A 1-Way Speed of Light

James Hasty
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TL;DR Summary
A proposal to test for differences in the speed of light for different directions relative to a gravitational field.
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy.

Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground. (Think of a "Ferris wheel".) Let points A and B align with the earth's radius, with A initially on top, and call that: 0-degree position. Then rotate the wheel clockwise so that A is on the right, diameter AB is perpendicular to the earth's radius, call that: 90-degrees orientation. Let an observer at point O be located along a line passing thru the wheel center and perpendicular to the wheel; O is an equal distance R from both A and B.

Synchronize clocks A and B: Send a "reset" signal (radio or light) from O to clocks A and B, to set the clocks to 0-time. If R is sufficiently distant, the angle <AOB will be very small, and lines OA and OB are effectively the same direction. Then both clocks receive the reset signal at the same instant and are synchronized.

Measure the 1-way speed of light in different directions: Begin in the 0-degree position, synchronize the clocks, and send a light signal from A at time t[A1] received at B at time t[B1]. Then send a light signal from B at time t[B2] received at A at time t[A2]. The speed of light in each direction is then:
c[AB]=L / (t[B1] - t[A1]) and c[BA]=L / (t[A2] - t[B2]) (1)
Repeat this test for line AB in the 90-degree position. Because the wheel is rotated, the reset signal must be sent again to synchronize A and B once stationary. Then the test is repeated and the speeds calculated again as per (1) above. Compare the results. If there is no difference, then the speed of light is the same in all directions.
 
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James Hasty said:
If R is sufficiently distant, the angle <AOB will be very small, and lines OA and OB are effectively the same direction. Then both clocks receive the reset signal at the same instant and are synchronized.
If the OWSOL is anisotropic then increasing the distance does not help. It gives a longer time for a smaller error to accumulate. The actual size of the desynchronization does not depend on the distance of R.

In any case, the difference in synchronization here is the same as the difference in time you are trying to detect later. It cannot work
 
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Speed isn't a well defined measure in curved spacetime. So the answer is yes or no, depending on your coordinate choice.
 
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What you are really measuring, if anything, is whether a (rigid) object retains the same proper length when rotated in a gravitational field.
 
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James Hasty said:
the angle <AOB will be very small
No, it won't. Ideally it will be 180 degrees, since A and B are on opposite sides of the wheel.

James Hasty said:
lines OA and OB are effectively the same direction.
No, opposite directions.

James Hasty said:
Then both clocks receive the reset signal at the same instant
No, they don't, not in any invariant sense. The two reception events are distinct events in spacetime and are not co-located. So any claim about the relative times of reception depends on your choice of coordinates.
 
James Hasty said:
Synchronize clocks A and B: Send a "reset" signal (radio or light) from O to clocks A and B, to set the clocks to 0-time. If R is sufficiently distant, the angle <AOB will be very small, and lines OA and OB are effectively the same direction. Then both clocks receive the reset signal at the same instant and are synchronized.
As others wrote, this trick does not work.
  1. The velocity of the "reset" signal (radio or light) from O to clocks A can be split into a component in direction of the axis of the wheel and a component perpendicular to the axis.
  2. The velocity of the "reset" signal (radio or light) from O to clocks B can be split into a component in direction of the axis of the wheel and a component perpendicular to the axis.
Relevant for the synchronization are only the two mentioned perpendicular components. They point in opposite directions. A definition of synchronicity of A and B requires a definition of the isotropy of the 1-way-speeds in these opposite directions.
 
James Hasty said:
TL;DR Summary: A proposal to test for differences in the speed of light for different directions relative to a gravitational field.

Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy.

Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground. (Think of a "Ferris wheel".) Let points A and B align with the earth's radius, with A initially on top, and call that: 0-degree position. Then rotate the wheel clockwise so that A is on the right, diameter AB is perpendicular to the earth's radius, call that: 90-degrees orientation. Let an observer at point O be located along a line passing thru the wheel center and perpendicular to the wheel; O is an equal distance R from both A and B.

