One way speed of light measurement proposal

[ValenceE]

Hello all,


Having read many threads and posts about light, it’s speed c, SR etc, the particular issue of not being able to measure the one way speed of light has always stood out and kept me very interested, so, after pondering about it for a while, I would like to propose an experimental setup in order to measure the one way speed of light.

The attached diagram shows the simple setup which has a source, two mirrors and one clock;

- Distance SC = distance SM2 = 10 meters
- mirrors M1 and M2 are positioned at a 90 degree angle from each other


Am I right in saying that for an observer at rest with the test setup;

1- the light pulse will start the clock (t1 = 0) and reflect from mirror M2 simultaneously.

2- the reflected pulse from M2 will stop the clock at t2 and, via Pythagoras, a calculation of the one way speed of light is given by the equation

SqrRoot ((M1M2)^2 + (M1C)^2) / (t2-t1) = SOL


Using a rotating gantry, this experimental setup could be repeated in any starting direction to gather data for additional validation.

Can this be a valid measurement of the one way speed of light?



Thank you for your comments and best regards,

VE

 

[Samshorn]

Am I right in saying that for an observer at rest with the test setup the light pulse will start the clock (t1 = 0) and reflect from mirror M2 simultaneously?

Only if you assume the speed of light is the same in perpendicular directions - but if you assume that, then the one-way speed equals the two-way speed by definition.

Can this be a valid measurement of the one way speed of light?

Regardless of whether you're trying to measure the one-way or the two-way speed of light (or anything else for that matter), you need to establish your basis of measurement, meaning a suitable measure of space and time intervals. Without establishing such a basis, you obviously can't measure anything. Once you have established such a basis, measuring the speed (one-way or two-way) is straightforward.

 

[Goodison_Lad]

Just to double-check that I’ve read your diagram correctly, your idea is that the light pulse that travels from S is split by the half-silvered mirror, M1. The part that is transmitted through M1 starts the clock C when it hits it (initially set at t1=0). The light reflected from M1 towards M2 then reflects off M2 towards the clock, stopping the clock when it arrives (t2).

If this is the case, your calculation is correct.

Regarding using a rotating gantry, what hypothesis are you interested in confirming with the additional data? Is it anything to do with the fact that many explanations in relativity employ the use of light reflected back to its source position?

(An alternative would be to use no mirrors, and two synchronised clocks, at the start and the end of the light path. Of course, one of the methods suggested for synchronising clocks involves reflecting light between them, so perhaps this isn't what you're after!)

 

[yoron]

You are still using two paths.

A strict 'one way' experiment should only use one, and, it wouldn't really work, would it??

I'm of two minds there as 'gravitational time dilations' should be everywhere in its path? Even though light is a constant it's so locally, but the observer will define its speed 'globally', over the whole path. In the two way experiment where it is reflected it should equal out, but in a one way experiment? And if you're a true believer in symmetries, as me is, joined by their 'complementary' Lorentz FitzGerald contractions.

0uch..

If someone has a good answer to that one I would be grateful.
It was a sweet idea with entanglements :)
==

The more I think of it the more confused I feel :
Da**

 

[ValenceE]

Hello all,


Samshorn, you wrote;

Only if you assume the speed of light is the same in perpendicular directions - but if you assume that, then the one-way speed equals the two-way speed by definition.

Indeed, this is the assertion I’m making here; that the light reaches both C and M1 simultaneously because the speed of light is the same in all directions, and I think this setup makes it possible to validate.

That it is equal to the two way speed by defenition is not an issue here, regardless of any predetermined parameters or assumptions, I think doing this simple experiment could be interesting, just to gather the data, just to see if the T2-T1 difference corresponds to the expected value in order to get c... I believe it will.



Goodison Lad, you wrote;

Regarding using a rotating gantry, what hypothesis are you interested in confirming with the additional data?

Well, no hypothesis in particular but rather compare all the different values of T2-T1 when the setup is rotated... I believe there will be no discrepencies.



Yoron, you wrote;

You are still using two paths.

Well, if this is using two paths, then I’m actually using three... the M2-C path is another one altogether.





Even if this is not a true one way measurement or validation, it still is only using one clock, no need for any synchronisation, and the distances between the instruments can be very accurately measured, or is 10m East not the same as 10m South ?

So, either T2-T1 will yield the current value of c or it won’t, it’s as simple as that... I believe it will, in any and all directions.


regards,
VE

 

[nitsuj]

one way speed is two clocks, spatialy seperated, and where simultanity creeps in.

two way speed is one clock.

I agree with Yoron that this is a there & back measure of c, and there being only one clock...

 

[Goodison_Lad]

Hello again, ValenceE

Well, if this is using two paths, then I’m actually using three... the M2-C path is another one altogether.

If you moved the clock closer to M1 so that it was right next to it (adjusting the orientation of M2 appropriately), then the clock would start at the same time as the light from S was transmitted through/reflected from M1, and stop again on reception of the reflection from M2.

Effectively, then, you would be measuring the round-trip time for the beam that went from M1 to M2 and back again. I think, then, that this set up is essentially the same as your original proposal, and is a round-trip measurement.

