There is something I don't quite get about relativistic velocities, which I am hoping to clarify with this question.

Suppose you have an emitter and receiver both located somewhere far away from one another, and far away from any gravitational field. The emitter and receiver do not need to be synchronized in any way. They are both in a reference frame such that they appear to be at rest with one another. For clarity, all "speeds," "velocities" and etc are with respect to this reference frame.

The emitter then transmits two electrons simultaneously toward the receiver. Each electron is propelled forward using some precisely determined amount of kinetic energy, such that we can compute either its Newtonian or relativistic speed just from its initial kinetic energy. These two kinetic energies are different, but sufficiently high that relativistic effects become important. The receiver then records the time differential between the two electrons arriving.

The basic idea is this: the Newtonian and relativistic equations for kinetic energy give different predictions of what this time differential is. In the relativistic framework, the same amount of kinetic energy leads to lower resulting speeds than the Newtonian one, and this changes the time differential. The amount to which these two predictions differ depends only on the value of $c$, such that as $c \to \infty$, the relativistic prediction agrees with the Newtonian one.

To put some numbers to this: suppose we sent the first electron with 1 GeV of kinetic energy, and the second with 10 MeV. Then Newtonian mechanics predicts a time differential of 144 ms, and SR predicts a time differential of only 1 ms. This is because in Newtonian mechanics, the 1 GeV electron is traveling 10x as fast as the 10 MeV electron, and thus gets there much faster. But in SE, both are traveling at approximately the same speed, which is getting asymptotically very close to $c$. Thus, the time between the two measurements is very small. Basically, we are measuring the extent to which increasing the kinetic energy of an electron beyond a certain point fails to increase its speed an appreciable amount. If we were to imagine that $c$ were higher, the measured time between the two would increase and tend toward the Newtonian value as $c \to \infty$. (I used these calculators for this: relativistic, Newtonian).

The problem: it would seem we even can use this to provide an estimate of $c$. Given knowledge of the initial kinetic energies, as well as how far the receiver is from the emitter, we can see which value of $c$ leads to the measured time differential. Basically, we are looking at the relationship between change in kinetic energy and change in apparent velocity, and measuring "how non-Newtonian" it is. But the problem is: we can rotate the setup and try again, and see if things are isotropic or not. If we measure something different in different directions, we can see which value of $c$ gives the measured result in each direction. Now we've measured the one-way speed of light in each direction, which is supposed to be impossible.

My question: What is going on? Is there something wrong with this experimental setup? How does this relate to the principle that there is no way to measure the "one-way speed of light," or differentiate between Lorentz's ether theory and SR?

(FWIW: I have searched a bunch of these "one-way speed of light" questions and none of them seem to address this, and frankly this seems much more subtle than most of those anyway.)

EDIT: I think this is related to the "Bertozzi experiments": https://en.wikipedia.org/wiki/Tests_of_relativistic_energy_and_momentum#Bertozzi_experiment

So does look like the suggested experimental setup is feasible, and something like it has even been done to calibrate particle accelerators (e.g. determine how much energy is needed to accelerate an electron to some speed). The "Bertozzi experiments" basically measure this relationship between kinetic energy and velocity. I guess I'm just wondering what is wrong with the naive idea to do this kind of experiment, then rotate, and do it again to measure anisotropy. Basically, what we are really measuring is the relationship between change in kinetic energy and change in apparent velocity for massive particles, and seeing if this relationship is the same in all directions; it doesn't seem like there are any clocks that need to be synchronized or any round-trip that needs to happen at all since we are only very indirectly measuring the speed of light via this particular relativistic effect. I mean, we don't even need to assume anything about the exact model that lets you infer $c$ from the time differential, since it may not even be philosophically valid to use the relativistic KE equation with anisotropic $c$. All we need to do is assume that some relativistic effect exists that becomes relevant at higher kinetic energies, and which would affect the time differential in some unknown way which would depend on the value of $c$, and then see if we measure the same time differential after rotating in different directions.