Synchronize clocks A and B: Send a "reset" signal (radio or light) from O to clocks A and B, to set the clocks to 0-time. If R is sufficiently distant, the angle <AOB will be very small, and lines OA and OB are effectively the same direction. Then both clocks receive the reset signal at the same instant and are synchronized.

Measure the 1-way speed of light in different directions: Begin in the 0-degree position, synchronize the clocks, and send a light signal from A at time t[A1] received at B at time t[B1]. Then send a light signal from B at time t[B2] received at A at time t[A2]. The speed of light in each direction is then:
c[AB]=L / (t[B1] - t[A1]) and c[BA]=L / (t[A2] - t[B2]) (1)
Repeat this test for line AB in the 90-degree position. Because the wheel is rotated, the reset signal must be sent again to synchronize A and B once stationary. Then the test is repeated and the speeds calculated again as per (1) above. Compare the results. If there is no difference, then the speed of light is the same in all directions.

Is clock syncrhronization actually important to your question? It's not obvious to me why it's relevant to your problem, and it'd be a lot easier to analyze the problem if you could get rid of it. I'd suggest about thinking why you need it, and the approach I'd take to that is to try and eliminate it and note any roadblocks to said elimination.

My general approach personally is to eliminate clock synchronization as much as is possible to any problem. It usually provides little physical insight, and it is painful to really communicate in a precise manner to be sure all parties in a conversation understand exactly how it is being done. I'd guess you are using Einstein synchronizaton along some path, but to be really clear we'd have to talk about what exact path you are using.

The second problem you'll have is coming up with a sufficiently rigid body. It's easy to talk about a circular wheel rotating, but an actual physical wheel will deform due to stress. If one imagine it's made out of steel, one can calculate the defomation due to the changing load on the wheel, for instance. I'd expect that you would find that this deformation will dominate any actual experimental result you hope for. Basically, physical objectgs are "jelly" compared to measurements made by the speed of light, because rigidity is in some sense proportional to the speed of sound in a material, and the speed of sound in any physical material is MUCH less than the speed of light. There's a formula, I believe, that the speed of sound is proportional to the square root of the bulk modulus divided by the density, or somesuch. See https://en.wikipedia.org/wiki/Speed_of_sound under "Equation". So without going into the detailed math, physical materials with a low speed of sound will just not have a high strength-weight ratio compared to synthetic concepts to measure distance based on the speed of light.

The way I'd personally think about the underlying problem is whether there exists a Born-rigid congruence for such a rotating wheel. But this is a somehwat advanced concept, so rather than talking about the theories of Born rigidity, I'll just mention it, and instead rely on my observations about materials with a high strength/weight ratio having a high speed of sound. This makes it reasonably clear why actual materials fall far short of the goal of what measurements using light can do, in a reasonably intuitive manner, which I think is about all I can reasonably expect to accomplish.

The final point is about the definition of the meter, and hence the definition of distance, which I think is important to your question. The SI definition of the meter is:

SI meter said:
The metre 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 ∆ν.

The wording of the definition was updated in 2019.

I am reasonably confident that you are NOT using this definition of the meter, but I have learned caution on guessing what people might mean by a meter when they aren't using the standard SI one. My _guess_ is that you're basing your definition of the meter on a prototype meter bar. But who knows? Well, you may know:). On the other hand, I find it's likely that you might not have actually thought about the issue, based on pass conversations. Anyways, The important point here is that you are assuming some sort of definition of distance, and that it's probably not based on the SI definition. And it's probably tied to the notion of rigidity. So both of those would be good topics to discuss to clarify your thinking.

Anyways, out of time, and I think I've probably already rambled on a bit here.
 
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pervect said:
Is clock syncrhronization actually important to your question?
Clock synchronization is inherent in any experiment to measure the one-way speed of light, which is the very title of this thread. So yes, it is.

Maybe there's a different question the OP could be asking that would not involve measuring the one-way speed of light, and in which clock synchronization would not be an issue. But that would be a different question for a different thread.
 
