Relativity Reviewed

How can we measure the length of a long, thin, straight bar of steel? Conventionally we compare the bar to some standard length that we have defined, such as the meter, and manifested in rules. If we say that the bar has a length of 9.45 meters, we mean that we can take nine meter sticks and a stick five centimeters shy of half a meter and lay them end to end next to the bar and the ends of that array will lie next to the ends of the bar.

If we can’t get close to the bar, how can we measure its length? What can we do if the bar is moving?

Prevented from getting close to the bar, we use a long-armed calipers to measure the bar’s length. On the near end of the calipers we have a finely-calibrated scale whose pointer shows the distance between the points at the opposite end of the device. Thus we place the points of the calipers next to the ends of the bar and read 945 centimeters off the scale.

Movement of the bar with some uniform velocity relative to our location makes the measurement more difficult and adds another complication. We must swipe the calipers past the bar as it flies by us. In doing so we are not so much measuring the length of the bar as we are measuring the distance between two events, each event consisting of one of the calipers’ points passing an end of the bar. The obvious caveat tells us that the points of the calipers must pass the ends of the bar simultaneously if we are to get a correct measure of the bar’s length. If the calipers’ points pass the ends at different times, the bar will move some distance in the interval between those times and the calipers will measure the wrong length.

Using a material calipers is clumsy at best. We want something better, especially when the bar moves at a substantial fraction of the speed of light. We need an immaterial calipers, one made of light. To that end we establish a Cartesian coordinate frame (spatial coordinates x, y, and z measured on straight, mutually perpendicular axes) and set up a screen that we have marked with a grid, setting up the screen such that it floats parallel to the x-axis and the z-axis and a certain distance from the origin of the frame in the negative y-direction. At some distance from the screen in the positive y-direction we set up a flash lamp and the necessary optics to transform its output into an arbitrarily thin plane wave. The light propagates in the negative y-direction and illuminates all points on the screen simultaneously with an arbitrarily brief pulse of light (Professor Heisenberg says that we can’t do that, but we’ll pretend that we can make the pulses as brief as we want).

Triggering the flash-lamp at the right instant sends out a plane wave that washes over the bar on its way to the screen: the bar is simply moving through our optical system at the right instant for this to happen. If the bar is oriented and moves parallel to the x-axis, the light will strike all points on its +y side simultaneously and the bar will cast onto the screen a shadow that’s just as long as the bar. We have set up a camera that makes a photograph of the screen at the right instant (we conduct our experiment in the dark and leave the shutter open), so we can determine the length of the bar by measuring the length of the shadow on the grid that we drew on the screen. Our optical calipers have given us, we are certain, the correct length of the bar.

A team of observers moving with the bar, at a speed that we calculate as v, disagrees. In their observation of the event, the light did not hit the bar simultaneously at all points: the light hit the fore end first and then, a short time later, hit the aft end. That statement stands true to Reality because the light is aberrated by the moving observer’s motion relative to its source (and relative to our frame of reference).

We know that light travels at the same speed for both teams of observers, stationary (us) and moving, and that its propagation vector as seen by the stationary observers is oriented parallel to the y-axis, pointing entirely in the negative y-direction. In the moving observers’ frame of reference the propagation vector makes an angle of arcsine(V/c) with a straight line parallel to the y-axis, in which V=-v represents the speed at which the moving observers see the flash-lamp moving through their frame. Light always propagates in the direction perpendicular to its wavefronts, so our aberrated plane wave must necessarily make the angle arcsine(V/c) with the x-axis. Thus that plane wave will strike the fore and aft ends of the bar at different times for the moving observers.

To prove and verify that statement, the moving observers have mounted clocks on the fore and aft ends of the bar and set them to run in synchrony with each other. They accomplish that latter feat by setting both clocks to show the same time and then starting them with a pulse of light that originates at the exact center of the bar’s surface. The plane wave from the flash-lamp will strike the fore clock and then strike the aft clock after crossing a distance equal to the length of the bar multiplied by the sine of the angle by which the plane wave is tilted away from the x-axis. Thus, between the plane wave striking and illuminating the fore clock and it doing the same to the aft clock, an interval of

(Eq’n 1)

elapses on both clocks (note that ΔX represents the length of the bar as measured by the moving observers). Thus, in the photograph made by the light of the plane wave both observers will see the aft clock displaying a time later than the fore clock shows by that amount of time.

