[guest4091]

It is written, "the measured speed of light is c", A. Einstein.
aka the 2nd postulate.

Let's elaborate on the simultaneity convention.

"That light requires the same time to traverse the same path A to M as for the path B to M is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity."
Relativity The Special and the General Theory
Albert Einstein 1961 Crown Publishers Inc. pg 23
aka, physics by decree.

[dutch]

You didn't.
But, unless there are circumstances where they are invalid then it doesn't matter that I made them.
Pointing out that I'm making an assumption is only worthwhile if that assumption is, or might be, wrong.

Generalizations of Lorentz transformations with anisotropic one-way speeds

A synchronization scheme proposed by Reichenbach and Grünbaum, which they called ε-synchronization, was further developed by authors such as Edwards (1963),[48] Winnie (1970),[17] Anderson and Stedman (1977), who reformulated the Lorentz transformation without changing its physical predictions.[1][2] For instance, Edwards replaced Einstein's postulate that the one-way speed of light is constant when measured in an inertial frame with the postulate:

The two way speed of light in a vacuum as measured in two (inertial) coordinate systems moving with constant relative velocity is the same regardless of any assumptions regarding the one-way speed.[48]

So the average speed for the round trip remains the experimentally verifiable two-way speed, whereas the one-way speed of light is allowed to take the form in opposite directions:

c± = c / (1 ± κ)   where κ is any value between 0 and 1.

As demonstrated by Hans Reichenbach and Adolf Grünbaum, Einstein synchronization is only a special case of a more broader synchronization scheme, which leaves the two-way speed of light invariant, but allows for different one-way speeds. The formula for Einstein synchronization is modified by replacing ½ with ε:[4]

t2 = t1 + ε (t3 - t1)

ε can have values between 0 and 1. It was shown that this scheme can be used for observationally equivalent reformulations of the Lorentz transformation, see Generalizations of Lorentz transformations with anisotropic one-way speeds.

As required by the experimentally proven equivalence between Einstein synchronization and slow clock-transport synchronization, which requires knowledge of time dilation of moving clocks, the same non-standard synchronizations must also affect time dilation. It was indeed pointed out that time dilation of moving clocks depends on the convention for the one-way velocities used in its formula.[17] That is, time dilation can be measured by synchronizing two stationary clocks A and B, and then the readings of a moving clock C are compared with them. Changing the convention of synchronization for A and B makes the value for time dilation (like the one-way speed of light) directional dependent. The same conventionality also applies to the influence of time dilation on the Doppler effect.[18] Only when time dilation is measured on closed paths, it is not conventional and can unequivocally be measured like the two-way speed of light. Time dilation on closed paths was measured in the Hafele–Keating experiment and in experiments on the Time dilation of moving particles such as Bailey et al. (1977).[19] Thus the so-called twin paradox occurs in all transformations preserving the constancy of the two-way speed of light.

https://en.wikipedia.org/wiki/One-way_speed_of_light#Generalizations_of_Lorentz_transformations_with_anisotropic_one-way_speeds

The formula is:

c± = c / (1 ± κ)   where κ is ANY value between 0 and 1.

as κ approaches 1 c- → ∞  which is nowhere near c. Infinity isn't "close to" c.

You didn't.
But, unless there are circumstances where they are invalid then it doesn't matter that I made them.
Pointing out that I'm making an assumption is only worthwhile if that assumption is, or might be, wrong.

It might be wrong, it might be right the point is we CANNOT measure the one-way speed of light independent of a choice of synchronization. You're essentially breaking the Lorentz Transformation into parts and you CANNOT do this. We measure the Lorentz Transform taken as a whole.

The following is true:

t'/t = f'/f = (1 - v/c) / γ     where γ = √(1 - (v/c)²)      (this is the Lorentz Time Transformation in simplest form)

However, the following is also true:

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.

We CANNOT measure time dilation τ by any known means. What we can measure in a lab is observed time t' and our time t. We can only measure values locally. No matter what assumptions we make t' and t are symmetric.

Everyone could agree on the two-way speed of light c relative to everything, everyone will agree on t'/t, everyone will agree on f'/f, and defining v by a specific f'/f there's agreement on v. People won't agree on τ and they won't agree on the one-way speed of light UNLESS you force a certain convention. Even then they will seem to contradict because both observers will think the others clock is running slow and both will think light is moving at exactly c relative to them (and from their perspective only for them).

