[Bored chemist]

The light must return for you to be aware of a distant event....
The fundamental problem is, you are present at the local events, emission and detection. You are not present at the distant reflection event.
...

Not really. I go to the camera later and look at the film. The camera was (sort of) "aware" of the event, and it recorded it as well as recording the local time of that event.

There is no "reflection event".

[guest39538]

Please tell me I am not going mad.  On another science forum the moderator said I was wrong and banned me. 

If r1=r2 and light travels both vectors simultaneously, both the detection points detect the light in the same time. 

He said no , he is quite mad I think.

[guest4091]

In the graphic, Martin has synchronized the master clock m with clocks s1 and s2, using the SR clock synch convention. producing the mathematical axis of simultaneity, s1-s2 (red). Martin wants to try the slow transport method with another pair of clocks.
The red hyperbola, referred to as the calibration curve by Max Born, in his book on Relativity, is a line of constant time where it intercepts the time lines of objects moving at varying speeds. I.e. the faster it moves, the longer it takes to indicate t on a moving clock. The pair is set to the m clock. In test 1 the pair separates from m at a fast speed, with a large difference in time dilation, as shown by the green line c1-c1. The slope is greater than s1-s2.
In test 2 the pair separates from m at a slower speed, with a small difference in time dilation, but a total time of 2t (not shown to avoid clutter, and v=half of first speed ), as shown by the green line c2-c2. The slope is much closer to s1-s2.
The slope of s1-s2 is actually tangent to the calibration curve at m.
Conclusion:
Slow clock transport more closely approximates the SR convention as the separation speed v decreases. The statement "it equals the SR convention at v=0" would be redundant, since it is equivalent to "if you don't move them, they will stay synchronized".
I think Dutch was referring to this aspect of synchronization.
https://app.box.com/s/59yk88qpfjodkwxsk3faii8eqwjyydwz

[guest4091]

Not really. I go to the camera later and look at the film. The camera was (sort of) "aware" of the event, and it recorded it as well as recording the local time of that event.

There is no "reflection event".

Assuming you think you are moving:
For the camera to record the background event requires a 2-way trip.
How do you know the difference in the clock there and your local clock?

[Bored chemist]

Not really. I go to the camera later and look at the film. The camera was (sort of) "aware" of the event, and it recorded it as well as recording the local time of that event.

There is no "reflection event".

Assuming you think you are moving:
For the camera to record the background event requires a 2-way trip.
How do you know the difference in the clock there and your local clock?

Did you actually read what I posted about the scenario I'm talking about?
It's the one where I'm at rest (at least in my frame of reference) and where there are two synchronised cameras which record a singe event from different distances.

[guest4091]

Did you actually read what I posted about the scenario I'm talking about?
It's the one where I'm at rest (at least in my frame of reference) and where there are two synchronised cameras which record a singe event from different distances.

Look at the right side of the graphic in #78. If someone in the theoretical 'fixed frame' U was observing your moving frame, with a flash from you to a distant object, on a long leg out and returning on a short leg, using SR conventon, you in your pseudo rest frame assigns equal legs, with the same round trip time. The answer will be the usual, light speed is c. You can check the Lorentz/SR coordinate transformations to verify that x'/t' = x/t = c, where the primed coordinates are your measurements. That is the very reason you can assume a pseudo rest frame. Pseudo because there is no real rest frame (that can be used as a common reference for measurements). It's not a conspiracy, it's because light propagation is constant and independent of source.

[guest39538]

Did you actually read what I posted about the scenario I'm talking about?
It's the one where I'm at rest (at least in my frame of reference) and where there are two synchronised cameras which record a singe event from different distances.

Look at the right side of the graphic in #78. If someone in the theoretical 'fixed frame' U was observing your moving frame, with a flash from you to a distant object, on a long leg out and returning on a short leg, using SR conventon, you in your pseudo rest frame assigns equal legs, with the same round trip time. The answer will be the usual, light speed is c. You can check the Lorentz/SR coordinate transformations to verify that x'/t' = x/t = c, where the primed coordinates are your measurements. That is the very reason you can assume a pseudo rest frame. Pseudo because there is no real rest frame (that can be used as a common reference for measurements). It's not a conspiracy, it's because light propagation is constant and independent of source.

There is a rest frame, the observed darkness background of space, space itself is not observed to move.  Relative to bodies , space is at rest and all bodies are moving relative to space.
All things move relative to the rest frame of space observably.

[Bored chemist]

Look at the right side of the graphic in #78.

