I Synchronizing clocks in an inertial frame if light is anisotropic

[James Hasty]

TL;DR Summary

Given the round-trip average speed of light reflected between any two points is a constant: c = 300,000 kilometers per second. The method described below may be used to synchronize two clocks in an inertial frame without the assumption that the speed of light is isotropic.

ASSUMPTIONS
1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both.
2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals.
3. The speed of light is anisotropic.
METHOD
1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A.
2. Clock B receives the signal from A at time t[B2] and sends a light signal back to A encoded with the time t[B2].
3. Clock A receives the signal from B and reads the time t[B2].
4. The time t[B1] can now be determined by A since it is given: t[B1] = t[B2] - 2L/c.
5. Then adjust clock A time to t[A(sync)]=t[A(now)]+Δt , where Δt=t[B1]-t[A1].

 

[PeroK]

TL;DR Summary: Given the round-trip average speed of light reflected between any two points is a constant: c = 300,000 kilometers per second. The method described below may be used to synchronize two clocks in an inertial frame without the assumption that the speed of light is isotropic.

ASSUMPTIONS
1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both.
2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals.
3. The speed of light is anisotropic.
METHOD
1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A.
2. Clock B receives the signal from A at time t[B2] and sends a light signal back to A encoded with the time t[B2].
3. Clock A receives the signal from B and reads the time t[B2].
4. The time t[B1] can now be determined by A since it is given: t[B1] = t[B2] - 2L/c.
5. Then adjust clock A time to t[A(sync)]=t[A(now)]+Δt , where Δt=t[B1]-t[A1].

We might as well assume that $t[A1] = t[B2] = 0$. When A receives the signal back from B, the clock reads $t = \frac {2L}{c}$. A resets its clock to $t = \frac{2L}{c} + 0 - \frac{2L}{c} = 0$.

Effectively, B sent a signal to A at time $t_B = 0$ and when A receives the signal it sets its clock to $t_A = 0$. That's one way to do it.

Then the measured speed of light from B to A is infinite/undefined; and the speed of light from A to B is $c/2$.

 

[Sagittarius A-Star]

it is given: t[B1] = t[B2] - 2L/c.

Why is this given?

 

[Dale]

1. At time t[A1] and time t[B1], clock A sends
...
4. The time t[B1] can now be determined by A

Umm, A can already determine t[B1] in step 1.

 

[Dale]

Also did you read all of the material on this topic that we posted last time you asked a question on this same topic:

A Post in thread '1-Way Speed of Light'

Aug 30, 2025

I have Google searched for "Anderson's kappa" but not readily found. What specifically is it?

This is the OWSOL convention used in Anderson, R.; Vetharaniam, I.; Stedman, G. E. (1998), "Conventionality of synchronisation, gauge dependence and test theories of relativity", Physics Reports, 295 (3–4): 93–180. doi:10.1016/S0370-1573(97)00051-3

Anderson is not particularly a seminal author in this field. I would say that is Reichenbach. But Anderson’s convention is easier to actually use for real calculations.

• Dale

A Post in thread '1-Way Speed of Light'

Aug 30, 2025

Possibly useful (for conventions and links to references to theories and experiments):
https://en.wikipedia.org/wiki/One-w...ansformations_with_anisotropic_one-way_speeds

• robphy

• 

A Post in thread '1-Way Speed of Light'

Aug 31, 2025

I have Google searched for "Anderson's kappa" but not readily found. What specifically is it?

Here is a similar thing, Reichenbach's $\epsilon$
https://sites.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/significance_conv_sim/index.html#epsilon

Here is another explanation:
https://www.mathpages.com/home/kmath229/kmath229.htm

• Sagittarius A-Star

• 

If so, then please do the analysis right (i.e. use Anderson’s $\kappa$ or Reichenbach’s $\epsilon$ and actually calculate the values in your experiment)

 

[Ibix]

4. The time t[B1] can now be determined by A since it is given: t[B1] = t[B2] - 2L/c.

Here's your assumption of isotropy.

Edit: I misread the OP - as PeroK says,this is a $c/2$ one way, $\infty$ the other way synchronisation convention, nit an isotropic one.

 

[James Hasty]

Why is this given?

In the introduction, "it is given" that the average speed of light is c for the 2WSOL . Then t[B2] - t[B1] = 2L/c.