 

[Goodison_Lad]

Afterthought: you would, of course, be justified in saying that I'm talking about a different experiment, because your original set-up was an ingenious way of basically clocking the travel time from M2 to C. This to me weakens the point I made in my last post!

So, if it is assumed that the velocity of light is independent of direction (as suggested by Samshorn you would have to), I'm starting to think it is a one-way measurement. Every piece of apparatus is static, so there's only one frame of reference involved.

Fickle, aren't I?

 

[Samshorn]

Samshorn, you wrote "Only if you assume the speed of light is the same in perpendicular directions - but if you assume that, then the one-way speed equals the two-way speed by definition." Indeed, this is the assertion I’m making here; that the light reaches both C and M1 simultaneously because the speed of light is the same in all directions, and I think this setup makes it possible to validate.

No it doesn't. It's well known that, if the speed of light depends on direction, it must be of the form C(theta) = c/[1 + k cos(theta)] for some constant k between 0 and 1, where theta is the angle of the direction of the light ray relative to some fixed reference. For example, let's take the line from your source to your clock to be the theta=0 direction, so the speed of light is c/(1+k) in that direction and c/(1-k) in the opposite direction. It follows that the speed in the perpendicular direction (from M1 to M2 in your setup) is simply c, and the speed in the "hypotenuse" direction is c/[1 + k/sqrt(2)]. Plug all these speeds into the setup, and you find that the elapsed time between the pulses reaching the clock is always exactly L sqrt(2)/c, regardless of the value of k. So you get the same result, regardless of whether c is the same in all directions or not (i.e., regardless of whether or not you have chosen a coordinate system in which c is the same in all directions).

I think doing this simple experiment could be interesting, just to gather the data, just to see if the T2-T1 difference corresponds to the expected value in order to get c... I believe it will.

Indeed it will, but it will do so regardless of whether or not the speed of light is the same in all directions. In fact, it's easy to show that any complicated set of light paths you can imagine (not just a simple triangle) will all give the same results, provided only that the speed of light has the directional dependence given by the ellipsoidal form noted above. You can read all about this in any good book on relativity.

 

[yoron]

You have a really god point in that you're using one clock. Maybe? In SR naturally. I'ts quite nice.
==
The thing is, every time I think of a 'one way experiment' I remember NIST, and get a headache :) But that is GR.

 

[Goodison_Lad]

This experiment has really got me – I’m arguing with myself over it in my sleep!

It occurs to me that, in using your setup to determine the travel time of the light from M2 to the clock, the implicit assumption is that the reflected light reaches M2 simultaneously with the transmitted light reaching the clock i.e. time at M2 is synchronised with time at the clock.

In ordinary two-way measurements, no such assumption is necessary, since the light leaves and returns to the same clock.

Since the measurement of the one-way speed of light requires two different locations, we need to know something about the synchronisation of time at those two points in order to determine the flight time of the light between them.

So the question becomes: how would we know, without assuming it, that time at M2 is synchronised with time at the clock?

 

[ghwellsjr]

There is no question but that this experiment will give a result that is consistent with Einstein's postulate that the one-way speed of light is equal to the two-way speed of light, in other words, c. If it didn't, it would prove Einstein's postulate to be wrong.

But this experiment and all others like it are designed to measure the ether wind, or the rest state of the ether, which is the same thing. If super stable clocks and super accurate rulers and super rigid structures were available at the time of MMX, they could have performed experiments on the one-way speed of light instead of just on the two-way speed of light, but they would have come to the same conclusion, that they couldn't measure any ether wind, and that would have thrown them into exactly the same theoretical quandary that they were thrown into with the null results from their two-way measurements and they would have come to the same conclusions, based on their persistent belief in an absolute ether rest state and absolute time and absolute space, that their super stable clocks and super accurate rulers and super rigid structures were in fact not so super stable, accurate and rigid as they thought, not that time and space were relative.

The point is, if you start with the concept of an absolute ether rest state in which light travels at c, then you will interpret the result of this experiment as not really measuring the one-way speed of light, but rather that your rulers contract along the direction of motion through the ether and that your clock is time dilated and your structure deforms as you rotate it to always yield the same measurement for the one-way speed of light.

Even in Special Relativity, with Einstein's convention for establishing a Frame of Reference, which is identical to the concept of being at rest in the absolute ether rest state, another identical experiment performed with a relative motion in that FoR is interpreted to not be measuring the defined one-way speed of light, but rather to be subjected to the same length contraction, time dilation, and deformation that would explain the null result without Einstein's convention.

So the bottom line is, there is no experiment that can determine the actual one-way speed of light, apart from a previous assumption about what that one-way speed of light is. Or to put it another way, as Einstein said, apart from defining the one-way speed of light, we can't know what it is, that is, we can't measure it.

 

[lugita15]

So the bottom line is, there is no experiment that can determine the actual one-way speed of light, apart from a previous assumption about what that one-way speed of light is. Or to put it another way, as Einstein said, apart from defining the one-way speed of light, we can't know what it is, that is, we can't measure it.