If I look at it, the first thing I see is that it's rather unclear- just a bunch of lines and letters.
The second thing I see is that it's  not a representation of the scenario I put forward.

Why is everyone so reluctant to answer the simple  question I asked?
If I get 2 clocks, sync them, separate them slowly enough that the time dilation is small and put them in front of 2 video cameras.
I the set off a flash- nearer to one camera than the other.
I look at the video and I note when the flash arrives at each camera.
I know the difference in paths and I know the difference in times.
So, what will I get if I divide the distance by the time?

[guest39538]

Look at the right side of the graphic in #78.

If I look at it, the first thing I see is that it's rather unclear- just a bunch of lines and letters.
The second thing I see is that it's  not a representation of the scenario I put forward.

Why is everyone so reluctant to answer the simple  question I asked?
If I get 2 clocks, sync them, separate them slowly enough that the time dilation is small and put them in front of 2 video cameras.
I the set off a flash- nearer to one camera than the other.
I look at the video and I note when the flash arrives at each camera.
I know the difference in paths and I know the difference in times.
So, what will I get if I divide the distance by the time?

speed?   It is the same way I did it with the graph using one detector.

[guest4091]

BC;

I've read your posts from #70 on, trying to get a picture of your setup. Your last post helped.
Is this correct?
You are at x=0 with a camera and clock1.
A 2nd camera and clock2 are at say x=10.
The flash occurs at x=6.
Both cameras record the flash and the local time.
You expect the clocks are synched to within an insignificant amount by slow transport.
If it is correct, then (6-4)/(6-4) = 1. (light speed)

[David Cooper]

So, what will I get if I divide the distance by the time?

You'll get the apparent speed of light for a round trip. Your synchronisation method tunes your apparatus to its speed of movement through space such that your measurement hides that movement.

[Bored chemist]

So, what will I get if I divide the distance by the time?

You'll get the apparent speed of light for a round trip. Your synchronisation method tunes your apparatus to its speed of movement through space such that your measurement hides that movement.

A trip "round" what?
Incidentally, from my perspective the apparatus isn't moving through space: no tuning needed.

[David Cooper]

A trip "round" what?

A round-trip is a 2-way trip.

Incidentally, from my perspective the apparatus isn't moving through space: no tuning needed.

That's the whole issue - you're assuming you're stationary up front, you're synchronising your clocks to fit with your idea that you're stationary, and then you're measuring the "one-way" speed of light on that false basis. You cannot move your clock infinitely slowly because infinite slowness won't move it at all, so you have to move it as some speed greater than zero, and while you are moving it there are time dilation effects which tune your clock synchronisation in such a way as to maintain the illusion that your system is stationary. Your measurement of the distance travelled by the moving and "stationary" clocks is an unknown because you can't tell how far they're really moving while you move one of your clocks away from the other. If the system is moving at very high speed, both clocks may have moved a lightyear and the time dilation difference between the "moved" and "unmoved" clocks will be more severe.

[Bored chemist]

A trip "round" what?

A round-trip is a 2-way trip.

Incidentally, from my perspective the apparatus isn't moving through space: no tuning needed.

That's the whole issue - you're assuming you're stationary up front, you're synchronising your clocks to fit with your idea that you're stationary, and then you're measuring the "one-way" speed of light on that false basis. You cannot move your clock infinitely slowly because infinite slowness won't move it at all, so you have to move it as some speed greater than zero, and while you are moving it there are time dilation effects which tune your clock synchronisation in such a way as to maintain the illusion that your system is stationary. Your measurement of the distance travelled by the moving and "stationary" clocks is an unknown because you can't tell how far they're really moving while you move one of your clocks away from the other. If the system is moving at very high speed, both clocks may have moved a lightyear and the time dilation difference between the "moved" and "unmoved" clocks will be more severe.

A round trip implies that you go somewhere and then come back.
At what stage in the process I described does anything come back?

I look at the "near" camera and see when the flash arrived, then I go and look at the "far" camera and see when the flash arrived. Then I do the arithmetic.
The only thing that "comes back" in any way is me and as I keep pointing out, I can make the relativistic effects there as small as I like.
"because you can't tell how far they're really moving while you move one of your clocks away from the other"
I forgot to mention- there's a railway track along the route I chose to move the clock.
I can count railway sleepers.
I can also check that each sleeper is the same distance as the last one- because I have a ruler.
As long as I go slowly, the length of the ruler is as near constant as I like.

"you're assuming you're stationary up front"
And, once again, from my point of view, it's not an assumption, it's an observation.

