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By this I mean say you had and entire galaxy to work with, and you could measure the exact wavelengths of the waves you send out and record it. You would shoot this light beam across the galaxy and measure the exact wavelength of the light you are transmitting, and receiving. Could you compare the difference in the wavelengths between the same light and calculate the one way speed of light from it?

benrg
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Baker Rentmeister
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  • How does "and entire galaxy" help? Why do you expect a difference in the wavelengths? – PM 2Ring May 11 '23 at 02:01
  • Or this https://physics.stackexchange.com/questions/590983/can-one-way-speed-of-light-be-instantaneous – Dale May 11 '23 at 02:06

1 Answers1

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Is it possible to measure the one way speed of light with [...]

No. The one-way speed of light is a statement about coordinates. You can always pick a non-standard coordinate system in which the speed of light is anisotropic - even infinite in one direction and finite in the other.

If you have a (1D, for simplicity) network of rulers and clocks which are all synchronized as per the standard Einstein synchronization convention, then all you need to do is set each clock backward by an amount $\Delta t = -x/c$, where $x$ is the clock's spatial coordinate. If you release a light pulse from the origin traveling to the right at 3:00 PM, then the clock offsets will make it so that the the light pulse passes every clock at 3:00 PM, and the one-way speed of light to the right will be infinite. For the same reason, the one-way speed of light to the left will be cut in half.

From a more mathematical perspective, by offsetting your clocks you are defining new coordinates $X=x$ and $T = t - x/c$. In this new coordinate system, the Minkowski line element takes the form

$$c^2 \mathrm d\tau^2 = c^2 \mathrm dt^2 - \mathrm dx^2 = c^2(\mathrm dT + \mathrm dX/c)^2 - \mathrm dX^2$$ $$= c^2 \mathrm dT^2 +2c \mathrm dT \mathrm dX = c^2\mathrm dT \big(\mathrm dT + 2\mathrm dX/c)$$

Light follows trajectories whereby $\mathrm d\tau^2 = 0$, which means that either $\mathrm dX/\mathrm dT = -c/2$ (the light pulse travels to the left at $c/2$) or $\mathrm dT = 0$ (the light pulse travels to the right with no lapse in $T$, i.e. $\mathrm dX/\mathrm dT \rightarrow \infty$.

The main takeaway is that this is a coordinate artifact. If you do your analysis in $(X,T)$ coordinates, then the velocity of a light pulse will be either $\infty$ or $-c/2$. If you do your analysis in $(x,t)$ coordinates, then the velocity will be $ \pm c$. Any experiment you can possibly imagine which indicates the latter has used (at some point) the assumption that you are in standard Minkowski coordinates.

J. Murray
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