# Transitivity condition for Einstein synchronization and the one-way light speed

1) Consider that one way light speed is anisotropic.
2) Use the Einstein method to synchronize all the watches by the watch located at (x,y,z)=(0,0,0).

Now, all the watches are synchronized by the watch at (0,0,0) but they are not necessarily synchronized with each-other (Consider watches that are not in one line). This means that the transitivity condition is not met by the synchronization. So the synchronization is not an equivalence relation. If so, the one-way light-speed is an observable quantity regardless of how we synchronize watches. Perhaps you are not agree with me, but could you spare me what I am missing? Or do you also want to accept that one way light speed is a measurable quantity?

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JesseM
Now, all the watches are synchronized by the watch at (0,0,0) but they are not necessarily synchronized with each-other (Consider watches that are not in one line).
How do you figure? If all the watches are at rest relative to one another, then if two pairs A,B and A,C meet Einstein's synchronization condition, then it's guaranteed that the pair B,C will meet it too.

DrGreg
Gold Member
1) Consider that one way light speed is anisotropic.
2) Use the Einstein method to synchronize all the watches by the watch located at (x,y,z)=(0,0,0).
These two conditions contradict each other. It's not possible for both to be true at the same time. If you use the Einstein method to sync all watches then you have forced the one-way speed of light to be isotropic. If you sync the watches in such a way that the one-way speed of light is anisotropic then you haven't used the Einstein method.

pervect
Staff Emeritus
Consider a non-rotating inertial frame of reference. Then you can synchronize all clocks with a central clock via Einstein's method, and find that they are all synchronized with each other.

Now, however, suppose you are on a rotating frame of reference, a rotating disk. Then if you synchronize all clocks via Einstein's method to the central clock, you will find that the clocks at the edge are not synchronized via the Einstein's convention.

So, transivity of Einstein synchronization tells us something interesting about rotation, but I don't see how it tells us anything about the one-way speed of light.

Consider a non-rotating inertial frame of reference. Then you can synchronize all clocks with a central clock via Einstein's method, and find that they are all synchronized with each other.

Now, however, suppose you are on a rotating frame of reference, a rotating disk. Then if you synchronize all clocks via Einstein's method to the central clock, you will find that the clocks at the edge are not synchronized via the Einstein's convention.

So, transivity of Einstein synchronization tells us something interesting about rotation, but I don't see how it tells us anything about the one-way speed of light.

Suppose that we are in an inertial frame. Then consider the triangle ABC.
a) Synchronise A and B with the Einstein method.
b) Synchronise A and C with the Einstein method.
c) Measure the light speed that travels from B to C.
d) Measure the light speed that travels from C to B.
The Einstein synchronisation implemented in "a" and 'b" does not guarantee that the results of "c" and "d" coincide to each other. This means measuring the one-way speed of light in my eyes.

If you use the Einstein method to sync all watches then you have forced the one-way speed of light to be isotropic.
Use the Einstein method to sync all watches for the observer sitting at the origin. Then this observer can not measure the one-way speed of light. The one way-light speed is isotropic for this observer. But you have synchronised all the watches, so please ask other observers to measure the one-way speed of light.

ghwellsjr
Gold Member
It is not possible for anyone to ever measure the one-way speed of light. We can only measure the "average" round trip speed of light using a mirror at some measured distance away and it always calculates to the constant c. Einstein postulates that each half of the round trip measurement takes the same amount of time and this enables us to use his method to synchronize all the clocks that are stationary with respect to each other. If we synchronize clocks A and B and we also synchronize clocks A and C, then we know that clocks B and C will also be synchronized. Remember, the process of "synchronizing clocks" is nothing more than "defining" the two directions of light travel time between two points to be equal. If, after applying this definition, we attempt to measure the one-way speed of light, of course we will get c, but that is because we defined it that way.

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It is not possible for anyone to ever measure the one-way speed of light. We can only measure the "average" round trip speed of light using a mirror at some measured distance away and it always calculates to the constant c. Einstein postulates that each half of the round trip measurement takes the same amount of time and this enables us to use his method to synchronize all the clocks that are stationary with respect to each other. If we synchronize clocks 1 and 2 and we also synchronize clocks 1 and 3, then we know that clocks 2 and 3 will also be synchronized.
I repeat myself. What is wrong with the algorithm below:
Suppose that we are in an inertial frame. Then consider the triangle ABC.
a) Synchronise A and B with the Einstein method.
b) Synchronise A and C with the Einstein method.
c) Measure the light speed that travels from B to C.
d) Measure the light speed that travels from C to B.
The Einstein synchronisation implemented in "a" and 'b" does not guarantee that the results of "c" and "d" coincide to each other. This means measuring the one-way speed of light in my eyes.

ghwellsjr
Gold Member

You need to edit your post again to be consistent since you have added "d" where I think you mean "b".

