Using Special relativity theory, of course. Can Lorentz transformations to "tell" something about it? "Wikipedia's" article: http://en.wikipedia.org/wiki/One-way_speed_of_light .

  • $\begingroup$ How about rotating by 180 degrees? I don't understand what could possibly be different in the 2-way speed, and the term "two way speed" is strange. So what is confusing exactly? $\endgroup$
    – Ron Maimon
    Commented Jun 1, 2012 at 19:32
  • $\begingroup$ @Ron The proposal is that the speed of light might be different on the outbound leg than on the return leg: say $c(\theta) = c/(1 + \epsilon \cos\theta)$. Obviously that is inconsistent with the principle of relativity and a Noetherian origin for the conservation of angular momentum, but it is hard to test experimentally. $\endgroup$ Commented Jun 1, 2012 at 21:43
  • $\begingroup$ For the fun of it: Measure period of revolution of a photon at Schwarzshild radius around a black hole $\endgroup$
    – IljaBek
    Commented Jun 2, 2012 at 19:37
  • 1
    $\begingroup$ @troyaner: 1.5 times the Schwarzschild radius is the circle orbit position. $\endgroup$
    – Ron Maimon
    Commented Jun 2, 2012 at 20:05
  • $\begingroup$ @RonMaimon: sry, correct $\endgroup$
    – IljaBek
    Commented Jun 6, 2012 at 13:53

3 Answers 3


The big misunderstanding about this "one-way speed of light" thing is that there is something to be proven experimentally. While it is just a matter of convention about

how to synchronize the clocks at the source and the detector

Let me explain with an example. Consider a following experimental setup:

  • We have three observers $O$,$A$ and $B$ at rest in our reference frame.
    Their positions are $x_O=0, x_A=-1$ and $x_B=1$.
  • First they synchronize their clocks following the standard procedure:
    1. The $O$ observer sends a signal at $t_O=-2$ both to $A$ and $B$.
    2. At the moment of the signal arrival, $A$ and $B$ set their clocks to $t_{A,B}=-1$
    3. $A$ and $B$ return those signals and $O$ receives them back at $t_O=0$
    4. Clocks are synchronized.
  • Now this guys measure something. For example there are two light flashes -- red and green, passing by. And observers are fixing the times of their passage.

Here is the space-time diagram for the process with the "standard" speed of light:

enter image description here
Such a diagram is drawn by the observers when they gather together at a meeting after the experiment was performed. Every observer has a list of events that locally happened at his place. And altogether they recreate "the big picture" of the experiment.

But now the observers have some doubts -- why do they presuppose that the speed of light is the same in both directions? So they agreed to change that and see what happens. Say that right-speed is three times as fast as the left-speed:

enter image description here

The description has changed. But it is just a change in the coordinates. It couldn't possibly affect any experimental predictions by any theory that observers have.

So it is just a convention. And there is nothing to be "proven" experimentally.

  • $\begingroup$ This is true, it is part of the issue regarding this, but the question was to do it experimentally, to determine if the back and forth motion in the interferometer frame is equal. You don't need two frames to check this, and you can determine this using Fizeau methods, or simply using the fact that frequency isn't changed upon reflection. $\endgroup$
    – Ron Maimon
    Commented Jun 3, 2012 at 18:44
  • $\begingroup$ @RonMaimon Well, my whole point is that there is nothing to "do experimentally". It is like asking to "experimentally prove" that there are 60 seconds in a minute. Just a convention... Regarding Fizeau experiments -- I'm pretty sure that those can be reduced to analysis of events in space-time, right? Then my argument holds. $\endgroup$
    – Kostya
    Commented Jun 3, 2012 at 19:54
  • $\begingroup$ I see your point--- the Einstein synchronization makes it obviously a coordinate issue. I was imagining adiabatic synchronization, synchronize two clocks when they are close, then slowly move one clock away to the other side of the room (in relativity, this is the same as Einstein synchronization, but we aren't granting relativity, since this is the whole point). Now ask, relative to adiabatically synchronized clocks, does the light travel with the same speeds back and forth. The methods I gave answer this question. I agree if you define synchronized Einstein's way, it's trivial. $\endgroup$
    – Ron Maimon
    Commented Jun 4, 2012 at 5:42
  • $\begingroup$ +1, BTW, yours is the best answer. I took the problem experimentally seriously, I didn't interpret it as a question about coordinate systems. This is probably the best interpretation, however. $\endgroup$
    – Ron Maimon
    Commented Jun 4, 2012 at 5:43
  • $\begingroup$ You are basically saying that if we synchronize the clocks in our experiments using light, we will not be able to detect any directional change in speed of light in the following experiment because we started with synchronizing our clocks using light. Suppose we can't see. Then we will synchronize our clocks using sound. Suppose there is eternal directional wind, so sound travels faster in one direction than the other. In that situation, you will again say that there is nothing to be proven experimentally for directional speed of sound when actually it can be proven using light as synchronizer $\endgroup$
    – Prem
    Commented Feb 11, 2016 at 15:36

Revision: use a stationary and moving dielectric

My first answer is wrong--- you can easily imagine a model in which light components are independently cancelling without being equal. The best way is by a version of Fizeau's experiment: put a dielectric along the ongoing leg of an interferometer and not the return leg, then shift the same dielectric to the return leg while at the same time removing it from the outgoing leg (you can do this instantly by a small motion of the two beams). You can do this easily with a diamond shaped dielectric, where the back and forth directions cut on opposite tapering sides. Then you slide the diamond back and forth, removing a portion from one leg and adding the same portion to the other, and look for fringes. You won't find any.

If you believe there are mysterious fitzgerald contractions making this experiment work out, you can do Fizeau's experiment--- make the interior of the dielectric a fluid moving in a given direction. If your model is consistent with both dielectrics and Fizeau, at all seasons of the year, then you will be forced to the relativity transformation for the moving frame.

I should point out that nothing prevents you from taking the time in the rest frame of the sun (say) to be the true global time, and then doing boosts by Galilean transformations. This won't preserve the metric, but it will give different laws according to the Galilean transformed metric. In these coordinates, the light travels at different coordinate speeds back and forth, and the cancellations in the two experiments above look like miracle.

Original experiment that doesn't work

This answer is updated. Originally, I said you should make an equilateral triangle of mirrors in a Michaelson interfermeter, and check for interference fringes as you rotate the apparatus slowly. I assumed that the "back and forth" speed can't cancel on back and forth paths and triangle paths simultaneously, but this is wrong--- in the Galilean boost model, the different components parallel to the velocity do the cancelling. But in more complicated models, you could get an effect here too.

This idea is sort of silly--- the light would have to be conspiratorial for something like this to work. You didn't give a definite model, if you did, it would be easier to rule it out.


A nice animation about Michelson and Morley's Experiment can be found on: http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/mmexpt6.htm

It clearly shows that the Aether doesn't influence the speed of light if you rotate the whole experiment about 90°. In fact Aether doesn't even exist.


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