B Measuring the One Way Speed of Light

[Sagittarius A-Star]

By changing the electromagnetic interaction the produces sound waves in a material. Suppose that you have a material whose speed of sound is $0.8 c$ under the isotropic convention. Then if you use the convention that $c(0)=\infty$ and $c(\pi)=0.5 c$ then clearly the speed of sound in the $\theta=\pi$ direction can no longer be $0.8 c$.

Yes. In this case, the speed of sound in the $\theta=\pi$ (negative x-)direction is $v_{-}' = \frac{4}{9}c$, and that in the opposite direction $v_{+}' = 4c$.

Proof: I use the following transformation, with $k = 1$:

$x' = x \ \ \ \ \ \ \ \ \ \ y' = y \ \ \ \ \ \ \ \ \ \ z' = z \ \ \ \ \ \ \ \ \ \ t' = t+ \frac{kx}{c}$
...
$\frac{c'}{c} = \frac{1}{1-k \cos(\theta)}$

Source:
https://www.mathpages.com/home/kmath229/kmath229.htm

From the transformation for $t'$ follows: $\frac{-x'}{0.8 c} = \frac{-x}{v_{-}'} + \frac{x}{c}$
and with $x' = x$ follows:
$$v_{-}' = \frac{4}{9} c$$

From the transformation for $t'$ follows: $\frac{x'}{0.8 c} = \frac{x}{v_{+}'} + \frac{x}{c}$
and with $x' = x$ follows:
$$v_{+}' = 4 c$$

 

[Flatland]

Can't you theoretically measure the one way speed of light using a black hole? You shoot a beam of light at the black hole at a geodesic path that curves the light beam back to your detector.

Also if the speed of light is directional wouldn't we see this in the CMB where the universe will appear younger in one direction as opposed to another?

 

[Nugatory]

Can't you theoretically measure the one way speed of light using a black hole? You shoot a beam of light at the black hole at a geodesic path that curves the light beam back to your detector.

That’s a two-way measurement that uses a massive gravitating body (actually, you will need more than one to get the path you want) instead of a mirror to send the light signal back to the source.

 

[Dale]

Can't you theoretically measure the one way speed of light

No. (None of the details are relevant)

 

[cmb]

I was thinking about this conundrum recently and didn't realize there was a recent thread going on about it.

Here are two thoughts/questions;
1) Would a viable experiment exist which deliberately introduces speed variation and then uses that as the reference calibration? Thus; Take a multi-km of optical cable and, when in a spiral in the lab, measure its propagation delay. Then lay it out straight and measure the differential time of signals one end to other compared with a beam of light. If the differential propagation delay is, say, 0.3c then 0.3 times instantaneous speed would be zero delay. Do that in both directions and if the delay is the same, would this not show light is isotropic in that axis, in those two tested directions?

2) Best to do this experiment vertically, because the most likely direction that light speed would be anisotropic seems to me to be when it is crossing a gravitational field gradient. If the speed of light varies according to gradient of the gravitational field it is passing through then this seems logical either as a consequence of relativity, or in an philosophical way relativity would be an effect of such anisotropy?

 

[Vanadium 50]

Measuring the "one way speed of light" is equivalent to solving one equation in two unknowns. No amount of Rube Goldebergery will change that.

 

[Dale]

Would a viable experiment exist

No. Again the details are irrelevant.

 

[Nugatory]

and if the delay is the same, would this not show light is isotropic in that axis, in those two tested directions?

It would show that the hypothetical anistropy affects light and the signal speed in the cable similarly.

 

[Vanadium 50]

the details are irrelevant.

Right. Just like in a perpetual motion machine. You can tart it up all you like, but energy is still conserved.
Let me repost what I wrote two weeks ago the last time this came up:

...people who say they have found a way to measure the one-way speed of light should be treated the same as people who claim they can solve one equation in two unknowns.

Specifically, the one-way speed of anything (not just light) going from A to B is:

v = \frac{x_B-x_A}{(t_B - \delta t)-t_A}

where δt is the difference in clock syncronization from A to B. In Newtonian physics, δt = 0. In Relativity, you tell me what one-way speed you want, and I'll tell you the clock syncronization convention to use. It's one equation in two - count 'em, two - unknowns.

 

[cmb]

Measuring the "one way speed of light" is equivalent to solving one equation in two unknowns. No amount of Rube Goldebergery will change that.

Adding in a calibrated delay is providing the second unknown.

Using calibrated delays to determine transmission speeds is quite a normal thing, I think?