You need to edit your post again to be consistent since you have added "d" where I think you mean "b".
I mean "d"

If we synchronize clocks A and B and we also synchronize clocks A and C, then we know that clocks B and C will also be synchronized.
I disagree with this point. If one-way light speed is anisotropic, synchronisation of A and C, and synchronisation of B and C does not necessarily imply synchronisation of A and B.

As a simplified example: Please draw a triangle. Assume a clockwise speed and an anti-clockwise one for light on the edge of the triangle. The light pulse that travels clockwise shall go faster or slower than the one on the anti-clockwise direction. This difference is independent of how you synchronise your watches on this special case placed on the corner of the triangle. This difference is a physical quantity, while you are arguing it is not.

JesseM
I repeat myself. What is wrong with the algorithm below:
Suppose that we are in an inertial frame. Then consider the triangle ABC.
a) Synchronise A and B with the Einstein method.
b) Synchronise A and C with the Einstein method.
c) Measure the light speed that travels from B to C.
d) Measure the light speed that travels from C to B.
The Einstein synchronisation implemented in "a" and 'b" does not guarantee that the results of "c" and "d" coincide to each other. This means measuring the one-way speed of light in my eyes.
It does guarantee it, at least if the two-way speed is still c. Try coming up with an example and doing the math, you'll see (for example, suppose that in your chosen coordinate system, light has a speed of c+v in one direction and c-v in the opposite direction)

ghwellsjr
Gold Member
I mean "d"
Sorry, I didn't notice the difference in your lower-case nomenclature vs your upper-case.
I disagree with this point.
What specifically do you disagree with? That you cannot measure the one-way speed of light OR that you cannot define the two halves of the round-trip measured speed of light to take the same time OR just that synchronizing different pairs of clocks does not guarantee that other pairs will also be synchronized OR something else?
If one-way light speed is anisotropic, synchronisation of A and C, and synchronisation of B and C does not necessarily imply synchronisation of A and B.
Very true but Einstein defined light speed to be isotropic not anisotropic. If you want to define light speed to be anisotropic, then why do you care about Einstein's clock synchronization method?
As a simplified example: Please draw a triangle. Assume a clockwise speed and an anti-clockwise one for light on the edge of the triangle. The light pulse that travels clockwise shall go faster or slower than the one on the anti-clockwise direction. This difference is independent of how you synchronise your watches on this special case placed on the corner of the triangle. This difference is a physical quantity, while you are arguing it is not.
Are you suggesting a three-segment speed of light test in which we have two mirrors and one source/detector? First you aim the light at one mirror which reflects it over to the second mirror which then reflects it back to the source/detector and measure the time it takes for the pulse to get around back to the source/detector and then you repeat with the light aimed at the other mirror so that the light travels the same path in the reverse direction? Are you saying that if you perform this experiment you will get two different answers for the time it takes for the light to go around the triangle in the two different directions? If I understand your experiment correctly, why do you need to synchronize any watches when you can do the whole experiment with just one watch located at the source/detector of the light? Please clarify.

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JesseM
Very true but Einstein defined light speed to be isotropic not anisotropic. If you want to define light speed to be anisotropic, then why do you care about Einstein's clock synchronization method?
Well, you can still analyze what will happen with clocks synchronized according to Einstein's method from the perspective of a coordinate system where the speed of light is anisotropic, you'll still get the same predictions about physical events like whether two clocks that have been previously synchronized with a third clock will be synchronized with each other (not 'synchronized' relative to this coordinate system, but 'synchronized' in the sense that if a signal is sent from one clock to the other and back, the midpoint of the readings when the signal left and returned to one clock is equal to the reading when the signal bounced off the other clock).