If all you can know is t' and t (really t'/t) you can certainly choose one equation to make your life easier. However, you CANNOT look at individual chunks of that equation like they have some reality beyond your choice of convention. This is the greatest misunderstanding of Relativity. You CANNOT break the Lorentz Transformation into component pieces without making major assumptions. We only measure the input t and the output t'.

It is written, "the measured speed of light is c", A. Einstein.
aka the 2nd postulate.

Let's elaborate on the simultaneity convention.

"That light requires the same time to traverse the same path A to M as for the path B to M is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity."
Relativity The Special and the General Theory
Albert Einstein 1961 Crown Publishers Inc. pg 23
aka, physics by decree.

That's Einstein admitting he had formed a convention of his own freewill. It's OK to have a convention as long as people admit it's a convention. He wanted A definition of simultaneity so he could make the math easier. It's a beautiful shortcut that works but we must still acknowledge that we can't measure the one-way speed of light  independent of a synchronization convention.

[dutch]

Showing how this:

t'/t = f'/f = (1 - v/c) / γ     where γ = √(1 - (v/c)²) 

Becomes this:

t' = ( t - v x / c²) / γ    (I used the reciprocal of what's normally γ in the last post so the math wouldn't look as messy)

t'/t = (1 - v/c) / γ

=  (t - t v/c) / γ

Assume the one-way speed of light is c and synchronize using Einstein Synchronization t = x/c (synchronize clocks with light).

=  (t -  x/c v/c ) / γ

= ( t - v x / c²) / γ

The equation f'/f = (1 - v/c) / γ  is also the Relativistic Doppler shift  but so is f'/f = γ / (1 + v/c)  and... (side note: the LT and RDS are highly related as they should be)

f'/f = √(1 - v/c)  / √(1 + v/c)    and  f'/f =  (1 - a/c)  / (1 + b/c)  γb/γa   for   v = (a + b) / (1 + a b /c²)   

Why is  t'/t = f'/f   when f is a frequency of light?

If I send a radio signal out at 10 MHz and someone else receives it at 20 MHz they observe my time at 2 times the normal rate. I can confirm with them and they can send a signal to me from an identical radio. I will receive 20 MHz from their 10 MHz radio keeping the observables symmetric. The ratios of frequency and ratios of time observed must be equal to conserve information. I don't know τ independent of a convention nor can I know the one-way speed of light independent of a convention.

https://en.wikipedia.org/wiki/Doppler_effect
https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

[Bored chemist]

"If I send a radio signal out at 10 MHz and someone else receives it at 20 MHz ... "
Then we know that you are not playing by the rules I specified where everyone is walking or cycling round on the surface of the Earth.

[dutch]

Then we know that you are not playing by the rules I specified where everyone is walking or cycling round on the surface of the Earth.

?

What are you even writing about? I was making a point about the math. The size of the frequency shift doesn't matter to the point whatsoever. You completely missed the point that t' and τ are completely different variables. I don't care if you're "walking or cycling" you still don't know the one-way speed of light without using a clock synchronization convention.

However, I'm glad you're getting your exercise.

I've got questions but let me set the questions up. Say there's a clock positioned at A and another at B when those clocks remain stationary on the surface of the Earth such that one is at sea level (A) and the other (B) 1,000 feet above sea level their reference frame feels acceleration and clock B will run faster than clock A.

Using the Equivalence Principle, I can accelerate two clocks connected by a 1,000 foot steel cable. If I accelerate in the direction of clock B then clock B will run faster when observed by clock A during the acceleration. I can accelerate slow enough such that when I inspect the cable later it will be the same length and in the same condition it originally was. I therefore conclude the length remains 1,000 feet between ships when finished (at least as observed by me). When the acceleration stops clock B will again run at the same rate as observed by clock A (this observed time is t').

If I assume Einstein Clock Synchronization (one-way speed of light is c for my frame) then clocks that were synchronized before the acceleration are now out of sync. However, my clocks remain synchronized to the original reference frame. Therefore, I decide of my own freewill to move either the rear clock's time reading forward or the front clock's time reading backwards to fix the "problem" I perceive based on my convention. I choose to re-synchronize the clocks because I assume the one-way speed of light is constant in my new reference frame. I had assumed the one-way speed of light was constant in my old reference frame but after I choose to re-synchronize my clocks the old assumption is no longer valid. I decided I didn't like my original choice to keep the math simpler.

Questions:

1) Why couldn't I just keep my original clock synchronization? Why do I have to re-synchronize?

