This is something that bothered me for a long time regarding special relativity. The Minkowski diagram is symmetric. That is, space and time look the same. But time is dilated, and length is contracted.

So I came up with this thought experiment. Imagine a very long Einstein passenger train with an effectively continuous line of LEDs strung from the front to the back of the outside of the train. These are set to flash simultaneously with a fixed period, as measured in the train's rest frame.

An observer in another train can only see directly out his window as the first train passes him at relativistic speed. At what frequency will he observe the flashes outside his window?

Edit to add Minkowski diagram. Notice that the rest frame time axis may be assigned to either the horizontal or vertical lines. The diagram itself is symmetrical in that sense. Of course that assumes we are only considering the one space dimension in which the relatively moving frame translates.

enter image description here

  • $\begingroup$ Seems a bit inconvenient. When he is straight in front of a flash it may be off. Can we replace them with Einstein - synchronized clocks? He can take readings of the clocks succesively then and compare readings. This way he will see that set of clocks is ticking $\gamma$ times faster then than his own clock. $\endgroup$ – Albert Jan 22 '18 at 17:55
  • $\begingroup$ See pages 16 and 17 of gutenberg.org/files/36276/36276-pdf.pdf "The time so specified is at all events independent of the position of the system of co- ordinates in our space of reference, and is therefore an invariant with respect to the transformation " $\endgroup$ – Steven Thomas Hatton Jan 22 '18 at 18:08
  • $\begingroup$ If you move in a train car along an embankment and there are many Einstein- synchronized clocks on it (the embankment), and you look at these clocks successively, one after one, you will see them running not slow, but very, very fast. That goes out straight from the Lorentz transformations. Single clock runs slower. Set of synchronized clocks runs faster "from the point of view" of a single clock. We consider that an observer makes observations in immediate vicinity, where delays are negligible. $\endgroup$ – Albert Jan 22 '18 at 18:26
  • $\begingroup$ 'These are set to flash simultaneously with a fixed period, as measured in the [first] train's rest frame.' $\endgroup$ – Steven Thomas Hatton Jan 22 '18 at 18:38
  • 1
    $\begingroup$ $\gamma$ times faster. This long lamp is a reference frame. Single moving clock runs slower from the point of view of a reference frame. Time in the reference frame is running faster. In SR observers change reference frame, that's why A is slower than B and B is slower than A. If we don't change frame, single clock runs slower than many clocks. but many clocks run faster from the point of view of a single clock. $\endgroup$ – Albert Jan 22 '18 at 19:12

Let’s long LED lamp is placed along an embankment. We can imagine, that each diode of that long LED lamp is a clock. All these clocks are Einstein – synchronized (one-way speed of light is c) and flash simultaneously in the Embankment’s frame.

An observer in a train car is moving along the Embankment and sees these diodes – clocks successively, in immediate vicinity.

It should be noted, that what clock rate the observer will measure, depends on how he measures it.

Fig.1 Fig.2

We can demonstrate time dilation of the SR in the following experiment (Fig. 1). Moving with velocity $v$ clocks measure time $t'$. The clock passes past point $x_{1}$ at moment of time $t_{1}$ and passing past point $x_{2}$ at moment of time $t_{2}$.

At these moments, the positions of the hands of the moving clock and the corresponding fixed clock next to it are compared.

Let the arrows of moving clocks measure the time interval $\tau _ {0}$ during the movement from the point $x_ {1}$ to the point $x_ {2}$ and the hands of clocks 1 and 2, previously synchronized in the fixed or “rest” frame $S$, will measure the time interval $\tau$. This way,

$$\tau '=\tau _{0} =t'_{2} -t'_{1},$$

$$\tau =t_{2} -t_{1} \quad (1)$$

But according to the inverse Lorentz transformations we have

$$t_{2} -t_{1} ={(t'_{2} -t'_{1} )+{v\over c^{2} } (x'_{2} -x'_{1} )\over \sqrt{1-v^{2} /c^{2} } } \quad (2)$$

Substituting (1) into (2) and noting that the moving clock is always at the same point in the moving reference frame $S'$, that is,

$$x'_{1} =x'_{2} \quad (3)$$

We obtain

$$\tau ={\tau _{0} \over \sqrt{1-v^{2} /c^{2} } } ,\qquad (t_{0} =\tau ') \quad (4) $$

This formula means that the time interval measured by the fixed clocks is greater than the time interval measured by the single moving clock. This means that the moving clock lags behind the fixed ones, that is, it slows down.

However, time in reference frame from the “point of view of the single clock” is running $\gamma$ times faster instead.

This way we can see, that if moving observer compares his clock readings successively with synchronized clocks of reference frame he moves in, he will see, that these clocks change readings $\gamma$ times faster.