 

[James Hasty]

Here's your assumption of isotropy.

That is for the 2WSOL (2-way round trip of light), which is a given because it is true. I am assuming the 1WSOL is anisotropic.

 

[Ibix]

I am assuming the 1WSOL is anisotropic.

No you aren't. Dividing the round trip time by two and assuming that is the appropriate clock offset is an assumption of isotropic light speed.

As Dale said, do the maths with $\kappa$ and see what comes out.

 

[James Hasty]

Umm, A can already determine t[B1] in step 1.

An observer can only know that if the 1WSOL is isotropic. I am assuming it is not.

 

[James Hasty]

No you aren't. Dividing the round trip time by two and assuming that is the appropriate clock offset is an assumption of isotropic light speed.

As Dale said, do the maths with $\kappa$ and see what comes out.

Let's be clear on definitions:
The 2WSOL can be calculated by timing a reflected light beam using 1 clock. For example: A sends a signal to point B, it is reflected back to A. Tee speed of light is calculated: c = 2L / ( t[A2] - t[A1] ). It is the same for all observers.
The 1WSOL cannot be calculated, unless two clocks are first synchronized which requires: (a) assuming the 1WSOL is isotropic, or (b) defining the speeds for A to B and B to A.

 

[Sagittarius A-Star]

In the introduction, "it is given" that the average speed of light is c for the 2WSOL . Then t[B2] - t[B1] = 2L/c.

No. At time t[B2] clock B receives light from clock A, that has moved only once over the distance L.

 

[Ibix]

I misread the OP. As @PeroK says in #2, you have a $c/2$ one-way speed in the A-to-B direction and an $\infty$ speed in the B-to-A direction.

Note that this has the odd effect that two simultaneous events can be causally connected, but only in one spatial direction.

 

[PeroK]

Let's be clear on definitions:
The 2WSOL can be calculated by timing a reflected light beam using 1 clock. For example: A sends a signal to point B, it is reflected back to A. Tee speed of light is calculated: c = 2L / ( t[A2] - t[A1] ). It is the same for all observers.
The 1WSOL cannot be calculated, unless two clocks are first synchronized which requires: (a) assuming the 1WSOL is isotropic, or (b) defining the speeds for A to B and B to A.

Not really. You just need a synchronization procedure. That defines the 1wsol.

There are no physical assumptions, per se, in defining a coordinate system.

 

[Dale]

An observer can only know that if the 1WSOL is isotropic. I am assuming it is not.

No. They know that at step 1 because that is the time in their own clock when they send the signal. They could be sending the signal by FedEx for all that matters.

 

[Dale]

Let's be clear on definitions

Indeed, to be clear post a spacetime diagram with each event labeled and each signal indicated. And perform your calculations using one of the standard notations provided to you earlier.

Also, use Latex for all of your math. When you have prepared that send me a DM and I will reopen the thread.

Edit: but I do think it is clearly unfeasible. You know the times the signal was emitted, received, and returned. You need to calculate the speed of light there, the speed of light back, the length, and the offset. Seems like too few equations in too many unknowns.

 

[Dale]

When you have prepared that send me a DM and I will reopen the thread.

@James Hasty has prepared the diagram and some careful thoughts so I am reopening the thread.

 

[PAllen]

I misread the OP. As @PeroK says in #2, you have a $c/2$ one-way speed in the A-to-B direction and an $\infty$ speed in the B-to-A direction.

Note that this has the odd effect that two simultaneous events can be causally connected, but only in one spatial direction.

I would think of that more like the case of lightlike coordinates. That a time coordinate is the same along a null path in some direction doesn't make it a possible set of simultaneous events. I would still require that a path be spacelike to make such claim.

 

[Dale]

a time coordinate is that same along a null path in some direction doesn't make it a possible set of simultaneous events. I would still require that a path be spacelike to make such claim.

There is disagreement in the literature on this point. But I tend to agree with you. Anderson, in contrast, does not restrict his $\kappa$ to be less than 1.

 

[Ibix]

I would think of that more like the case of lightlike coordinates.

But I tend to agree with you.

I also agree. To call what you get with a homogenous but anisotropic two way speed a "simultaneity plane" you need to require that $c/2<c_\pm<\infty$. Going further than that turns one of your spacelike coordinates null or timelike, which is a fine coordinate system but doesn't have a natural interpretation in terms of "stationary objects" and "simultaneous events".