2) How is my synchronization not based on convention? 

3) If I did attempt to measure the one-way speed of light using my original synchronization it would be different in different directions. If I'm lazy and I keep my original synchronization why is this less valid then re-synchronizing? Why do I have to reset spatially separated clocks after every acceleration? Does nature know that it needs to re-synchronize its clocks after accelerations?

Say you're in a lab that changed velocity by .8c a while ago and the scientists in charge of the lab re-synchronizes the clocks. You measure the one-way speed of light and get exactly c. You get excited thinking you just measured the one-way speed.  You also don't realize that the scientist reset the clocks... You attempt to justify the re-synchronization. However, careful analysis of slow clock transport and light synchronization you find that those are identical. To your surprise you go back into the system and restore the original clock settings and the one-way speed of light is 1.8γ² c in one direction and .2γ² c in the other (where γ = √(1 - .8²) = .6 the gamma is due to time dilation/length contraction with respect to the original frame chosen as a reference; also remember v = d/t).

c± = c / (1 ± κ)     If   κ = .8 then c - = c / (1 -.8 ) = 5c = 1.8γ² c      and c+ = c / (1 +.8 ) = 5/9c = .2γ² c  (predicted value without resync)   
 
Why is slow clock transport the same as fast clock transport? Take a light clock where every bounce of the light between two mirrors is one tick. As you slowly transport the clock from A to B you realize something amazing... you just sent a light beam from A to B just like light synchronization does. You sent the light in a zigzag pattern but careful analysis of the math shows that it doesn't matter because the light still goes through a net horizontal distance from A to B. ALL clocks will give the exact same result because they are all more alike than people think. We are after all made out of a bunch of sub atomic particles interacting via the fundamental forces. If you don't believe this than careful analysis of experiments proves it (it's an experimentally verified fact). Hell, Quantum Field Theory predicts that particles like electrons, quarks, weak bosons, etc would move at light speed if not for their interaction with the Higgs Field and/or the EM field and/or strong field (forming a rest mass). This interaction doesn't slow down the fundamental two-way speed of their "fields" it just gets them "stuck in the mud" via very complex interactions with other fields. This is often called "self energy." Gravity, the EM field, and the strong force all have a two-way speed c. Careful analysis shows that you quickly start to run out of "slow clock transports" that aren't really "light synchronizations" in disguise. Experiments and Relativity also verifies this fact.

They discuss QFT in this series.

Finally, FYI even tiny changes in velocity like walking or cycling changes the speed of light if you don't re-sync the clocks. We do measure this difference as the speed of light is only c locally in General Relativity (small but measurable). Einstein stipulates that you need to re-synchronize your clocks locally. That is you need to keep re-syncing your clocks after every acceleration so they are temporarily localized. In a gravity field or when under acceleration the speed of light is only c in a small enough volume around every point (spatially localized; how local this needs to be depends on the acceleration/curvature). This is the Law of Physics called Local Lorentz Covariance (the heart of GR and included in all modern mainstream theories). We always can assume a convention where we can measure c locally in all directions.

[guest39538]

To test the present speed of light one way we can simply use a strobe set to flash once per second.   We can then have a detector (radiometer) a set distance away to detect the light.

How are you going to time it? Are you going to use one clock or two?

If you use one clock and place it by the strobe unit, how long does it take for the information to get back from the detector to the clock once the light has reached the detector? That signal returns at the speed of light, so you're actually timing the two-way speed of light (i.e. a round trip).

If you use two clocks and have one by the strobe and the other by the detector, you need to synchronise the two clocks, and how do you synchronise them? Any viable method of synchronisation that you use will automatically lead to you measuring the two-way speed of light instead of the one-way speed of light.

I am not using any clocks to time it in the above.  The strobe flashes 1 flash per second, the detector then detects and the detection of the pulses should be in correlation to the strobe at distance x, if it isn't , then c is not c as it stands at the moment and Maxwell would be wrong . i.e the detects should be 1 second apart.

[dutch]

I am not using any clocks to time it in the above.  The strobe flashes 1 flash per second, the detector then detects and the detection of the pulses should be in correlation to the strobe at distance x, if it isn't , then c is not c as it stands at the moment and Maxwell would be wrong . i.e the detects should be 1 second apart.

I already answered this. You're measuring t', t' (and f') are symmetric as required by Relativity as shown below:

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.

