many experimental verifications of time dilation to differing levels of accuracy. The 1971 Hafele–Keating experiment using an airplane and atomic clocks is conceptually problematical, and the time increments measured are on the nanosecond scale (STP2 p. 133). Experiments with high speed unstable particles that exhibit lengthened half-lives in Earth frame have been very successful—these include the muons measured by Rossi and Hall 1941. A test originally performed on hydrogen ions by Ives and Stilwell (1938, 1941) used the transverse Doppler effect. The Ives-Stilwell result was markedly improved by G. Saathoff et al. in 2003 (published paper cited below, but see also their website summary). They used laser spectroscopy measuring Doppler shifted frequencies on 7Li+ ions in the Heidelberg Heavy-Ion Test Storage Ring (TSR) at an ion velocity v/c = 0.064, and obtained extremely good agreement with theory, finding that the parameter α was < 2.2x10-7 (where a value of α = 0 is expected by special relativity). By comparison, the Ives-Stilwell experiment yielded a value of α < 1x10-2, thus a much less precise test.) The TSR experiment is apparently the most precise confirmation of special relativity time dilation to date.
This problem is presented in STP2 (p. 117), the article by Boughn (1989) referenced below. It is not purely a problem in special relativity, because it involves accelerations, so that I have to take somewhat on faith the validity of its statements. Boughn states that this paradox illustrates the "problem of clock synchronization in special relativity and a simple example of how acceleration from one inertial frame to another renders two initially synchronized clocks unsynchronized"—again, not a feature of special relativity. The essence of the paradox is that although Jane and Dick are twins and thus were born in Earth frame S at the same time, Jane ceases accelerating in frame S' before Dick, and Jane's birthday occurs Δt' = γvx0 /c2 before Dick's in the moving frame S' that they end up in. (Here, x0 is their original separation just before they start their trip, and v is the final velocity attained.) If their initial separation is 12 light years (sic!) in S, and they attain 0.75c in S, Δt' = 13.61 years... (Obviously, these are extreme conditions for human twins.)
(Problem 4-1 in STP2 p. 135). James departs Earth at Earth time t=t'=0, and travels at v = 0.75c (with respect to Earth frame) to Sirius, L0 = 8.7 ly away (again as measured in Earth frame S). During travel, he is moving relative to Earth frame, at rest in frame S'. Once he stops near (not on!) Sirius, he waits 7 years (again at rest in Earth frame S), then returns to Earth at the same speed as before but in the opposite direction and effectively in a new frame S''.
According to Earth frame S observers:
4-1a. His arrival at Sirius occurs at tSir = distance/speed = L0/v = 8.7 ly / 0.75 = 11.6 years.
4-1b. The rocket leaves Sirius after an Earth-frame time delay DSir = 7 years, so it departs Sirius at tSirDep = L0/v + DSir = 11.6 + 7 = 18.6 years.
4-1c. The rocket travels back to Earth at the same speed as outgoing though in the opposite direction, so Sirius-Earth travel time is again 11.6 years. Therefore, he has been gone a total of ttot = 2L0/v + DSir = (2 * 8.7 ly / 0.75) + 7 = 30.2 years.
According to James's observations in his rocket frame in which he is always at rest:
4-1d. He travels in his rocket frame S' and arrives at Sirius at t'Sir = (1/γ) distanceEarth/speedEarth = (1/γ) L0/v where γ = 1/(1-v2 /c2) =1.511858. Note that γ > 1 so 1/γ < 1 and his time is dilated (larger) in Earth frame S relative to his moving frame while traveling, S'. Therefore, in his moving frame S', he arrives at t'Sir = 8.7 ly / (1.511858•0.75) = 7.67268 years. This is the time his body has biologically aged at the time he reaches Sirius.
4-1e. During his stay at Sirius, he and his rocket frame is at rest with Earth frame, so the time added to his Rocket frame is also simply D'Sir = DSir = 7 years, thus he departs Sirius at t'SirDep = 7.67 + 7 = 14.67268 years in his S' frame. This is the time his body has biologically aged at the time he leaves Sirius.
4-1f. Upon returning to Earth, he again experiences a time lapse of t'Sir = (1/γ) L0/v = 7.67268 additional years, so that he experiences a total time lapse round trip of:
t'tot = 2γL0/v + DSir = 22.34536 years. This is the time his body has biologically aged for the round trip.
4-1g. Using a string of lookout stations outgoing from Earth accompanying James's flight, what is the distance he measures between Earth and Sirius in S'. (Each lookout station is moving at his speed and each has a clock synched to his own—in other words, there is a standard lattice of lengths and clocks in S' with John's position always at x' = 0 and his position now coinciding with Sirius, and in which all clocks in S' are synchronized to read the same time as defined above). This problem is asking to compute Lorentz contraction. The proper length between Earth and Sirius L0 is the length with both objects at rest (in S), namely 8.7 ly. However, when he is moving at speed 0.75c, he measures in his frame S' the contracted length L', specifically L' = L0 / γ = 5.75451 ly.
4-1h. One of these lookout stations in S', call it Q, passes Earth when James reaches Sirius. What is the time t'Q that Q reads in S' at this event? Recall that James's clock reads t'Sir = 7.67268 years. Then t'Q = t'Sir = 7.67268 years (because the clocks of S' are synchronized to read the same value in the sense above).
What time tQ does an Earth clock read for the same event when Q passes by Earth? This is given by the Lorentz transformation, where coordinates are specified relative to James who is now on Sirius:
tQ = γ(t'Q + vrelx'/c2)
but c = 1, vrel = 0.75, and x' = -5.75451 ly (because Earth's x' coordinate in S' is - 5.75451 ly relative to James at x'=0 on Sirius), so
tQ = 1.511858•(7.67268 - 0.75•5.75451) = 5.07500 yr
4-1i. As he travels back to Earth, James is effectively moving in a new frame S'', and is accompanied by a string of lookout stations incoming toward Earth moving along the direction of his motion (the x'-axis), each with a clock synched to his clock in S''. Assume that at the time of departure, the S'' time scale is chosen at Sirius so that t''SirDep = t'SirDep = 7.67268 + 7 = 14.67268 years. One of these stations, call it Z, passes Earth just when he leaves Sirius (i.e., simultaneously in S").
What time does Z's clock read when passing Earth?
In S'' frame, this clock when passing Earth reads the same as James's clock when departing Sirius, namely t''Z = t''SirDep = t'SirDep = 7.67268 + 7 = 14.67268 years.
What time does a clock on Earth read at the same event when Z passes Earth? (Note that from James's point of view in S'', the Earth is coming toward him while he is stationary. The distance to Earth appears Lorentz contracted in S.)
Using the LT, the time that elapses in S on Earth clocks during the return trip is given by
tZ = γ(t''Z + vrelx''c2)
where c = 1, vrel = -0.75 (negative now that Earth is incoming to S" and James), t''Z is the travel time in S'' back to Earth (7.67268 years), and x'' = L0 /γ, so
tZ = 5.07500 years
Therefore, when James departs Sirius, the Earth clock seen from Z must read Total trip time - Travel time from Sirius = 30.2 years - 5.07500 = 25.125 years.
Euclidean 3-Vectors: Points or positions in Euclidean 3-space are expressed by ordered coordinates (x, y, z). Euclidean 3-vectors for various vector quantities Q are defined by means of starting coordinates and 3 ordered components (ΔQx , ΔQy, ΔQz). Together, these specify the direction and magnitude of traditional vector quantities such as displacement, velocity, force, and acceleration in 3-dimensions. The 3-vector representing the net spatial displacement between two events is termed the displacement vector. Its existence and magnitude are independent of frame, but the individual displacements are frame-dependent (STP2 p. 192).
The unit displacement vector for a 3-vector quantity expresses direction and has a scalar magnitude of (Δx2 + Δy2 + Δz2)1/2 = 1. A specific 3-vector such as a velocity may be defined by a starting position where the vector arises (in this example, where the velocity is measured from), a unit displacement vector providing direction for the vector (in this case the direction of movement), and a scalar magnitude expressing the length of the vector quantity (in this case the speed of movement).
4-Vectors: In contrast, a four-vector (4-vector) is a vector defined in a four-dimensional real vector space called Minkowski space or spacetime. Events in spacetime are positioned at points with coordinates specified by the position 4-vector, namely X(τ) ≡ (ct, x, y, z). The displacement 4-vector expresses movement along a worldline —or the "arrow" expressing a distance between 2 events—and is given by ΔX(τ) ≡ (cΔt, Δx, Δy, Δz). It represents the net spacetime displacement between two events, say E1 at (ct1, x1, y1, z1) and E2 at (ct2, x2, y2, z2). Although this 4-vector also has a spacetime direction or an arrow linking the two events, this 4-dimensional arrow cannot easily be visualized. It has an existence and magnitude that are independent of frame, but the individual component displacements are frame-dependent, including cΔt. The 4-vector displacement vector is defined by the coordinates of the starting event E1 and the 4 net spacetime displacements (cΔt, Δx, Δy, Δz) needed to reach E2, where Δx = x2 - x1, etc. (The position vector X(τ) represents the displacement vector with respect to the origin where x = y = z = t = 0.) As previously discussed, the magnitude of the spacetime displacement 4-vector is given by
Δs= (c2Δt2 - Δr2)1/2 where Δr2 = Δx2 + Δy2 + Δz2, or
Δs= (Δt2 - Δr2)1/2 for units chosen so that c = 1.
4-vectors in spacetime include 4-displacement, unit 4-displacement, 4-velocity, 4-acceleration, 4-force, 4-momentum, the momentum-energy 4-vector (momenergy), 4-current, electromagnetic 4-potential, 4-frequency, etc. In the momenergy 4-vector, momentum and energy are components of the 4-vector but are not in themselves 4-vectors.
Unlike ordinary vectors, all 4-vectors can be transformed by Lorentz transformations.