For the sake of discussion, let's assume I start by destroying the rest of the universe.

[guest4091]

BC;
I wasn't making your example difficult. If you provide distances for all the objects involved as measured from a common location it's easier to visualize. General terms like foreground, near, etc. are vague.
Your example is just a single 1-way trip divided into 2 parts, i.e. set the origin at the location of the flash.
Let's put your example in a realistic perspective. One light second is just short of 1.25 ls, the time for light to travel between the surface of the earth to the surface of the moon. Thus your  clock should measure ns (nano seconds), and the distance unit 1 foot or .3 m.
Set clock 2 (c2) and a camera at 3 m distance, send a flash across that distance to c2.
Assuming your clocks are synched to less than 1 ns, c2 reads 3 ns.
If you had sent a flash at t=0, that reflected from a mirror at c2, it would return at t=6.
The SR synch convention would have set c2 to indicate 1/2 of the round trip,6/2=3.
Just as in the MMX, regardless of speed, the observer perceives the td time and lc distance, as if he is not moving, i.e. x/t=c.

[Bored chemist]

I've read your posts from #70 on, trying to get a picture of your setup.

If you provide distances for all the objects involved as measured from a common location it's easier to visualize.

Only if you read the distances I provided; you didn't.
Feel free to try reading from the start and see if that helps.

[David Cooper]

A round trip implies that you go somewhere and then come back.
At what stage in the process I described does anything come back?

Your clocks function by counting round-trip cycles of something moving, and while you move one of your clocks, those round-trips are affected by that movement regardless of how slowly you move it.

and as I keep pointing out, I can make the relativistic effects there as small as I like.

No you can't. All you can do is minimise the big error while making no difference to the residual error, but the only way you're going to see how these errors come into play is by producing a table of data to show it all clearly, so I'm going to help you build such a table.

I can also check that each sleeper is the same distance as the last one- because I have a ruler.
As long as I go slowly, the length of the ruler is as near constant as I like.

You can't tell how much length-contraction might be acting on the track, and you don't know how far you've moved your moving clock through space either (or how far you've moved your "stationary" clock).

"you're assuming you're stationary up front"
And, once again, from my point of view, it's not an assumption, it's an observation.

If you're doing this on one planet and your cousin is doing the same thing on another planet moving relative to yours, both of you are making the same claim to be stationary, and at least one of you is being fooled.

For the sake of discussion, let's assume I start by destroying the rest of the universe.

That's not going to help you do anything other than help you deceive yourself. Let's just build the table and do the maths. We want to generate numbers for two cases, one where we assume the system is stationary and the other where we assume it's moving at very high speed. We then want to generate numbers for each case using a range of numbers for moving the clock at different speeds, so I'll show you a method for doing the calculations which you can apply yourself to extend the table if you want to use slower speeds than the ones I've chosen for relocating your clock.

Let's start with case 1, where we assume the system is stationary. If we move the clock at the speed of light (1c) and move it a distance of one lightsecond, it will not tick at all while it's being moved, so it will end up ticking one whole second behind the stationary clock. If we move the clock at 50% the speed of light instead (0.5c), we can work out how much the clock will be slowed by finding the inverse sine of 0.5 and then finding the cosine of that answer. That comes out as 0.866 ticks per second, meaning that the moved clock has lost 0.134 of a second while it was being relocated. Because we're assuming that the system is stationary, we can always calculate how to correct the moved clock to make up for the time it has lost, so we can ensure that our clocks are perfectly synchronised for the frame of reference in which the system is stationary regardless of how fast we move the clock. Let's now try 10% the speed of light (0.1c): cos(arcsin(0.1)) = 0.995, so the clock has lost 0.005 of a second. So, here's the first column of our table (speed; relative tick rate for moving clock compared with stationary clock; lost time for moved clock compared with stationary clock; correction value; corrected value):-

1c, 0, 1, 1, 0
0.5c, 0.866, 2 x 0.134, 2 x 0.134, 0
0.1c, 0.995, 10 x 0.005, 10 x 0.005, 0

You can extend that as much as you like, but remember that we can remove the big error in each case simply by working out how much time the clock must have lost and adding it back on. As we use slower and slower clock transport, the uncorrected results tend towards the corrected results (which are all zero), but they never quite get there, so you should always make the adjustment if you want to maximise the accuracy.