Just to add another perspective to this, while it is certainly true that the one-way speed of light cannot be defined independent of simultaneity convention, there are some simultaneity conventions which unlike Einstein synchronization do not define the one-way speed of light in advance, but rather the one-way speed of light under such a convention has to be determined experimentally (and such an experiment can constitute a test of SR). See this long discussion.

 

[ghwellsjr]

Just to add another perspective to this, while it is certainly true that the one-way speed of light cannot be defined independent of simultaneity convention, there are some simultaneity conventions which unlike Einstein synchronization do not define the one-way speed of light in advance, but rather the one-way speed of light under such a convention has to be determined experimentally (and such an experiment can constitute a test of SR). See this long discussion.

Does ValenceE's "One way speed of light measurement proposal" qualify as an experimental determination of the one-way speed of light since it did not define that speed in advance?

 

[lugita15]

Does ValenceE's "One way speed of light measurement proposal" qualify as an experimental determination of the one-way speed of light since it did not define that speed in advance?

No, of course not. You can't measure the one-way speed of light, tautologically or otherwise, with only one clock. You and I are in complete agreement for the purposes of this thread.

 

[ghwellsjr]

No, of course not. You can't measure the one-way speed of light, tautologically or otherwise, with only one clock. You and I are in complete agreement for the purposes of this thread.

But you can with two clocks?

 

[ValenceE]

Hello ghwellsjr and lugita15,


Thank you both for your comments…


I do agree that in this setup, we don’t know the SOL for paths M1-C and M1-M2, we don’t know if they are equal and, if not, which is ahead or behind, we only know that the paths are of equal lengths and that the M2-C length is sqrt(200), the only measured value being T2-T1.

Let’s say you do this experiment 360 times, rotate the apparatus 1 degree at a time, and record the result of [sqrt(200)/ (T2-T1)], I’d really like to know if it would match the accepted value of c… my belief is that it would on all 360 readings.

Again, there is only one clock, no to-from measurements, nothing assumed…

So I ask you, given what we know about SR;

- what can we make of the results for the ones that do match?, those that don’t ?
- what if they match for all 360 readings ?
- what can we make of the results if none match?


Regards,

VE

PS: As far as ether is concerned, I don’t believe in an absolute ether rest state but I have ideas about a dynamic one, but that is another subject altogether, although related…

 

[Samshorn]

So I ask you, given what we know about SR;
- what can we make of the results for the ones that do match?, those that don’t ?
- what if they match for all 360 readings ?
- what can we make of the results if none match?

You do realize that all your questions were answered in post #9, right?

 

[ValenceE]

hello Samshorn,


actually, no I don't... I wanted to respond to your post #9 earlier but got sidetracked.

You say that;

"it must be of the form C(theta) = c/[1 + k cos(theta)] for some constant k between 0 and
1"

and

"...so the speed of light is c/(1+k) in that direction and c/(1-k) in the opposite direction."

From this it follows that if K=1 then you get c/2 in one way which to me seems awfully slow, and how about c/0 in the other way ? is it undefined or infinite?

Sorry, but I'm just not comfortable with those values...


regards, QE


edit: sorry again, but I must leave now... will be back later today, thx

 

[lugita15]

But you can with two clocks?

We've covered this territory pretty thoroughly, but to repeat you can synchronize two clocks according to slow transport, and then if you do a measurement of the one-way speed of light with respect to these clocks, you cannot predict the result in advance, i.e. without knowing what universe you're in. Whereas if you synchronize them with respect to Einstein synchronization, you immediately know that the one-way speed of light with respect to these clocks will be isotropic, even if you don't know what universe you're in or what the laws of physics are.

 

[Samshorn]

You say that "it must be of the form C(theta) = c/[1 + k cos(theta)] for some constant k between 0 and 1" and "...so the speed of light is c/(1+k) in that direction and c/(1-k) in the opposite direction." From this it follows that if k=1 then you get c/2 in one way which to me seems awfully slow, and how about c/0 in the other way ? is it undefined or infinite?

Infinite, meaning the time for light to travel any distance in that direction is zero. So, if you assume k=1, the speed from M1 to the Clock in your setup is c/2, and the speed from M1 to M2 is c, and the speed from M2 to the Clock is c/(1 + 1/sqrt(2)). With these speeds, the time difference T2 - T1 will be sqrt(2)L/c, which of course is exactly the same as it would be for any other value of k. This is just 4th grade algebra.

For another example, suppose k=-1, so the speed of light is infinite from M1 to the clock, and the speed from M1 to M2 is (again) just c, and the speed from M2 to the clock is c/(1 - 1/sqrt(2)). In this case the time difference T2 - T1 comes out to be (surprise) sqrt(2)L/c.

You get the same value of T2 - T1 for ANY value of k. Do you understand this?

Sorry, but I'm just not comfortable with those values...

Your comfort isn't the issue. The point is that regardless of how asymmetric you assume the speed of light to be (even the ridiculously extreme asymmetry implied by k=1), and regardless of how you orient the device, the value of T2 - T1 always comes out to be sqrt(2)L/c.