The one-way speed of light could be measured with values from zero to infinity with the following (depending on synchronization used):

c± = c / (1 ± κ)       where  κ = 0 to 1

If v = 0 between two objects then f'/f = 1 and t'/t = 1 but so what? No one disputes this fact. Both observers will think the other clock is running at the same rate as their own clock. Receiving light pulses every second IS a light clock... so you are using a clock.

To measure the one-way speed of light independent of a synchronization convention you MUST:

1) Universally prove what now is at two different locations beyond some arbitrary convention.

2) Get clocks A (start) and B (finish) ticking at the same rate in an absolute sense.

For over 100 years and with countless geniuses thinking about the subject the first condition still has not been met. The second condition can be met (have no acceleration/minimal curvature) but it's nowhere near sufficient to measure the one-way speed independent of a convention.

You may know the pulses are 1 second apart BUT you don't know what time the pulses were emitted without relying on a synchronization convention that let's you sync two spatially separated clocks. You MUST start the stop watch at the same absolute time at position A and at position B to measure the speed in an absolute sense. We do NOT have a sense of absolute time. In Relativity absolute time simply doesn't exist so we use a convention to form "planes of simultaneity." In LET absolute time exists but we have no way to identify it because while a rest frame exists in principle no frame is preferred in the math in any experimentally distinguishable way.

[David Cooper]

I am not using any clocks to time it in the above.

Let me get this straight - you're timing something without using any kind of clock?

The strobe flashes 1 flash per second, the detector then detects and the detection of the pulses should be in correlation to the strobe at distance x, if it isn't , then c is not c as it stands at the moment and Maxwell would be wrong . i.e the detects should be 1 second apart.

So, you've got a flash every second at the strobe and a flash being detected every second at the detector, so how do you get from there to a measurement of the time taken for the light to get from the strobe to the detector?

[guest39538]

I am not using any clocks to time it in the above.

Let me get this straight - you're timing something without using any kind of clock?

The strobe flashes 1 flash per second, the detector then detects and the detection of the pulses should be in correlation to the strobe at distance x, if it isn't , then c is not c as it stands at the moment and Maxwell would be wrong . i.e the detects should be 1 second apart.

So, you've got a flash every second at the strobe and a flash being detected every second at the detector, so how do you get from there to a measurement of the time taken for the light to get from the strobe to the detector?

At the detector end the received packets should be an equal space apart on the readout, equal to the distance the light as travelling .plus 1 second.

My thoughts are still a bit incomplete, but I can visual ''see'' the idea in my head.   Maybe I will draw the idea later.

The distance between the strobe and detector would have to be a fraction of the distance of the speed of light, the readout would have to scroll off at a second rate, so many lines per second, then we could do the maths.  I am using the scrolling of the readout as my second clock.

added- ok I am going to visualise the experiment in my head, I am know longer looking at the screen so forgive me if my tyoing is rubbish, I now see the strobe and see the first photon emitted, the photon travles across the distance tot he detector which is already in motion, the photon striokes the detector and puts a mark on the graph, The second one comes and strikes a mark oon the graph, I can now use scaled reading to give a result.


detect.jpg (31.83 kB . 1152x648 - viewed 5247 times)

added - let us start with the first detect and in example we will say 0.1s, the second detect 1.1s, the third 2.2s and so on .

let us use 3.26cm of distance on the printout to equal 1.s

printout speed = 3.26cm/1.s


We are using an equivalent to measure time and by the comparison can measure speed .

added-  No need to worry about synchronous starts either. Turn on the detector , turn on the strobe, get measuring lol


added- obviously if we could get the distance of 299792458m between emit and detect it would be much easier to calculate.

Each flash should be recorded 3.26cm apart , this would equal the present prediction and Maxwell had it right. If not, we can then use the distance difference to get a true speed. Quite elementary really.

<3.26cm = faster

=3.26cm= equal 299792568m/s

>3.26cm = slower

In short we are setting up the experiment to the exact prediction of Maxwell. All results should conform to this, if not then it is quite obvious we are wrong about the speed.


3.26cm /s  =   299792458m/s    in our scaling.


A photon travels 299792458m compared to a line that travels 3.26cm in the same time.

added- only the first 3 detects are important to give a result.

[guest39538]

Ok, I have brought my head back to reality after some deep thought and drew a working model .  However you science geniuses now need to scale down the model or re-scale the model to suit your needs and make it simpler.


detect 1.jpg (31.06 kB . 1152x648 - viewed 5313 times)

p.s Just elementary my dear Watson , what next Watson?  that was too easy ....

p.s don't forget to make your printout machine variable in speed so you can adjust the speed to get a true speed of c if the prediction fails.