Now we need to start to build our second column, but it takes a bit more work. We need to choose a speed for the system to move through space at (the system being the "stationary" clock and anything that remains at rest relative to it), and you can add more columns of your own to cover any speed you fancy if you don't like my choice. I'm going with 0.866c. At this speed, our 1 lightsecond distance has been length-contracted to 0.5 of a lightsecond. (You can calculate the length-contraction for any other speed you want to work with in the same way as you calculate the time dilation.) The relative speeds of travel are also affected by the speed of travel of the system because of relativistic mass complications, so we have to do some extra calculations to convert to their real speeds of travel through space. For example, we naively think that light moves at 1c faster than the system if we're at rest with the system, but because the system is actually moving at 0.866c, for light to be moving at 1c relative to the system it would have to be moving through space at 1.866c which is clearly impossible. We know that light always travels through space at c though, so it must be moving relative to the system at 0.134c.

We have to correct our other speeds too, because with our 0.5c speed for moving a clock relative to our "stationary" clock which is moving at 0.866 we can't just add the two speeds together as again it would mean moving a clock through space faster than the speed of light. The formula for adding velocities correctly involves adding the two together (as you might expect), but then dividing the answer by 1 + the two velocities multiplied by each other, so that's (0.866 + 0.5) / (1 + 0.866 x 0.5). The answer is 0.953c. The time dilation for that speed is 0.3022. We now need to work out how long it will take for the clock to move to it's destination, and that's going to involve a longer distance travelled than the 0.5 lightsecond length of the course because the system (including the course) is moving along at 0.866c. So, we need to know how long it will take for an object moving at 0.953c to close in a distance of 0.5 lightsecond on something that's moving away from it at 0.866c. (Note: I stored all the values in memories on a calculator with 9 memories so as to be able to reuse them more easily on subsequent steps without losing any accuracy, and I also used sin 60 to get a more precise value than 0.866 which is just a close approximation of the speed which dilates time and contracts lengths by exactly 0.5x, so every time I write 0.866c I expect you to type in and use sin 60 instead.) The closing speed is 0.953 - 0.866 = 0.0872, so we can divide the 0.5 lightsecond figure by this to get the time taken for the clock to move to its destination, and that gives us the time 5.732s. (The distance the clock actually moves through space during its relocation is therefore  0.953 x 5.733 = 5.46 lightseconds, but we don't need to use that figure in building our table.) We know that the clock is moving at 0.953c for 5.733 seconds with a time dilation of 0.3022 applied to it, so during that 5.733 seconds it is only measuring 5.733 x 0.3022 seconds as having passed for it while our stationary clock is measuring 5.733 x 0.5 seconds as having gone by. The difference between these two numbers is the actual amount of time lost by the moved clock, and that value is 1.1339.

Let's now do the same for 0.1c: (0.866 + 0.1) / (1 + 0.866 x 0.1) = 0.889c. The time dilation for that speed is 0.4578. We need to know how long it will take for an object moving at 0.889c to close in a distance of 0.5 lightsecond on something that's moving away from it at 0.866c. The closing speed is 0.889 - 0.866 = 0.023, so we divide the 0.5 lightsecond figure by this to get the time taken for the clock to move to its destination, and that gives us the time 21.732s. We know that the clock is moving at 0.889c for 21.732s with a time dilation of 0.4578 applied to it, so during the 21.732s it is only measuring 21.732 x 0.4578 seconds as having passed for it while our stationary clock is measuring 21.732 x 0.5 seconds as having gone by. The difference between these two numbers is the actual amount of time lost by the moved clock, and that value is 0.91615.

You can repeat that process for any other speeds you like, but to complete my part of the table we still need to work out what happens if the clock is moved at the speed of light, so I'll do that bit now. The closing speed this time is 1c - 0.866 = 0.134, so we divide the distance 0.5 by this to get the time taken for the clock to get into position, and that comes to 3.73s. During this time, the stationary clock gains time over the moving clock, but as it's ticking at 0.5 the rate of the time of the stationary reference frame, that means the moved clock loses 1.866s

(Note that the values that we correct the end result by are the same as in the previous table because they will still be calculated by the naive experimenters on the basis that the system is not moving, so these are the corrections they will make - we are describing the exact same events so we can't change the history as to which correction values they apply.) So, here's the new column for the table (speed, relative tick rate for moving clock compared with "stationary" clock; lost time for moved clock compared with "stationary" clock, correction value, corrected value):-

1c, 0, 1.866s, 1, 0.866s
0.5c, 0.6044, 1.1339s, 2 x 0.134, 0.8659s
0.1c, 0.9157, 0.91615s, 10 x 0.005, 0.866s

So, what have we learned from this? Look at the corrected times on the right. They are all the same (or would be if we used more accurate correction values rather than the approximations 0.134 and 0.005). That is how much the moved clock's time lags behind the "stationary" clock, and it's the same lag regardless of the speed at which the clock is moved. You can copy the way I did the calculations for moving the clock slower and slower, but if the system's moving at 0.866c, the residual error will always be 0.866.