 

[cmb]

It would show that the hypothetical anistropy affects light and the signal speed in the cable similarly.

Only if the delay was a constant addition of delay rather than a factor addition.

 

[cmb]

No. Again the details are irrelevant.

Could we possibly spend a moment to discuss, rather than write it off without thought?

If it was found that there was a 50% delay factor in the propagation speed down an optic fibre, and one sets up an experiment with a photon emitter timed to traverse a straight 10,000 feet, then at 1ft/ns (approx c) it'd take 10us for the light to get there and 15us for the delayed path, being uncoiled and laid out straight along the 10,000 test path.

The speed of light is then the distance divided by difference of the two arrival times and times the delay factor. Without any delays in that one direction, this would then be (10,000'/5us) * 50% = 1'/ns, as expected.

If there was a whole-sale slowing down of 25% such that it'd take 12.5us for light and 18.75us for the delay, thus (10,000'/6.75us) * 50% = 0.74'/ns.

The delay factor is the 2nd unknown.

The delay factor being calibrated in a lab with a coil in which the direction is an average of both.

The experiment does not, and probably does not need, to be trying to prove/disprove the whole thing in one go, but if one does this experiment in each direction and one gets a different 'absolute' delay it would disprove that light behaves the same in both directions.

 

[Dale]

Could we possibly spend a moment to discuss, rather than write it off without thought?

Can you spend a moment to read the previous material that has already answered your question over and over and over. We have already spent over 100 posts in this thread alone plus many other posts in many other threads. The details of your scenario are absolutely 100% irrelevant. It is impossible as a matter of definition.

 

[cmb]

We have already spent over 100 posts in this thread alone plus many other posts in many other threads. Can you spend a moment to read the previous material that has already answered your question over and over and over. The details of your scenario are absolutely 100% irrelevant.

I did, and I noted nothing that excludes the matters in my last post.

If you might simply identify the error of physics/maths, then may I please propose that this would be preferable rather than jumping to an instant bias that there is no possible answer.

With a calibrated delay, which is calibrated in a manner which is not biased towards either one way or two way speed measurements because it takes an average, I submit one can then use thsi calibrated relative delay to measure the speed of light.

 

[Dale]

I noted nothing that excludes the matters in my last post.

Then read again.

This is impossible as a matter of definition. The definition of the one way speed of light is the distance that light travels (in a single straight line path) divided by the time that it takes for the light to travel that distance. Since that time is measured by clocks at two different locations then the time depends on your synchronization convention. Let me repeat that for emphasis:

The one way speed of light depends on your synchronization convention by definition.

Your choice of synchronization convention determines the one way speed of light. In other words, BY DEFINITION, the one way speed of light requires that you make an assumption and that assumption determines the one way speed of light. You cannot avoid making that assumption because it is part of the definition, and once you make that assumption you have determined the one way speed of light.

Your experimental details are irrelevant. This is not a matter of clever experimental design.

 

[cmb]

Then read again.

This is impossible as a matter of definition. The definition of the one way speed of light is the distance that light travels (in a single straight line path) divided by the time that it takes for the light to travel that distance.

I have used a different definition of speed, avoiding the conundrum that you say I have not read. I propose that I have just avoided it, but if I am wrong and not avoided it at all then surely it is so obviously a fallacy? If so, please just tell me where the logic breaks down.

Say;
D = known test distance
V1 = unknown speed of 1st measurand
V2 = unknown speed of 2nd measurand
average{V1} = k * average{V2}, by local calibration of each measurand following an identical loop circuit whose test radius << D {and using the same clock}

I propose there no longer a need for any time synchronisation between two distant points.

Instead, let dt = observed time interval between 1st measurand and 2nd measurand passing the end of test distance D, having started off together, and only time-measured at end of D, not requiring any prior synchronisation reference to any other point {and using the same clock}.

then V1 = (D/dt) * (k-1)

'IF' I can perform that calibration, then thereafter this definition of velocity is only using a time measurement made at one singular point and not synchronised to any other clock or any other time reference anywhere else.

There may be some fallacy embedded in the calibration procedure, but I am not seeing it as it does not favour any particular velocity direction prior to the actual one way measurement.

I can do the test run along D, one way, with nothing more than one singular time piece. I do not need two timepieces, thus there is no synchronisation error.

 

[Sagittarius A-Star]

The delay factor is the 2nd unknown.

The delay factor being calibrated in a lab with a coil in which the direction is an average of both.

