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lets say we have two person A and B. Person A is holding a light source (lets call this light source 'S1') which is pointed towards north. Now person A starts moving at speed 0.86c WRT (with respect to) Person B in the direction of north. Now Person A turn on the S1. what will be the speed of light emitted from S1:

1- WRT Person A.

2- WRT Person B.

This question can also be asked as an excited electron traveling at 0.86c emits a photon in the same direction in which it is moving. what will be the speed of that photon WRT to the electron and WRT to the observer (here the observer is the one from whom's respect electron is moving at 0.86c).

Please consider this hypothetical experiment that may help under the question better.
I have a metal cylinder which is 1m long. it absorbs all the light. now on its both ends i have light sensors which gives the signal to a very accurate micro-controller that can record the signal and detect the time between two events. An event is described as light being detected by a sensor. I then place this cylinder in my home horizontally and put a light source on one end. light first hit one sensor and after some time (i guess) the other sensor. My micro controller records this time as 1/299792458 seconds accurately. now i put the same experiment on a train that is moving at 0.86c. the cylinder is now placed in the direction of the train. lets say its end 1 face the engine and end 2 faces the tail and i put the light source at end 2. now i run the experiment again. now what time will be recorded by my micro controller. In this case i think it should be greater then 1/299792458 seconds. In both cases, the one at my home and the one in the train uses the same clocks and have same distance between the two sensors so this dictates that speed of light should remain same (299792458 m/s). But as both sensors are moving at speed near the speed of light, after hitting one the other is farther apart (i think due to doppler effect) the speed would change.

what will be actual time i will find in this experiment and can i use it to find my speed relative to the observer standing on the platform?

PS the above apparatus is in a vacuum chamber.

ALK007
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    The photon has the same speed with respect to both observers. That is the point of relativity. – Bob Knighton Jun 04 '17 at 19:09
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    I don't understand the downvote. OP is just using his common sense to understand a physical situation. The point is that common sense doesn't work in this case, but I don't see it worth a downvote. If you never heard about relativity or you're still assimilating it I think this is a legitimate doubt – RenatoRenatoRenato Jun 04 '17 at 19:41
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    Duplicate: https://physics.stackexchange.com/q/79331 or https://physics.stackexchange.com/q/7446 Related and near duplicates: https://physics.stackexchange.com/q/288327 https://physics.stackexchange.com/q/114523 https://physics.stackexchange.com/q/2230 https://physics.stackexchange.com/q/242281 and so on. The problem isn't that there is anything wrong with the question, its that every physicist has been asked this question in its many variation over and over again and is bored with it. – dmckee --- ex-moderator kitten Jun 04 '17 at 20:42
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    Possible duplicate of If I run along the aisle of a bus traveling at (almost) the speed of light, can I travel faster than the speed of light? – Wrichik Basu Jun 04 '17 at 21:07
  • I thinks this question has not been asked before as there is no 'source of light' movement in the questions that has been asked. as photons also behave as particles i think they should have initial speed WRT Person B much like a bullet and a gun example i.e. if very fast moving gun shot a bullet what will be the speed of bullet wrt the object it hit. In all other examples the light source is very far e.g. sun from which all the objects are 'taking their light'. So this question is being asked from the relative position of source of light which in other question is fixed i think. that the Que.? – ALK007 Jun 04 '17 at 21:55

2 Answers2

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Well, a photon in the vacuum always travels with the same speed c. But if you want to check, it's possible to use the relativistic velocity addition formula:

$$ v_A=\frac{u+v_B}{1+\frac{uv_B}{c^2}} $$

where $v_S$ is the velocity measured by observer S and $u$ is the velocity of B relative to A. Now, the movement of the light source doesn't interfere with the speed at which each photon travels, so $v_B=c$.

Plugging this in, you'll find that $v_A$ is consistently also $c$.

Renan Nobuyuki Hirayama
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    You do understand that your formula is only true if Einstein procedure is used to synchronise clocks, don't you? And that this procedure ensures that the one-way speed of light is $c$. Your answer is therefore a tautology. –  Jun 04 '17 at 20:33
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    @LucJ.Bourhis, Einstein's work does not prove that the speed of light in any reference frame always is $c$. Einstein started with that as a postulate. It was Maxwell who came up with the idea---his theory of electromagnetic waves. And, it was Maxwell who calculated the value of $c$ based on other measurable constants. Einstein's contribution was to discover the consequences (time dilation, length contraction, $c$ as a universal speed limit) if Maxwell's theory really was true. – Solomon Slow Jun 05 '17 at 01:47
  • @jameslarge As I understand it, the problem is that the postulate leads to an ambiguous definition of speed, because you no longer have a unique way to synchronize clocks, and thus to measure/define time for spatially separated events. https://plato.stanford.edu/entries/spacetime-convensimul/ –  Jun 05 '17 at 02:12
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Your question is ill-posed. You are obviously talking about a one-way speed, i.e. a light signal is emitted at a point $M$ at time $t_M$ and received at a point $N$ at time $t_N$ and then the speed is $v=\frac{MN}{t_N-t_M}$. But then this formula is obviously meaningless until you specify how the clocks at $M$ and at $N$ have been synchronised. Otherwise $t_N - t_M$ could take any value.

In special relativity, Einstein's procedure is the golden standard. $M$ emit a light signal at $t_M$, which is reflected by $N$ toward $M$, and $M$ gets the signal back at time $t'_M$. Then $N$ sets its clock so that the time of arrival of the light signal was $t_N=\frac{t'_M - t_M}{2}$. Now since $c=\frac{2 MN}{t'_M-t_M}$, the round-trip speed of light, is a constant independent of how $M$ and $N$ are placed, as proven to amazing precision by one century of experimental endeavours, you can see that the one-way speed of light $v=\frac{MN}{t_N-t_M}$ is also equal to $c$.

But the reasons are completely different: the round-trip of light is a constant because of a law of nature whereas the one-way speed of light is a constant because of the convention we used to synchronise clocks.

  • For anyone interested in further reading, see, for example, the One-way speed of light article at Wikipedia: "The "one-way" speed of light from a source to a detector, cannot be measured independently of a convention as to how to synchronize the clocks at the source and the detector. What can however be experimentally measured is the round-trip speed (or "two-way" speed of light) from the source to the detector and back again." – Alfred Centauri Jun 04 '17 at 23:16
  • Alternatively, just read Einstein 1905 paper: https://www.fourmilab.ch/etexts/einstein/specrel/www/ My answer just parrots Einstein section I.1 –  Jun 05 '17 at 07:58