Your claim was that if the one-way speed of light was not the same in all directions, your setup would reveal this by giving different values for T2-T1 in different directions, but this simple 4th grade algebra proves that you were mistaken. Do you understand this?

 

[ValenceE]

hello again Samshorn,


you write;

Your claim was that if the one-way speed of light was not the same in all directions, your setup would reveal this by giving different values for T2-T1 in different directions, but this simple 4th grade algebra proves that you were mistaken. Do you understand this?

I’ve never made such a claim, quite the contrary... I’m trying to find a way to show that it IS the same in all directions, but that’s not what I’m after. My thought for this setup was to find out if T2-T1 would yield the current value of c, 299,792.458 km/s


in Post#9;

For example, let's take the line from your source to your clock to be the theta=0 direction, so the speed of light is c/(1+k) in that direction and c/(1-k) in the opposite direction. It follows that the speed in the perpendicular direction (from M1 to M2 in your setup) is simply c, and the speed in the "hypotenuse" direction is c/[1 + k/sqrt(2)]. Plug all these speeds into the setup, and you find that the elapsed time between the pulses reaching the clock is always exactly L sqrt(2)/c, regardless of the value of k.

It is true that I might not grasp this 4th grade algebra, so can you tell me what you mean by “plugging these speeds into your setup” and also what the L term represents; is it the Lorentz factor, is it a length ?... then maybe I could figure out if using L sqrt(2)/c also yields 299,792.458 m/s





regards,

VE

 

[Samshorn]

Samshorn, you write "Your claim was that if the one-way speed of light was not the same in all directions, your setup would reveal this..." I’ve never made such a claim, quite the contrary... I’m trying to find a way to show that it IS the same in all directions...

That makes no sense. You're trying to show that it IS the same in all directions by performing an experiment whose outcome depends (you believed) on whether or not the speed of light is the same in all directions. Otherwise how could the experiment possibly tell you anything about whether it is the same in all directions? And why even go to the trouble of arranging the convoluted path for the light if you accept that the speed of light is the same in all directions?

...but that’s not what I’m after. My thought for this setup was to find out if T2-T1 would yield the current value of c, 299,792.458 km/s

Again, that makes no sense. If you aren't concerned about the possible difference between the one-way speeds of light in different directions (as you now claim), then all you're really "after" is the two-way speed of light, but that contradicts the subject title of the thread you started (and many of your previous - and some of your subsequent - statements).

Can you tell me what you mean by “plugging these speeds into your setup” and also what the L term represents; is it the Lorentz factor, is it a length ?... then maybe I could figure out if using L sqrt(2)/c also yields 299,792.458 m/s

Sure, L is the distance from M1 to M2, which is also the distance from M1 to the clock. (Remember, you defined those distances to be the same.) If the one-way speed of light is c in all directions, then obviously T2 - T1 would equal sqrt(2)L/c.

The question is, if the round-trip speed of light is always c, but the one-way speed of light is different in different directions, would we expect to get a different value of T2-T1? The answer is no, we would get exactly the same value. So the experiment is worthless.

When I say "plug these speeds into your setup" I simply mean compute the value of T2-T1 you would expect for your setup, given the speeds corresponding to any assumed value of k. You just compute the times required for the pulses to reach the clock by the two different paths. You can easily do this knowing the distances traveled and the speeds. When you do this, you find that the elapsed time between the pulses reaching the clock is always exactly L sqrt(2)/c, regardless of the value of k.

 

[ValenceE]

Ok, I found a good explanation of the directional dependence you are talking about, where I see the equations you have posted.

http://mathpages.com/home/kmath229/kmath229.htm

What I read from it is that this applies to a round trip path, and yes, as you mentionned, any fancy path you can devise could always be reduced to a simple closed path like a triangle. But the proposed setup is not such a path...

The light pulse does not return to the source in a round trip, it is split up at the beginning, the clock is not at the source and is not started and stopped from the same path. So I’m not sure that this set of equations apply here...


VE

 

[Samshorn]

Ok, I found a good explanation of the directional dependence you are talking about, where I see the equations you have posted.
http://mathpages.com/home/kmath229/kmath229.htm
What I read from it is that this applies to a round trip path, and yes, as you mentionned, any fancy path you can devise could always be reduced to a simple closed path like a triangle. But the proposed setup is not such a path... The light pulse does not return to the source in a round trip, it is split up at the beginning, the clock is not at the source and is not started and stopped from the same path. So I’m not sure that this set of equations apply here...

You mis-read. The directional dependence applies to the one-way speed of light. That is the form that the one-way speed of light must have in order for the round-trip speed of light to always be c. So, if you accept that the round-trip speed of light is always c (which it empirically is), then the one-way speed of light must be of that form for SOME value of k.

Now, for any value of k you like, plug in the corresponding speeds for your setup, and compute the value of T2-T1 you would expect. What do you get? As you know, the answer is sqrt(2)L/c, regardless of the value of k. So your setup doesn't provide any information beyond what a round-trip test would provide.