A key 4-vector in Relativistic Mechanics is the momentum-energy 4-vector that is used to describe mass and energy in motion. I have not identified a single symbol that is universally used to represent the momenergy 4-vector. The expression sometimes used, (E, p), is potentially confusing as the momentum components are not standard 3-vector momentum components. The Momentum-Energy 4-Vector appears to be the same as the momentum 4-vector. Wheeler and Taylor (STP2 p. 191) have chosen to adopt the neologism "momenergy" to designate the momentum-energy 4-vector. Although this word does not appear to have gained common usage, it is succinct and unique, and I will also adopt it here.
Momenergy can be defined for a single particle or for a system of particles. The Momentum-Energy 4-Vector ("Momenergy") of a particle or system at a particular point or event on the worldline is defined by a scalar magnitude (m) multiplied by a unit 4-vector, thus yielding another 4-vector having magnitude m and the same spacetime direction as the unit 4-vector.
Magnitude m is called the mass, the invariant mass, the system mass, or especially applicable for single particles the rest mass.
The unit displacement 4-vector, which gives the spacetime direction along the worldline at a particular point or event, is given by a quotient:
• the numerator of the quotient is
the spacetime displacement 4-vector Δs or ds = (cdt , dx , dy , dz)
• the denominator of the quotient is
the proper time increment Δτ or dτ required for that displacement
Recall that total proper time of a displacement along a worldline is a scalar which gives the measure or magnitude of this displacement. Since the quotient is a vector divided by a scalar representing the vector's magnitude, the quotient yields a unit 4-vector, where "unit" indicates magnitude = 1 (regardless of actual units by which mass are measured).
The momenergy 4-vector may be expressed by
momenergy 4-vector = (mdt/dτ , mdx/dτ , mdy/dτ , mdz/dτ) = m(dt/dτ , dx/dτ , dy/dτ , dz/dτ) , where
m is the magnitude (mass) expressed above, and
(dt/dτ , dx/dτ , dy/dτ , dz/dτ) is the unit displacement 4-vector.
Properties of momenergy and calculation of its magnitude (m2 = E2 - p2 for c = 1) are discussed in detail here.
In momenergy and other discussions, here are some additional definitions which apply:
Isolated System: an Isolated System of particles is a system in which no matter, energy, radiation, or external forces enter into the system from outside or escape to the outside from the system. (Such a system is generally a fiction or an approximation of real systems in the universe.)
Invariant and Invariance: Invariant and Invariance describe a quantity (such as the momenergy magnitude or the magnitude of a Lorentz interval) which remains the same when observed in various inertial frames of reference (STP2 usage). This is not the same as conservation during an interaction. (However, some authors use it loosely in this way. Moreover, Noether's Theorem states that any law of physics with continuous differentiable symmetry, such as the invariances we are dealing with in spacetime, has a corresponding conservation law, so this provides an opportunity for overlapping usage. The term covariance is apparently broader and to be preferred in the tensor mathematics of GR.)
Conserved and Conservation: Conserved describes a quantity (such as momenergy of an isolated system) which remains the same before and after some type of interaction such as a collision or disintegration. It is not the same as invariance, which deals with differing inertial frames.Constant: Constant describes a quantity or property (such as the speed of light) which does not change with time. Some quantities may be constant in a particular frame (such as a particle's velocity) but not invariant when viewed from other inertial frames.
Mass proves to be a more complex topic in relativity theory than might be expected. The quantity m = "mass", which is the momenergy magnitude, is the best usage of the word in relativity theory according to Taylor and Wheeler's text. Because momenergy mass in invariant in all frames, it is also usefully and more meaningfully called the invariant mass, a term applicable to single particles as well as to entire isolated systems. (Other terms purported by some to be synonyms and sometimes used include intrinsic mass and proper mass—however, these appear to be ambiguous and/or problematic.)
Rest / Invariant Mass m: For a single particle observed in a frame in which the particle is at rest (though its innards may not be!), the particle's linear momentum is zero and momenergy mass is the same as the particle's Energy. Therefore, for single particles their momenergy mass may be said to be the particle's rest mass, also called the rest energy. This term rest mass is trickier to apply and perhaps inappropriate when referring to two or more particles that are in relative motion, because there is no frame in which they can both be at rest simultaneously. In classical mechanics, all isolated systems of particles have a center of mass which moves with a steady velocity. In relativistic mechanics, for any isolated system it is also possible to define and solve for a center of momentum frame. In the center of momentum frame, the center of momentum is stationary, the total momentum is zero, and the system—or at least its center of momentum—may be thought of as not moving and therefore "at rest". The invariant mass of the system as a whole in this frame is equal to the total system energy. This invariant mass is therefore the minimal possible energy which the system may be observed to have compared to other frames that are moving with respect to the center of momentum, and for which both total momentum and total energy increase. Therefore, the system mass may be thought of as the minimal possible energy, perhaps in a limited sense a rest mass, but probably better the invariant mass. The invariant-system-rest mass is also conserved for interactions such as collisions and decays in isolated systems, but the terms invariant and conserved are not entirely synonymous (see definitions above).
Relativistic Mass M: The term "mass" in current preferred relativity usage generally refers to rest or invariant mass, the magnitude of the momenergy 4-vector. In earlier usage including Einstein's own (and some current authors, see UR p. 212), relativistic mass was also used to refer to the total mass or relativistic total energy of a particle, having the value mγc2. This makes sense in relation to expressing acceleration as the ratio of Force to Mass. However, the relativistic mass includes kinetic energy K and is velocity- and frame-dependent and is therefore not invariant. This usage has generally fallen out of favor compared to using [relativistic] total energy E and momenergy in problem solving, and is deprecated by the authors of STP2 (p. 250).
In general, the simple arithmetical sum of individual rest / invariant masses of particles in an isolated system is not a conserved quantity in relativity—mass is not additive (STP2 p. 224).
Matter: Mass does not express the "amount of matter" in relativity, and the latter is not readily defined. Even if we can count the number of atoms, the mass of matter will vary with temperature, degree of bonding, radioactive decay, and particles can also come into existence spontaneously. (STP2 p. 248)
In 3-D Euclidean space, velocity is simple ds/dt, distance traveled (a vector quantity) per unit coordinate time.
In 4-D spacetime, coordinate velocity is the same traditional 3-vector ds/dt, distance traveled per unit coordinate time.
However, proper velocity for an object moving with respect to frame S in the positive x-direction is the ratio between the x-distance traveled in S and the proper time τ elapsed on the clocks of S' for this distance, where S' is the frame in which the object it is at rest. Thus,
♦ Proper velocityx = dx/dτ
and more generally
♦ Proper velocity = ds/dτ,
which is a 3-vector consisting of the space-like components of the object's 4-vector velocity.
For a particle and for any speed,
Proper velocity = p / m [momentum p (which is frame dependent) divided by its invariant (rest) mass m].
At low non-relativistic velocities, | proper velocity | = | coordinate velocity |.
Proper velocity is also sometimes called celerity.
LT of relativistic velocity: The LT of relativistic velocity is given in UR p. 110 and 238.
In 4-D spacetime, momentum is retained as a 3-vector spatially defined quantity, and represents the "space part" of momenergy.
In Newtonian mechanics, the classical momentum of a particle p = mv = m(dv/dt), where v is a 3-D vector, m is the (unvarying) mass of the particle, the units of momentum are for example kg-m-s-1, and t is the universal time. Individual components of Newtonian momentum are given by:
px-Newton = mvx = m(dx/dt), py-Newton = mvy = m(dy/dt), pz-Newton = mvz = m(dz/dt)
In Relativistic Mechanics, Momentum is modified but still a 3-D vector and still conventionally (but confusingly) represented by the same symbol p. When expressing momentum p of a particle at rest in its own frame S' but moving with respect to a frame S at relative velocity v (for which the magnitude or speed = v), m now represents the rest /invariant mass of the particle. Further, the units of speed v are often expressed as light-related unitless ratios vconv /c (or alternatively phrased, as meters traveled per meters of light travel time required to travel this distance).
When v is unitless:
♦ |p| = mvγ = mv / (1- v2)1/2 and |p| is given in the same units of mass as m.
(This relationship is proven in UR p. 215)
Momentum magnitude may also be expressed in conventional units, where vconv is in m/s and c is in m/s, etc.
♦ |pconv| = mvconv (1- (vconv2/ c2)1/2 where |pconv| is now given in kg m s-1 etc.
Note that relativistic momentum magnitude mvγ reduces to Newtonian momentum mv for low values of v for which γ approaches 1. But at high speed where vconv /c is close to 1, γ and therefore the relativistic momentum can increase without bounds whereas Newtonian momentum even at speed c is merely mc.
The individual spatial scalar components of relativistic momentum are defined relative to the increment in proper time and are given by:
px = m(dx/dτ)
py = m(dy/dτ)
pz = m(dz/dτ)
but dτ = (dτ2)1/2 = [dt2 - dx2 ]1/2 = [dt2 - vdt2 ]1/2 = (dt) (1 - v2)1/2 = dt / γ (where c=1), so
♦ dt/dτ = γ
and since vx ≡ dx/dt (the x-direction speed as measured in the S frame)
px = m(dx/dt)(dt/dτ) = mvx γ or
♦ px = mvx γ = mvx (1 - v2)-1/2 for c=1, vx is unitless, and px is given in the same units of mass as m
♦ py = mvy γ = mvy (1 - v2)-1/2 for c=1, vy is unitless, and pY is given in the same units of mass as m
♦ pz = mvz γ = mvz (1 - v2)-1/2 for c=1, vz is unitless, and pZ is given in the same units of mass as m
again with units of mass or the energy equivalent (e.g., kg, eV, J, etc.)
In conventional units with vx expressed in m/s, these may be given equivalently as
♦ px-conv = mvx-conv γ = mvX (1 - (vconv /c)2)-1/2 where px-conv is now given in kg m s-1 etc.
♦ py-conv = mvy-conv γ = mvY (1 - (vconv /c)2)-1/2 where py-conv is now given in kg m s-1 etc.