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In my opinion arguments over or about the one way speed of light are pretty much a dead end, but rereading the post I see Peter was right in commenting that this was the original question. Somehow I thought the OP was interested in whether the radial speed of light was the same as the transverse speed of light - this question can be asked about the two-way average speed of light rather than about the one way speed of light.

As far as the later question goes, I can point out that coordinates do exist around a massive single body where the coordinate speed of light is the same in all directions. These are the isotropic version of the Schwarzschild coordinates. Of course, just because these coordinates exist doesn't mean one is obligated to use them, and it' worth noting there are common coordinate choices, for instance standard Schwarzschild coordinates (without the isotropic modifier) that do not have the property that the radial coordinate speed is equal to the transverse coordinate speed. It's unclear how this relates to the proposed experiment though, which I find interesting for other questions it raises which probably aren't relevant.

I think some of the other general points I made would still be useful for the OP to think about. As I said previously, the way they ask the question strongly suggests they are using some definition of distance other than the SI defintion of the meter. Which is perfectly fine, but some discussion of what they are replacing it would be helpful in even attempting to answer the question. The existence or lack t hereof of an appropriate rotational symmetry in the Schwarzschild geometry is also an interesting question, it's likely that the OP is assuming that the geometry of space is Euclidean, and that's not true around a massive object.
 
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pervect said:
it's likely that the OP is assuming that the geometry of space is Euclidean, and that's not true around a massive object
More precisely, the geometry of the surfaces of constant Schwarzschild coordinate time is not Euclidean, and that is the coordinate chart that is adapted to the time translation symmetry of the spacetime. Or, to put it in coordinate-free terms, the geometry of the spacelike 3-surfaces that are orthogonal to the integral curves of the timelike Killing vector field of the spacetime is not Euclidean.
 
  • #11
Worth noting that you'll get different two way speeds of light in the vertical configuration depending on which clock you use. If you fire a pulse from the bottom clock and reflect it off the top clock you will get a higher speed than if you fire from the top clock and reflect off the bottom one, due to gravitational time dilation. So different one way speeds would not surprise me, although I suspect you could rig your synchronisation to get equal one way speeds.

If you drop the apparatus and carry out the two way speed measure while it's in free-fall you would get equal two-way speeds, because there the device is at rest in a local inertial frame.
 
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Thank you all for your comments! Very much appreciated. I do hope that my reply here will be seen by all of you (Dale, Ibix, PeroK, Peter Donis, Saggitarius A-Star, pervect).
1. Desynchronization can occur between A and B because of the perpendicular components of O's reset signal. This requires a definition.
2. Gravitational time dilation will make synchronization difficult in the vertical configuration.
 
  • #13
James Hasty said:
Desynchronization can occur between A and B because of the perpendicular components of O's reset signal.
And its value is independent of R. It does not become small for large R
 
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Yes. And this is why Einstein defined time as: the time it takes for light to travel from A to B is the same as it takes light to travel from B to A (in Special Relativity). So, in General Relativity, is the speed of light defined differently, and is it not constant for all observers.
 
  • #15
James Hasty said:
Yes. And this is why Einstein defined time as: the time it takes for light to travel from A to B is the same as it takes light to travel from B to A (in Special Relativity). So, in General Relativity, is the speed of light defined differently, and is it not constant for all observers.
General Relativity entails general curved geometries, but locally the geometry is Minkowski - technically, it's a Minkowski tangent space at every point. This means that SR applies locally. The speed of light is locally invariant. Globally, speed is not a well-defined concept - especially if you have a dynamic spacetime, where the proper distance between two points changes with time.

The classic example is light from a distant galaxy. There is the initial distance between the galaxy and Earth; the current distance (greater as the universe has been expanding); and the proper distance travelled by the light (which is ##c## multiplied by the time between emission and detection).

It's essentially a postulate that light travels locally in vacuum at an invariant speed ##c##.
 