[Bored chemist]

Then we know that you are not playing by the rules I specified where everyone is walking or cycling round on the surface of the Earth.

?

What are you even writing about? I was making a point about the math. The size of the frequency shift doesn't matter to the point whatsoever. You completely missed the point that t' and τ are completely different variables. I don't care if you're "walking or cycling" you still don't know the one-way speed of light without using a clock synchronization convention.

However, I'm glad you're getting your exercise.

I've got questions but let me set the questions up. Say there's a clock positioned at A and another at B when those clocks remain stationary on the surface of the Earth such that one is at sea level (A) and the other (B) 1,000 feet above sea level their reference frame feels acceleration and clock B will run faster than clock A.

Using the Equivalence Principle, I can accelerate two clocks connected by a 1,000 foot steel cable. If I accelerate in the direction of clock B then clock B will run faster when observed by clock A during the acceleration. I can accelerate slow enough such that when I inspect the cable later it will be the same length and in the same condition it originally was. I therefore conclude the length remains 1,000 feet between ships when finished (at least as observed by me). When the acceleration stops clock B will again run at the same rate as observed by clock A (this observed time is t').

If I assume Einstein Clock Synchronization (one-way speed of light is c for my frame) then clocks that were synchronized before the acceleration are now out of sync. However, my clocks remain synchronized to the original reference frame. Therefore, I decide of my own freewill to move either the rear clock's time reading forward or the front clock's time reading backwards to fix the "problem" I perceive based on my convention. I choose to re-synchronize the clocks because I assume the one-way speed of light is constant in my new reference frame. I had assumed the one-way speed of light was constant in my old reference frame but after I choose to re-synchronize my clocks the old assumption is no longer valid. I decided I didn't like my original choice to keep the math simpler.

Questions:

1) Why couldn't I just keep my original clock synchronization? Why do I have to re-synchronize?

2) How is my synchronization not based on convention? 

3) If I did attempt to measure the one-way speed of light using my original synchronization it would be different in different directions. If I'm lazy and I keep my original synchronization why is this less valid then re-synchronizing? Why do I have to reset spatially separated clocks after every acceleration? Does nature know that it needs to re-synchronize its clocks after accelerations?

Say you're in a lab that changed velocity by .8c a while ago and the scientists in charge of the lab re-synchronizes the clocks. You measure the one-way speed of light and get exactly c. You get excited thinking you just measured the one-way speed.  You also don't realize that the scientist reset the clocks... You attempt to justify the re-synchronization. However, careful analysis of slow clock transport and light synchronization you find that those are identical. To your surprise you go back into the system and restore the original clock settings and the one-way speed of light is 1.8γ c in one direction and .2γ c in the other (where γ = 1/√(1 - .8²) = 1.67 the gamma is due to time dilation with respect to the original frame chosen as a reference; also remember v = d/t).

c± = c / (1 ± κ)     If   κ = 2/3 then c - = c / (1 -2/3) = 3c = 1.8γ c      and c+ = c / (1 +2/3) = 1/3c = .2γ c  (predicted value without resync)   
 
Why is slow clock transport the same as fast clock transport? Take a light clock where every bounce of the light between two mirrors is one tick. As you slowly transport the clock from A to B you realize something amazing... you just sent a light beam from A to B just like light synchronization does. You sent the light in a zigzag pattern but careful analysis of the math shows that it doesn't matter because the light still goes through a net horizontal distance from A to B. ALL clocks will give the exact same result because they are all more alike than people think. We are after all made out of a bunch of sub atomic particles interacting via the fundamental forces. If you don't believe this than careful analysis of experiments proves it (it's an experimentally verified fact). Hell, Quantum Field Theory predicts that particles like electrons, quarks, weak bosons, etc would move at light speed if not for their interaction with the Higgs Field and/or the EM field and/or strong field (forming a rest mass). This interaction doesn't slow down the fundamental two-way speed of their "fields" it just gets them "stuck in the mud" via very complex interactions with other fields. This is often called "self energy." Gravity, the EM field, and the strong force all have a two-way speed c. Careful analysis shows that you quickly start to run out of "slow clock transports" that aren't really "light synchronizations" in disguise. Experiments and Relativity also verifies this fact.

They discuss QFT in this series.