♦ pz-conv = mvz-conv γ = mvz (1 - (vconv /c)2)-1/2 where pz-conv is now given in kg m s-1 etc.
LT of Momentum: See here for the LT for momentum (and also STP2 p. 215, UR p. 227).
Conservation of Relativistic Momentum and Its Individual Components: relativistic total p, px , py , and pz are each conserved for an isolated system for any and all interactions when viewed in a particular frame. That is, (∑pxj) , (∑pxj), and (∑pzj) are conserved, as are their vector sum direction and vector sum magnitude ( (∑pxj)2 - (∑pxj)2 - (∑pzj)2 )1/2 (STP2 p. 212, and more completely in UR p. 217). Of course these quantities are not invariant.
In 4-D spacetime, energy is a one-component scalar quantity, and represents the "time part" of the momenergy 4-vector.
Newtonian Energy: In Newtonian mechanics, a stationary particle has no rest energy and with acceleration to a speed v it acquires a kinetic energy KE = mv2/2. If other forms of energy are present (such as radiations emitted, heat, etc.), they must be separately considered and factored in to yield total energy for an isolated system. Classical kinetic energy is conserved only for elastic collisions whereas total energy is conserved for isolated systems provided all possible forms of energy are included.
Total and Rest Energy in Relativistic Mechanics: is modified and conventionally (but somewhat confusingly) represented by the same symbol E. However, this is now taken to be the total energy in all forms for the particle, so E = Etot. The energy of a particle with velocity v is given by
♦ E = Etot = m(dt/dτ) = mγ = m(1 - v2)-1/2 for c=1, and where as usual γ = (1 - v2)-1/2, or
♦ Econv = Etot-conv = mc2γ = m(1 - (vconv/c)2)-1/2 (expressed in J = kg m2 s-2, or other energy units),
Here, the quantity m is termed the rest energy (rest / invariant mass) of the particle, also expressed as Erest, sometimes shown as m0, so that
♦ Erest = m
Note that m is equal to the energy for a particle only when it is observed in a frame for which it is at rest, and that in all other frames with relative motion, E > Erest.
In conventional units, we also have the famous equation for rest energy of a particle:
♦ Erest-conv = mc2 (expressed in J = kg m2 s-2, or other energy units).
To reiterate once again, this formula is only valid in a frame for which the particle is at rest.
Equivalence of Mass and Energy: Sartori provides a clear summary of Einstein's thought experiment first deriving the E=mc2 result (UR p. 203, original 1905 in English translation here). In Einstein's analysis, a particle at rest having mass m0 emits 2 "light waves" each of energy L/2 in opposite directions. He showed that the particle then has a lesser mass m1, where m0 - m1 = L/c2, verifying the equivalence between mass and energy. (This is not the same equivalence as the equivalence principle discussed above.) Einstein concludes in 1905 "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c2. The fact that the energy withdrawn from the body becomes energy of radiation evidently makes no difference, so that we are led to the more general conclusion that the mass of a body is a measure of its energy-content; if the energy changes by L, the mass changes in the same sense by L / (9 × 1020), the energy being measured in ergs, and the mass in grammes." [107 ergs = 1 J, 103 g = 1 kg, so this is the same as 9 x 1016 J/kg]
STP2 problem 8-5 provides a detailed discussion of Einstein's alternate proof of the E=mc2 relationship, by showing that a photon emitted from one end of a closed container transfers energy equivalent to mass m = E to the other end, etc.
Relativistic Kinetic Energy: A moving particle has E = Etot = Erest + K where K is the relativistic kinetic energy, a term given by
♦ K = Etot - Erest = mγ - m = m(γ -1)
In conventional units:
♦ Kconv = mc2(γ -1) (expressed in J = kg m2 s-2 or other energy units)
Note that for any given particle with rest/invariant mass m, at high speeds v the relativistic K is considerably larger than Newtonian KE and in fact can increase without bounds as v approaches 1. In contrast, KE at light speed (v=1) would be only m /2. (Of course, Newton did not know of v = c = 1 as representing the speed limit for matter.)
A massless particle such as a photon can have Etot = K even though the rest/invariant mass is zero.
Conservation of Relativistic Total Energy: E = Etot is conserved for an isolated system for any and all interactions, including both "inelastic" as well as "elastic" collisions, and for all forms of energy contributing to E (STP2 p. 206 and 212), when viewed in a particular frame. However, E and K are not invariant (STP2 p. 246). Moreover, relativistic K is not conserved in general. (In classical Newtonian mechanics, KE is conserved but only for elastic collisions—not for inelastic collisions—and this relationship remains valid in relativity only for low-speed interactions—STP2 p. 206)
Conversion of Energy To Conventional Units: In general, the units of energy (rest, kinetic, and/or total) are the units of mass or the energy equivalent (e.g., kg, eV, J, etc.) To convert energy (rest, kinetic, or total) calculated with c = 1 to conventional units with c expressed as m/s (resulting in J = kg m2 s-2), multiply by c2 (m2/sec2)
Interrelationships of E, m, v, and p for an object:
|v| = mγ |v| / mγ , so
♦ |v| = |p| / E where |v| is expressed relative to c (also less accurately written as v = p/E)
or in conventional terms,
♦ |vconv| = |pconv | c2 / E.
By the same formula,
♦ |p| = |v| E , also less accurately written as p = vE
Note that in the limiting case with v/c = 1, E = cp, a relationship which holds true for the photon.
Because m2 = E2 – p2 (see below), we have
♦ m = (E2 – p2)1/2 for c = 1
♦ E = (m2 + p2)1/2 for c = 1
♦ |p| = (E2 – m2)1/2 for c = 1
In conventional units, we have m2c4 = Econv2 - |pconv|2 , so
♦ m = [ (Econv2 - |pconv|2)/c4 ]1/2
♦ E = (m2 + p2)1/2 for c = 1
♦ |p| = (E2 – m2)1/2 for c = 1
Low Velocity Relativistic Energy Limit: The formula for relativistic E = (m2 + p2)1/2 can be approximated at low velocity by the binomial expansion:
Etot = m(1 + p2/m2)1/2 = m(1 + p2/2m + ... + much smaller higher order corrections) so at low velocity , the relativistic Etot reduces to
♦ ETot-LowV ≈ m + p2/2m (expressed in terms of momentum), or
ETot-LowV≈ m + mv2 /2 (expressed in terms of velocity), or
ETot-LowV / m ≈ 1 + v2/2
♦ ETot-LowV≈ mrest + KEclassical
Quantitation Of The Tiny Discrepancy Between Classical vs. Relativistic Energy At Low Velocity:
If a jet travels at v = c10-6 so that v = 300 m/s and has rest mass m, the fractional increase in relativistic kinetic energy K with increasing rest mass is given by
dE/dm = dK/dm = d/dm (mγ - m)) = γ -1 , so for v = 10-6,
Relativistic dE/dm = γ - 1 =
But the classical dE/dm = v2 /2 =
0.0000000000005000000000000000000 = 5 x 10-13
so the discrepancy is only
0.0000000000000000000000093750000 = or 9.375 x 10-24.
Therefore the ratio of the discrepancy in dE/dm to the value of dE/dm is only 2 x 10-11 , and thus negligible at such low speed.
LT of Energy: See here for the LT for energy (and also STP2 p. 215, UR p. 227).
At the risk of redundancy, some statements are repeated or pounded in mercilessly here in several ways to assure clarity of understanding:
Formula for momenergy = (mdt/dτ , mdx/dτ , mdy/dτ , mdz/dτ) = m(dt/dτ , dx/dτ , dy/dτ , dz/dτ)
Direction of Momenergy: Momenergy for a particle is a 4-vector directed along and pointing in the direction of a particle's worldline—that is, in precisely the same direction as the direction of instantaneous motion of the particle in spacetime (STP2 p. 193 and 211) This direction is specifically given by the direction of the spacetime displacement or its unit 4-Vector (spacetime displacement 4-vector / proper time displacement). The direction of the momenergy 4-vector is the same as the direction of the spacetime displacement between a reference event and a very nearby event. This direction is independent of the momenergy magnitude (they are separate quantities).
Invariance of Momenergy Direction: The momenergy 4-vector "arrow" or direction exists and has a particular direction that is independent of frame (STP2 p. 196). However, as we select different frames in which to view momenergy, the component values of E and px change. An attempt to graphically display the momenergy 4-vector from the viewpoint of different frames and plotted in Euclidean space on paper leads to differing appearing momenergy vector directions and magnitudes, but the momenergy "direction" and magnitude in Minkowski spacetime are invariant (STP2 p. 198—this is genuinely difficult to understand, but STP2 p. 210 helps a little with the analogy of a tree vs. its shadow.)
Invariance of Proper Time In The Quotient: Displacements of spacetime (in the numerator of the defining quotient) are measured in relation to proper time (the denominator of the quotient in the definition), and that total proper time is a frame-independent invariant quantity which is used to "clock" the particle. (STP2 p. 194) Recall that:
Total Proper Time τ = ∫P [dt2 - dx2/c2 ]1/2 = ∫P [dt2 - dx2]1/2 for c = 1
Invariance of Momenergy 4-Vector: Perhaps because momenergy is defined as a scalar (mass m) multiplied by the quotient of two invariant quantities, momenergy is also invariant. (However, when we break momenergy down into energy and 3-D momentum components with values established in a particular frame, the energy and momentum components are not invariant—see STP2 p. 196.)
Components and Units: Momenergy consists of energy and momentum components, both of which are expressed in units of mass or the energy equivalent (e.g., kg, eV, J, etc.). These components are not invariant. The units of the Momenergy magnitude are also expressed as mass or the energy equivalent (e.g., kg, eV, J, etc.)
Calculation of Energy Components of Momenergy: When computing momenergy for 2 or more particles considered together in an isolated system, their energies are additive as scalar quantities.