What happens then if you measure your "one-way" speed of light? In case 1 your clocks are correctly synchronised because the system is stationary, and because your clocks are ticking in sync, it will take one second for light to move one lightsecond. In case 2, your clocks are out of sync. Light leaves one clock and has to chase after another clock which is 0.5 of a lightsecond ahead of it and which is moving away from it at 0.866c, so its closing speed is 0.134c. We divide 0.5 by this and get 3.73s. During this time, our clocks are ticking half as slowly, so they advance by 1.866 seconds, and that gives the experimenters a timing for the light of 1 second to cover the distance which they mistakenly think is 1 lightsecond long while they also mistakenly think their clocks are ticking in sync.

[dutch]

I was moving my parents to a new house out of state and myself to a new house (in the same city) so it took me awhile to answer.

I have spoken of two errors, one of which is larger the faster you move one clock while not moving the other clock (which you assume to be at rest). That error is real and predictable, so it can be corrected for.

There is no "error." Even from a LET framework all reference frames still work just fine even if ε=1/2 only for one of them. Furthermore, LET doesn't give you the option to know a true rest frame so you can't "correct" for the “error.” You first must do one thing... CHOSE A CONVENTION. Everything else you show is a result of this choice including time dilation. Unless you're God or you know how to break Lorentz Invariance, there is no error just choice of convention. It's a convention because we just arbitrarily agree to it and THEREFORE arbitrarily agree to the one-way speed of light. Time dilation has no objective reality we can test unless there are round trips (like the Twins Paradox). The one-way speed of light is set BEFORE we even consider time dilation by choice of convention. Time dilation only shows how we can’t distinguish between our infinite choices.

The whole business of trying to measure the one-way speed of light necessarily brings in the idea of an absolute frame if you imagine that light can travel relative to you at a speed other than c, so of course we have to explore this on that basis.

No, Einstein just chose ε =1/2. He chose it because it made the math easier, but he could have chosen ε =1/4, ε =1000, or ε =1/1000. You could have a convention where you always choose ε = 1/4 for your point of view. You could even make a block universe where ε =1/4 for your own viewpoint. What LET suggests is a unique true ε (from a "God's" view) exists for every single velocity (forcing a system to exist for how ε changes with velocity and gravity). This may be true but it adds considerable more complexity when applying it to General Relativity and doesn't work with the extremes of GR (perhaps GR is broken at the extremes). GR is already complex and this increases the complexity by a significant factor.

That ε can be arbitrarily chosen is good enough. I don't need to have the extra assumption that ε is different in an absolute sense for every frame. SR believers simply do not believe this at all. I know you like LET which is fine but you're opening an entire can of worms that is unnecessary to prove that we can't measure the one-way speed of light WITHOUT referring to an arbitrary convention.

Now you can personally think that one convention is "true" but you cannot prove it.

Now I personally think it's weird that people automatically assume ε=1/2 for them no matter what (often in an absolute sense) forcing them to look at other frames with ε ≠ 1/2 at least from their chosen perspective. However, I can't stop them from doing this because it mathematically works. Shoving LET down their throats doesn't work. I can point out that ε =1/4 etc would also work and that ε=1/2 is arbitrarily chosen in SR.

If we assume that the

You mean if you assume a convention? Stop, that's all you need to say to prove the one-way speed of light is unknown without a convention.

Now, here's the problem. If the clocks are moving so slowly that there is no error of the second type creeping in (due to different time dilations applying to different clocks while they're being moved apart), they must stay in sync in BOTH frames. They cannot possibly be ticking simultaneously in both frames though, so how can they get out of sync in the absolute frame where we watch the system moving past us?

You're invoking an absolute frame again when all anyone must do is show that ε is arbitrarily chosen. I don't care if you use LET or SR both set ε arbitrarily. Both assume a framework we can't verify. One by forming a block universe where ε=1/2 everywhere locally and the other via an absolute rest frame where ε=1/2 absolutely for only one frame. The underlying reason behind our ability to choose is we have no way to tell what ε equals because whatever we choose it doesn't alter experiments.

