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1 - Has anyone ever measured the one way speed of photons traveling perpendicular to the Earth at the Earth's surface?

2 - Given our current understanding of Physics is there any way both the upward and downward speed would not be $c$?

3 - If the measurement were made and the downward speed were found to be considerably faster than $c$, would there be any plausible explanation given our current understanding of Physics?

4 - If it hasn't been done in the past, how would one do so and how difficult would it be to make such a measurement in both the upward and downward directions with less than 1 km/s error bars?

EDIT: Someone helped me find my error on another thread. Thanks again for all help.

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What is your reasoning for it being faster or slower? I think it should be red-shifted going up and blue-shifted going down. As long as it follows the same path the overall time it takes should be the same. – Brandon Enright Apr 10 '13 at 16:39
If the (local) speed of light was different up and down, the wavelength of the light would be different up and down, and we'd see shifts in fringe positions as you rotated an interferometer. I don't know what the experimental limits on this are, but I'd have guessed a speed change of 10$^4$m/s would be easily visible. – John Rennie Apr 10 '13 at 16:49
The speed of light will never be measured to be greater than c. Having a BA in physics, surely you've had a course in SR? – Dmitry Brant Apr 10 '13 at 17:00
People have measured the gravitational redshift (Rebka and Pound did it first) which is essentially the inverse problem. The redshift agrees with the assumption of constant $c$. This is based on the same thought that @John is talking about but also accounting for the gravitational time dilation (because $c = \lambda f$). – dmckee Apr 10 '13 at 17:09
up vote 5 down vote accepted

GPS satellites orbit at an altitude of around $15\times10^6 m$. Sometimes they are directly overhead, sometimes they are nearer the horizon. An error of 11187 m/s in the speed of light, varying with direction, would show up as an enormous error in estimating position, rendering the currently used GPS computations completely useless.

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Ok, my calculation is that the difference is at the surface and will diminish with altitude. But, your point is still well taken. I'll try to calculate the variance from 15x10^6 through the surface and see where that leads me. – aepryus Apr 10 '13 at 21:38

Everyone seems to be misunderstanding the question. The one-way speed of light cannot be measured even in principle. Einstein knew about this and even proclaimed that the one-way speed of light is not a feature of nature but rather a human preference. What we know as the speed of light c is actually 1/2 of the "Two-way speed of light." The one-way speed of light is simply DEFINED to be c for the sake of simplicity but this is not necessarily so. The one-way speed of light can be DEFINED to be ANY constant number within a range of values as long as the opposite direction of light compensates so that the Two-way speed of light is c. Many gets confused by this because this is not mentioned in basic relativity textbooks. This, I believe, for the sake of simplicity but at the expense of knowing a very peculiar and interesting part of Relativity.

To directly answer the questions:

1) No, the one-way of speed of light cannot be measured whether downward, upward, or any direction. (for a general reference see:

2) Yes. Special Relativity allows you to DEFINE a constant, one-way speed of light as long as the two-way speed of light remains c. For instance, you can define the downward speed of light to be greater than c but you would then have the upward speed of light to be less than c such that you will still get c as the two-way speed of light.

3) Our current understanding of Physics (Special Relativity) allows you to define the downward speed of light to be greater than c. Infinity is also possible (this means it takes 0 time for distant light to arrive to an observer) and have the upward speed of light to be 1/2 c. (The total time of travel of light back and forth would be 2t corresponding to two-way speed of light to be c).

3) See answer 1.

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In my subsequent research of this question, I had come to understand this point. However, (correct me if I'm wrong) it also seems possible to tease out the difference using a variation of the Michelson-Morley experiment. (Excepting for possible logistical difficulties) one could imagine a vertical variation of the Michelson-Morley experiment that would be able to detect a 11,187 m/s discrepancy. – aepryus Nov 16 '13 at 17:18

Can you measure one way speed of light? I think you can: Let’s have 5 points A, B, C, D, E equally spaced (distance AB=BC =CD=DE. Let the point C be exactly in the middle. We can indirectly synchronize the clocks at B and D by sending the signals from B and D towards C and calculating back the time on B and D clocks so the signals from B and D would arrive at C at the same time. The clocks at B and D would be perfectly synchronized if the speed of light were the same in both directions. If not, there would be a difference. But the difference will be nullified if we send the light the same distance but in opposite direction. So let’s send the signal from B to A and from D towards E at the same time as indicated on clocks B and D. Regardless of the unidirectional speed of light, the signals from B to A and from D to E will arrive at exactly the same time. So now we will have perfectly synchronized clocks at A and E. Measuring unidirectional speed of light now would be quite trivial.

Let us assume for example that the light needs 1s to travel from B to C and 3s from D to C. If we synchronize the clocks at B and D to be 0 at the moment the light was emitted (if the light signals arrive at C at the same time), the real time at the time tC as indicated on the clock at C will be as follow: tB =tC+1s;

Now if we send the signals from B to A and from D to E (at the same time as indicated on clocks at B and D) we will have the time on the clocks at A and E as follows: tA=tB+3s=(tC+1s)+3s =tC+4s tE=tD+1s = (tC+3s)+1s=tC+4s So the clocks at A and E will be perfectly synchronized.

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