Calculation of Momentum Components of Momenergy: When computing momenergy for 2 or more particles considered together in an isolated system, their momenta are additive as vector quantities. For example, two electrons moving at 100 keV but in the opposite direction along x have px = mvxγ and px = -mvxγ, respectively, so that their net px = 0. However, two electrons moving at 100 keV in the same direction along x have px = mvxγ and px = mvxγ, respectively, so that their net px = 2mvxγ.
Calculation of Momenergy Magnitude: Momenergy Magnitude is symbolized by m. The Momenergy Magnitude m for an isolated system is given by
♦ m2 = E2 - |p|2 where c = 1,
or in conventional terms:
♦ m2c4 = Econv2 - |pconv|2
More specifically (for c=1):
♦ Momenergy Magnitude m =
[ (sum ∑ of individual total energies)2 - (sum ∑ of magnitudes of individual momentums)2 ] 1/2 , or
♦ Momenergy Magnitude m2 = (∑Ej)2 - (∑pxj)2 - (∑pxj)2 - (∑pzj)2 where the individual pxj values etc. may be positive or negative but all Ej values are positive. Note that It is incorrect to calculate the momentum contribution by px12 + px22 +px32 + py12 + py22 +py32 ... as this prevents negative and positive values from canceling each other out.
System Rest/invariant mass is not usually additive: Momenergy magnitude (system invariant rest mass) is invariant but not usually equal to the sum of the rest/invariant masses of the individual parts or particles in the system. For example, certain particles with mass of 8 each may be combined into a "system", even before they have any interaction, in which the resulting system mass is 20. But after inelastically colliding, their system mass is also of 20. Added energy contributes to the added mass (STP2 p. 224-225)
Invariance of Momenergy Magnitude: When viewed from different frames, both momentum and energy vary for a particle, but the Momenergy Magnitude is invariant (STP2 p. 198). In other words, for frames S and S',
m2 = (∑Ej)2 - (∑pxj)2 - (∑pxj)2 - (∑pzj)2 = (∑E'j)2 - (∑p'xj)2 - (∑p'xj)2 - (∑p'zj)2
Momenergy "Hyperbola" and Rest Energy/Mass: When momentum is confined to the x-direction (only px ≠ 0) and px2 is plotted on paper (in Euclidean space!) against energy E2, the paired values of E and px trace out a hyperbola given by m2 = E2 - px2, where m2 (mass2) is Momenergy Magnitude2, an invariant quantity. The vertex of this hyperbola is at m2 = E2, the point where there is no net momentum and the momenergy m = Erest = mrest, the lowest possible energy state for the particle given by its "rest energy" or "rest mass". (STP2 p. 198) Note that although this hyperbola is drawn as if formed by arrows of varying lengths, the varying Euclidean magnitudes of these arrows are given by (E2 + px2)1/2, which is a meaningless quantity. In contrast, the momenergy magnitude (E2 - px2)1/2 = m2 is invariant but does not explicitly appear in STP2 Figure 7-4.
Multiples Of Momenergy Magnitude: A particle A having exactly the same direction of motion and velocity as another particle B but having 5 times the Momenergy Magnitude (rest mass) of B, has 5 times the Momenergy Magnitude (rest mass) as that of B—i.e., mA = 5mB (STP2 p. 193).
Conservation of Momenergy Direction and Magnitude: The momenergy 4-vector direction and its magnitude (expressed as a system mass) are conserved for all isolated systems, so that no interaction of the constituent particles (such as a collision or disintegration) can change the Momenergy Magnitude=mass of the system or its direction. (Of course, constituent particles can change in a reaction, and even if the particles are unchanged, energies and momentum components for the individual particles are not generally conserved in an interaction.)
Utility of Relativity Energy/Mass Balance Analysis At Low Non-Relativistic Speeds: Even when speeds are relatively low and therefore "non-relativistic", such as in many nuclear reactions, the use of the energy-mass equivalence and the balance among E, p, and m in relativity are useful or essential for analyzing the changes in energies and masses resulting from these nuclear reactions, etc. (UR p. 219)
Given frame S' moving at velocity v in positive direction along x-axis compared to frame S, for c=1, and applying LT for incremental displacements and addition of velocities formula, etc. (see STP2 p. 215, UR p. 227, and partial proof here):
♦ E = γ(E' + vp'x)
♦ px = γ(p'x + vE')
♦ py = p'y
♦ pz = p'z
The inverse transformation are:
♦ E' = γ(E - vpx)
♦ p'x = γ(px - vE)
♦ p'y = py
♦ p'z = pz
Note that as expected, the frame S', in which the x-velocity of a particle is effectively reduced compared to in S, has lower E' and px' than the values for the lab frame S.
¶ Calculate Momenergy Magnitude m from E, px , and py (STP2 sample problem 7-1):
Given E = 6.25 kg, px = 1.25 kg, py = pz = 2.5 kg
Calculated Momenergy Magnitude m2 = E2 - (px)2 - (py)2 - (pz)2 = 39.0625 - 1.5625 - 2*6.25 = 25 , so
Calculated Momenergy Magnitude m = 5 kg mass
¶ Calculate Momenergy Magnitudes m in different frames (STP2 Fig. 7.3):
GIven E = 101 and px = -99, the Momenergy Magnitude m = (1012 - (-99)2 ) 1/2 = (400) 1/2 = 20 units (e.g., kg)
GIven E = 25 and px = 15, the Momenergy Magnitude m = (252 - (15)2 ) 1/2 = (400) 1/2 = 20 units (e.g., kg)
GIven E = 52 and px = 48, the Momenergy Magnitude m = (522 - (48)2 ) 1/2 = (400) 1/2 = 20 units (e.g., kg)
As we select different frames in which to view momenergy, the component values of E and px change. (so that an attempt to graphically display the momenergy 4-vector appears to lead to different vector directions), but the momenergy magnitude does not.
¶ Calculate Energy, Momentum, and Kinetic Energy from Momenergy magnitude m (Rest energy) and velocity v (STP2 problem 7-2):
Given m = 3 kg, Δx = 8 m, Δt = 10 m dtl, Δy = Δz = 0, calculate kinetic energy:
v ≡ Δx/Δt = 8/10 = 0.8c
γ = 1.666667
E ≡ Etot = mγ = 5 kg
p = px = mvx γ = 5 * 0.8 = 4 kg
Erest ≡ m = 3 kg
Erest-conv = mc2 = 2.69627 x 1017 kg m2 s-2 = = 2.69627 x 1017 J (because 1 J = 1 kg m2 s-2)
K (Kinetic Energy) ≡ ETot - Erest ≡ E - m = 2 kg
Kconv-units = 2 kg x c2 = 1.79751 x 1017 kg m2 s-2 = 1.79751 x 1017 J
Note that for the same v and m, the classical Newtonian KE would be mv2 /2 = 0.32m = 0.96 kg. Therefore, at such high speeds, relativistic KE is considerably larger than the value predicted by Newton.
Given E and p, note that
|v| = |p| / E = 4/5 = 0.8 as already calculated.
¶ Calculate Energy from Rest Energy m and Kinetic Energy Or Momentum
in lab frame for a particle with mass m (STP2 problem 7-3):
• a. If the particle is moving along x-direction and K = 3m:
K ≡ ETot - Erest , so
ETot = Erest + K = m + 3m = 4m
E2 - px2 = m2 , so
px2 = E2 - m2 =16 - 1 = 15 or
px = (15)1/2 m = 3.873m
Then the momenergy 4-vector components are expressed as (4m, (15)1/2m, 0, 0), which has a momenergy magnitude of m.
• b. If the same particle is observed from a rocket frame in which K = m
K ≡ ETot - Erest = m, so
ETot = Erest + K = m + m = 2m
E2 - px2 = m2 , so
px2 = E2 - m2 =4 - 1 = 3 or
px = (3)1/2 m
Then the momenergy 4-vector components are expressed as (2m, (3)1/2m, 0, 0), which has a momenergy magnitude of m.
• c. Instead, the particle moves in the y-direction in the lab frame with py = 2m
py = 2m , px = 0, pz = 0
So p2 = 22 m2
E2 = m2 + p2 = 1 + 4 = 5 m2
Then the momenergy 4-vector components are (51/2 m, 0, 2m, 0), which has a momenergy magnitude of m.
• d. Instead, the particle moves in the negative x-direction in the lab frame with E = 4m
px2 = E2 - m2 = 42m2 - m = 15m2
py = 0 , pz = 0
Then the momenergy 4-vector components are (4m, -(15)1/2m, 0, 0), which has a momenergy magnitude of m.
• e. Instead, the particle moves in the lab frame with px = py = pz and with K = 4m
K ≡ ETot - Erest
4m = ETot - m
ETot = 5m
p2 = E2 - m2 = 25m2 - m2 = 24m2
3px2 = 24m2
px= py= pz = (8)1/2m
Then the momenergy 4-vector components are (5m, (8)1/2m, (8)1/2m, (8)1/2m), which has a momenergy magnitude of m.
¶ Momentum and Energy for System Of 2 particles in Elastic Collision (STP2 p. 207)
• a. GIven Particle A has E = 13 units of energy, px = -5, Particle B has E = 17, px = +15, and all y and z momentum components are 0. Then
Momenergy Magnitude for A = mA = (132 - (-5)2)1/2 = (169 - 25)1/2 = (144)1/2 = 12 units
Momenergy Magnitude for B = mB = (172 - (15)2)1/2 = (289 - 225)1/2 = (64)1/2 = 8 units
Momenergy Magnitude for A + B = msys = ((13+17)2 - (-5+15)2)1/2 = (900 - 100)1/2 = (800)1/2 = 28.284 units
Note that the Momenergy Magnitude for A + B = msys ≠ mA + mB.
Thus momenergy magnitude for the system is not simply the sum of the individual rest/invariant masses.
