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I'm a little confused by the following classic example about the speed of light being constant for all observers (paraphrasing):

Jack and Jill are both travelling at a significant portion of the speed of light; Jack is travelling behind Jill. Jack shines a laser pointer at Jill. Both observe the light emitted from it as travelling at the speed of light, $c$.

I understand that the speed of the emitter of the light doesn't influence the speed of light; the photons always travel at the maximum speed they can. If Jack were moving backwards, the photons would "instantly accelerate" to $c$; Jack moving backwards wouldn't detract from their speed. Jack moving forwards isn't adding to their speed, because they can't move faster than $c$.

However, the observer's perspective is harder to understand for me. Assuming Jill is observing the incoming laser by counting the frequency of incoming photons, shouldn't that rate change depending on how fast Jill is moving herself? Assuming Jill would be travelling at the speed of light (ignoring the practical impossibility), shouldn't she not be receiving any light at all? Scaling that down to her travelling at half $c$, shouldn't her incoming photon rate be half as much as if she was "standing still"?

(This is very related to Seeing light travelling at the speed of light, however it's about the other side of the same setup.)

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I think your point of view needs a little correction.

You are thinking in terms of 3D space but it was shown by relativity that space and time are actually related and make a 4D world. The relative speed between two persons also alters the rate of change of time in their frames hence the photons do not accelerate or deaccelerate to 'maintain their speed'.

However they do accelerate or deaccelerate (i.e. gain or loose energy) and this is manifested in the Doppler shift of the light. If two sources are moving away from each other they will see red shifted (observed wavelength is longer than emitted wavelength) light, and if they are moving towards each other they will see blue shifted light.

Hypothetically if two persons are moving away with relative velocity c then the light emitted by person1 will be seen by person2 as 0 frequency i.e. DC (and he practically do not see any light). However if the relative velocity is anything less than c person2 will definitely see emission at non-zero frequency.

In this case if emitter is moving with velocity v and observer is still or if observer is moving with velocity -v and emitter is still does not make any difference, the end result will be same.

I hope it will help to clear your confusion.

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  • $\begingroup$ I was afraid that the interconnectedness of spacetime would be involved in the answer... :) Are you saying that a) due to relativistic effects and time dilation (excuse me if I'm misusing these terms) Jill would observe the light as plain old c; or b) that Jill would observe the light as half c, but know to correct her POV by compensating for redshift and after this compensation conclude that the speed is c? $\endgroup$ – deceze Jun 8 '16 at 10:00
  • $\begingroup$ you are right option a) is correct the light will remain as plain old c. This experimental observation is actually the basis of the origin of spatial relativity. $\endgroup$ – hsinghal Jun 8 '16 at 10:07
  • $\begingroup$ Alright, that sets me off in the right direction then. I understand it in the abstract, I think, but can't really wrap my head around it entirely just yet. Thank you! $\endgroup$ – deceze Jun 8 '16 at 10:09
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WHEN jack is moving backwards at velocity C/2 THEN Jill will still receive the light at velocity C BUT at half the frequency of the transmitter (doppler). As long as the relative velocity between jack (transmitter) and Jill (receiver) ,assuming they are going away from each other , is less than C THEN SPEED OF LIGHT IS UNAFFECTED but the freq received is decreased ).Now if jack is moving backwards at velocity C ,then doppler leads to that the received freq is zero which as equation C= F * WAVELENGTH LEADS TO THAT NO LIGHT IS RECEIVED BY jILL .

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  • $\begingroup$ In other words, the individual photons will still travel at c, but you'll receive half as many in the same amount of time? $\endgroup$ – deceze Aug 8 '16 at 7:04
  • $\begingroup$ Jack fundamentally cannot "move backwards at $c$". $\endgroup$ – pela Aug 8 '16 at 7:31
  • $\begingroup$ pls use the formatting features in the editor instead of writing in capitals, it makes it hard to read your answer. $\endgroup$ – Wolpertinger Aug 8 '16 at 7:51
  • $\begingroup$ Welcome on Physics SE and thank you for your answer :) Please refrain from USING CAPSLOCK BECAUSE IT SEEMS VERY AGGRESSIVE and see the help section for information about the built-in TeX-editor to format formulas. $\endgroup$ – Sanya Aug 8 '16 at 7:55

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