Finally, FYI even tiny changes in velocity like walking or cycling changes the speed of light if you don't re-sync the clocks. We do measure this difference as the speed of light is only c locally in General Relativity (small but measurable). Einstein stipulates that you need to re-synchronize your clocks locally. That is you need to keep re-syncing your clocks after every acceleration so they are temporarily localized. In a gravity field or when under acceleration the speed of light is only c in a small enough volume around every point (spatially localized; how local this needs to be depends on the acceleration/curvature). This is the Law of Physics called Local Lorentz Covariance (the heart of GR and included in all modern mainstream theories). We always can assume a convention where we can measure c locally in all directions.

In the limit of a low speed, low acceleration, low gravity situation, we can measure the one-way speed of light to an arbitrarily high accuracy.

If you have someone moving WRT us so fast that they perceive our 1 MHz clock as 20MHz then that's a different scenario.

You are plainly better at the maths than I am.
So, calculate the answer to the question I set earlier. If the one way speed of light is the same as C then m will be 10,000.
What value do you calculate (for simplicity assume that C is exactly  300,000,000 m/s, the frame rate is (locally) exactly 1 million frames per second and the separation is exactly  million metres.)
I'd expect m to be 10,000.

[dutch]

In the limit of a low speed, low acceleration, low gravity situation, we can measure the one-way speed of light to an arbitrarily high accuracy.

No, not independent of a clock synchronization which makes the one-way speed of light measurement a two-way speed of light measurement.

If you have someone moving WRT us so fast that they perceive our 1 MHz clock as 20MHz then that's a different scenario.

This simply doesn't matter. Where did I say this mattered? I made an example before to show you how the math works and you completely misunderstood it.

You are plainly better at the maths than I am.

If you don't know the math then you most likely don't understand the argument. From what you've written it appears that you're missing the points and the math. You probably should understand those before asserting anything.

So, calculate the answer to the question I set earlier. If the one way speed of light is the same as C then m will be 10,000.
What value do you calculate (for simplicity assume that C is exactly  300,000,000 m/s, the frame rate is (locally) exactly 1 million frames per second and the separation is exactly  million metres.)
I'd expect m to be 10,000.

OK. I don't think your math here is right even if I make your assumptions. The separation should be 3 million meters with your assumptions not 1 million meters. Also you're making the exact same mistake Thebox is making with some of his setups to measure the one-way speed of light. The 10,000 is a frequency measured in Hz. You have a signal of 10 KHz.

Both position A (emitter) and position B (receiver) if their relative velocity is 0 will measure this same frequency of 10 KHz. OK fine, this doesn't tell me when position A emitted a particular flash it only tells me the flashes occur 10,000 times per second. This isn't helpful in measuring the one-way speed of light at all.

Let's slow way down and look at the Classical Doppler Shift:

fmo = f (1 - v)        This is the equation for a moving observer.

fms = f / (1 + v)     This is the equation for a moving source. Please note v is a value between -1 and 1 (velocity negative is towards observer and positive is away from observer).

for a moving source and observer    fmso = f (1 - a) / (1 + b)   (combining the two equations)

Say you've got a long train and you've got a locomotive in the back and front (pusher and puller). These locomotives have train whistles and the train is moving at a velocity of  .1.

If f = 1,000 then what frequency would the locomotive in the back see?

Well to get this value it's simple. The emitting locomotive in the front is a source moving away and the locomotive in the back is the observer moving towards:

fmso = f (1 - a) / (1 + b) = 1000 ( 1 - (-.1))/(1 + .1) =1000 * 1.1/1.1 = 1000

The rear locomotive is moving at the same velocity as the front locomotive so the frequency does NOT change.

Likewise the following occurs if the locomotive in the front observes the whistle from the rear.

fmso = f (1 - a) / (1 + b) = 1000 ( 1 - .1)/(1 + (-.1)) =1000 * .9/.9 = 1000

Again the frequency does not change.

However, if the rear train whistle emits a noise it moves forward at a velocity of .9 because the train is moving at .1. The other sound moves rearward at 1.1.

The sound moving rearward reaches the rear locomotive FASTER than the sound moving toward the forward locomotive. However, the frequency does NOT shift.

How do we know the sounds travel at different speeds going forwards and backwards? Well, we can because sound isn't the fastest thing we could use to send a message. We can use light signals to notify the front locomotive of when the rear locomotive fired its whistle. We can then use light to more accurately set our clocks than sound. Light is simply much closer to instantaneous than sound so its a better option to synchronize clocks (light moves so fast it gives us an existence where we think we know what "now" is at different locations). We can definitely tell that the rear whistle must fire much earlier than the front whistle in order for the observers to receive the sounds at the same time as synchronized by light. The rear whistle must fire earlier if a light signal can be sent informing the front locomotive that the rear locomotive fired its whistle.