EDIT 2: this answer gives an interesting summary of Reichenbach's research on this. Basically, there is (what the author terms) a "conspiratorial anisotropy" effect that takes place, where all kinds of physical quantities distort in very strange ways as a result of changing the one-way speed of light such that the experimental results are always the same. It makes sense in principle; I guess I'm just curious regarding the details of "how the magic happens" in this particular situation. I deliberately chose a very simple experimental setup that avoids having to synchronize multiple clocks (or at least I think, anyway), and just looks at this KE-velocity relationship in different directions, and I am curious what kind of bizarre stuff has to happen to somehow make this all appear isotropic at the end.

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    $\begingroup$ If the speed of light isn't anisotropic you do need to use different equations for KE, etc. Or just transform to a system with $\varepsilon=\frac12$. ;) See sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/… $\endgroup$
    – PM 2Ring
    May 20, 2022 at 5:32
  • $\begingroup$ Thanks @PM2Ring. I would guess you are saying the modified equations for KE would be those that make the KE effect isotropic in all directions? Even though $c$ is anisotropic. I had thought of that but it seems there'd be other issues with that... $\endgroup$ May 20, 2022 at 5:39
  • $\begingroup$ I'm also not quite sure if the stuff Reichenbach is talking about is quite the same... I get the basic idea of Einstein's synchronicity convention and how it is arbitrary, and the difficulty of directly measuring the one-way speed of light as a result. But measuring anisotropy in the kinetic energy/velocity relationship seems to be a different thing and I'm not sure how it relates... $\endgroup$ May 20, 2022 at 5:41
  • $\begingroup$ From physics.stackexchange.com/a/591436/123208 for any other value [of epsilon] the one way speed of light is anisotropic but in a very specific way that is sometimes called "conspiratorial anisotropy". It is anisotropic, but in a way that does not affect any physical measurement. Instead this synchronization convention causes other things like anisotropic time dilation and even anisotropic stress-free torsion which conspire to hide the anisotropic one way speed of light from having any experimental effects. $\endgroup$
    – PM 2Ring
    May 20, 2022 at 5:44
  • $\begingroup$ Thanks @PM2Ring, I have seen some of this before and I have even heard of the "conspiratorial anisotropy" thing, I guess I am just curious of the details regarding how that happens in this particular situation. For this to be true the kinetic energy/velocity relationship should also be anisotropic but then you get some truly strange things. For instance, suppose that you have $\epsilon = 1$, meaning the speed of light is $c/2$ in one direction and infinite in the other direction. Then suppose you send off a proton in addition to the two electrons. (1/2) $\endgroup$ May 20, 2022 at 5:55

2 Answers 2


This is a very interesting question. Indeed, as a one-way-speed-of-light (OWSOL) experiment the answer is a duplicate of all of the other myriad OWSOL questions: there is no possible experiment which can measure the OWSOL. However, this question is unique in that it is the only question I have ever seen that does not use light in the experiment and instead relies on relativistic kinematics to make the argument. I am not sure how to treat questions that are not duplicates with answers that are duplicates, so I will go ahead and answer.

Indeed, as you point out there is no way to measure the OWSOL, so this experiment is no exception. To measure the OWSOL it must be treated as a variable and then the equations of the experiment need to be solved to find that variable. Here, I will use Anderson's convention and use units such that the two-way speed of light is $c=1$. Anderson's $\kappa$ is related to the more famous Reichenbach $\epsilon$ by $\kappa = 2 \epsilon -1$, but I find Anderson's approach more convenient to work with. I will use lower case letters for variables in a standard Einstein synchronized frame (Minkowski metric with isotropic OWSOL) and upper case letters for variables in the Anderson synchronized frame (anisotropic OWSOL).

The transformation between the frames is: $$t=T+\kappa X$$ $$x = X$$ and the metric in the Anderson frame is $$ds^2=-dT^2+\left(1-\kappa^2 \right) dX^2-2\kappa \ dT \ dX$$

So in an Einstein frame if electron 1 is emitted with velocity $v_1$ and electron 2 is emitted with velocity $v_2$ then in the Anderson frame the velocities are $$V_1=\frac{v_1}{1-\kappa \ v_1}$$ $$V_2=\frac{v_2}{1-\kappa \ v_2}$$ Then to travel a distance $L$ requires a time $T_1=L/V_1$ and $T_2=L/V_2$. Substituting and simplifying we get $$T_1=L\left(\frac{1}{v_1}-\kappa\right)$$ $$T_2=L\left(\frac{1}{v_2}-\kappa\right)$$ $$\Delta T = T_2-T_1 = L\left(\frac{1}{v_2}-\frac{1}{v_1} \right) $$ So indeed a measurement of $\Delta T$ provides no information on $\kappa$, the OWSOL asymmetry parameter, and this experiment will yield the same answer regardless of the OWSOL.