However, if the observer changes frame and places two spatially separated clocks of his frame $S'$ at points $x'_{1}$ and $x'_{2}$ then every single diode that moves from $x'_{1}$ to $x'_{2}$ will flash $\gamma$ times slower.

The animation below vividly demonstrates that:

enter image description here

Small note in regard of Transverse Doppler effect (diagram is below). Transverse Doppler effect reflects rate of clock. We can imagine, that observer moves along the row of clocks and looks at these clocks through a telescope. Let all clockfaces of these clock are highlighted in green monochromatic light.

What color of the clocks the observer will see? If he think, that he is at rest, he will direct his tube at right angle to the line that connects clocks (long LED lamp). Then he will see, that clock rate of every single clock is $\gamma$ times slower. Color of each clockface will be red, or $\gamma$ times redshifted. However, during reversal, when he moves at half – circle around the last clock, he will not see any light, because he annot acribe himself state of rest then. Due to aberration, he must always keep his telescope at angle $\sin \alpha = v/c$

He must remember, that he himself moves in the reference frame of the LED lamp. He must take aberration into account and turn his telescope forward at relativistic aberration angle $\sin \alpha = v/c$. He will always, all the time will see clocks and each clock will be of blue color, or $\gamma$ times BLUESHIFTED due to Transverse Doppler effect. This way he will see, that each individual clock is ticking $\gamma$ times faster, since his own clock dilates, and the set of clocks, if he looks at them successively, runs faster. He can explain that by dilation of his own clock, since he moves himself in the reference frame $S$.

Actually it is Twin paradox, simply explained, no acceleration involved. Don't change frame and there is no paradox.

See Transverse Doppler Effect in Wikipedia.

enter image description here Transverse Doppler blueshift comes out straight from the Einstein’s 1905 paper, chapter 7.

Einstein gives this Doppler shift formula for moving observer:

$$f_0= \frac {(1+\cos\theta_s\cdot v/c)}{\sqrt {1-v^2/c^2}}f_s(2)$$

That means that moving observer sees increase of source's clock rate, since square root resides in denominator. If source emits at right angle in its frame ($\cos\theta_s = 0$) frequency will be $\gamma$ times blueshifted. Observer who makes U - turn around the source will always see blueshift of frequency, i.e. that source's clock is running faster. Source always emits at right angle. He cannot see any other frequency shift, only blue one.

Actually, an observer in $S'$ can measure higher rate of a single clock of $S$ by means of two spatially separated clocks at points $x'_{1}$ and $x'_{2}$, but he must synchronize these clocks by Reichenbach, not Einstein, or simple take time of reference frame $S$. One - way speed of light measured with these clocks will be different from c, though speed of light back and forth (round trip, measured with single clock) will still be c. Then single moving clock of $S$ will appear ticking $\gamma$ times faster than two clocks at $x'_{1}$ and $x'_{2}$.

  • $\begingroup$ I like your work. In my opinion, this idea of "time-contraction" is an important, overlooked aspect of SR. Another way to think about the situation is to consider that the embankment observers (all along the rail-bed) conclude that the LEDs are not synchronized in their inertial frame. $\endgroup$ – Steven Thomas Hatton Jan 24 '18 at 19:07

First of all Minkowski spacetime is not symmetric as for space and time, in fact the metric signature is different between the temporal and the spatial coordinates.
Then, the thought experiment assumes a standard configuration for the two reference frames, that is S and S' with aligned spatial coordinates and S' moving in the +x direction relative to S with speed $\beta$ = $v/c$. Note that in SR (special relativity) what is meaningful is the relative velocity between the two frames, so we can assume the long passenger train with the LED's as S' and the other train as S.
Your question is twofold, as it refers both to the time dilation and the relativistic Doppler effect as measured by the observer in S.
Time dilation:
The observer in S will measure a time dilation vs. the proper time in S', hence the period of time between one flash and the next one will be longer.
Relativistic Doppler effect:
A peculiarity of the relativistic Doppler effect is that it occurs also when the distance between the source in S' and the observer in S is at a standstill, that is at a minimum. It is the so called transverse Doppler effect, where the frequency (wavelength) measured is lower (longer) as a consequence of the time dilation. Therefore the light of the flashes will be redshifted.

  • $\begingroup$ Look at the Minkowski diagram I added to the original question. Time and space are interchangeable. Choose time to be vertical. Now, imagine the moving train to be flashing at unit time intervals. Each line of simultaneity in that reference frame coincides with a flash of the train. The "stationary" observer is sitting at the origin. He sees a flash of light every time the dashed red space-like line intersects his world-line. His world line being the heavy green vertical line. This is exactly the same geometry as length contraction. The period is shorter so the light is blue-shifted. $\endgroup$ – Steven Thomas Hatton Jan 24 '18 at 17:45

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