HOWEVER, light does not move instantaneous and it's actually infinitely far from instantaneous (light speed is finite). It's true it's much better than sound to synchronize times but it's still highly limited. The Relativistic Doppler Shift doesn't add anything more to this argument.

We can tell one locomotive emitted a sound earlier than another within our setup because we can use something faster than sound to synchronize. However, what can we use that's faster than light? NOTHING (ever seen). The math I wrote already explained the concepts behind all of this.

[Bored chemist]

OK, Sorry about any typos.
Let's clarify what I consider to be synchronisation.
I get two "clocks"
They are boxes with a button marked "zero" and  a display on the front that is a number. It continuously counts "microseconds since someone pushed the zero button".
They are both "perfect" clocks. If you set them up next to each other the difference between the two displays will always be the same.
If you press the "zero" buttons on both at the same time then the displays will count up in synchrony.
At the start of the experiment I set both clocks to zero by pressing the "zero" buttons on both at the same time.
(I have to say that I thought we could assume that, rather than having to spell it out).

Since the two cameras take a picture a million times a second, each frame is effectively "numbered" by the picture of the clock.
Since each pair (a clock and a camera) are always kept together and they are "perfect" they will remain synchronised to one-another.
Now, just as a sense check, I can do a thought experiment where I repeat the "clock on a plane" trial.
I take one clock and it's camera and fly them round the world on a jet plan.
When they get home, I can compare them and I see that the moving clock no longer agrees with the clock that stayed at home (from my POV).

Do you agree so far that there's a difference between what the  two clocks say?
(for a trip on a plane round the world the difference is too small to see on a microsecond scale, but that's OK, I can set the clocks and cameras to record nanoseconds instead.  I lke thought experiments- they are so much cheaper).
If I repeated the original Hafele–Keating experiment the disparities measured would be much the same as they measured.
-59 ns one way and 273 ns the other.

Do we agree so far?

OK.
Now the clocks are back home I re-synchronise them (locally) by pressing both buttons at the same time.
And I get on my bike with one pair and ride off 3000 km  into the distance (I'm assuming the Earth is flat for this  experiment).
I put the clock + camera down and I ride home again. (Again, I'm glad this is just a thought experiment).

For the sake of having some numbers, I set off, and accelerate to 3 m/sec linearly over the course of 10 seconds then I maintain exactly the same speed until I get near the end, and I then decelerate linearly over 10 sec to zero .

I set of a flash bulb (it's a very short  very bright flash).
I stop the nearby clock and camera and look at the recording.
The flash appears on frame n.
I get back on my (t)rusty bike and set off to the other camera.
I look through that recording and I find that the flash appears on frame n+m

What  value do you predict for m?
For convenience of arithmetic I's saying that C is exactly 300,000,000 m/s so the time it takes to cover the distance of 3 km is 1/100 seconds.
And with a  million frames a second, m should be 10,000
But obviously, that's ignoring any relativistic effects.

How big are those effects?
By how much does m differ from 10,000 ?

(I don't mind if it's a fraction- in principle, we can just speed the clocks up and make it a big enough integer to count reasonable accurately)

And I'm going to keep asking the same question until I get an answer of the form "m=10000.0001 " or whatever or a request to clarify some aspect of the experimental procedure.

[David Cooper]

If I'm on a spaceship moving through space at high speed in a direction which we'll call "north" (for want of a better word), I can go to the middle of the ship and synchronise three clocks there. I can then send two of the clocks to the ends of the ship (one to each) such that we now have three synchronised clock spaced apart with one perhaps 50m to the north of the middle one and another 50m to the south of the middle one. By moving the clocks in this way though, their synchronisation has actually changed - the southern clock's time is running ahead of the other two, and the northern clock is lagging most (but they are back to ticking at the same rate as each other). If I now send a signal from the middle clock to the other two, both those clocks will record the same time of arrival even though the signal takes longer to reach the northern clock.

[Because a clock's ticking rate slows more dramatically the faster it is moved through space, moving the clocks more quickly to the ends of the ship will lead to both of them lagging further behind the middle clock, but the lag will be the same for both. You will therefore get different times from them by changing the speeds of relocation, but despite this, both the northern and southern clock will always agree on the time taken for a signal to reach them from the middle clock.]