 

[ghwellsjr]

We've covered this territory pretty thoroughly, but to repeat you can synchronize two clocks according to slow transport, and then if you do a measurement of the one-way speed of light with respect to these clocks, you cannot predict the result in advance, i.e. without knowing what universe you're in. Whereas if you synchronize them with respect to Einstein synchronization, you immediately know that the one-way speed of light with respect to these clocks will be isotropic, even if you don't know what universe you're in or what the laws of physics are.

Can those clocks be light clocks?

 

[lugita15]

Can those clocks be light clocks?

Yes, just like Einstein synchronization, slow transport synchronization works for any clocks including light clocks. And yes, just like Einstein synchronization, it is frame dependent. That is not where their difference lies.

 

[ghwellsjr]

Yes, just like Einstein synchronization, slow transport synchronization works for any clocks including light clocks. And yes, just like Einstein synchronization, it is frame dependent. That is not where their difference lies.

The difference between what and what?

 

[lugita15]

The difference between what and what?

The difference in the properties of the two synchronization conventions.

 

[ghwellsjr]

The difference in the properties of the two synchronization conventions.

So it doesn't matter the orientations of the light clocks or their sizes?

 

[lugita15]

So it doesn't matter the orientations of the light clocks or their sizes?

It is an experimental fact that in our universe, but not all possible universes, Einstein synchronization yields exactly the same result as slow transport synchronization, so to find out whether two clocks are synchronized according to one it's sufficient to answer the same question for the other. I don't know what else to tell you.

For an example of a universe in which the two synchronization schemes would yield markedly different results, consider Newton's emitter theory of light, in which the speed of light was like the speed of a bullet, dependent on the motion of the source.

 

[DrGreg]

If I can step in as interpreter between ghwellsjr and lugita15, I would guess the point that ghwellsjr is trying to make is that if you "don't know what universe you're in or what the laws of physics are" then you can't know that a light clock will accurately measure time. In principle a similar objection could be raised for other types of clocks. Whenever we talk of "clocks" in relativity we mean ideal time-measuring devices which work accurately under whatever conditions we put them in. If you were performing a real-world experiment you would first of all have to be convinced that whatever sort of clock you were using would indeed accurately measure time (to the precision your experiment needs) under the conditions that you were going to use it.

 

[lugita15]

DrGreg, in the thread I linked to earlier, you made an excellent post on the subject that ghwellsjr and I are disagreeing about:

Mansouri & Sexl^[1] consider a "test theory" of relativity in which the transformation between two frames is postulated to be\begin{align}<br /> t &amp;= a(v)\,T + \epsilon(v)\,x\\<br /> x &amp;= b(v)\,(X - vT)<br /> \end{align}where a, b and ε are unknown functions to be determined by experiment. (Note: the first equation intentionally contains x, not X.) Special relativity is a special case of this test theory for a particular choice of these three functions. Experiments to test the validity of relativity can be performed from which the values of a(v), b(v) and ε(v) can be estimated. If the experimental values match the values predicted by SR, this is a confirmation of SR.

Mansouri and Sexl point out that the function ε(v) depends on the clock sync convention chosen, whereas a(v) and b(v) are both independent of sync convention. Under these assumptions, they go on to prove a result (pp.506–508) that slow clock transport and Einstein synchronisation are equivalent if and only if a(v) takes the value predicted by SR, viz<br /> a(v) = \sqrt{1 - v^2/c^2}<br />To avoid any misunderstanding, the term "slow clock transport" is defined to mean in the limit as the speed of clock transport tends to zero (as others have pointed out).

Thus, if you sync clocks by slow clock transport and then measure the one-way speed of light, if you get an answer of c regardless of direction, you have experimentally confirmed that a(v) takes the value predicted by SR.Reference
[1] Mansouri, R and Sexl, R U (1977), "A Test Theory of Special Relativity: I. Simultaneity and Clock Synchronization", General Relativity and Gravitation 8 (7), pp.497–513, Bibcode: 1977GReGr...8..497M, DOI: 10.1007/BF00762634Further reading
Test theories of special relativity, Wikipedia

I think this summarizes my point of view pretty well.

 

[ValenceE]

Hello Samshorn,


you wrote;

When I say "plug these speeds into your setup" I simply mean compute the value of T2-T1 you would expect for your setup, given the speeds corresponding to any assumed value of k. You just compute the times required for the pulses to reach the clock by the two different paths. You can easily do this knowing the distances traveled and the speeds. When you do this, you find that the elapsed time between the pulses reaching the clock is always exactly L sqrt(2)/c, regardless of the value of k.

Now, for any value of k you like, plug in the corresponding speeds for your setup, and compute the value of T2-T1 you would expect. What do you get? As you know, the answer is sqrt(2)L/c, regardless of the value of k. So your setup doesn't provide any information beyond what a round-trip test would provide.