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James Hasty said:
Yes. And this is why Einstein defined time as: the time it takes for light to travel from A to B is the same as it takes light to travel from B to A (in Special Relativity). So, in General Relativity, is the speed of light defined differently, and is it not constant for all observers.
The two way speed of light is only guaranteed to be invariant for inertial observers. Your apparatus is not inertial.

The difference between SR and GR here is that in GR velocity is not well-defined for things that are not co-located with the observer. In a relatively small region of spacetime like that occupied by this experiment that doesn't actually matter because the curvature is small. The issue is the non-inertial coordinates you are using.
 
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  • #17
James Hasty said:
So, in General Relativity, is the speed of light defined differently, and is it not constant for all observers.
As both @Ibix and @PeroK said, the difference between SR and GR in this respect is that speed can only be compared locally. In both SR and GR the speed of light is invariant in inertial frames, and may not be invariant in non-inertial frames.

Another difference (which is related to the fact that relative velocity is only defined locally) is that in SR inertial frames are global, but in curved spacetime they are only local.

By the way, this is not really the main point of the discussion, but “constant” and “invariant” are different concepts. “Constant” means it doesn’t change over time. “Invariant” means that different frames agree on the value. The speed of light is both constant and invariant, but the reason that we are interested in it, the reason all of the relativistic effects are related to it, is because it is invariant. Not because it is constant.
 
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  • #18
Dale said:
As both @Ibix and @PeroK said, the difference between SR and GR in this respect is that speed can only be compared locally. In both SR and GR the speed of light is invariant in inertial frames, and may not be invariant in non-inertial frames.
I'd add the qualifier coordinate speed in both SR and GR is invariant in inertial frames since the 4-velocity of light, like any 4-vector, is a geometric quantity/object therefore invariant by its very definition (like the light cones in any spacetime).
 
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  • #19
OK. Good points. If I understand you correctly, my "experiment" is a "local, weak gravitational" case, which approximates an "inertial frame". My coordinates to 1st approximation are Minkowski. Let the center of the wheel be point C, then with respect to O, the reset signals sent are two closely related vectors: OA=OC+CA and OB=OC+CB, where CA=-CB and magnitudes |CA|=|CB|. Then for any orientation of the wheel, i.e., any direction of the line AB with respect to the vertical, these relationships would hold true. So, it looks to me like the clocks would be synchronized at least initially, but gravitational time dilation in most orientations would take over to slowly desynchronize the clocks. Do I have that right so far?
 
  • #20
James Hasty said:
which approximates an "inertial frame".
No it doesn't! That's the point!
James Hasty said:
My coordinates to 1st approximation are Minkowski.
No, you're approximately using Rindler coordinates.
 
  • #21
James Hasty said:
Einstein defined time as: the time it takes for light to travel from A to B is the same as it takes light to travel from B to A (in Special Relativity).
He defined time in a particular kind of coordinate chart (an inertial chart with orthogonal axes) that way.

But that definition is specific to that kind of coordinate chart.

James Hasty said:
in General Relativity, is the speed of light defined differently
In GR, the "speed of light", except when measured within a single local inertial frame, is not a meaningful quantity at all. The meaningful quantity is the light cone structure of spacetime.

James Hasty said:
is it not constant for all observers.
The coordinate speed of light depends on your choice of coordinates. This is true whether spacetime is flat (SR) or curved (GR).

The light cone structure of spacetime is invariant; it's the same for all observers. That's why GR focuses on that.
 
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  • #22
OK. If my "experiment" were performed on the International Space Station (ISS), which is a "falling reference frame", that would satisfy as a "local inertial frame". The gravity field is "eliminated". But we could still test for the 1-Way speed of light in all directions, as the wheel is turned and the ISS orbits the earth.
 
  • #23
James Hasty said:
OK. If my "experiment" were performed on the International Space Station (ISS), which is a "falling reference frame", that would satisfy as a "local inertial frame". The gravity field is "eliminated". But we could still test for the 1-Way speed of light in all directions, as the wheel is turned and the ISS orbits the earth.
The question then boils down to how local is local? If you place two objects inside the space station, one a short distance above the other, then the objects initially appear to remain the same distance apart - because the rest frame inside the space station is approximately inertial. But, the objects eventually are seen to drift apart, because it's only approximately inertial.