We CHOOSE a plane of simultaneity. Sure, clocks moving in other reference frames time dilate moving in one direction but ONLY after we assume our choice of convention.

t'/t = f'/f = (1 - v/c) / γ = γ / (1 + v/c) =  (1 - a/c)  / (1 + b/c)  γb/γa  where  v = (a + b) / (1 + a b /c²)
 
Time dilation could be :  τ = t / γ  or  τ = γ t   or  τ = γb/γa t  these are all different. Given a t, τ could be 0 to infinity.

LET says we can't locate the absolute rest frame so it also chooses a convention based exactly on the math above. SR does the same thing based on the math above.

I have spoken of two errors, one of which is larger the faster you move one clock while not moving the other clock (which you assume to be at rest). That error is real and predictable, so it can be corrected for. Even if you move the clock extremely slowly, an error of that kind will remain - all you can do is decrease its size until it becomes small enough that you have less need to bother making a correction. In denying the existence of that error, you are only going to confuse people who can see that it must exist.

Why do you keep bringing in "errors of the second type?" If an absolute rest frame exists such that ε=1/2 only absolutely relative to that reference frame, then clocks moving at v still work normally. They may run slower but that's not an "error" as the clocks are working as they should. There aren't two sources of errors there's time dilation and that's it. As an object moves at v with a specific time dilation then the clocks would get out of sync. How much they get out of sync one-way depends on distance at a specific v AFTER we choose the "rest frame."

t' =  (t - v x/c²) / γ   Let t = 0  then we have  t' =  v x/c² / γ    →  t' is proportional to x  as v → c γ → 0

The above isn't a "second error." Time dilation gets clocks out of sync regardless. You're assuming a "first error" relative to an arbitrarily set absolute rest frame (at least from our point of view) and a second error relative to another arbitrary frame moving at velocity v. Both “errors” are relative to whatever frame of reference/convention we choose. You can account for this “error” exactly but ONLY after assuming a convention. You can make the “error” arbitrarily small (slow clock transport) but ONLY with respect to a chosen convention. Other conventions would have the “error” significantly larger (in terms of the speed of light it could be infinitely off with another convention; simultaneity on the other hand is bounded by the two-way speed of light).

The one-way speed of light is literally chosen by convention in our mathematics. If it does have a definite value for a specific reference frame like both SR AND LET claim then we simply have no way to measure it. 

[David Cooper]

I have spoken of two errors, one of which is larger the faster you move one clock while not moving the other clock (which you assume to be at rest). That error is real and predictable, so it can be corrected for.

There is no "error."

If a clock runs slow because it's moved, it fails to record all the time that has passed for it, and that results in an error in its timing. The faster you move it, the bigger that error is.

Even from a LET framework all reference frames still work just fine even if ε=1/2 only for one of them. Furthermore, LET doesn't give you the option to know a true rest frame so you can't "correct" for the “error.”

I don't know if you understand how reasoning works, but it's generally conditional using IF and THEN clauses. E.g. IF the system is stationary, THEN a clock moving through the system runs slow. While we are discussing things under that condition ("IF the system is stationary"), everything said is governed by that condition. When we switch to a different condition such as "IF the system is moving at 8.866c", everything we then say is governed by that new condition. The fact that we can't tell which of the cases is true does not negate the truth (conditional truth) of what is said about that case on the basis that it's true if the condition is true and may not be true if the condition is false.

You first must do one thing... CHOSE A CONVENTION.

No, you choose a frame of reference. Choosing a convention then gives you the same result no matter which valid convention you go for.

Everything else you show is a result of this choice including time dilation.

Everything I show is a result of which frame you choose to use.

Unless you're God or you know how to break Lorentz Invariance, there is no error just choice of convention.

You are choosing a frame, and you are almost always choosing the wrong one. Your results will seem just as valid as any others because you can't tell who's right, but the underlying mechanism must be a rational one and must hold a truth about what is happening which it doesn't reveal to us. That doesn't stop us looking at the way the mechanism works though by working through things on a conditional basis.

The one-way speed of light is set BEFORE we even consider time dilation by choice of convention.

It's the choice of frame that selects a proposed one-way speed of light relative to the system.

The whole business of trying to measure the one-way speed of light necessarily brings in the idea of an absolute frame if you imagine that light can travel relative to you at a speed other than c, so of course we have to explore this on that basis.