Ok, I got it, so here we go…

the directional dependency equations for light speed being c(theta) = c / 1 + k cos(theta)

for simplicity, let’s make;

paths M1-C and M1-M2 = 1 m
path M2-C = sqrt(2) m = 1.4142136 m
k= 1
c= 1m/s

we then get the following speeds;

in the M1-C direction = 1 / 1 + cos(0) = 0.5 m/s
in the M1-M2 direction = 1 / 1 + cos(270) = 1 m/s
in the M2-C direction = 1 / 1+ cos(45) = 0.58578643… m/s

so, plugging those in the setup we see that;

- from M1, it takes 2 seconds for the light pulse to reach C
- from M1, it takes 1 second for the light pulse to reach M2
- from M2, it takes 2.4142135… seconds for the light pulse to reach C

from this we see that;

- while the light pulse reaches and reflects from M2, we’re only halfway towards C
- there is still 0.5 m to go at 0.5 m/s to reach C and start the clock at T1
- in the 1 second this will take, the reflected pulse from M2 will have traveled 0.585784643… m towards C, leaving 0.82842713… m to go at 0.58578643… m/s until it reaches C and stops the clock at T2

- thus T2 - T1 = 0.82842713… / 0.58578643… = 1.4142135623… seconds

which is the same as if the speed of light would be equal to 1 m/s in all directions


regards,
VE

 

[ghwellsjr]

So it doesn't matter the orientations of the light clocks or their sizes?

It is an experimental fact that in our universe, but not all possible universes, Einstein synchronization yields exactly the same result as slow transport synchronization, so to find out whether two clocks are synchronized according to one it's sufficient to answer the same question for the other. I don't know what else to tell you.

Since you didn't deny that orientation or size matters when it comes to the slow transport of light clocks, I invite you to consider the following scenario:

We make a light clock on a large rigid structure with the two ends labeled A and B with mirrors facing each other. This light clock is a little different in that it has two light pulses bouncing back and forth between the mirrors, a green light pulse and a red light pulse. Every time a green light pulse reflects off the mirror at A, it increments a counter to the next odd number. Every time a red light pulse reflects off the mirror at A, it increments the counter to the next even number. The time interval between green-red and red-green is the same so that the counter increments in a steady manner.

Now what we want to do is have another counter placed at B so that it counts synchronously with the counter at A. We decide to use the slow clock transport method. Someone comes along and says, well we don't have to actually transport a clock from A to B because the light clock itself stretches all the way from A to B so we can just use the reflections of the green and red flashes but we will change what they do so that a red flash increments the new counter at B to the next odd number and a green flash increments the counter to the next even number. All he has to do at location B is look at the counter at A at the time the local counter at B increments and set it so that it reads one count more and this will synchronize the two counters so that they always read the same time.

Does this make sense to you?

 

[lugita15]

ghwellsjr, that sounds like Einstein synchronization, not slow clock transport.

 

[ghwellsjr]

ghwellsjr, that sounds like Einstein synchronization, not slow clock transport.

Well then, let's tell the guy he's all wet and actually build a second identical light clock next to the first one making sure the two A counters increment in lock step and then slowly move the second one so that it's A counter and mirror are adjacent to the B counter and mirror of the first one. Will the B counter of the first setup be in lock step with the A counter of the second one?

 

[lugita15]

Well then, let's tell the guy he's all wet and actually build a second identical light clock next to the first one making sure the two A counters increment in lock step and then slowly move the second one so that it's A counter and mirror are adjacent to the B counter and mirror of the first one. Will the B counter of the first setup be in lock step with the A counter of the second one?

Yes in our universe, but not in all possible universes.

 

[ghwellsjr]

Yes in our universe, but not in all possible universes.

Then we shouldn't have told the guy he was all wet?

 

[lugita15]

Then we shouldn't have told the guy he was all wet?

Well, we should have told the guy he was all wet when he was originally performing Einstein synchronization but claiming it was slow transport synchronization. But then, now that he slowly moved one of the light clocks past the other, he has managed to genuinely slowly transport the A counter of one clock to the B counter of the other clock. Now if he happens to live in a universe like ours in which the two synchronization schemes give the same result, then he will find that the two counters he has now aligned are perfectly in sync. But if he lives in another universe in which special relativity is not true, like the universe that Newton believed that he lived in, then he will find that the A counter of one and the B counter of the other are not in sync at all.

Let me make my point another way. Suppose we synchronized two clocks using a method analogous to Einstein synchronization, except using sound other than light. Then we will find that "sound synchronization" yields very different results than slow transport synchronization, unless you happen to be in a frame that is at rest with respect to the air. And Newton would have agreed with this fact. But Newton would be flabbergasted to learn that slow transport synchronization and Einstein synchronization yield the same results.

 

[ghwellsjr]

Even in Newton's universe where Special Relativity is not true, and by that, I assume you mean there is no time dilation or length contraction, the two synchronization schemes will give the same result. In fact, in Newton's universe, they are both the same no matter what speed the clock is moved at. This is true even if you are not at rest with respect to the medium, whether it be air or ether. Even in our universe, the two synchronization schemes give the same result even if you are moving in the assigned reference frame or even if you assume LET to be the way the universe works and you are moving with respect to the ether.

So Newton would not be flabbergasted that slow transport synchronization and Einstein synchronization yield the same results but he might be flabbergasted to learn that fast transport synchronization yields a different result than slow transport synchronization (and Einstein synchronization) in our real universe.

