# Einstein's train experiment-Do waves from lightning flashes meet after the same number of cycles for all observers? [closed]

In Einstein's thought experiment about moving and stationary observers and two lightning flashes, let's say that we replace the lightning flashes with a pulses of mono-chromatic light, with both sources being the same wavelength in the frame of the sources. For a stationary observer exactly half-way between the light pulses, the pulses will complete the same number of cycles by the time they reach the observer. Is that also true for any observer who is half way between the pulses when they flash, but moving relative to them?

• Are the flashes simultaneous in the lamp frame or in the observer frame? Currently it's unclear what you're asking. – WillO Aug 28 at 3:30
• Wait, if the pulses are not simultaneous wrt the moving observer, what is meant by saying that he is halfway between the pulses when they flash? Are you assuming that they are simultaneous wrt the moving observer? I am asking to confirm because this thought experiment is usually done in a set-up where the flashes are simultaneous for the stationary observer (and thus, not so for the moving observer). – Dvij Mankad Aug 28 at 4:27
• What I was thinking was that the observer in the moving train would be facing the observer (the same x) in the relative non-moving frame at the moment that the lights flash simultaneously for the observer in the relative non-moving frame. @DvijMankad – Joseph Hirsch Aug 28 at 22:28

For a stationary observer exactly half-way between the light pulses, the pulses will complete the same number of cycles by the time they reach the observer. Is that also true for any observer who is half way between the pulses when they flash, but moving relative to them?

The number of pulses is directly proportional to the phase of the wave, $$\phi$$, specifically it is $$\phi/2\pi$$. Fortunately, the phase is a relativistic invariant which is rather easy to calculate in terms of four-vectors. I will use units where c=1 for convenience and $$\cdot$$ indicates the Minkowski product between two four vectors, not the usual dot product. Then for the phase of a wave emitted at position $$x_i$$ with a wave k-vector of $$k$$ we have: $$\phi_i = R\cdot K = (t,x-x_i)\cdot (|k|,k)$$

So if we have two coherent sources at different locations emitting their waves towards each other then the phases will be equal at $$\phi_1=\phi_2$$ $$(t,x-x_1)\cdot(k,k)=(t,x-x_2)\cdot(k,-k)$$ $$k t - k(x-x_1)=k t + k(x-x_2)$$ Which you can solve for $$x$$ to obtain $$x=\frac{1}{2}(x_1+x_2)$$ So this shows that only the observer who remains at the mid point between the sources in the frame where they are coherent will see equal phases from each.

Alternatively we can find $$\Delta\phi=\phi_2-\phi_1$$. After simplification that gives us $$\Delta \phi=k(2x-x_1-x_2)$$ This quantity is positive for $$x>(x_1+x_2)/2$$ meaning that the phase from the source on the right is greater if you are to the right of the midpoint.

• Does the moving observer see more phases from light that that they are moving towards or away from? Obviously the one they are moving toward has a higher frequency, but will arrive at the observer first, while the other will have its frequency reduced but will arrive later. I thought at first that the two effects would cancel out. – Joseph Hirsch Aug 28 at 22:34
• More phase from the one that they are moving towards. I have updated the answer to explain that. The reason you have more phase from the one you are moving towards is that at any moment the light you are receiving from the closest was emitted later than the light you are receiving from the furthest. – Dale Aug 29 at 0:26
• how can the light you are receiving from the closer source have been emitted later that the light you are receiving from the farther source? Don't you see the light from the closer source first? Also, I know it's off the topic, but since more cycles are passing you in the direction you are moving than in the direction you are coming from, doesn't that slow you down? (It should vary by your mass but it seems like more energy is hitting your front than back.) Are there any implications of that? – Joseph Hirsch Aug 29 at 0:36
• “how can the light you are receiving from the closer source have been emitted later that the light you are receiving from the farther source?” Just work out the numbers. The speed of light is about 1 ft/ns. Suppose that the sources are 6 ft apart and at t = 10 ns you are 2 ft to the right of the center. Then you are at that moment receiving the right light from t = 9 ns and the left light from t = 5 ns. So the light from the right was emitted later. – Dale Aug 29 at 1:32
• Oh I see that now. – Joseph Hirsch Aug 29 at 1:41

Is that also true for any observer who is half way between the pulses when they flash, but moving relative to them?

Well there's a lot to consider. One thing is that the flashes can't be simultaneous in the moving observer's frame, if these flashes are the same flashes that the non-moving observer observes to be simultaneous.

Let's call the moving observer Joe. Let's say there are "buoys" lined up on the path of the light pulses. The buoys bounce up and down when an EM-wave passes them. Let's say Joe is far from the line of buoys like this:

Beacon......buoy1......buoy2.....byoy3......buoy4......Beacon .

.

.

.

.

                     Joe


Joe sees the beacons moving to the right,the buoys co-move with Joe. First Joe sees buoy1 bounce 10 times rapidly (the pulse consists of 10 wave-crests). Later Joe sees buoy4 bounce 10 times slowly (Doppler-shift is the cause of the slowness). If Joe sees buoy3 and buoy4 to bounce very slowly then Joe would see those buoys bouncing in sync, which would cause Joe to say that the buoys are on the same wave-crest.

Now Joe might say: "I saw 20 waves on the left side and 10 waves on the right side".

What Joe really saw at the left side was:

first he saw EM-field waving 10 times at the position of buoy1
then he saw EM-field waving 10 times at the position of buoy2

Joe did not see any buoy waving 20 times, so he did not see any EM-field waving 20 times either. The "number of waves" was 10, the same number as the "number of waves" at the right side.