• b. Given that after colliding, we now have
Momenergy Magnitude for A = mA = (202 - (16)2)1/2 = (400 - 256)1/2 = (144)1/2 = 12 units (unchanged, thus conserved)
Momenergy Magnitude for B = mB = (102 - (-6)2)1/2 = (100 - 36)1/2 = (64)1/2 = 8 units (unchanged, thus conserved)
Momenergy Magnitude for A + B = msys = ((20+10)2 - (16-6)2)1/2 = (900 - 100)1/2 = (800)1/2 = 28.284 units (unchanged, thus conserved)
To summarize, Momenergy Magnitude for each particle (corresponding to their individual rest/invariant masses) and also for the system is conserved, but the momenergy magnitude for the system is not simply the sum of the individual rest masses.
¶ Conversion Of Kinetic Energy To Mass In An Inelastic Collision (STP2 problem 7-3)
Two trains A and B each have mass 5x106 kg, and collide while each traveling at 42 m/s in opposite directions, coming to rest. The isolated system in this problem includes the tracks and roadbed, and we ignore the radiation and sound produced that exits the system.
• a. The Newtonian KE of each train is mv2 /2 = 4.41x109 J = 4.91x10-8 kg = 4.91x10-2 mg. Note that at these low speeds, the Newtonian formula for KE is adequate.
• b. With the inelastically crashed trains now at rest, how much mass does the train wreck increase by?
Net momentum pre and post collision is zero.
Momenergy msys-pre = Momenergy msys-post , (by conservation of msys) , so
((Precollision Energy components)2 + 02) = ((Postcollision Energy components)2 + 02) , so
ΔEtot = (Precollision Total Energy components) – (Postcollision Total Energy components) , so
ΔEtot = EA-rest--pre + KA-pre + EB-rest-pre + KB-pre + Erest-track and roadbed-pre + Ktrack and roadbed-pre
– EC-rest-post + KC-post + E-rest-track and roadbed-post + Ktrack and roadbed-post , so
Ktrack and roadbed-pre = Ktrack and roadbed-post = KC-post = 0
ΔErest = (EC-rest-post + E-rest-track and roadbed-post ) - (EA-rest-pre + EB-rest-pre ) = - (0 - KA-pre + KB-pre ) = -2 x (-4.91x10-2 mg) =+ 9.81x10-2 mg
Conclusion: there is no longer any K energy, and this former K energy has been added to rest energy Erest of the system particles (wreck, track, and roadbed) in the form of heat and deformation strain, etc. However, the momenergy magnitude msys is unchanged and conserved!:
msys = ((EA-rest--pre + KA-pre + EB-rest-pre + KB-pre + Erest-track and roadbed-pre - 0 )2)1/2 = ((EC-rest-post + E-rest-track and roadbed-post )2)1/2
¶ Weight of Heat Energy: Efforts to weigh the increase in mass resulting from adding heat created for instance in collisions have not yet succeeded (STP2 p. 223). Heat is a system property. Here are some fractional increases in system mass produced by heating (STP2 p. 228):
• Heat bath water from 15 °C to 40 °C: Fractional increase is 10-12
• Separate all molecules of Earth to infinite separation: Fractional increase is 10-9
• Ionize hydrogen atom to infinite separation of proton and electron: Fractional increase is 10-8
• Split deuteron into proton and neutron: Fractional increase is 10-3
¶ Calculate Momenergy components (E, px , py , pz ) (STP2 problem 7-1):
• a. If a particle is moving along x-direction with E = 5m:
px2 = E2 - m2 =(25 - 1)m = (24)1/2 m or
Momenergy 4-vector = (5m, (24)1/2 m, 0, 0)
• b. Same but observed in particle's rest frame
px2 = 0 = E2 - m2 so E = m
Momenergy 4-vector = (m, 0, 0, 0)
• c. If a particle is moving along z-direction with p = 3m
pz2 = p2 = E2 - m2
E2 = p2 + m2 = (9+1)m2
E = (10)1/2 m
pz 2 = p2 = 32m2
Momenergy 4-vector = ((10)1/2, 0, 0, 3m)
• d. If a particle is moving along negative y-direction with K = 4m
E = m + K = m + 4m = 5m
py2 = p2 = 52m2 - m2 = 24m2
py = -(24)1/2
Momenergy 4-vector = (5m, 0, -(24)1/2m, 0)
• e. If a particle has E = 10m and px : py : pz are in ratio 1:2:3
p2 = E2 - m2 = 102m2- m2 = 99m2 so p = (99)1/2
By Pythagorean theorem
14px2 =99 so px2 =99/14 and px = 2.659m, pY = 5.318 m, pZ = 7.978m
Momenergy 4-vector = (10m, 2.659m, 5.318m, 7.978m)
¶ Particles collide elastically in isolated system (STP2 problem 7-2)
Given 2 Particles having initial EA = 13, pxA = -5, EB = 17, pxB = 15
Initial individual invariant masses: mA = 12; mB = 8
The starting system invariant mass is
msys = ( (13+17)2 + (-5+15)2)1/2 = 28.28 units
This is not the sum of the individual particle invariant masses (20).
After elastic collision, particles have final EA = 13, pxA = -5, EB = 17, pxB = 15.
msys = ( (20+10)2 + (-6+16)2)1/2 = 28.28 units
Conclusion: The invariant mass is conserved in the collision and does not change.
¶ Relativistic Protons: Proper Speed, Momentum, Energy, γ, proper and lab time (STP2 problem 7-4)
Constant velocity protons emit flashes of light each 1 meter of light travel time (proper time) Δτ. Rest/invariant mass of proton m = 0.938 GeV. Do not solve for v.
|Lab Distance Δx traveled
by proton between flashes
= mrest dx/dτ
= (mrest2 + p2)1/2
|Lab time Δt
|0 meter||0 GeV||0.938 GeV||1||1 meter|
|0.1 m||0.0938 GeV||
|1 m||0.938 GeV||1.33 GeV||1.414||1.414 m|
|5 m||4.69 GeV||4.78 GeV||5.099||5.099 m|
|10 m||9.38 GeV||9.43 GeV||10.05||10.05 m|
|103 m||938 GeV||938 GeV||1000||1000 m|
|106 m||938000 GeV||938000 GeV||1000000||1000000 m|
Conclusion: It is apparent that as speed increases, momentum and energy become equal because mrest becomes insignificant by comparison.
¶ Fast Electrons at SLAC: E, v and proper length (STP2 problem 7-6)
Linearly accelerated electrons go from 0 to 47 GeV in 3000 m. For electron, rest/invariant mass mrest = 0.511 MeV.
• a. What is energy gain per meter (ΔE/Δx)?
Gain = 47 GeV / 3000 m = 15.6667 MeV/m
For Newtonian KE, at light speed v = c and KENewt = 1/2 mc2 = 1/2 * 0.511 MeV = 0.256 MeV, so
Distance to achieve this KENewt = ΔE / (ΔE/Δx) = 0.256 MeV / 15.6667 MeV/m = 0.016 meter
• b. Estimate 1 - v:
Actual final speed v is given by E = mγ so
γ = E/m = 47 GeV/0.511 MeV = 91980
γ = 1/(1-v2)1/2 = 91980 so 1/(1-v2)= 919802 so 1/(1-v)(1+v)= 919802 but v≈1 so
1/2(1-v)= 919802 so
v = 1 - 1/2*919802 = 1 - 5.91x10-11 = 0.9999999999409
In traveling 12740 km through Earth, light requires t = 0.0425 s, so discrepancy in distance traveled by the electron vs. light over this distance is
- 5.91x10-11 x 0.0425 s X c (m/s)= 0.753 mm less
• c. Estimate proper length of SLAC to a 47 GeV electron: L = L0 /γ = 0.033 m
¶ Formation of Deuterium from Hydrogen and Neutron: (UR p. 222)
The deuterium isotope (1H-2) has m = 2.014102 u, whereas the ordinary hydrogen isotope (protium 1H-1) has m = 1.007825 u and the neutron has m = 1.008665 u. Therefore, the mass defect in forming deuterium is 0.002388 u. The missing mass (mass deficit or defect) δ = 2.22 MeV represents the binding energy Eb for the deuterium nucleus (deuteron) that must be provided to separate the two nucleons, where δ = Eb /c2. This binding energy has been experimentally confirmed.
If we attempt to create a deuteron from the collision of a proton and neutron, the reaction must give off a gamma ray in order to balance the momentum conservation requirement (according to UR p. 235). This is called radiative capture.
¶ Radioactive Decay: E, p, m, and mass deficit or defect (STP2 problem 7-8)
A lone particle A that is at rest with invariant mass mA = 20 decays into particles D and C moving along the x-axis and having mC = 2 and EC = 5 [there is no B].
• a. What is EA of A? 202 = EA2 - pA2 and pA=0 so
EA = 20
• b. What is ED of D? EA = EC + ED so
ED = EA - EC = 20 - 5 = 15
• c. What is pc ? EC2 - pC2 = mC2 so pC2 = EC2 - mC2 = 25 - 4 so
pC = -(21)1/2
• d. What is pD? pD = - pC by cons. of momentum so
pD = (21)1/2 (the 2 products have equal but opposite momenta)
• e. What is mD ? ED2 - pD2 = mD2 so mD2 = ED2 - pD2 = 152 - 21 so
mD = (204)1/2 = 14.283
• f. Compare pre and post collision invariant masses:
mA = system invariant mass predecay = 20.
mC + mD = 2 + 14.283 = 16.283 postdecay (so mA > mC + mD)
Interpretation: The missing mA mass/energy or mass deficit or defect Δm has gone into Kinetic energy K in the products (assuming there is no radiation or transfer of heat outside the system). Increased K indicates the reaction is exoergic. That is, Kf = Ki + Q where Ki is initial kinetic energy, Kf is final kinetic energy, and Q = -(Δm)c2 is the Q value of energy for this reaction, Q being a positive value when Δm is negative. (UR p. 220)
System invariant mass postdecay = (202 - 02)1/2 = 20 predecay, unchanged
¶ Relativistic Particle Decay
If a particle "A" at rest (such as a K+ ) decays into particles "B" and "C" (such as a π+ and π- respectively), the resulting particles B and C travel in opposite directions along some axis in order to preserve the momentum = 0. According to UR p. 224, the velocities of C and D are given by their γ's with respect to A's rest frame as follows:
mBγB = MA/2 + [ (mB2 - mC2) / 2mA ]
mCγC= MA/2 + [ (mC2 - mB2) / 2mA ]
If A is moving initially in lab frame, do the calculation in the rest frame of A and then do a LT to obtain the final results in lab frame. However, B and C cannot be expected to move along the x-axis defined by A's movement in lab frame.