source/detector? First you aim the light at one mirror which reflects it over to the second mirror which then reflects it back to the source/detector and measure the time it takes for the pulse to get around back to the source/detector and then you repeat with the light aimed at the other mirror so that the light travels the same path in the reverse direction? Are you saying that if you perform this experiment you will get two different answers for the time it takes for the light to go around the triangle in the two different directions? If I understand your experiment correctly, why do you need to synchronize any watches when you can do the whole experiment with just one watch located at the source/detector of the light? Please clarify.
Thank you. We don't need synchronisation. Synchronisation is not needed in closed path. But there seems to be an unhappy/redundant marriage between one-way speed of light and synchronisation in mind of people.
Thank you. It is exactly the method that directly measures the one-way speed of light. This is the Trimmer experiment. But I have faced a large inertia which keeps repeating that the one-way light speed is not a measurable quantity. But the Trimmer method measures anisotropy in the one way light speed, in fact it has measured it about 40 years ago.

JesseM
Thank you. We don't need synchronisation. Synchronisation is not needed in closed path. But there seems to be an unhappy/redundant marriage between one-way speed of light and synchronisation in mind of people.
Thank you. It is exactly the method that directly measures the one-way speed of light. This is the Trimmer experiment. But I have faced a large inertia which keeps repeating that the one-way light speed is not a measurable quantity. But the Trimmer method measures anisotropy in the one way light speed, in fact it has measured it about 40 years ago.
But I'm pretty sure the only way this experiment could fail to give equal times for the path in different directions would be if the two-way speed along any leg of the triangle (or any other straight-line path) was not equal to c. If that's correct, then I would say this experiment is not measuring anisotropy in the one way speed, rather it is just a particular type of test of the two-way speed.

But I'm pretty sure ..
I would highly appreciate it if you first would possibly put your "assurance" on a test on the simplified case I described above, repeated below:

Consider a triangle wherein the light speed in not the same for clockwise direction and anticlockwise one. In this example, the two-way speed of light is isotropic but it is the one-way speed which is not isotropic.

JesseM
I would highly appreciate it if you first would possibly put your "assurance" on a test on the simplified case I described above, repeated below:

Consider a triangle wherein the light speed in not the same for clockwise direction and anticlockwise one. In this example, the two-way speed of light is isotropic but it is the one-way speed which is not isotropic.
You're right, I was thinking of the type of anisotropic speed that results from doing a non-Lorentzian coordinate transformation on an inertial frame where the speed of light is isotropic (an example would be the Mansouri-Sexl transformation), but more generally it would be possible to have a theory where there is no frame in which the speed of light is isotropic.

An issue here is that if you have a theory with an anisotropic speed of light that isn't physically equivalent to SR (i.e. you can't do a coordinate transformation to get an SR inertial frame where the speed of light is isotropic), then even if you have one frame where observers at rest in that frame will measure the two-way speed to be c, I have my doubts that it'd be possible to have such a theory where observers moving at arbitrary constant velocities in this frame will also measure the two-way speed to be c regardless of which direction they aim the beam. And if there is only one preferred frame where the two-way speed is c, then experiments like the Michelson-Morley experiment should not be expected to show the two-way speed is constant when the experiment is performed at different points in the Earth's orbit.

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Dale
Mentor
Consider a triangle wherein the light speed in not the same for clockwise direction and anticlockwise one. In this example, the two-way speed of light is isotropic but it is the one-way speed which is not isotropic.
That sounds like a test of rotation, not anisotropy. I.e. the Sagnac effect. You are correct that the Einstein synchronization is not transitive in a rotating reference frame.

That said, there has been some research in this area:
http://ajp.aapt.org/resource/1/ajpias/v31/i7/p482_s1?isAuthorized=no [Broken]

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That sounds like a test of rotation, not anisotropy. I.e. the Sagnac effect. You are correct that the Einstein synchronization is not transitive in a rotating reference frame.
I agree that it can be used to test rotation too. But in this set-up, we do not need to keep something rotating. We just display the triangle a bit. Make it at rest with an inertial frame and perform the measurement. This is somehow similar to Sagnac effect, but it is not this effect.

Dale
Mentor
Can you demonstrate mathematically what the difference is? It sounds exactly the same to me.

Also, the triangle cannot be at rest in an inertial reference frame since an inertial reference frame by definition is one in which the one-way speed of light is isotropic and equal to c. So you are by definition considering non-inertial reference frames. From your description it sounds like the specific non-inertial reference frame you are considering is a rotating reference frame (i.e. same metric as a rotating frame).