The above analysis is a purely kinematical analysis and applies irrespective of the dynamics. Regardless of the dynamics, the experiment provides no information on the OWSOL. However, it is also instructive to delve into the dynamics regarding the energies of the electrons, since that was a key element in distinguishing Newtonian from relativistic physics.

From the metric the Lagrangian of a free massive particle in the Anderson frame is $$\mathcal{L}=m\sqrt{\dot T^2-\left(1-\kappa^2 \right) \dot X^2+2\kappa \ \dot T \ \dot X}$$ which is cyclic in $T$. This leads to a conserved total energy of $$E=m\frac{1+\kappa \ V}{\sqrt{1-\left(1-\kappa^2\right) V^2+2 \kappa \ V}}$$ So the relationship between velocity and kinetic energy is also anisotropic in the Anderson coordinates unless $\kappa=0$ and we recover the standard relativistic formula. I leave it as an exercise to the interested reader to show that the above formula for energy leads to the correct velocity in each direction shown above. However, note that the Taylor series expansion of the total energy is $$E = m + \frac{m}{2} V^2 - \kappa m V^3 + \left(\frac{3m}{8} + \frac{3m \kappa^2}{2}\right) V^4 + O(V^5)$$ which includes the rest energy $m$, the Newtonian KE $\frac{1}{2}mV^2$ and the usual lowest order relativistic energy correction term $\frac{3}{8}mV^4$. However, note that the lowest order anisotropy correction term $-\kappa m V^3$ is larger than the relativistic correction term.

  • $\begingroup$ A nice answer. Separately, @dale, I have seen people muse on the possibility that one might test the anisotropy of the speed of light by an arrangement in which light travels out and is returned by reflection, but one leg of the journey is through a medium with a higher refractive index than the other. Does the 'conspiratorial anisotropy' defeat such a scheme? I assume so, but an explanation of how would be super. $\endgroup$ May 20, 2022 at 17:25
  • $\begingroup$ @MarcoOcram Sounds like an answer to be made on a separate question $\endgroup$ May 20, 2022 at 18:26
  • $\begingroup$ @MarcoOcram you can use the velocity formula above to derive a formula for the index of refraction. It is anisotropic in exactly the needed amount to make the outcome independent of $\kappa$. If you need more details then ask, but ask simply about the index of refraction In anisotropic OWSOL coordinates without asking about measuring the OWSOL $\endgroup$
    – Dale
    May 20, 2022 at 18:39
  • $\begingroup$ @dale I will do so! $\endgroup$ May 20, 2022 at 19:08
  • $\begingroup$ Thanks @Dale, that is a very interesting answer! The only remaining question l have is, is there some physical interpretation of this altered quantity? For instance, with other experiments that claim to measure the OWSOL, it is easy to show how things go wrong. For instance, if you try to synchronize two clocks and slowly move them apart, there will be an anisotropic time dilation and this leads to the same predictions as if c were isotropic. Is there some similar way to interpret this result here, with maybe some fictitious force arising and etc, due to the way the axes have been shifted? $\endgroup$ May 21, 2022 at 5:06

Photons are bound by relativity. Even if a really fast vessel. It is o ly when an object approaches 1/10th c that dilation becomes heavily noticeable. One aspect of Special Relativity speed of light perpetuation...namely the photon, is constant for all frames of reference. Since the clock that is synchronized with the ship is hours or days behind...time is only perceptible.... we send light speed signals to craft in space all the time.... Voyager, pioneer, new horizons. At light speed from earth to ship it would take its time to travel. Ship to craft....would appear similar. If you have a ship going 70% would take 9 extra hours.

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    $\begingroup$ I think your answer is missing the crux of the OP's question. $\endgroup$
    – PM 2Ring
    May 20, 2022 at 5:33

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