¶ Particles collide inelastically in isolated system (STP2 problem 7-9)
Given 2 Particles A and B that collide in lab frame to form C which is at rest, and with initial mA = 2, EA = 6, mB = ?, EB = ?, mC = 15, EC = ?
• a. What is EB of B?
mC = 15 = (EC2 - pC2)1/2 = EC so
EC = 15
By conservation of energy, EA + EB = EC so
EB = 15 - 6 = 9 units
• b. What is pA of A?
EA2 - pA2 = mA2 so
pA = (62 - 22 )1/2 = √32
Momentum is conserved, and is 0 after collision, so pA = -pB , so
pB = - √32
• c. What is mB of B?
EB2 - pB2 = mB2 so
mB = (92 - 32)1/2 = 7
• d. Compare mC to mA + mB:
Precollision msys = ((∑E)2 - (∑p)2)1/2 = [ (6+9)2 - (0)2 ] 1/2 = 15
Note that these E components consist of rest energy plus kinetic energy subcomponents for each particle. Also,
mA + mB = 2 + 7 = 9
Postcollision msys = ((E)2 - (p)2)1/2 = [ (EC )2 - 0 ] 1/2 = EC = 15 (by conservation of invariant system mass), and so since mC= msys
mC = 15
Conclusion: Although invariant system mass msys is unchanged by the collision, the sum of the rest masses of each particle considered separately mA + mB = 9 is < the rest mass of the final combined particle mC. The smaller precollision rest mass compared to postcollision rest mass is accounted for by the precollision kinetic energy that is converted to additional system rest energy = mass.
¶ Balls, only one moving, collide inelastically in isolated system (STP2 problem 7-10)
Given putty ball A of rest mass mA with kinetic energy KA sliding on ice, collides with identical ball B at rest, and they stick together and continue moving as putty ball C in the positive x-direction. Let m = mA-rest = mB-rest
• a. What is ETot-sys prior to the collision?
ETot-sys = EA-rest + KA + EB-rest = m + KA + m = 2m + kA = ETot-C = EC-rest + KC (by energy conservation)
• b. What is momentum pA of A?
EA2 - pA2 = m2 so
pA2 = EA2 - m2 = [ (EA-rest+ KA)2 - m2 ]
pA = [ (m + KA)2 - m2] 1/2 = [ m2 + 2mKA + KA2 - m2] 1/2 = [ 2mKA + KA2] 1/2 in the positive direction
psys-tot-pre = pA because pB = 0
pC = psys-tot-pre = pA = [ 2mKA + KA2] 1/2
• c. What is mC ?
ETot-sys = ETot-C = (kA + 2m)2 - 2mKA + KA2
mC2 = (KA + 2m)2 - pA2 = KA2 +4mkA + 4m2 - 2mKA - KA2 = (2m)2 + 2mk = (2m)2 (1 + KA /2m)
• d. What are limiting values of final mass at high and low velocity?
For low velocity limit, k/m << 1 and mC2 = (2m)2 or mC= 2m (as expected, the Newtonian result)
For high velocity relativistic limit, k/m >> 1 and mC2 = (2m)2 (K/2m) = 2mK or mC= (2mK)1/2
Therefore for high velocity, mC can increase without limits as v approaches c and K rises without limit.
¶ Relativistic Elastic Scattering In 2 Dimensions (Special Case of Symmetric Scattering) (STP2 p. 240)
Given Particle A with mass m, energy Ea, momentum pa traveling along the x-axis It collides with stationary particle b with mass = Eb = m and pb = 0. Assume in this case that both a and b scatter through equal angles θ/2 with respect to the x-axis (so that the angle between their scatter paths is θ). In other words, assume symmetrical scattering results—a special case—and also that alternative processes such as creation of new particles do not occur.
• a. What is final kinetic energy of the particles?
Ea = m + Ka and the masses of the 2 particles are conserved
Because of conservation of Energy,
Ea + Eb = Ec + Ed
K before reaction is Ka
K after reaction is Kc + Kd = Ka so
Kc = Kd =Ka/2 (K has become evenly divided between the two scattered particles due to the symmetry assumed)
• b. What is final momentum of the particles?
By conservation of momentum, the total momentum was simply pa and there was initially no py so the y-axis components of final momenta must cancel. Then
ptot = pa = pc cos θ/2 + pd cos θ/2 = 2pc cos θ/2 , or
pc = pa / 2(cos θ/2)
• c. What is final angle θ between the particles?
Using momenergy, E, and p magnitudes:
pa = (E2 - m2)1/2 = ((Ka + m)2 - m2)1/2 = (Ka2 + 2Kam + m2 - m2)1/2 = (Ka2 + 2Kam )1/2 so
cos2 (θ/2) = (K + 2m) / (K + 4m) or
but since cos2 (θ/2) ≡ (cos θ + 1)/2
cos θ = (K/m) / [ (K/m) + 4]
Conclusion: The angle between the symmetrically scattered particles depends on the ratio of K/m.
• When K/m is small and approaches 0 as with low velocity non-relativistic impact, cos θ approaches 0 radians so θ = 90 degrees, the Newtonian limiting case. Note that in classical Newtonian elastic scattering between particles of equal mass where one that is traveling at low velocity impacts the other at rest, as with billiard balls, the overall scattering angle θ formed between the elastically scattered objects is always 90 degrees (though the individual scattering angles may not be equal with respect to the original direction of travel.)
• When K/m is high (indicating higher velocity of impact), the value of cos θ approaches 1 so θ = 0 degrees. In other words, at increasing relativistic velocities, the overall scattering angle is smaller and smaller and the scattered particles scatter into nearly the same x-axis direction when impacting. (This forward concentration of scattering at relativistic velocities seems reminiscent of beaming seen with relativistic aberration.) The angle of scattering (for example as seen in early cloud chamber experiments reported for electrons by FC Champion, see below) could therefore be used to estimate initial kinetic energy.
Note: A more general formula for unequal scattering angles formed in "relativistic billiards" is derived in Rindler's text cited below (p. 86-7), namely:
♦ tan θ • tan φ = 1/γ2 ,
where θ and φ are the respective scattering angles in the lab frame with respect to the incident particle path and γ is computed for the COM frame. Note that when the combined angles are 90 degrees, tan θ tan φ = tan θ • cot θ = 1 (the classical case), but since γ > 1, the product tan θ • tan φ decreases with increasing γ and therefore collision velocity.
Photons have no rest mass, but do have E and p, and travel at light speed. To recap properties listed above:
ν (Hz) = frequency = c / λ
λ (m) = wavelength = c / ν
E (kg m2 s-2) = energy = hν = hc / λ
m = mass = 0
p (kg m s-1) = momentum p = hν / c = E / c
p = E (kg, when c = 1)
The quantum particle nature and momentum of the photon was verified beyond reasonable doubt by the demonstration of Compton scattering of X-rays by electrons by Arthur Compton reported in 1923. The scattered photons have altered wavelength that is a function of the scattering angle and the mass of the electron me, given by
♦ λpost - λpre = h/mec (1 - cos θ) = 2.43 x 10-12 (1 - cos θ)
a result that does not vary much for a variety of materials (consistent with similar orbital electrons being the objects causing the scattering). A proof of this relationship is given here.
¶ Examples of photon properties include (STP2 p. 229):
• AM Radio photon at 1020 KHz: f = 1.02 x MHz; λ = 294 m; E = p = 6.86x10-28 J = 4.22 x 10-9 eV
• Infrared at λ = 10-4 m (100,000 nm): f = 3 x 1012 Hz; λ = 10-4 m; E = p = 1.99 x 10-21 J = 1.24 x 10-2 eV
• Ultraviolet at λ = 2.5 x 10-7 m (250 nm): f = 1.2 x 1015 Hz; λ = 2.5 x 10-7 m; E = p = 5 eV
• Gamma rays from proton+antiproton annihilation: f = 2.3 x 1023 Hz; λ = 1.32 x 10-15 m; E = p = 0.938 x 109 eV
The spacetime Lorentz interval for events produced by photons is always 0. Its momentum and energy are equal, so invariant mass (rest mass) for a photon is always 0. Because they have momentum and energy, they can alter the invariant system mass of a system having matter, or even a system consisting only of photons.
¶ Photon Backscattering Example: an orbital electron (which may be considered essentially stationary compared to speed of light c) with E = mrest = 1 is collided by a photon with E = p = 2 and m = 0. Together they make up a system with msys = ((2+1)2 + 22)1/2 = √5. After backscattering 180° by Compton scattering, the photon still has mass 0, and for the photon -p = E = 0.4, and the electron has m = 1, E = 2.6, and p = +2.4. The msys = (32 - 22)1/2 = √5, unchanged (STP2 p. 231).
¶ Photon Systems & Collisions (STP2 p. 232):
• Electron stationary with mass m = me , photon collides with E = 3me:
Then msys = (16 - 9)1/2 = √7
• Photon with E = 3, 2nd photon traveling same direction with E = 1:
Then msys = (16 - 16)1/2 = 0
• Photon with E = 3, 2nd photon traveling opposite direction with E = 1:
Then msys = (42 - (3-1)2)1/2 = √12
• Photon with E = 3, 2nd photon traveling at right angles with E = 1:
Then msys = (42 - 10 )1/2 = √6. Here we must consider the 3 components of the vector momentum.