Likewise, two clocks on the space station would eventually show different times if their orbits are at slightly different radii.
 
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  • #24
James Hasty said:
But we could still test for the 1-Way speed of light in all directions, as the wheel is turned and the ISS orbits the earth.
And the answer would depend on your assumptions, as always.

Your error is in (a) assuming that the clocks are initially synchronised, which is only true if the one way speed of light is the same in both directions between the clocks; and (b) assuming that there is no non-negligible clock drift when you rotate your wheel, which is only true with an isotropic speed of light. The only complication from GR in a free-falling frame is that you need to add the word "locally" a couple of times.
 
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  • #25
PeroK wrote: two clocks on the space station would eventually show different times if their orbits are at slightly different radii.
Ibix wrote: Your error is in (a) assuming that the clocks are initially synchronised, which is only true if the one way speed of light is the same in both directions between the clocks;
(I hope I am using this posting thread correctly.)
1. On the ISS, clocks A and B will be synchronized initially when O transmits a reset signal because of symmetry in the inertial frame. Let the O signal be a radio wave traveling in all directions. Distance |R|=|OA|=|OB|. The c velocity components of the radio wave with respect to O are: c[R]=c[OC]+c[CA]= and c[R]=c[OC]+c[CB]. Since |c[R]|=constant in the inertial (ISS) frame, then the magnitudes of the velocities along the line AB will be the same |c[CA]|=|c[CB]|, according to O.
2. The clocks will desynchronize over time, so the light signals from A to B: c[AB] and from B to A: c[BA] must be sent immediately after initial synchronization.
 
  • #26
James Hasty said:
1. On the ISS, clocks A and B will be synchronized initially when O transmits a reset signal because of symmetry in the inertial frame.
They will be initially synchronised if you assume an isotropic speed of light. If you don't assume that then they aren't synchronised.
 
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  • #27
I realize that assuming a radio wave travels at the speed of light in all directions to initiate synchronization of the clocks is a flaw in logic, since we are trying to prove light speed from A to B is the same as from B to A. That is why in my original post I was trying to avoid that flaw by making R large so the angle <AOB is very small, so that effectively R is one direction. R can be very large if O is far away, say on earth while A and B are on the ISS.
 
  • #28
James Hasty said:
so that effectively R is one direction.
The difference between "effectively" and "actually" will always lead to a desynchronization that exactly counters the effect you hope to measure.
 
  • #29
James Hasty said:
I realize that assuming a radio wave travels at the speed of light in all directions to initiate synchronization of the clocks is a flaw in logic, since we are trying to prove light speed from A to B is the same as from B to A. That is why in my original post I was trying to avoid that flaw by making R large so the angle <AOB is very small, so that effectively R is one direction. R can be very large if O is far away, say on earth while A and B are on the ISS.
Just for the record, one way to synchronize two clocks would be for one clock to send out a signal at ##t_1 = 0## and when that signal is received by the second clock, it is set to ##t_2 = 0##. With that set-up, the speed of light is infinite/undefined in one direction and ##c/2## in the opposite direction.

More generally, you are not obliged to use the Einstein synchonization convention. Other coordinate systems with a different time-like coordinate are routinely used in SR and especially in GR. Many students struggle with the concept that there is more than one way to synchronise clocks. In fact, if you accept that different synchronization conventions are possible, then all talk of an unambiguous one-way speed of light is seen to be pointless.

The critical point is that your choice of time-like coordinate/clock synchronization determines the one-way speed. But, not, of course, the two-way speed, which involves only a single clock.
 
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  • #30
James Hasty said:
we could still test for the 1-Way speed of light in all directions, as the wheel is turned and the ISS orbits the earth.
You cannot ever test for the one way speed of light with any experiment in any location at any time (without assuming the answer at some point). The OWSOL is an arbitrary convention that you choose by setting your coordinates, not a feature of the universe that can be physically measured.