That temporal offset makes perfect sense to the moving observers, but the stationary observers know that in their own frame the clocks were illuminated simultaneously. Yet the aft clock appears, by the time it displays, to lie in the future relative to the fore clock. The stationary observers must thus infer that the elapse of time in the moving frame is distorted relative to the elapse of time in the stationary frame. But that’s not all there is to the distortion of time between inertial frames.

For convenience the stationary observers have set up a pole parallel to the y-axis and marked it for use as a ruler. Nicely enough, both observer teams can use that pole to measure distances oriented perpendicular to the direction of relative motion between them and they will get the same result. That statement stands true to Reality because, if the moving observers have their own measuring pole parallel to the y-axis, then both teams of observers must see all of the points with the same y-coordinates pass each other at the same instant: that fact necessitates that it be the same pairs for both teams, which means that the observers don’t see any distortion of distance in directions perpendicular to the direction of relative motion.

If the stationary flash-lamp emits a plane wave at regular intervals, the observers will see a series of bright spots on the stationary pole. The stationary observers will see those spots move down the pole at the speed of light, because the plane waves are propagating parallel to the pole. Because they see the light aberrated, the moving observers will see the spots moving more slowly. In the moving frame the propagation vector of the plane wave is the hypotenuse of a right triangle whose X-side is V. The Y-side of that triangle, according to Pythagoras, has the magnitude

(Eq’n 2)

If the spots on the stationary pole have a spacing of y=Y, then the stationary observers will see two successive spots pass a given mark in a time interval

(Eq’n 3)

The moving observers will see the same spots pass the same mark in the time interval

(Eq’n 4)

That statement means that the stationary observers put fewer microseconds between the events as do the moving observers, who state that the stationary observers are using dilated time to measure the intervals between pairs of events.

If the moving observers transmit plane waves of light in the Y-direction, the stationary observers will see them aberrated by an angle arcsine(v/c). Those waves make spots of light on the moving pole (which is stationary in the moving observers’ frame) and by timing the movement of those spots the stationary observers infer that the moving observers are using dilated time such that

(Eq’n 5)

The symmetry between that equation and Equation 4 tells us our designation of one observer as moving and the other as stationary is purely arbitrary.

The symmetry between Equations 4 and 5 confounds many people: it seems absurd that the clocks should run slower than each other. That symmetry, naively accepted and interpreted, is the basis for the infamous twin paradox. But in order to determine whose clock ran slower than the other’s, the observers must ensure that their clocks were together at the beginning of the measured interval and that the clocks are brought back together at the end of that interval. So one observer must turn around and return to the other observer: because that observer has occupied two different inertial frames, theirs is the clock that shows less elapsed time.

That paradox and its resolution tell us that when we measure the time of events either the clocks must attend the events or we must apply a suitable correction for distance between the clock and the event. Imagine that two instantaneous events (such as the popping of firecrackers) occur at a single point in the stationary observers’ frame. The stationary observers need only one clock to measure the times of those events, but the moving observer needs two clocks.

The moving observers’ fore clock attends the first event and the observers record the time at which that event occurred. Then the aft clock attends the second event and the time of that event is recorded. By subtracting the first recorded time from the second the moving observers calculate the time T that elapsed between the events. The stationary observers’ clock attends both events and records the times at which they occurred, enabling the stationary observers to calculate the time t that elapsed between the events.

On first impression the two observer teams disagree on the amount of time that elapsed between the events. But they can reconcile their calculations through a simple procedure. The stationary observers know that the moving observers’ clocks are offset from each other in accordance with Equation 1, so that much time must be subtracted from T, because, in essence, the aft clock is running that much fast relative to the fore clock. That, of course, is dilated time, so the stationary observers must use Equation 5 to make the appropriate correction. When the moving observers give the stationary observers their data, the stationary observers calculate

(Eq’n 6)

That’s the fourth (the temporal) equation of the Lorentz Transformation.

What does that all have to do with measuring the length of a moving bar? We have a photograph of the shadow that the bar casts upon our gridded screen. That should be sufficient, shouldn’t it? No, in fact, it’s not sufficient and Equation 1 gives us the reason why it’s not sufficient.

Our stationary observers see what they are told are well-synchronized clocks offset from one another: the aft clock on the bar appears to be advanced into the future relative to the fore clock. The fact that the bar is moving leads to the expectation that the aft clock will, at least in some sense, partly overtake the fore clock. That effect is prorated along the full length of the bar, which leads to a uniform shrinkage of the bar. The stationary observers multiply the offset between the clocks by the speed of the bar’s motion and subtract the result from the proper length of the bar to calculate the shrunken length as

(Eq’n 7)

In that equation I have put the calculated length in sneer quotes because it’s wrong. The subtraction is correct, insofar as it goes, but the equation lacks a factor.