In fact, slow transport and Einstein synchronization are exactly the same. Neither one validates the other and neither one is an experimental verification of the other. There is no difference between them except the common misconception that they are different.

One thing that needs to be made clear is that even if the time it takes for light to propagate from A to B is different than the time it takes for light to propagate from B to A, then Einstein will call these unequal times equal and the slow clock transport will also end up with the clock being "out of sync" with the stationary clock by exactly the same amount, but we make the claim that they are in sync and we simply define the time to be equal and the clocks to be in sync.

I urge you to analyze what actually happens with slow transport of a light clock and see that what I say is true before you react negatively.

 

[Alfie]

Just a thought about slow transport synchronisation.

What if, instead of moving one of the clocks away very slowly, each clock underwent acceleration programs, equal in all respects other than direction, so that they could be moved apart at arbitrarily high velocities, before being brought to rest with respect to each other? Would this qualify as a valid way of synchronising clocks?

 

[nitsuj]

sure would but the challanges remain the same.

 

[ghwellsjr]

Just a thought about slow transport synchronisation.

What if, instead of moving one of the clocks away very slowly, each clock underwent acceleration programs, equal in all respects other than direction, so that they could be moved apart at arbitrarily high velocities, before being brought to rest with respect to each other? Would this qualify as a valid way of synchronising clocks?

No, they would remain synchronized with each other but not with a clock that remained at their starting location. The whole idea of Einsteinian synchronization is to build a coordinate system where their is a synchronized clock at every location and the three fixed coordinates for that location plus the changing time on its coordinate clock becomes the four spacetime coordinates for that location.

 

[lugita15]

Even in Newton's universe where Special Relativity is not true, and by that, I assume you mean there is no time dilation or length contraction, the two synchronization schemes will give the same result.

No, they wouldn't. For instance, in a non-Lorentz ether theory like the one Maxwell believed in, they would give dramatically different results if you're not in the rest frame of the ether, just like in our universe slow transport synchronization and sound synchronization (Einstein synchronization with sound rather than light) give dramatically different results if you're not in the rest frame of air.

Even in our universe, the two synchronization schemes give the same result even if you are moving in the assigned reference frame or even if you assume LET to be the way the universe works and you are moving with respect to the ether.

Yes, Lorentz ether theory also says that the two synchronization methods give the same results, because LET and SR make the same experimental predictions. But other aether theories of light, and emitter theories of light, will not make the two methods equivalent.

So Newton would not be flabbergasted that slow transport synchronization and Einstein synchronization yield the same results but he might be flabbergasted to learn that fast transport synchronization yields a different result than slow transport synchronization (and Einstein synchronization) in our real universe.

I definitely agree that Newton would have expected slow transport synchronization and "fast transport" synchronization to produce the same, "correct" result. But he would have believed that both Einstein synchronization and sound synchronization produce "incorrect" results.

In fact, slow transport and Einstein synchronization are exactly the same. Neither one validates the other and neither one is an experimental verification of the other. There is no difference between them except the common misconception that they are different.

Do you disagree with the proof given in the Mansouri and Sexl paper DrGreg referred to in the old thread? (For your convenience, I've attached the relevant pages.) They say "We thus arrive at the important result that Einstein proceudre in general differs from synchronization by clock transport. The equality of both procedures is neither trivial nor logically cogent." And then they say "Thus clock synchronization by clock transport and by the Einstein procedure agree if and only if the time dilatation factor is given exactly by the special relativistic value." In other words, the procedures are not logically equivalent, but special relativity makes them equivalent.

One thing that needs to be made clear is that even if the time it takes for light to propagate from A to B is different than the time it takes for light to propagate from B to A, then Einstein will call these unequal times equal and the slow clock transport will also end up with the clock being "out of sync" with the stationary clock by exactly the same amount, but we make the claim that they are in sync and we simply define the time to be equal and the clocks to be in sync.

I'm not sure what you're saying here.

I urge you to analyze what actually happens with slow transport of a light clock and see that what I say is true before you react negatively.

I don't think what you say is true, and a very easy way to demonstrate that is, replace the red and green light beams with high-pitch and low-pitch sound waves. Then I think you should agree that if initially the A counters of the two "sound clocks" were in sync and the B counters were in sync, if we slowly transport one of the sound clocks so that the A counter of one clock is in sync with the B counter of the other, we'll find that they're not in sync as long as you're in a frame different than the rest frame of air.

 

[ghwellsjr]

I definitely agree that Newton would have expected slow transport synchronization and "fast transport" synchronization to produce the same, "correct" result.

I haven't got time to respond to all your points but let me just ask you right now, what is the difference between "fast transport" where we take the limit of the clock as it approaches the speed of light (just like we take the limit in slow transport as the speed approaches zero) and Einstein's synchronization?

EDIT: Just to make it clear: I'm talking about in Newton's world where Special Relativity is not true, that is, where there is no time dilation and no length contraction.

 

[lugita15]

I haven't got time to respond to all your points but let me just ask you right now, what is the difference between "fast transport" where we take the limit of the clock as it approaches the speed of light (just like we take the limit in slow transport as the speed approaches zero) and Einstein's synchronization?