I should have put "simultaneous" in scare quotes in #13.

 

[pervect]

TL;DR Summary: Given the round-trip average speed of light reflected between any two points is a constant: c = 300,000 kilometers per second. The method described below may be used to synchronize two clocks in an inertial frame without the assumption that the speed of light is isotropic.

ASSUMPTIONS
1. Two identical clocks A and B in the same inertial frame are stationary relative to each other a fixed distance L apart. Time passes at the same rate for both.
2. Both clocks are able to send/receive light signals and to write/read the send/receive times into signals.
3. The speed of light is anisotropic.
METHOD
1. At time t[A1] and time t[B1], clock A sends a light signal to clock B. The clock B time is unknown to A.
2. Clock B receives the signal from A at time t[B2] and sends a light signal back to A encoded with the time t[B2].
3. Clock A receives the signal from B and reads the time t[B2].
4. The time t[B1] can now be determined by A since it is given: t[B1] = t[B2] - 2L/c.
5. Then adjust clock A time to t[A(sync)]=t[A(now)]+Δt , where Δt=t[B1]-t[A1].

As long as you don't believe that your clock synchronization has any physical significance, in particular as long as you don't address the issue of the laws of motion in your supposedly inertial frame, I don't see any issues.

If you were to make further assumptions like assuming that your clock synchronization is compatible with Newton's laws in the low speed limit, then you probably do have an issue. I'm guessing is that you are making such assumptions, just not stating them explicitly.

To illustrate the problem, consider two identical masses, moving at the same measured velocity. If the clocks are synchronized fairly, then when the two identical masses with the same velocity collide, they will come to a stop.

There is only one clock synchronization scheme that will make this happen exactly, assuming you compute the velocity by dividing the distance travelled between time difference between your two clocks, syncrhoized according to your scheme of choice.

What you have failed to demonstrate, and most likely failed to consider, is whether the particular clock scheme you have proposed is this unique clock synchronization scheme (usually called a fair clock synchronization and also called an isotropic clock syncrhonization) that makes two identical masses moving at identical velocities come to a stop when they collide inelastically.

It's easiest to see this in a non relativistic example. Suppose you synchronize clocks by the position of the sun overhead. Then with a fast enough plane, you can make a trip instantaneously in one direction, while the trip in the other direction will take twice the amount of time, where the time difference is computed by subtracting the clock readings. This infinite velocity, if you apply p=mv from newton's laws, would give an infinite momentum.

THis example is deliberately exxagerated to show the effect. It demosntrates the principle, though - the principle being that if we use Newton's laws, there is an implied clock syncrhonization that makes those laws work in the applicable limit.

The idea that the plane physically moves at an infinite velocity is just wrong in principle. The plane does not actually have an infinite momentum, as can be seen by looking conceptually at what would happen if they collided inelastically. (Note that this is a thought experimeint, real planes tend to disintegrate).

To describe the situation loosely, I would say that the "infinite" velocity doesn't have any direct physical significance. But this isn't too clear as to what I mean by "physical significance". So it's better to spell it out in more detail.

Another clue that something is wrong is that the "instantenous" trip in one dirtection takes a finite amount of time as measured by the passenger on the plane. This observation works for anything but light to define isotropy and fairness of clock syncrhonization, but since light doesn't experience time, we can't even conceputally use the idea of comparing the trip time as measured by two clocks to the trip time as measured by one onboard clock.

THis isn't really that unexpected an issue. If you are timing footraces, for instance, using a clock at the finish line and a clock at the start line, you probably already realize that you need to synchronize the clocks fairly for the race time to count for a world record.

 

[pervect]

I'd like to expand on my earlier point.

There is only one clock synchronization scheme that will make this happen [[make Newton's laws work exactly]] , assuming you compute the velocity by dividing the distance traveled between time difference between your two clocks, synchroized according to your scheme of choice.

Suppose we have some frame in which Newton's laws work. Two equal masses moving in opposite directions at the same velocity as measured by the two-clock system of measuring velocity collide, and come to an exact stop.

Now, let's change the clock synchronization. We haven't changed the physics at all, so the two masses still collide, and they still stop. But - we've changed how we measure the velocities, by changing the way we synchronize clocks.