To measure the OWSOL you would need to construct an experiment whose measurement depends on Anderson’s kappa. There is no such possible experiment because your choice of coordinates cannot change any measurement.
 
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  • #31
James Hasty said:
That is why in my original post I was trying to avoid that flaw by making R large so the angle <AOB is very small, so that effectively R is one direction. R can be very large if O is far away, say on earth while A and B are on the ISS.
R doesn’t matter. The distance can be as large or small as you like. It doesn’t change the amount of desynchronization if the OWSOL is not isotropic.

You should work out the math. The desynchronization depends on Anderson’s kappa, but not R.
 
  • #32
James Hasty said:
the angle <AOB is very small
No, it isn't. Read my post #5.
 
  • #33
PeterDonis said:
No, it isn't. Read my post #5.
I don't think O is in the plane of the wheel - it's meant to be on the axis but far outside the plane, so AOB is small. Not that it makes any difference to this problem.
 
  • #34
@James Hasty - here's why all attempts to measure the one way speed of light fail.

This is a Minkowski diagram - it shows the position of objects at different times. If you came across displacement-time graphs in high school physics, this is the same thing except we usually draw position horizontally and time vertically.
1756492846975.webp

This one shows two clocks, one red and one blue. Initially (at the bottom of the diagram) they are at ##x=-2## and ##x=+2## respectively. At ##t=-4## a light pulse (yellow) is emitted from ##x=0## in both directions. It arrives at the clocks at ##t = -2## an they immediately set themselves to zero and begin moving, swapping places. They complete the manoeuvre at ##t=3##, come to a stop and immediately emit light pulses back to the origin, which arrive there simultaneously at ##t=5##.

The interesting point about Minkowski diagrams is that we take them pretty literally as maps of spacetime (at least, one space dimension and the time dimension). So the clocks (which are just point-like objects on this scale) are really the lines - points extended in time.

Note the assumption that the speed of light is equal in both directions in the diagram above. Can we consider an anisotropic one way speed of light? Yes - like this:
1756493120126.webp

Notice how the horizontal grid lines are now slanted? That means that the initial light pulse leaves at ##t=-4## but arrives at one clock slightly before ##t=-2## and the other slightly after ##t=-2## - the speed of light is slightly higher in the ##+x## direction than in the ##-x## direction. And a similar thing happens at the top of the diagram with the returning light pulses which are not emitted at the same time but arrive at the origin at the same time.

But notice that none of the red, blue or yellow lines has changed. The only thing that's changed are the grey lines which are not there in reality. They are just things I added to the diagram to make interpretation easier. So there is no actual physical difference between the isotropic and anisotropic cases - the actual physical measurements will always be the same.

And that is the important bit. It does not matter how many bells and whistles you add to your experimental design, nor whether you have a 2d or 3d setup, whether you use wheels or bananas or whatever. The difference between the isotropic and anisotropic cases is always and only in the shape of the grid you imagine drawing on spacetime. This has no physical consequences.
 
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  • #35
As @Ibix showed, the orthogonal grey grid lines in the diagram actually correspond/represent straight lines orthogonal in spacetime (note that the notion of straight line in spacetime is a geometric notion). The isotropy of coordinate one way speed of light corresponds to pick orthogonal grid lines in spacetime (or in the spacetime diagram itself).
 
  • #36
Very good points, thanks all of you for your comments. I know it took time out of your days to respond back to me, and I appreciate it. (I have learned along the way.) Thanks again. Live long and prosper.
 
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  • #37
Dale said:
R doesn’t matter. The distance can be as large or small as you like. It doesn’t change the amount of desynchronization if the OWSOL is not isotropic.

You should work out the math. The desynchronization depends on Anderson’s kappa, but not R.
I have Google searched for "Anderson's kappa" but not readily found. What specifically is it?
 
  • #38
James Hasty said:
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.
 
  • #39
##\kappa## is the thing that controls how slanted the lines in my second diagram are, IIRC.
 
  • #42
Dale said:
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.
Thank you.
 
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