That calibration does not work. You find a similar scenario in my above posting #101. You need only to replace your speed of light in an optical cable ($0.5 c$) by the hypothetical speed of sound in a material ($0.8 c$) in the isotropic case. In my scanario you can easily calculate, that for example on a distance of 300,000 km in x-direction, the difference of arrival time between light and sound will be $0.25 s$, in both, the isotropic and the anisotropic example.

 

[iVenky]

I have a question. If the velocity of light from a point source of light depends on the direction, doesn't the intensity of light (which is a one-way measurement) change with direction? Clearly, we don't observe this, which means velocity is constant as well regardless of the direction.

 

[Ibix]

'IF' I can perform that calibration, then thereafter this definition of velocity is only using a time measurement made at one singular point and not referenced to any other clock or any other time reference anywhere else.

I find it difficult to work out what you think you are doing. However, I think you are firing two light pulses simultaneously, one through free space and one through an optical fibre. This is essentially synchronising clocks except that the "receiving" clock has an offest of $D/c$ compared to Einstein synchronisation. It is therefore subject to the same synchronisation problems as any other one way speed measure.

I have a question. If the velocity of light from a point source of light depends on the direction, doesn't the intensity of light (which is a one-way measurement) change with direction?

No. Why would it?

 

[Dale]

I have used a different definition of speed

Then it is not what anyone else is talking about when we say “one way speed of light”

You don't get to redefine it. Here on PF we use the standard definition as accepted by the professional scientific community.

There may be some fallacy embedded in the calibration procedure, but I am not seeing it as it does not favour any particular velocity direction prior to the actual one way measurement.

Your calibration is a two way measurement. You then simply assume that the two way speed is the same as the one way speed, which is the Einstein synchronization convention.

 

[iVenky]

I find it difficult to work out what you think you are doing. However, I think you are firing two light pulses simultaneously, one through free space and one through an optical fibre. This is essentially synchronising clocks except that the "receiving" clock has an offest of $D/c$ compared to Einstein synchronisation. It is therefore subject to the same synchronisation problems as any other one way speed measure.

No. Why would it?

If I consider light as a stream of photons, then the amount of photons striking a light sensitive element should change based on velocity of the photons. The total energy for a given time should be different based on velocity, right? Is there anything wrong with my understanding?

 

[cmb]

That calibration does not work. You find a similar scenario in my above posting #101. You need only to replace your speed of light in an optical cable ($0.5 c$) by the hypothetical speed of sound in a material ($0.8 c$). In my scanario you can easily calculate, that for example on a distance of 300,000 km in x-direction, the difference of arrival time between light and sound will be $0.25 s$, in both, the isotropic and the anisotropic scenario.

I am not following your point then. I thought you were showing there that there would be no difference in a path with a changing angle?

In a 'calibration' phase, I am proposing that taking the average path (two, or multiple, angles back to the same point) gives an average propagation delay, which I thought you were confirming. It would not be possible to pull apart, say, two variables if that delay was different in two directions.

If there is more to what you put, I am sorry I did not understand it.

I am proposing once we have a known delay, thus measured (and unaffected by anisotropy), it might then be used for a one-way measurement.

 

[Sagittarius A-Star]

In a 'calibration' phase, I am proposing that taking the average path (two, or multiple, angles back to the same point) gives an average propagation delay, which I thought you were confirming.

Yes.

It would not be possible to pull apart, say, two variables if that delay was different in two directions.

But as you can see in my posting #101, the "delay" (refraction-index $n$ of the optical cable) is different in two directions in the anisotropic case.

 

[Nugatory]

Only if the delay was a constant addition of delay rather than a factor addition.

I'm not sure what you mean by that? If the anistropy delays both signals by the same constant factor per unit length, it will not appear in your setup?

But that's all beside the point when the reply starts with the words "Only if..." because that is excluding one particular form of anistropy by assumption. And the point of this discussion is that anything that purports to be a measurement of the one-way speed of light requires some unverifiable assumption

It is true that anisotropy that would be consistent with all experiments and observations would be somewhat perverse and implausible. That's why we choose to assume that it doesn't exist - but resonable though it is, that's still an assumption.

 

[Dale]

If I consider light as a stream of photons, then the amount of photons striking a light sensitive element should change based on velocity of the photons.

No, why would it?

 

[Nugatory]

If I consider light as a stream of photons...

Light is not and cannot be accurately modeled as a stream of photons. Photons only appear in quantum mechanical treatments of electromagnetism, and there the picture bears no resemblance to what you're imagining here.

