Is there an experiment where they tested if c is constant to any observer moving at any speed in any direction? Am I correct that to check the speed of light for two observers (and letting both observers know each other's results) we would need to exchange information in between the two observers, and that the exchange of information can only be done also with max c speed? So how would we let the other observer know of our results if we can send information only with max c speed?
4 Answers
There have been countless studies on the topic. The easiest for me to understand are the interferometry studies. In these studies, you shine a laser at a semi silvered mirror which splits the beam into two beams at right angles. These each bounce off a mirror and recombine, interfering with each other. You can then move this apparatus in one axis and not the other or rotate it and show that the interference pattern does not change (intuition, such as that gained by thinking about throwing a ball on a train, leads to a different answer).
The most famous of these was the Michaelson-Morley experiment, which showed that the speed of light did not vary with direction or velocity by using the earth's own motion and rotation.
These interference experiments could be though of a a single observer showing that difference in reference frame velocities do not change the velocity of light. If you want it proven with two observers, it may be easier to instead focus on proving that time dilation occurs. That is another expected outcome of relativity. For that, you need to go no further than your local GPS receiver in your phone. Your phone actually has to account for relativistic effects such as time dilation when it is calculating your location. It can be shown that, without that correction, your GPS would be more inaccurate than it is today!
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$\begingroup$ so basically there has not been an experiment with two observers? $\endgroup$ Oct 10, 2016 at 17:09
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$\begingroup$ @ÁrpádSzendrei That depends on your definitions. We have done two observer studies which demonstrate time dilation. Time dilation is a far more unintuitive result than "the speed of light is constant," and the equations for time dilation easily show that the speed of light is constant. There may also be a direct study, as you are looking for. The real question is why do you care? The Michaelson-Morley experiment is so incredibly close to being a 2-observer test, that you almost might as well treat it as a 2 observer test. $\endgroup$ Oct 10, 2016 at 17:43
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$\begingroup$ If the time dilation that you can prove with GPS and Michaelson-Morley are not sufficient experimental evidence, then you may need to be more specific about what effect you are looking for. What experiment you think you want to see may simply be infeasible, and nobody did it because there were much more feasible ways to prove the same thing. $\endgroup$ Oct 10, 2016 at 17:45
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$\begingroup$ (Also, you may find an experiment matching your particular desires in the links Robin Ekman provided) $\endgroup$ Oct 10, 2016 at 17:46
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$\begingroup$ the reason I am asking is because I understand that time dilation is proven and that observers experience a different time rate and measure distances differently. so obviously distance/time=speed for the same photon will always be constant c. What i am asking is to prove that two different observers will see the same photon moving at the same speed at the same moment. They would obviously use Emwaves (that is information) to measure information (Em waves) themselves. So how can we measure speed of light that is the speed of information with itself? $\endgroup$ Oct 10, 2016 at 19:06
There exists a comprehensive list of experimental tests of special relativity, including experiments testing the invariance of the speed of light.
If the light emitted from a source moving with velocity $v$ toward the observer has a speed $c+kv$ in the observer's frame, then these experiments place a limit on $k$.
The best result reported as far as I can see is $k < 5\cdot 10^{-9}$.
Nature has given us the best test. Go out on a starry night and look at the stars and planets. The points of light are arriving on Earth at a velocity of c. Yet, they all are moving in their own proper motion, and you, on the Earth, are moving as well. With all that motion of reflecting and emitting bodies we see a constant c.
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1$\begingroup$ While I do see where you are coming from, I do believe that to the answer the question fully using this response, you would also need to be able to measure the speed of the incomming light from the stars and planets. Simply stating that they travel at a velocity of $c$ does not prove that $c$ is indeed constant. $\endgroup$ Jun 13, 2019 at 3:07
As the moving observer measures the frequency shift, he actually measures the shift in the speed of the light relative to him. When the initially stationary observer starts moving towards the light source with speed v, the frequency he measures becomes
f' = (c+v)/λ
http://www.hep.man.ac.uk/u/roger/PHYS10302/lecture18.pdf "Moving Observer. Now suppose the source is fixed but the observer is moving towards the source, with speed v. In time t, ct/λ waves pass a fixed point. A moving point adds another vt/λ. So f'=(c+v)/λ."
On the other hand, the speed of the light relative to the moving observer is
c' = λf'
Combining the two formulas gives
c' = c+v
in violation of Einstein's relativity.
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$\begingroup$ The lecture you quote is discussing the classical theory of waves in a medium which is (still and always) a different situation from that of light. The relativistic expression is $f_o/f_s = (1 \pm \beta)/(1 \mp \beta)$ where $\beta$ is the normalized relative speed of approach (using upper sign) or recession (using lower sign). $\endgroup$ Oct 10, 2016 at 22:28
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$\begingroup$ "The lecture you quote is discussing the classical theory of waves in a medium which is (still and always) a different situation from that of light." No. Later Barlow introduces relativistic corrections (time dilation) and the frequency measured by the moving observer becomes f' = γ(c+v)/λ. The speed of the light relative to the moving observer is, accordingly, c' = λf' = γ(c+v), in violation of Einstein's relativity. $\endgroup$ Oct 10, 2016 at 23:03