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.