EDIT: Just to make it clear: I'm talking about in Newton's world where Special Relativity is not true, that is, where there is no time dilation and no length contraction.

Well, let me ask you a question back. What's the difference between sound synchronization (Einstein synchronization with sound waves) and fast transport if we take the the clock's speed approaches the speed of sound? I'm asking this in either Newton's world, or our world where the speed of sound is small relative to the speed of light. It's kind of ambiguous, because the speed of sound is different in opposite directions, just like in Maxwell's aether theory the speed of light is different in opposite directions.

Anyway, I look forward to your response to the rest of my post.

 

[nitsuj]

No, they would remain synchronized with each other but not with a clock that remained at their starting location.

The whole idea of Einsteinian synchronization is to build a coordinate system where their is a synchronized clock at every location and the three fixed coordinates for that location plus the changing time on its coordinate clock becomes the four spacetime coordinates for that location.

If the topic of the thread is still the context (measure one way speed of light), Yes this method of "symmetricaly" seperating the clocks would work equally to slow clock transport. It acheives the same goal of keeping the two clocks syncronized.

But maybe I am jumping in the middle and the detail was important

 

[ValenceE]

Indeed it has veered off from my original question about a one way speed of light test setup validation using only one clock… and unfortunately, although my belief is that the one way speed of light is the same in all directions, the proposed experiment can’t distinguish between same or different speeds.

However, even if it's not the main subject of these other exchanges, there is mention of aether and that is interesting to me, but there are plenty other threads about it, so I don't mind reading on and don't mind it being locked as it doesn’t add any more to my enquiry for the moment…


Regards,

VE

 

[ghwellsjr]

No, they wouldn't. For instance, in a non-Lorentz ether theory like the one Maxwell believed in, they would give dramatically different results if you're not in the rest frame of the ether, just like in our universe slow transport synchronization and sound synchronization (Einstein synchronization with sound rather than light) give dramatically different results if you're not in the rest frame of air.

Keep in mind, I have been talking about comparing the synchronization via Einstein and slow transport for a particular instance of a large light clock and I maintain that as long as there is no time dilation and no length contraction and the speed of the light waves (or water waves or air waves) is a constant with respect to some fixed frame, then those two synchronization schemes are identical, even if the light clock is moving with respect to the fixed frame, not just in their outcome, but they physically are carrying out the same process with either synchronization scheme. I'm not saying that this could be used to synchronize all clocks in all orientations because the tick rate of one of these clocks could be different than another one at right angles to it, except in a universe like ours where SR is in effect.

Yes, Lorentz ether theory also says that the two synchronization methods give the same results, because LET and SR make the same experimental predictions. But other aether theories of light, and emitter theories of light, will not make the two methods equivalent.I definitely agree that Newton would have expected slow transport synchronization and "fast transport" synchronization to produce the same, "correct" result.

With the stipulations that I presented above, even emitter theories, as long as the light has constant speeds in each direction, will be the same synchronization process for both schemes.

But he would have believed that both Einstein synchronization and sound synchronization produce "incorrect" results.

Whether you call the results "correct" or "incorrect" they will be identical for any given instance of the type of clock I described.

Do you disagree with the proof given in the Mansouri and Sexl paper DrGreg referred to in the old thread? (For your convenience, I've attached the relevant pages.) They say "We thus arrive at the important result that Einstein proceudre in general differs from synchronization by clock transport. The equality of both procedures is neither trivial nor logically cogent." And then they say "Thus clock synchronization by clock transport and by the Einstein procedure agree if and only if the time dilatation factor is given exactly by the special relativistic value." In other words, the procedures are not logically equivalent, but special relativity makes them equivalent.

Yes, but like I said, they are covering a broader ground than I am. I'm simply pointing out that for a given instance of a light clock in a particular orientation, Einstein's synchronization is the same process as slow transport.

I'm not sure what you're saying here. I don't think what you say is true, and a very easy way to demonstrate that is, replace the red and green light beams with high-pitch and low-pitch sound waves. Then I think you should agree that if initially the A counters of the two "sound clocks" were in sync and the B counters were in sync, if we slowly transport one of the sound clocks so that the A counter of one clock is [STRIKE]in sync with[/STRIKE] adjacent to the B counter of the other, we'll find that they're not in sync as long as you're in a frame different than the rest frame of air.

[I edited your response to fit what I think you meant.]

It depends on what you mean by "in sync". If you mean that someone looking at the counters via light sees if they count synchronously, then of course the slowly transported A counter will not be in sync with the stationary B counter. But if we could synchronize clocks with some higher speed communication than light provides, we'd have a whole different world.

But if by "in sync" you mean that the stationary B counter increments to an odd count when the low frequency signal bounces off it (while the stationary A counter increments to an odd count when the high frequency signal bounces off it), then, yes, the slow transport will result in its A counter in sync with the stationary B counter when they get together.

Since you disagree, I would suggest that you work through how the sound waves propagate at two different speeds in each direction and yet the slow transport of the sound clock will produce the same synchronization as the method I described which you called Einsteinian synchronization (and I agree, it is).