No, Einstein just chose ε =1/2. He chose it because it made the math easier, but he could have chosen ε =1/4, ε =1000, or ε =1/1000. You could have a convention where you always choose ε = 1/4 for your point of view. You could even make a block universe where ε =1/4 for your own viewpoint.

You choose a frame of reference and thereby conditionally assert that the speed of light is the same relative to that frame in all directions. That makes the speed of light not equal to c relative to objects moving through that frame in most directions and provides a different theory as to the reality of the situation to any of the theories based on using different frames (all of those theories being in contradiction of each other as only one of them can be true).

That ε can be arbitrarily chosen is good enough. I don't need to have the extra assumption that ε is different in an absolute sense for every frame. SR believers simply do not believe this at all. I know you like LET which is fine but you're opening an entire can of worms that is unnecessary to prove that we can't measure the one-way speed of light WITHOUT referring to an arbitrary convention.

Your choosing a convention is simply selecting a frame to act as an absolute frame and thereby setting yourself up to measure the speed of light relative to that frame as c.

Now you can personally think that one convention is "true" but you cannot prove it.

There can only be one truth for the universe regardless of whether we can identify it or not.

Now I personally think it's weird that people automatically assume ε=1/2 for them no matter what (often in an absolute sense) forcing them to look at other frames with ε ≠ 1/2 at least from their chosen perspective. However, I can't stop them from doing this because it mathematically works. Shoving LET down their throats doesn't work. I can point out that ε =1/4 etc would also work and that ε=1/2 is arbitrarily chosen in SR.

If you restrict yourself to ε=1/2, you are selecting the frame of reference in which the system is at rest, and that sheds no light on anything. It's only when you work things through using at least two different reference frames that you can gain a real understanding of what's going on.

If we assume that the

You mean if you assume a convention? Stop, that's all you need to say to prove the one-way speed of light is unknown without a convention.

It's much easier to understand that you're choosing a particular frame of reference, and that's all your convention actually is, so why are you objecting to me calling a spade a spade instead of burying it in obfuscation?

Now, here's the problem. If the clocks are moving so slowly that there is no error of the second type creeping in (due to different time dilations applying to different clocks while they're being moved apart), they must stay in sync in BOTH frames. They cannot possibly be ticking simultaneously in both frames though, so how can they get out of sync in the absolute frame where we watch the system moving past us?

You're invoking an absolute frame again when all anyone must do is show that ε is arbitrarily chosen. I don't care if you use LET or SR both set ε arbitrarily. Both assume a framework we can't verify. One by forming a block universe where ε=1/2 everywhere locally and the other via an absolute rest frame where ε=1/2 absolutely for only one frame. The underlying reason behind our ability to choose is we have no way to tell what ε equals because whatever we choose it doesn't alter experiments.

What I've done is point out that infinitely slow transport (which can't be done because infinitely slow will never separate the clocks at all, but if it could by magic be done...) would lead to the separated clocks remaining in sync in all frames (which is impossible) rather than just one, but you don't seem to have taken that on board.

We CHOOSE a plane of simultaneity.

Yes - a frame of reference.

Sure, clocks moving in other reference frames time dilate moving in one direction but ONLY after we assume our choice of convention.

All frames of reference apply to every scenario, so most of them need time dilation to act on your "infinitely slowly moving" clocks, which means they can't be moving infinitely slowly relative to the "stationary" clock (or relative to another "infinitely slowly moving" clock "moving" in the opposite direction) if the time dilation differences are able to act. The idea of moving them infinitely slowly and having no time dilation act on them at all requires the synchronisation to remain in place for all frames, and that's impossible - that's why you should be banned from using it as an any kind of explanation of what's going on.

t'/t = f'/f = (1 - v/c) / γ = γ / (1 + v/c) =  (1 - a/c)  / (1 + b/c)  γb/γa  where  v = (a + b) / (1 + a b /c²)
 
Time dilation could be :  τ = t / γ  or  τ = γ t   or  τ = γb/γa t  these are all different. Given a t, τ could be 0 to infinity.

LET says we can't locate the absolute rest frame so it also chooses a convention based exactly on the math above. SR does the same thing based on the math above.

It's nothing more than choosing a frame of reference.

I have spoken of two errors, one of which is larger the faster you move one clock while not moving the other clock (which you assume to be at rest). That error is real and predictable, so it can be corrected for. Even if you move the clock extremely slowly, an error of that kind will remain - all you can do is decrease its size until it becomes small enough that you have less need to bother making a correction. In denying the existence of that error, you are only going to confuse people who can see that it must exist.

Why do you keep bringing in "errors of the second type?"