[Bored chemist]

If I'm on a spaceship moving through space at high speed in a direction which we'll call "north" (for want of a better word), I can go to the middle of the ship and synchronise three clocks there. I can then send two of the clocks to the ends of the ship (one to each) such that we now have three synchronised clock spaced apart with one perhaps 50m to the north of the middle one and another 50m to the south of the middle one. By moving the clocks in this way though, their synchronisation has actually changed - the southern clock's time is running ahead of the other two, and the northern clock is lagging most (but they are back to ticking at the same rate as each other). If I now send a signal from the middle clock to the other two, both those clocks will record the same time of arrival even though the signal takes longer to reach the northern clock.

[Because a clock's ticking rate slows more dramatically the faster it is moved through space, moving the clocks more quickly to the ends of the ship will lead to both of them lagging further behind the middle clock, but the lag will be the same for both. You will therefore get different times from them by changing the speeds of relocation, but despite this, both the northern and southern clock will always agree on the time taken for a signal to reach them from the middle clock.]

That's a nice story.
Do you  have an opinion on the value of m in the experiment I suggested?

[guest39538]

If I'm on a spaceship moving through space at high speed in a direction which we'll call "north" (for want of a better word), I can go to the middle of the ship and synchronise three clocks there. I can then send two of the clocks to the ends of the ship (one to each) such that we now have three synchronised clock spaced apart with one perhaps 50m to the north of the middle one and another 50m to the south of the middle one. By moving the clocks in this way though, their synchronisation has actually changed - the southern clock's time is running ahead of the other two, and the northern clock is lagging most (but they are back to ticking at the same rate as each other). If I now send a signal from the middle clock to the other two, both those clocks will record the same time of arrival even though the signal takes longer to reach the northern clock.

[Because a clock's ticking rate slows more dramatically the faster it is moved through space, moving the clocks more quickly to the ends of the ship will lead to both of them lagging further behind the middle clock, but the lag will be the same for both. You will therefore get different times from them by changing the speeds of relocation, but despite this, both the northern and southern clock will always agree on the time taken for a signal to reach them from the middle clock.]

no, you are measuring time by counting at different speeds which is wrong.

[David Cooper]

That's a nice story.
Do you  have an opinion on the value of m in the experiment I suggested?

I would have if I could be bothered doing the maths, but it's more important for you to realise that if you use the same method to put an identically synchronised clock+camera the same distance away in the opposite direction, the timing you get from it will always match up with the other measurement you make in the first direction, so you clearly aren't going to be able to measure the one-way speed of light unless you happen to be stationary relative to the fabric of space by luck.

[David Cooper]

no, you are measuring time by counting at different speeds which is wrong.

I'm discussing synchronising three clocks at a single location, then moving two of them away in opposite directions and using these synchronised clocks to time a signal sent at the speed of light from the middle clock to the outer two. The clocks are all counting at the same speed as each other, but they weren't doing so while two of them were being relocated.

It's important to point out though that if you repeat the process many times moving the synchronised clocks from the centre into position more and more slowly, the lag between them will get smaller and smaller each time you do it. This tends towards zero lag (assuming that the central clock is at rest relative to the fabric of space). You can get directly to zero lag by better synchronisation methods, so we're really discussing an extremely poor method of clock synchronisation here.

[guest39538]

no, you are measuring time by counting at different speeds which is wrong.

I'm discussing synchronising three clocks at a single location, then moving two of them away in opposite directions and using these synchronised clocks to time a signal sent at the speed of light from the middle clock to the outer two. The clocks are all counting at the same speed as each other, but they weren't doing so while two of them were being relocated.

The clocks are never out of synchronisation if the clock is being used correctly i.e 1=1

I understand the distance contractions/expansions would cause an issue in timing, sorry I thought you were trying to still say there was a time dilation.

[David Cooper]

The clocks are never out of synchronisation if the clock is being used correctly i.e 1=1

If you move one clock relative to another, you will necessarily push them out of synchronisation - the only issue is how much, with faster speeds of movement pushing them out of sync by more.

I understand the distance contractions/expansions would cause an issue in timing, sorry I thought you were trying to still say there was a time dilation.

Moving a clock slows its ticking. If you want to drag length-contraction in too, then it reduces the slowing of the ticking (by making it the same no matter which way the clock is aligned), but you still have slowed ticking while the clock's moving.