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We have two observers, A who is stationary and B who is moving. We have two points C and D that are equidistant from A. At the exact moment B passes A (from A's perspective, if relevant) moving towards C (and at such a distance that in the moment that B passes A, BC = AC in terms of line length, again from A's perspective), light is emitted from C and D. B will encounter the light from C first, then from D, while A will encounter both beams simultaneously (as I understand things). My question comes about what A observes about B, especially if B 'reacts' to C (say by sneezing). The events I'm trying to get in order, from each perspective, are as follows:

  1. B is beside A
  2. C emits light
  3. D emits light
  4. A encounters C
  5. A encounters D
  6. B encounters C
  7. B encounters D
  8. B sneezes
  9. A observes B encountering C
  10. A observes B encountering D
  11. A observes B sneezing
  12. B observes A encountering C
  13. B observes A encountering D

Attempting to see if I have this right, then: From A's perspective it goes:

  1. B is beside A
  2. C emits light, D emits light
  3. B encounters C
  4. B sneezes
  5. A encounters C, A encounters D
  6. B encounters D
  7. A observes B encountering C
  8. A observes B sneezing
  9. B observes A encountering C, B observes A encountering D
  10. A observes B encountering D

From B's perspective:

  1. B is beside A
  2. C emits light
  3. D emits light
  4. B encounters D
  5. B sneezes
  6. A encounters C, A encounters D
  7. B encounters C
  8. A observes B encountering C
  9. A observes B sneezing
  10. A observes B encountering D
  11. B observes A encountering C, B observes A encountering D

... Is that close to it?

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2 Answers 2

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When dealing with simultaneity in special relativity, it is always a good idea to draw a spacetime diagram1. In the diagram below, I have scaled the axes (hence the $ct$ instead of just $t$) such that light (represented by yellow lines) is at a 45°-angle (this is a convention that you will often see in spacetime diagrams) and I chose the moment where B passes A to be $t=t'=0$; $x=x'=0$. In this case, I chose the velocity of the train (and thus observer B) to be about $0.364c$ which corresponds to a 20°-angle with the $ct$-axis.

enter image description here

I also added the worldlines for observers A and B

For a general process to find simultaneous events in different frames using a spacetime diagram, start by drawing the events in one observer's coordinate system (the "stationary" one). Then, you just have to think what the coordinate system will look like for a moving observer. Since the coordinate axes for this observer are tilted, simultaneity changes but you will be able to simply read it from the diagram.

We arrive at the following order for observer A (each line represents a simultaneity line, i.e. events that are simulatneous for A):

  1. B passes A (AB), D emits light (D), C emits light (C)
  2. B encounters the light ray from C (BC), B sneezes
  3. A encounters the light rays from D (DA) and C (CA), A observes B sneezing
  4. B encounters the light ray from D (DB)

For observer B the coordinate system is different (The dotted line shows the time coordinate of each event in B's frame $-$ if you wanted to find the $x$ coordinates, you would draw lines parallel to the $ct$-axis and see where they intersect with the $x$ axis):

enter image description here

So the simultaneity is as follows:

  1. C emits a light ray (C)
  2. B passes A (AB)
  3. D emits alight ray (D)
  4. B receives the light ray from C
  5. A receives both light rays (DA/CA)
  6. B receives the light ray from D

You will have noticed that I left out the events where B passes D and C, and A observes this. I will leave it to you to figure this out for each frame using your a method of your choice.


1 I don't know how familiar you are with spacetime/Minkonwski diagrams. They are extremely useful in special relativity and thus worth learning. They can be intimidating at first (one seemingly confusing thing is that $x$ and $t$ axes are interchanged $-$ $x$ is horizontal and $ct$ is vertical instead of the other way around), but once you spend some time thinking about them, they will stop to look confusing and be a very valuable tool. I recommend you to learn them.

Here are some ressources on spacetime diagrams:

2 Be warned though that there are some errors later in the paper. See for example my question How can it be that Doppler Factor k=1/k?

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  • $\begingroup$ Lots confusion in SR comes from ppl thinking events (e.g., D,C) are stationary in one reference frame...which you imply by drawing world lines for them. Events are stationary in all reference frames, so take out the vertical world lines and the answer is a 100. $\endgroup$
    – JEB
    Mar 25, 2021 at 16:34
  • $\begingroup$ @JEB I meant to draw the world lines for the "locations" (flashlights or whatever they are) fpr C and D. Wouldn't this be correct? (Though I will edit my answer since this was apparently confusing) $\endgroup$
    – jng224
    Mar 25, 2021 at 18:47
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    $\begingroup$ Unfortunately, the original-poster used symbols that can lead to the confusion as @JEB describes. I prefer events labeled by a capital letter, like P, (like a point in a geometry book). Lines/rays/segments would be specified by an ordered pair of capital letters (like PQ)... and maybe even using names... like Alice is along PQ... or a lightray GH. (An alternate notation is to label lines by lower-case letters and describe points as intersections of lines by specifying the pair of lines.... similar to what you have done, following the notation given by the original-poster.) $\endgroup$
    – robphy
    Mar 25, 2021 at 20:49
  • $\begingroup$ With the above notation... Using a pair of points, like PQ, to describe a segment also can be used to symbolize the proper time along that segment (technically one would say mPQ (measure of PQ).... but one could drop the m since i think it would be clear from the context). $\endgroup$
    – robphy
    Mar 25, 2021 at 20:53
  • $\begingroup$ An event exists at one $(t,x)$, which is $(t',x')$ in another frame..it's a point in $M^4$. When you extend it to later $t$ you hold $x$ fixed, but then why can't you hold $x'$ fixed and extend it to later $t'$? Many ppl struggling with SR don't understand the difference, and how it implies a preferred frame when fixing someone's $x$. $\endgroup$
    – JEB
    Mar 26, 2021 at 1:03
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I think you are wrong in at least one point. A should encounter C (your 5.) simultaneous with A observing B encountering C (your 7.).

Consider if C emits two red photons simultaneously. Let B detect one of these photons (B encounters C), and instantaneously emit a single blue photon towards A. The remaining red photon and the new blue photon will move in step towards A. And A will detect the red photon (A encounters C) and the blue photon (A observes B encountering C) simultaneously.

There are other problems too: For example you state "At the exact moment B passes A (from A's perspective, if relevant) moving towards C (and at such a distance that in the moment that B passes A, BC = AC in terms of line length, again from A's perspective), light is emitted from C and D." so from A's perspective B passing and the emission of light from both C and D are simultaneous. But then you say that in A's point of view: Attempting to see if I have this right, then: From A's perspective it goes:

  1. B is beside A

  2. C emits light, D emits light

  3. ...

That is inconsistent

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