What? In that bit you've quoted which is all about the first type? The second type of error is the unidentifiable one that's left over once you think you've corrected for the first type.

If an absolute rest frame exists such that ε=1/2 only absolutely relative to that reference frame, then clocks moving at v still work normally. They may run slower but that's not an "error" as the clocks are working as they should.

If they are attempting to measure time, they are failing to record all of the time that has passed for them, so they are building up an error. They are counting cycles and they are correctly counting the number of cycles that have run through, so in that sense they are not producing an error. It therefore depends on what you expect of them as to whether you call it an error or not, and I call it an error on the basis that they are failing to measure all the time that's passed for them. That is a perfectly sound basis to work upon.

There aren't two sources of errors there's time dilation and that's it. As an object moves at v with a specific time dilation then the clocks would get out of sync. How much they get out of sync one-way depends on distance at a specific v AFTER we choose the "rest frame."

What I have done in dividing things up into two errors is fully sound. The first type of error is the component which you attempt to correct for on the basis that the system is not moving, and I separated it away from the total error in order to show how all the different speeds of travel for moving the "moving" clock can easily be corrected for to provide the same synchronisation as each other. If you move the clock fast, you make a big correction, and if you move it very slowly you make a tiny correction, but the result is the same. The point of doing this was to show that all speeds of movement for the clock are equivalent to just sending a signal at the speed of light to synchronise two clocks that are already separated and making a correction to that in the same way.

t' =  (t - v x/c²) / γ   Let t = 0  then we have  t' =  v x/c² / γ    →  t' is proportional to x  as v → c γ → 0

The above isn't a "second error." Time dilation gets clocks out of sync regardless. You're assuming a "first error" relative to an arbitrarily set absolute rest frame (at least from our point of view) and a second error relative to another arbitrary frame moving at velocity v. Both “errors” are relative to whatever frame of reference/convention we choose.

My two errors can be regarded as a two parts of a single error if you like, in which case we can rename the first kind of error as one component of the error (the part we can attempt to correct on the basis that the system isn't moving) and another component of the error which would only be corrected for if we then adjust to the absolute frame. There can be situations where the attempt at correcting the first component actually takes it away from the correct value to a wrong one. But the whole point of dividing the error into these two categories is to correct for the first one to produce the same end value regardless of which speed was used to move the moving clock away from the stationary clock and have the same degree of synchronisation for the infinitely slow clock transport as for sending a signal at c. It's all about showing the equivalence of these methods so that it becomes clear that no matter how slowly you move your clock, you might as well just send a light signal to set up the clock timings and then use those clocks to time another flash of light over the same course.

You can account for this “error” exactly but ONLY after assuming a convention.

Which is an obfuscated way of saying that you can account for it exactly only after choosing a specific frame of reference to work with.

You can make the “error” arbitrarily small (slow clock transport) but ONLY with respect to a chosen convention. Other conventions would have the “error” significantly larger (in terms of the speed of light it could be infinitely off with another convention; simultaneity on the other hand is bounded by the two-way speed of light).

Which is an obfuscated way of saying, using other frames of reference as your base would lead you to use different correction values. All you're really doing here is objecting to my clear use of language and trying to impose your obfuscated language on me instead, but you're not going to get your way. I prefer to call a spade a spade because it leads to greater understanding. And once the first component of the error has been corrected for (usually on the basis that the system is stationary), there remains an error which is unidentifiable to us depending on which frame is the actual frame in which the speed of light is c relative to it in all directions.

The one-way speed of light is literally chosen by convention in our mathematics.

= The one-way speed of light is assumed to be c relative to the frame of reference chosen for the calculations.

If it does have a definite value for a specific reference frame like both SR AND LET claim then we simply have no way to measure it.

And I have made no claim that it can be measured by anything (other than the universe itself).

[Bored chemist]

If you're doing this on one planet and your cousin is doing the same thing on another planet moving relative to yours, both of you are making the same claim to be stationary, and at least one of you is being foo

led.

No, I'm stationary from my PoV and he's stationary from his.
But, since these viewpoints are different there's no contradiction.
Nobody is being "fooled"
As I have pointed out before, I know that other people who are moving WRT me will see things differently- so it's no surprise that they will interpret the experiment I'm doing differently and will think the one-way speed of light I measure is different.
I know that.
But, as I pointed out before, they won't even agree with me about the speed of the bus into town, so I don't expect them to agree with me about the one way speed of light.
I'm only interested in measuring it in my frame of reference.
ymmv