¶ Pair Production From Photons: Photons such as cosmic rays having a suitable high energy (> 2 x 0.511 MeV) can create mass, specifically another electron plus a positron (pair production), when passing adjacent to a stationary electron [pair production more typically occurs with an atomic nucleus]. For example (STP2 p. 233), a photon has E = p = 4em and of course m = 0, whereas the electron has mass me = 1. In the pair production interaction, the photon is annihilated and a "polyelectron" (is this the positronium negative ion Ps- ?) is created with 2 e- and 1 e+, having p = 4, E = 5, and msys = (52 - 42)1/2 = 3. The relativistic velocity of the polyelectron is given by v = |p| / E = 4/5 = 0.8c. A substantial portion of the photon energy ends up as K in the polyelectron.
Because of momentum It is impossible to produce a pair from a solitary photon in empty space in the absence of additional matter to interact with (STP2 ex. 8-11): In the center-of-momentum frame for the alleged pair produced, the total linear momentum of the system is by definition zero (the pair particles diverge in opposite y-directions, so net y-momentum remains 0, and their net momentum in x is by definition also 0. Yet the incoming photon in this COM frame still has net x-momentum regardless of the frame's relative velocity. Therefore the pre-collision momentum cannot equal the post-collision momentum and the reaction is impossible.
¶ Solar Photon Pressure: Because photons carry momentum, light exerts a pressure when totally absorbed, and double that pressure when reflected back 180 degrees (because the change in momentum is doubled). The power of light falling on the Earth's atmosphere is expressed by the solar constant of 1366 W m-2. The resulting pressure from changing photon momentum is calculated as follows (after STP2 exercise 8-3):
Force F (kg m s-2 ≡ N) = Pressure (N m-2) x Area (m2) = Change of momentum Δp per unit area x Area force is exerted over
But Δp = ΔE/c when photons are fully absorbed but not reflected.
Then Force for totally absorbing surface on 1 m2 is:
FTotAbs = P x A = 1366/c N m-2 x 1 m2 = 1366/c N = 1366/299792458 N = 4.6 x 10-6 N
Solar pressure for total absorption = FTotAbs / Area = 4.6 x 10-6 N / m2 = 4.6 x 10-6 N / m2 ≡ Pa
For totally reflecting surfaces:
FTotRefl = 2 x 1366/c N m-2 x 1 m2 = 9.1 x 10-6 N m-2
Solar pressure for total reflection = FTotRefl / Area = 9.1 x 10-6 N / m-2 ≡ Pa
The force is independent of the wavelength and only depends on the rate of energy absorption per unit area.
¶ Gravitational Redshift of Photons: (STP2 exercise 8-6) The Newton gravitational force is given by F = GM1 M2 / r2. The work required to raise a test mass M2 against this force is from r to r + dr is
dWconv = GM/c2 dr/r2 = M* dr/r2 where M* is the mass of the center of attraction translated to units of meters. For Earth, M* = 4.44 x 10-3 m. The work required to move a test particle from r to infinity is simply M* / r. The fractional energy loss for a photon that rises from a spherical astronomical object of mass M (kg) or M* (m) ≡ GM/c2 , beginning at radius r, is given by
♦ z = Δf / f = -M*/r = - GM/c2 r.
An exact formula for the Gravitational Redshift derived from GR is presented here as follows for a nonrotating uncharged mass M:
♦ z = Δf / f = [1 / (1 - (2GM/rc2) ) ] - 1
Note that frequency decreases and is thus said to be "redshifted" whether or not in the visible spectrum, analogous to Doppler redshift.
• For the Earth, the calculated redshift to lift photons from the surface to infinity is z = - 4.44 x 10-3 m / 6371000 m = -6.96908 x 10-10. Light at 500 nm would be redshifted by -3.5 x 10-7 nm, which would be difficult to detect (except by Mössbauer spectroscopy, see below). However, gravitational redshift affects the timing signals of the Global Positioning System, and would cause a significant error of 38 microseconds per day if not corrected for.
• For the Sun, z = -2.13 x 10-6 and the shift in wavelength for 500 nm light would be -0.001 nm, still difficult to detect.
Gravitational redshift (actually blueshift) on Earth was demonstrated by the Pound-Rebka experiment, performed in 1959, using Mössbauer spectroscopy on 14 keV gamma rays emitted by iron-57. Their frequency shifts were on the order of 10-15 due to a much shorter path (74 feet) over which the shift in the descending gamma rays was measured.
¶ Gravitational Time Dilation: The gravitational redshift is a GR effect related to gravitational time dilation: Clocks close to a massive body run slow compared to ones that are farther away. This fact makes it difficult to synchronize clocks in a strong gravitational field, and furthermore causes the measured value of the speed of light c to vary according to the strength of the gravitational field. (UR p. 262) The effect on the clocks of the GPS navigational system may be described as resulting from this phenomenon.
¶ Photon Splitting: If it were possible for a photon to split into 2 photons, could the resultant photons be traveling in any other direction than the direction of the parent photon? Call this direction the x-axis and the angles formed by the new photons B and C with respect to the x-axis is θB and θC (STP2 ex. 8-10).
The incoming photon has EA = pA (true for all photons).
For the outgoing photons B and C, the component of momentum in the y-direction must add to 0, thus pB sinθB + pC sinθC =0 so
pB sinθB = - pC sinθC
The sum of the components of momentum in the x-direction must equal that of the incoming photon:
pA = pB cosθB + pC cosθC
Because of energy conservation and identity of E and p for photons:
EA = pA = EB + EC = pB + pC
pB cosθB + pC cosθC= pB + pC
This can only be true for cosθB = cosθC = 1 so θB= θC= 0 degrees. (This solution might need refinement.)
¶ Colliding Photons: Two gamma ray photons (1 and 2) collide, having unequal energies E1 = p1 and E2 = p2 and produce an electron-positron pair (particles 3 and 4). (STP2 ex. 8-12). The rest energy of 3 and 4 is 0.511 MeV x 2 = 1.022 MeV, and these particles also have kinetic energy. The incident gamma particles must have the minimum total energy = 1.022 MeV to end up producing stationary products 3 and 4. (I did not find time to complete this solution.)
Relativistic Doppler Along X-Axis By Photon LT: The LT can be used to derive the previously given formula for relativistic Doppler using photon energy E (STP2 p. 263). The photon energy in rocket frame E' moving at speed v along x-axis in positive direction away from the lab frame is given compared to lab frame value E by:
E = γE' + vγE' = γ(1+v) E' = (1+v)/(1-v2)1/2 E' = (1+v) / [ (1-v)(1+v )]1/2 E' = (1+v)1/2 / (1-v)1/2 E' or
♦ E = [(1 + (v/c)) / (1 - (v/c) )]1/2 E'
Since E = hf,
♦ f = f' [(1 + (v/c)) / (1 - (v/c) )]1/2 (for movement away from the observer, in positive x-direction) , or
♦ f = f' [(1 - (v/c)) / (1 + (v/c) )]1/2 (for movement toward the observer, in negative x-direction)
It is also possible for a proton impacting another proton to create an additional proton plus an antiproton (STP2 p. 236). For instance, a proton with mass = 1 has energy = 7, and momentum p = √48. The velocity v = |p| / E = (48)1/2 / 7 = 0.9897. It collides with another proton at rest, E = m = 1. The system invariant mass msys = ((7 + 1)2 - 48)1/2 = 4. The resulting particles (the original 2 protons plus a newly created proton and antiproton) have a relativistic velocity in the same direction as the impacting proton and stay together when the impacting proton is just at a threshold energy value. The final velocity is given by |p|/E for the bound 4 particles = |p| / E = (48)1/2 / 8 = 0.866, which is considerably slower than the impacting proton.
Stable and Unstable Nuclei: Nuclear stability is a relative thing. The stable isotopes 28Ni-62, 26Fe-58, and 26Fe-56 have the most tightly bound nuclei, with binding energies of about 8.8 MeV / nucleon. (26Fe-56 has the lowest mass per nucleon at 0.998838 amu/nucleon, slightly lower than 28Ni-62 with 0.998844 amu/nucleon, but 28Ni-62 has the highest binding energy of any isotope at 8.7946 Mev / nucleon compared to 8.7922 Mev / nucleon for 26Fe-56. This difference is because Ni-62 has proportionately more neutrons than Fe-56 and they are more massive—see also here and here and graph STP2 p. 238). This means that 8.8 MeV of energy must be input to split out a single nucleon from one of these isotope's nucleus. Isotopes that have smaller or larger numbers of nucleons compared to iron, and much lower binding energies than iron, are more likely to release energy, either by fission or fusion reactions. For instance, the deuterium nucleus has a binding energy of only 1.11 MeV /nucleon. (By way of comparison, the energy to chemically ionize the single electron in the n=1 ground state from the 1H-1 atom is only 13.6 eV.)
(Note that I am unable to show both superscripts and subscripts in the approved manner with both preceding the element abbreviation—instead, I have shown the number of protons as a preceding superscript, and number of neutrons + protons is shown to the right of the element abbreviation. E.g., for isotope 3Li-6, 3 is the number of protons (atomic number), and 6 is the total number of neutrons and protons.)
Nuclear Fusion reactions on Earth and in stars preferentially involve small nuclei, especially 1H-1 (hydrogen), 1H-2 (deuterium), 1H-3 (tritium), 2He-3, 3Li-6, 5B-11, but in stars Be and larger nuclei including 12C, 13C, 13N, 14N, 15N, and 15O are also formed, and larger and much larger nuclei synthesized in nucleosynthesis of supernovas and other cosmic reactions). Fusion reactions are initiated at high temperatures (e.g., 800 million K or more) and release large amounts of heat and radiation (generally exothermic for small elements < iron) or absorb energy (generally endothermic for large elements > iron). The mass deficit or deficit resulting from combining 2 n and 2 p to form a 2He-4 nucleus is approximately Δm= 0.0304 u which gives a binding energy of 28.3 MeV—this is the amount of energy that is released as heat and radiation to allow the binding to occur. The energy release per nucleon of fusion reactants is about 7 MeV/nucleon, substantially larger than for nuclear fission but smaller than for nucleon annihilations. I have summarized a history of thermonuclear fusion weapon development here.