But when we're talking about the behavior of light in relativity, there's no need to invove quantum mechanics. Classical electrodynamics works just fine.

 

[weirdoguy]

Besides, light is not a stream of photons the way river is a stream of water molecules. If you are not talking about quantum "situations" you should stick to classical picture of electromagnetic waves.

 

[Ibix]

If I consider light as a stream of photons,

Risky, unless you are genuinely planning on doing a proper anslysis using quantum field theory.

then the amount of photons striking a light sensitive element should change based on velocity of the photons.

If you have a state with a well-defined number of photons then the number of photons received per unit time must be the same as the number of photons emitted per unit time. Otherwise photons are disappearing or appearing somewhere.

Notice that the velocity of light does not appear anywhere in this.

 

[iVenky]

I see, thanks for the answers. In that case, if I consider it as an electromagnetic wave and integrate the Poynting vector over a period of time, shouldn't the energy (over a period of time) be different depending on the velocity of the EM wave?

 

[iVenky]

Besides, light is not a stream of photons the way river is a stream of water molecules. If you are not talking about quantum "situations" you should stick to classical picture of electromagnetic waves.

Yes, the flow of river was the analogy that led me to this question.

 

[Ibix]

I see, thanks for the answers. In that case, if I consider it as an electromagnetic wave and integrate the Poynting vector over a period of time, shouldn't the energy (over a period of time) be different depending on the velocity of the EM wave?

No. The same argument I made earlier applies - the energy that leaves the source in some time interval must arrive at the source in the same interval. Otherwise the energy must be escaping or being added. This is independent of the velocity of light.

 

[Nugatory]

shouldn't the energy (over a period of time) be different depending on the velocity of the EM wave?

the energy that leaves the source in some time interval must arrive at the source in the same interval.

There is, however, a pitfall lurking behind that phrase "the same interval".

If I point a one-watt laser at someone, turn it, let it go for one second according to my local clock, then turn it off I have sent one Joule (one watt for one second) in their direction. Eventually they will receive one Joule; how it long it takes for the leading edge of the pulse to reach them depends on the distance between us and the speed of light. But depending on our relative speeds and possible gravitational time dilation effects, the time measured on their clock between the arrival of the leading edge of the pulse and the trailing edge may be more or less than one second - the total energy delivered is one Joule but the power, which is energy per unit time, may be more or less than one watt.

So the power at the source and destination may not be the same. But that will be unrelated to the speed of light; it's a combination of relativistic Doppler and possible gravitational effects.

 

[Ibix]

So the power at the source and destination may not be the same. But that will be unrelated to the speed of light; it's a combination of relativistic Doppler and possible gravitational effects.

Indeed - I was assuming that source and receiver were at mutual rest in an SR situation. Otherwise you need to be very careful.

You can also consider light traveling through a medium with a time-varying index if you want to add to your list of caveats.

 

[Nugatory]

Indeed - I was assuming

I'm sorry - that's a perfectly good assumption for this thread - I forgot which thread I was in. My comment may be considered to be correct but irrelevant

 

[PeterDonis]

Eventually they will receive one Joule

No, they won't. Energy is affected as well as the time taken for the pulse. The gravitational case was discussed by Einstein: if you, lower in a gravity well, could send one Joule and I, higher in the gravity well, received the full one Joule, we could construct a perpetual motion machine (I could convert the one Joule to mass, drop it to you, and you could take the extra kinetic energy that the mass gained as it fell and convert it, plus the mass itself, into more than one Joule of energy, so you would have a closed loop that produced a net energy gain).

 

[Nugatory]

No, they won't. Energy is affected as well as the time taken for the pulse. The gravitational case was discussed by Einstein: if you, lower in a gravity well, could send one Joule and I, higher in the gravity well, received the full one Joule, we could construct a perpetual motion machine (I could convert the one Joule to mass, drop it to you, and you could take the extra kinetic energy that the mass gained as it fell and convert it, plus the mass itself, into more than one Joule of energy, so you would have a closed loop that produced a net energy gain).

Ah - yes, you are right - different than the relative motion case.

 

[PeterDonis]

different than the relative motion case

Note that the energy at the receiver also changes in the relative motion case. But the "perpetual motion machine" logic isn't the same in that case.

 

[cmb]

I absolutely and genuinely do not understand what the push-back on my question is?