An example will illustrate this. Consider the right-moving mass. It starts at the left, and arrives at the right. Suppose it has a velocity of 1. It started out at t=0, and arrived at t=1, covering a 1 unit distance course in one second.

But now we change the clock synchronization between the two clocks. There are several ways we could do this, but we'll assume that we keep the left clock reading the same, and add or subtract time to the right clock. You can do the reverse, it doesn't matter.

Now, the trip time for the mass moving to the right has changed. This means it's velocity is no longer 1. If we add time to the right clock, it might take 1.2 seconds to make the journey, making it's velocity according to our two-clock measurmenet lower. And if we subtract time, it's velocity would be higher.

If we assume that Newton's laws work, we are restricted in how we can change our clock synchronizations and keep the law working.

In the example of two colliding planes (or masses), we actually have three clocks to worry about. If we add equal adjustments to all clocks, we don't change the synchronization or the measured velocities. We can, technically, change the synchronization of the midpoint clock without messing things up. But changing the synchronization between the left clock and the right clock will inevitably lead to the velocities, which used to be the same, no longer being the same. The physics doesn't care about that - if the colliding masses came to a stop before, they'll still come to a stop. But calculating the momentum from Newton's formula, p=mv, will no longer give the correct answer. Because we've changed how we measure velocities.

Thus, if we have some clock synchronization scheme that makes Newton's laws work (in particular, equal masses moving at equal velocities except in opposite directions come to a stop when they collide), that clock synchronization is unique. Newton's laws need a certain type of clock synchronization to work, we call this a "fair" or "isotropic" clock synchronizaton.

Einstein's postulate is basically saying that the exact same clock synchronization that works for matter is the same clock synchronization that makes the speed of light the same in both directions.

 

[Sagittarius A-Star]

Thus, if we have some clock synchronization scheme that makes Newton's laws work (in particular, equal masses moving at equal velocities except in opposite directions come to a stop when they collide), that clock synchronization is unique. Newton's laws need a certain type of clock synchronization to work, we call this a "fair" or "isotropic" clock synchronizaton.

Einstein's postulate is basically saying that the exact same clock synchronization that works for matter is the same clock synchronization that makes the speed of light the same in both directions.

Einstein modified Newton's definition $p_x=mv_x$ to $p_x=m {dx \over d\tau}$. This does not depend on the clock-synchronization scheme.

Newton's laws do hold in an inertial coordinate system based on anisotropic one-way-speeds, if the underlying definitions are modified to the relativistic ones.

 

[pervect]

Einstein modified Newton's definition $p_x=mv_x$ to $p_x=m {dx \over d\tau}$. This does not depend on the clock-synchronization scheme.

Newton's laws do hold in an inertial coordinate system based on anisotropic one-way-speeds, if the underlying definitions are modified to the relativistic ones.

Yes, I personally think it would be simpler to introduce relativity by means of proper velocity, or as you write it $dx/d\tau$. Do you have an attribution for where Einstein actually wrote that? I was trying to make a case for introducing relativity by means of proper velocity, which as you point out do not require clock synchronization at all, but I got a lot of pushback on my attempts, so showing that it was, in fact, Einstein's original formulation all along would be very helpful.

Another advantage to using proper velocity as opposed to the standard "two clock" velocity, besides the simple relation between it and momentum and it's coordinate independence, is that ranges from zero to infinity, better matching our ideas of their not being any upper limit to velocity. One can, in theory, travel any arbitrarily large distance in an arbitrarily short of experienced, proper time, the "speed of light" limit does not change this.

The one thing that convinced me that one DOES eventually need to deal with clock synchronization at all (rather than simply choosing to sidestep it by using proper velocities instead of the two-clock method) is the need to appreciate the SI definition of the meter. This is, again, in a convention, but it's another very important one.

The metre, symbol m, is the SI unit of length. It is defined by taking the fixed numerical value of the speed of light in vacuum c to be 299 792 458 when expressed in the unit m s–1, where the second is defined in terms of the caesium frequency

Since accepting the SI definition of the meter makes any discussion of the one-way speed of light pretty irrelevant, (as we've just literally defined it as a constant), it's fairly clear to me that people who like to think and talk about the issue don't accept it.