Can you demonstrate mathematically what the difference is? It sounds exactly the same to me.
The mathematical framework for the triangular system is that of Trimmer experiment, extended to resonators in section II of
http://arxiv.org/abs/1010.2057

Also, the triangle cannot be at rest in an inertial reference frame since an inertial reference frame by definition is one in which the one-way speed of light is isotropic and equal to c.
This is your assumption. If you assume that there exists a frame in which the Lightspeed is isotropic, you confine yourself to theories similar to that of Robertson-Mansuori-Sexl model. In these subset of theories, one-way light speed is isotropic.

So you are by definition considering non-inertial reference frames. From your description it sounds like the specific non-inertial reference frame you are considering is a rotating reference frame (i.e. same metric as a rotating frame).
If so, please assume that this sound of my writing is accidental. I did not mean it.

If we measure the one-way light speed and its anisotropy's profile coincides to that predicted by being in a rotating frame we than conclude that our frame is rotating. The one-way light speed anisotropy chosen by nature, may not necessarily coincide to that of any rotating frame.

ghwellsjr
Gold Member
When you described your triangular experiment in earlier posts, you didn't mention a piece of optical medium in one leg to slow down the speed of light which is what the Trimmer experiment had. Why? Do you consider it insignificant?

And from everything I could find about the Trimmer experiment, it yielded a null result, just like MMX. Why are you suggesting otherwise?

I looked at the paper that you linked to and it does not describe the results of an experiment but rather a proposed new experiment. Why did you reference it?

Finally, I cannot find enough information on the Trimmer experiment to understand what they actually did so could you please explain it?

When you described your triangular experiment in earlier posts, you didn't mention a piece of optical medium in one leg to slow down the speed of light which is what the Trimmer experiment had. Why? Do you consider it insignificant?
Imagine an arbitrary deviation from one-way anisotropy. The glass is there, to measure just one particular deviation. Measuring the rest of deviations, do not need a glass.

And from everything I could find about the Trimmer experiment, it yielded a null result, just like MMX. Why are you suggesting otherwise?
I am not suggesting otherwise. The Trimmer experiment returned null results for one-way light speed. The MMX returns null results for anisotropy of two-way light speed.

I looked at the paper that you linked to and it does not describe the results of an experiment but rather a proposed new experiment. Why did you reference it?
You asked a mathematical framework for Sagnot and Trimmer experiment. I pinpoint to the mathematical framework for the Trimmer experiment. I hoped that this would helped you realise the distinction between Trimmer and Sagnot experiment.

Finally, I cannot find enough information on the Trimmer experiment to understand what they actually did so could you please explain it?
They considered a triangular interferometer. They measured how the phase difference between light moving clockwise and anticlockwise on the perimeter of the triangle, as function of the configuration of the triangle to the constant starts. They find no deviation for anisotropy of one way light speed with the precision of cm/s.

The whole aim of this post, is to rise the question and debate why now, today, some of us say and keep repeating that the one-way light speed can not be measured.

ghwellsjr
Gold Member
When I keep repeating that the one-way speed of light cannot be measured, I am merely pointing out what Einstein said in his 1905 paper. He described the experimental method of measuring the round-trip speed of light using a source/detector and a mirror a measured distance away and timing the round trip and calculating the speed. He then explained that this does not say anything about the one-way speed of light and then made the bold statement that in any inertial reference frame, we are free to define the two halves of the round-trip measurement to take the same amount of time. This is the essence of SR.

Many people don't realize that this is what he is saying because they think the statement that the speed of light is constant in all inertial reference frames is referring to something that has been measured and confirmed (which is true for the round-trip but not the one-way). If we could measure the one-way speed of light, we could identify an absolute rest frame and relativity would be out the window.

When I keep repeating that the one-way speed of light cannot be measured, I am merely pointing out what Einstein said in his 1905 paper.
Einstein wrote his paper correctly. In page 3, of the english translation of his paper, here
www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf
After defining synchronisation he writes that
We assume that this definition of synchronism is free from contradictions,
and possible for any number of points; and that the following relations are
universally valid:
1. If the clock at B synchronizes with the clock at A, the clock at A syn-
chronizes with the clock at B.
2. If the clock at A synchronizes with the clock at B and also with the clock
at C, the clocks at B and C also synchronize with each other.​
A general anisotropy of the one-way light speed contradicts these assumptions (provided that you use the Einstein method to synchronise all watches with the one at the origin, as this method is assumed in the paper). Nonetheless, the Einstein's work that introduces SR is right by itself.

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