Solar Fusion Energy Output (STP2 p. 242 and extended): The Sun's total energy received at the Earth's outer atmosphere in a plane perpendicular to the direction of the Sun is variously estimated at about 1366 W m-2 by satellite measurement. This quantity is called the solar constant, though it is not actually constant. The amount of solar radiant energy falling on the Earth's cross-sectional area of 1.28 x 1014 m2, not all of which is absorbed, is 1366 x 1.28 x 1014 m2 = 1.74 x 1017 W = = 1.74 x 1017 J s-1. This amounts to a mass conversion rate of 1.94 kg/s for all solar radiation received at Earth. Total solar radiation emission can be estimated by multiplying by the ratio of area presented by Earth to the total spherical area at 1 A.U. = 1.4960×1011 m. The ratio of total spherical area at 1 A.U. to area intercepted by Earth = 2.2 x 109 so the solar mass conversion rate translating into power output is 4.3 x 109 kg s-1. Since the solar mass is 1.989 x 1030 kg, this conversion rate is only 2 x 10-21 of the solar mass converted each second = 7 x 10-5 of the total mass converted per Gy at the current rate.
Nuclear Fission: Fission reactions occur with large nuclei such as U, Pu, Th, etc. A nuclear fission reaction used in fission weapons is
n + 92U-235 ⇒ 92U-236 ⇒ 37Rb-95 + 55Cs-141 + several n + heat and radiation such as gamma and X-rays
where n is a neutron. The large fission fragments and smaller species including neutrons together have the same number of neutrons and protons as the reactants. Typical fission events release about 200 MeV of energy for each fission event. The energy release per nucleon of fission reactants is about 0.8 MeV, considerably smaller than for fusion reactions. For comparison, chemical oxidation events release only a few eV per event.
Annihilation: These reactions involve matter-antimatter pairs such as protons-antiprotons, electrons-positrons, etc. Annihilation reactions release the most energy per mass of reactants because 100% of the matter is converted to energy (photons, heat, etc.) For example, the energy release in the annihilation of a proton-antiproton pair is 1.88 GeV, or about 0.9 GeV per nucleon of reactants. The annihilation of an electron-positron pair releases about 1.0 MeV which is about 1.8 GeV per atomic mass unit of reactants. Electron-positron pairs must emit at least 2 photons when annihilating (typically 2 photons in opposite directions if the starting reactants are essentially stationary), so that the reactants and the products add to the same net momentum vector. (STP2 p. 238)
¶ Example (STP2 p. 242): When a positron of mass m and K = m traveling along the axis collides with an electron of mass m and they annihilate forming 2 high energy photons, the sum of the energies (mass + kinetic energy) is conserved as is the total momentum. The vertical net momentum component must be zero as there was no component in this direction initially. Assume one photon is scattered at 90 degrees, and the other photon is scattered at angle θ with respect to the x-axis?
These relationships yield:
pa = Ec cos θ
Ed = Ec sin θ
Beta Decay: While most reactants produced by nuclear reactions and annihilations are readily detectable, the electron neutrinos and antineutrinos produced by beta decay are not. (There are also Muon and Tau neutrinos, not discussed here.) Neutrinos were originally postulated by Wolfgang Pauli in 1930 when energy and momentum for beta decay reactions were otherwise not conserved. Beta decay reactions consist of the following net reactions:
n (neutron) ⇒ p+ (proton) + e- (electron) + electron antineutrino , and
energy + p+ (proton) ⇒ n (neutron) + e+ (positron) + electron neutrino
When performing energy and momentum balance calculations for beta decay reactions involving neutrinos, the energy and momentum carried away by the neutrinos must be accounted for. The mass of the electron neutrino is currently thought to be nonzero but < 2.2 eV. (STP2 p. 239 again states that neutrino mass is zero and that p = E for neutrinos, but more recent work apparently indicates these statements are only approximately true.) A typical beta decay is:
6C-14 ⇒ 7N-14 + e- + antineutrino
in which the energy of the electron created varies considerably depending on how much is imparted to the antineutrino.
¶ Nuclear Excitation By a Gamma Ray
(Solution revised 10/2014 per notes of Michael Kiewe, University of Wisconsin - Madison, link no longer working)
A nucleus of mass m at rest absorbs an incoming gamma ray with energy Ep
and is excited to a higher energy state with mass 1.01m = M. (STP2 Ex 8-8)
In the process, the nucleus acquires a velocity v of in the rest frame.
This idealized problem ignores the electrons of the atom.
The Ep of the incoming photon is given by energy conservation as follows:
m + Ep = Mγ
m + Ep = M / (1 - v2)1/2
Then v = [1 - (M2/(Ep + m)2) ]1/2
By momentum conservation:
Ep = Mγv
Ep = (m + Ep)v
Ep = (m + Ep)[1 - (M2/(Ep + m)2) ]1/2
Ep2 = (m + Ep)2[1 - (M2/(Ep + m)2) ]
Ep2 = (m + Ep)2 - M2
Ep2 = m2 + 2mEp + Ep2- M2
0 = m2 + 2mEp - M2
Ep = (M2 - m2) / 2m = [(1.012 - 1) / 2 ] m = 0.01005m
The change in mass of the nucleus is 0.01 m but the photon had to have energy 0.01005m. The difference is due to conversion of some of the energy into kinetic energy of the nucleus.
¶ Nuclear Emission of a Gamma Ray: A moving radioactive nucleus having system mass M emits a gamma ray in the forward direction of motion, stopping instantly in its forward motion and dropping to a known mass of m at the same time. What is the initial energy of A in terms of M and m? (STP2 Ex 8-9)
The systems mass is M before and [(m + EC)2 - EC2 ]1/2 and these are conserved and therefore equal.
M2 = m2 + 2mEC2 + EC2 - EC2 = m2 + 2mEC2
but EC = EA - m so
M2 = m2 + 2m(EA - m)2 = m2 + 2mEA - 2m2 = 2mEA - m2 , so
EA = (M2 + m2) / 2m
(This phenomenon is alluded to in STP2 p. 160 and problem 7-7, but updated here to more recent data.) The highest energy yet detected as of 2000 for an "Ultra-high-energy cosmic ray" (defined as > 1020 electronvolts or eV) is reported to be
• Etot = E = 3.2 × 1020 electronvolts (eV)
for the so-called "Oh-My-God particle". (This total energy E includes both the rest energy of the particle as well as its kinetic energy.) This event was detected on 15 October 1991 by the High Resolution Fly's Eye Cosmic Ray Detector, and was consistent with a hadron-initiated event, probably a proton. This energy can be converted to Joules J by 1 eV = 1.60217653(14)×10−19 J, so that its energy was about
• Etot = 51 J,
which is "roughly the kinetic energy of a well-pitched baseball". The rest mass of a proton is 1.67262×10−27 kg which can be converted as follows:
• Eproton-rest = mproton-rest c2 = 1.50 x 10-10 J = 9.38 x 108 eV = 938 MeV.
(MeV are sometimes shown as MeV/c2 to keep units consistent when applying the mass-energy equation). Therefore, the Oh-My-God particle had an energy Etot expressed as a multiple of the proton rest energy of
• Etot = 3.2 × 1020 eV / 9.38 x 108 eV = 3.4 x 1011 times mproton-rest.
Now because Etot = γmproton-rest, this ratio is simply γ (the time stretch factor):
• γ (time stretch factor) = 3.4 x 1011 (no units)
The speed of this particle in Earth frame can be calculated by v/c = (1 - (1/γ)2 )1/2. However, if one attempts to use a medium precision calculator or spreadsheet program to compute this value directly from this formula, the result will be simply 1 because of the inadequacy of available precision. Instead, use the binomial expansion approximation, in which v/c is given by v/c ≈ 1 - 1/2γ2 + ... (much smaller terms). Therefore, v/c = 1 − (4.3×10−24), or in other words, the deviation from v/c=1 is only 4.3×10−24, and
• v = 0.9999999999999999999999957c
Similar calculations on the entertaining Web page by John Walker use the slightly different value for E, namely E = 3.0, and give
v = 0.9999999999999999999999951c
The deviation of the particle's speed from light speed is only Δv = (4.3 x 10−24) x 299792458 m/s = -1.288 x 10-15 m/s slower than c.
From the viewpoint of the particle just prior to its collision here on Earth, the rapidly approaching Earth diameter would be dramatically Lorentz contracted to a mere
• LEarth Diameter Contracted = LEarth Diameter Proper / γ = 1.2742 x 104 m / 3.41 x 1011 = 3.7 x 10-8 meters.
After traveling one year (31,557,600 s) in Earth frame, the particle would lag behind a photon emitted at the same time and starting place as the particle by a distance of only
• Δs = 4.06 x10-8 m or 41 nm
which is a distance the photon covers in
• Δt = 4.06 x10-8 m / 299792458 m/s = 1.4 x 10-16 s = 0.14 femtoseconds
During this year of traveling in Earth frame, the particle would have aged in its rest frame by only
• Δτ = Δt/γ = 31557600 s / 3.4 x 1011 = 9.25 x 10-5 s.
For this remarkably short proper time Δτ, a photon would propagate a distance of only
• distance = Δτc = 9.25 x 10-5 s * 2.99792458 x 108 m/s = 2.77 x 104 m = 27700 m.
Here is another interesting time comparison: Although it would take 2.44 x 109 years in Earth frame for light photons to travel the 2.44 x 109 light years to quasar 3C273, the proper (wristwatch) time Δτ required for the Oh-My-God particle to make this journey in its own rest frame would be only 2.44 x 109 years / 3.16 x 1011 ≈ 2.6 days.
The following primary references are of interest in addition to the hyperlinked references provided above, but were found on the Web only in restricted locations (that is, available without charge only through an academic proxy, or as partially displayed book contents in Google, etc.):• Boughn SP, "The case of the identically accelerated twins" Am J Phys 57:793 Sept. 1989