If two horses set off from a start line together, and you are at the finish line of a known length of course, and it is also known that one horse runs twice as fast as the other, are you actually telling me that there is no way to tell their speeds from the difference in time that they cross that finish line you are standing next to?

Course length; 1 mile
Horses; A and B
Interval of time between horses A and B crossing the line; 2 minutes
Relative average speed of horses; A = 2 * B

Absolute speed of horses? ... impossible to say, apparently?

I accept that this hinges on the reliability of the 'relative average speed of horses' factor. However, if, in respect of measuring light, what is the measurement dilemma in measuring this average speed in a loop of material? I don't see the problem with that.

The other thing to bear in mind that I did not aim to start out trying to achieve an absolute measurement by these means, but to see if there was anisotropy. I was not seeking to see if there was isotropy which is a different objective.

If, by these means, this process is done twice and in each run this measured interval (between horse arrivals) differs, one should surely come to the conclusion that the course is softer one way than the other?

It does not provide a measurement of 'course softness', but if they are measured to be different intervals between the horses' arrivals, then one can then conclude that the speed of the course is different, whether or not it is the same course backwards or another course altogether. One needs not know, or attempt to calculate, what the actual horse speeds are to say 'these are different'?

Is this not so?

 

[Ibix]

If two horses set off from a start line together, and you are at the finish line of a known length of course, and it is also known that one horse runs twice as fast as the other, are you actually telling me that there is no way to tell their speeds from the difference in time that they cross that finish line you are standing next to?

You seem to be assuming their speeds, which you haven't established a method to measure. When you do establish such a measure, the result will depend on your clock synchronisation convention, and hence you can get different speeds.

However, if, in respect of measuring light, what is the measurement dilemma in measuring this average speed in a loop of material? I don't see the problem with that.

I really don't understand what experiment you think you are doing. Are you just sending a light pulse one way in an optical fibre and the other way in free space, then rotating the experiment 180° and repeating it?

The other thing to bear in mind that I did not aim to start out trying to achieve an absolute measurement by these means, but to see if there was anisotropy. I was not seeking to see if there was isotropy which is a different objective.

Did you mean to include the "not" in the last sentence? I think you didn't. The thing is, measuring the anisotropy gives you the one-way speed if you know the average speed (you can solve $\Delta c=c_+-c_-$ and $\frac 2c=\frac 1{c_+}+\frac 1{c_-}$ simultaneously for $c_+$ and $c_-$). So a measurement of anisotropy is equivalent to a measurement of one-way speed. If one is impossible, so is the other.

 

[cmb]

You seem to be assuming their speeds, which you haven't established a method to measure. When you do establish such a measure, the result will depend on your clock synchronisation convention, and hence you can get different speeds.

There is only one clock, the one that is stationary with me at the finish line. (This remained stationary throughout the race, and is also stationary wrt start line too.)

How do you synchronise a clock with itself? A few have said this and I do not know what it means to synchronise a clock with itself. I simply cannot comprehend what that means.

The horses at the start line commence the trip once the start wire is raised, which is stationary with them and at the same location, thus again no synchronisation and not even a time piece to set.

I press my stop watch the moment the first horse crosses the line and press it again when the second passes.

If I know this, and the relative (proportional) speeds of the horses, then I can determine their speed had they commenced their run from a known distance away.

If I don't even know the relative (and/or proportional) speeds of the horses, it doesn't even matter so long as I can do a second run in the opposite direction; if the timing interval between the two horses when the race is done the other way is different to the timing interval in the first direction, I know something has changed. If they are the same, then it doesn't tell me much that is fully conclusive, but at least I have taken a look to see if they are different.

Different arrival intervals at the finish line = something is different.

 

[Dale]

what is the measurement dilemma in measuring this average speed in a loop of material?

There is no dilemma whatsoever. The problem is that that measurement is a two way measurement not a one way measurement. You cannot go from that to a one way speed without making an assumption.

There is only one clock,

This fact is characteristic of a two way measurement.

 

[Nugatory]

I accept that this hinges on the reliability of the 'relative average speed of horses' factor. However, if, in respect of measuring light, what is the measurement dilemma in measuring this average speed in a loop of material? I don't see the problem with that.

There’s no dilemma, but you’re just restating the assumption that the one-way speed (that is, the speed in any particular segment of the loop, which we cannot measure without synchronizing two clocks at the two ends of that segment) is equal to the two-way speed (the average across the entire round trip through the loop, which we can measure with one clock). We aren’t disagreeing with that assumption - it is a foundational postulate of special relativity - but it remains an assumption and not a provable fact.