We can determine the value of that missing factor through a simple calculation. Our observers know that they see each other moving at the same speed albeit in opposite directions (v=-V). Speed is simply the ratio of distance crossed to time elapsed as an object goes from one point to another, so our observers can write x/t=X/T. The four variables in that equation represent measurements that our observers make of, let’s say, the moving observers’ fore clock passing the edges of the stationary observers’ gridded screen. Applying Equation 5 to the result produces

(Eq’n 8)

which necessitates that

(Eq’n 9)

Using that result to modify Equation 7 gives us

(Eq’n 10)

which describes the Lorentz-Fitzgerald contraction of the bar in the stationary observers’ frame. That equation tells us that we must divide the length of the shadow on our gridded screen by the square root factor to calculate the proper length of the bar, the length that we would measure if the bar were not moving. More generally we have Equation 9 as

(Eq’n 11)

the first (the x-spatial) equation of the Lorentz Transformation.

If we add to Equations 6 and 11 the statements that y=Y and z=Z, we get the full Lorentz Transformation. With those equations observers in one inertial frame of reference can take measurements of distance and duration made between two events by observers in another inertial frame and translate them into the measurements that they would make between the same two events. And we got it through a simple exercise in basic optics.

Appendix: The Relativistic Doppler Shift

If we want to use optics alone to devise the full theory of Relativity, we must address the Doppler shift and see how it plays out between inertial frames of reference. We make the simple postulate that light has the nature of a wave, that it propagates at a speed that is the same for all observers (in accordance with Einstein’s second postulate of Relativity) and that its wavecrests pass any given point at some frequency that we can measure. As usual, we shall employ axiomatic-deductive logic through an imaginary experiment.

On a vast flat plane Stationary Stanley and Stationary Steven communicate with each other via beams of monochromatic light. The two men do not move relative to each other or to the plane. Somewhere between them Mobile Mary moves away from Stanley and toward Steven at some uniform speed V. She can use a pickoff mirror to take some light out of the beam that Stanley is sending to Steven and can read the message. She doesn’t want the men to know that she’s spying on them, so she recreates the light that she stole and sends it on to Steven. Thus Steven detects no diminution of the intensity of the beam and, thus, does not suspect that Mary is spying on him.

Stanley has sent the light out with a frequency of F=N/T, which means that in the time interval T a number N of wavecrests will pass a given point. Steven receives the light at the same frequency; that is, in the interval T his detector will receive N wavecrests. Because she is moving away from Stanley, Mary infers that in a time interval t she will count n wavecrests passing her in accordance with f=n/t=F(c-V)/c. That’s the classical Doppler shift. She then assumes that Steven will receive the recreated light at the frequency f(c+V)/c, which is wrong: it does not equal F. She calculates the Steven will receive her coverup radiation at the frequency

(Eq’n A-1)

and, thus, that her spying will be discovered.

But that inference cannot be right. If Mary has received the light at the frequency f and re-emits it at the same frequency, then it must come to Steven at the same frequency at which it left Stanley: in essence, nothing has changed in the light, even though Mary sees it differently from how the men see it. So there must be something wrong in how Mary thinks she sees the light; that is, the Doppler shift isn’t what she thinks it is.

If we write the Doppler shift as f=FD(-V) and F’=fD(+V) then we must have as true to Reality D(-V)D(+V)=1. The blueshift must exactly cancel the effect of the redshift. The geometric factors that Mary used above are legitimate, so she must multiply each of them by a corrective factor which is the same for both of them. Thus she writes

(Eq’ns A-2)

She notices that each Doppler coefficient equals the reciprocal of the other, so when they are multiplied together they equal unity, as required. Now she has to figure out what the square root means.

Frequency equals a number of wavecrests divided by a time interval. The number of wavecrests that exist in the beam won’t change due to relative motion between observers, so the time interval must change; that is, time intervals measured by different observers must differ from each other. When Mary calculates f=n/t, the geometric term calculates the number of wavecrests that pass a given point in the given time interval, so she has two separate factors,

(Eq’ns A-3)

Those equations tell her that Stanley’s clock puts more microseconds between two given wavecrests than her clock does. But that’s the calculation that Stanley would make if he knew about Mary’s espionage, which means that as Stanley sees the situation, Mary’s clock is running slower than his. Likewise, in the reverse calculation Mary infers that Steven’s clock runs slower than hers does. She’s discovered time dilation, one of the primary phenomena of Special Relativity.

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