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Experiment seen from two different inertial frames.

Consider two very short light pulses emitted from the centre (C) of two mirrors A and B (as shown in the diagram). From the point of view of the lab frame, the apparatus is all moving to the left at velocity v. Imagine there is also an electron near the centre of the apparatus, which is stationary in the apparatus frame and therefore also moving with velocity v to the left according to the lab frame.

The short light pulses (much shorter than the apparatus length) bounce off mirrors A and B and return and strike the electron. This situation has similarities with the Michelson-Morley experiment.

According to the frame moving with the apparatus, the pulses take an equal time to bounce off the mirrors and arrive back at C. Therefore the EM waves cancel and there is no net radiation pressure exerted on the electron.

According to the lab frame, the light pulse emitted to the left has less distance to travel overall and so arrives at C before the pulse that was emitted to the right. Therefore the first pulse accelerates the electron by exerting a radiation pressure on it.

Does the electron accelerate or not? :)

(I'm looking for derivations/proofs showing both frames' interpretations)

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  • $\begingroup$ a very, very slightly related topic in terms of conservation of momentum: space.com/26713-impossible-space-engine-nasa-test.html $\endgroup$
    – user108787
    Commented Jun 1, 2016 at 17:05
  • $\begingroup$ The trick is to realise that the pulses are not emitted simultaneously in one frame (or both frames: they may be simultaneously emitted in at most one frame). $\endgroup$
    – user107153
    Commented Jun 1, 2016 at 19:14
  • $\begingroup$ @tfb If they are not emitted simultaneously in both frames then doesn't the same paradox apply to the device that is the source of the emitted light? Either the pulses are emitted simultaneously and the source device undergoes no net acceleration or they are not emitted simultaneously and the source undergoes acceleration... $\endgroup$
    – czechmea
    Commented Jun 1, 2016 at 19:33
  • $\begingroup$ @Faraday7000 In any frame, if they arrive at a given point simultaneously but the path lengths as measured in that frame are different then, trivially, they were not emitted simultaneously (as measured in that frame). On the other hand if the path lengths are the same then, equally trivially, they were emitted simultaneously. $\endgroup$
    – user107153
    Commented Jun 1, 2016 at 20:28
  • $\begingroup$ @tfb I think I agree with what you've said, but I don't see how that helps resolve the paradox in which the electron accelerates or does not accelerate depending on whether pulses are emitted/received simultaneously or not. $\endgroup$
    – czechmea
    Commented Jun 1, 2016 at 20:39

4 Answers 4

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In the lab frame both pulses arrive at C at the same time. The reason is that the distances traveled are the same (they do not reach A and B simultaneously). The distances of paths CA and BC are equal, the same happens with the paths AC and CB. The distance CAAC is equal to CBBC.

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  • $\begingroup$ I'm confused by the term "distance of path". CA is a distance, and the pulse follows a path. Are you saying that the distances are the same, or that the path lengths traversed by the pulses are the same, or both? If so, how do you reconcile this with the standard analysis of the Michelson-Morley experiment in which the path lengths are not the same? (If I understand you correctly you are carrying out all analysis in the lab frame). $\endgroup$
    – czechmea
    Commented Jun 1, 2016 at 20:30
  • $\begingroup$ @Faraday7000 He is saying that in the lab frame, the two paths the photons take have equal lengths. Just like in the apparatus frame $\endgroup$
    – Jahan Claes
    Commented Jun 1, 2016 at 21:02
  • $\begingroup$ It refers to the paths lenghts, I believe, I'l write explicit equations to show it $\endgroup$
    – user83548
    Commented Jun 1, 2016 at 21:06
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If $t_{CA}$ refers to the time it takes in the lab frame for the light to reach C from A, and the same with $t_{AC}$, $t_{CB}$ and $t_{BC}$ then we have:

$t_{CA}=\frac{L/2+v t_{CA}}{c}$

$t_{AC}=\frac{L/2-v t_{AC}}{c}$

$t_{CB}=\frac{L/2-v t_{CB}}{c}$

$t_{BC}=\frac{L/2+v t_{BC}}{c}$

Thus $t_{CA}=t_{BC}$ and $t_{AC}=t_{CB}$

Finally: $t_{CA}+t_{AC}=t_{CB}+t_{BC}$

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Let $E$ be the event "pulse 1 arrives back at point $C$" and let $F$ be the event "pulse 2 arrives back at point $C$". Then $E$ and $F$ either are or are not two names for the same event, and this has nothing to do with choices of frames. Therefore if $E$ and $F$ have the same coordinates in one frame (e.g. the apparatus frame), then they must have the same coordinates in any other frame (e.g. the lab frame).

In other words: It's obvious in the apparatus frame that the pulses arrive at $C$ simultaneously. Therefore the pulses arrive at $C$ simultaneously, and all observers must agree on this.

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  • $\begingroup$ Yeah, we know they must both agree, I was looking for a detailed derivation/proof of this, something with lines/lengths/times etc explicitly worked out :) (Adjusted my question to ask for this). $\endgroup$
    – czechmea
    Commented Jun 1, 2016 at 17:16
  • $\begingroup$ S0 put the event where the pulses leave $C$ at the origin. In the apparatus frame, they hit the walls at, say, $t=1,x=1$ and $t=1,x=-1$, and come back together at $t=2,x=0$. Now Lorentz transform these coordinates to the lab frame. Where are you getting stuck? $\endgroup$
    – WillO
    Commented Jun 1, 2016 at 17:52
  • $\begingroup$ I'm looking for an analysis done in both frames to show they are consistent, not just starting in one and using Lorentz transforms to get the other... $\endgroup$
    – czechmea
    Commented Jun 1, 2016 at 18:07
  • $\begingroup$ Write down whatever you consider to be an "analysis" in the apparatus frame; then Lorentz-transform all times and locations to the lab frame, and you'll have an analysis in the lab frame. What more are you looking for? $\endgroup$
    – WillO
    Commented Jun 1, 2016 at 18:27
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    $\begingroup$ It should be possible to analyse the experiment in the lab frame and come to the same conclusion. If I am in the lab frame, I don't need to guess that the other frame is correct and use boosts to get back to myself - I should be able to use my own measuring instruments and come to my own conclusion about what happened. $\endgroup$
    – czechmea
    Commented Jun 1, 2016 at 18:57
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Because I like setting the $x$ axis point towards the right, reverse the speed on your figure. In the lab frame, the light moving towards the mirror A reaches the mirror in a time $\Delta t_1$ determined from $v \Delta t_1 + c \Delta t_1 = L/2$ giving the $\Delta t_1 = \frac{L/2}{c + v}$. It is then reflected by this mirror. Though in the lab frame the electron has moved it is still a distance $L/2$ from the A mirror. The light reflected from $A$, according to an observer in the lab frame, will reach the electron in the time $\Delta t_2$ determined by $c \Delta_2 = L/2 + v \Delta_2$ so giving $\Delta t_2 = \frac{L/2}{c - v}$. This gives the total time as $\Delta t_A = \frac{L c}{c^2 - v^2}$.

If you look to see what is happening with the pulse of light moving towards the B mirror, then the time to reach the mirror is $\frac{L/2}{c-v}$ and once reflected the electron has also moved forward so that it is still a distance $L/2$ when the pulse of light begins its reflected journey from the B mirror, but now it takes a time $\frac{L/2}{c + v}$ (because the electron is moving towards the reflected beam).

This gives the same time in the lab frame, specifically because when the light is reflected at mirror A it still leaves the mirror with speed $c$ and similarly for the light being reflected by the mirror B, the speed of light is constant.

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