Actually, I'm OK with that for the purposes of discussion, as long as the moderators don't complain. It unfairly shuts down thought and understanding to insist too strongly on following the current standards. However, I've found that very few people are interested in talking or even thinking about the issue. At first, I tried using my own personal, older definition of the meter, based on the prototype meter standard, as being likely what they were thinking. (And also likely what was used historically, when people wrote about measuring the speed of light. I've forgotten the details of the history of the meter, but I think it's likely that most discussions of measuring the speed of light were written at a time when the prototype meter bar was THE standard, and hence measuring the speed of light was a perfectly sensible thing to do).

But it became clear that people weren't necessarily thinking of it in those terms. Fair enough - but - what ARE they thinking then? When I ask I usually get silence, not even a vague hint, so I have to guess.

As I said, I've gotten a lot of silence, but my thoughts are that distance is actually something "in people's heads", usually along with Euclidean geometry, and they just don't want to or can't write about it. But it could just be a lack of interest.

 

[PeterDonis]

my thoughts are that distance is actually something "in people's heads"

I don't think distance itself is something in people's heads. But definitions of units of distance are. As you note, the SI definition of the meter depends on adopting an isotropic clock synchronization convention. (It also makes an implicit assumption that one is working in an inertial frame.) But that is a definition of a unit of distance, not a definition of the concept of distance itself.

One way of seeing that distance itself isn't just something in people's heads is to remember what someone (was it Wheeler?) said about time and space: "Time is what keeps everything from happening at once. Space is what keeps everything from happening to me." The facts that not everything happens at once, and that not everything happens at the same place, are real attributes of real events, not just things in people's heads, and those facts are what our intuitive concepts of "time" and "space" (and "distance" is really just another name for "space", for the fact that not everything happens in the same place) try to capture.

The issues arise when one tries to formulate a model that can incorporate these concepts and make accurate predictions. There are indeed many pitfalls and traps for the unwary lurking there.

 

[Nugatory]

The one thing that convinced me that one DOES eventually need to deal with clock synchronization at all (rather than simply choosing to sidestep it by using proper velocities instead of the two-clock method) is the need to appreciate the SI definition of the meter.

Does that force us to deal with clock synchronization? Start with a source of light signals and a mirror, move the mirror around until the reflected signal is back at the source in 2/299792458 seconds and we have a one-clock standard meter.

I’m thinking that we need clock synchronization for a more practical reason: it’s just too convenient in too much of life to not use it. I can’t even decide when to leave home to pick up my child from daycare without assuming synchronized clocks.

 

[Dale]

Suppose we have some frame in which Newton's laws work. Two equal masses moving in opposite directions at the same velocity as measured by the two-clock system of measuring velocity collide, and come to an exact stop.

Now, let's change the clock synchronization. We haven't changed the physics at all, so the two masses still collide, and they still stop. But - we've changed how we measure the velocities, by changing the way we synchronize clocks.

Indeed, if I am reading my notebook correctly the Lagrangian for a free particle in an Anderson synchronized frame is $$L=-(\dot T+(\kappa-1)\dot X)(\dot T +(\kappa+1)\dot X)+\dot Y^2 + \dot Z^2$$ This leads to a conserved energy $$E=\frac{m(1+V_X \kappa)}{\sqrt{1-V_Y^2-V_Z^2+2V_X\kappa+V_X^2(\kappa^2-1)}}\approx m+\frac{1}{2}m V^2-m\kappa V^3$$ and a conserved momentum $$p=\left( \begin{array}{c}
\frac{m(\kappa+V_X(\kappa^2-1))}{\sqrt{1-V_Y^2-V_Z^2+2 V_X\kappa + V_X^2(\kappa^2-1)}}\\
\frac{m V_Y}{\sqrt{-V_Y^2-V_Z^2+(1+V_X(\kappa-1))(1+V_X(1+\kappa))}}\\
\frac{m V_Z}{\sqrt{-V_Y^2-V_Z^2+(1+V_X(\kappa-1))(1+V_X(1+\kappa))}}
\end{array} \right)\approx -m \kappa + m V$$

Notice that for energy it is not a huge effect, coming in at 3rd order in V. But for momentum it is a 0th order effect in V. Indeed, I think that would make it fair to say that Newton's laws don't work in that frame even though technically you could modify stuff so that some version of it would technically work.