 

[Ibix]

There is only one clock,

Then how are you working out that one horse travels at twice the speed of the other? The experiment you did to determine that is the one where you need clock synchronisation.

If I don't even know the relative speeds of the horses, it doesn't even matter so long as I can do a second run in the opposite direction; if the timing interval between the two horses when the race is done the other way is different to the timing interval in the first direction, I know something has changed. If they are the same, then it doesn't tell me much, but at least I have taken a look to see if they are different.

Let's consider an easy case. We measure the two-way speeds of the horses and find that one is three times faster than the other. Now we send out the horses, measure the difference in arrival times, and then send them back and measure again. But the fast horse has time to kill while waiting for the slow horse - just enough to get to the start point and return to the finish. I'll draw that, assuming isotropy:

Time is horizontal, the black lines represent the start and finish, the fast horse is represented by the red line, and the slow one by the blue line.

Hopefully you can see that if the difference in arrival times is equal to the two-way trip time for the fast horse. So the difference in arrival times in either direction must be the same unless the two-way speed of the horse is also anisotropic.

Just to hammer the point home, let's try drawing the above if we assume anisotropy of the one-way speed. The rules for drawing the diagram are that

1. Upward sloping red lines must be parallel, as must downward sloping ones (constant one-way speeds)
2. Durations of round-trips must be the same as above (require the same two-way speed)
3. If the horses leave at the same time in the original diagram, they must leave at the same time in the new one

Let's make it really anisotropic and allow the red lines to be vertical upwards. Rules 1 and 2 then require us to draw this diagram:

Rule three let's us add the blue horse, because its arrival and departure times are dictated by those of the red one:

Still, the difference in arrival times is the two-way trip time of the faster horse. Maybe you can also now see where the clock synchronisation comes in - the third diagram is the first one, sheared. That is, the difference is just where the $t=0$ points on the two black lines lie.

Finally, you might argue that this is trickery based on using horses with this 1:3 speed ratio. But for any given speed ratio you can add a custom course that the faster horse has to complete after one leg of the main course, choosing its length so that the horse returns to the finish at the same time as the slower one arrives. Again, the difference is set to some two-way time for the faster horse, and must be invariant.

 

[cmb]

Hopefully you can see that if the difference in arrival times is equal to the two-way trip time for the fast horse. So the difference in arrival times in either direction must be the same unless the two-way speed of the horse is also anisotropic.

Yes!

Exactly!

I call that progress of a sort.

IF there is isotropy in both the calibration and two tests then the arrival time intervals (for the two tests) would be the same.

IF there is isotropy in at least one of the calibration and two tests then the arrival time intervals (for the two tests) might be the same.

but ...

IF the arrival time intervals (for the two tests) are not the same, then there must be anisotropy somewhere.

Is this controversial?

What I have proposed may not be a solution to measuring one way speed, just a first step to see if there is a difference between what should be two symmetrical outcomes.

The assumption in the calibration is that the two-way trip delivers a result being the average speed in both directions*. The two test runs are then compared to 'that' singular average speed result. Asymmetry in the results of the two tests would support an anisotropy hypothesis.

*(The alternative, that the calibration result is NOT an average of the two-way speeds axiomatically dictates that the result shows anisotropy.)

(If they are exactly the same, it doesn't say anything as the two might perfectly balance out. If they do not perfectly balance out, then I can't see how the speeds in each direction can still be isotropic.)

 

[Ibix]

You seem not to have understood what I wrote.

I said that if the two-way speed of light is isotropic (which it is - see Michelson-Morley) then the result of your experiment is that the arrival times are the same in either direction regardless of any anisotropy in the one-way speeds. So your experiment adds nothing in light of the results of Michelson-Morley.

 

[cmb]

You seem not to have understood what I wrote.

I said that if the two-way speed of light is isotropic (which it is - see Michelson-Morley) then the result of your experiment is that the arrival times are the same in either direction regardless of any anisotropy in the one-way speeds. So your experiment adds nothing in light of the results of Michelson-Morley.

Sorry, no I don't understand what you wrote.

Once you have your assumed-isotropic two-way calibration measurement, you can go do two one-way tests as I described. Are you saying it is then impossible to get two different results from those two tests, or are you saying there can be two different results even with isotropy?

 

[cmb]

Do you guys recall how the scientific mainstream simply accepted as read that there was no way to test quantum entanglement?

Until John Bell came along with a simple test .. (also involving 3 legs (polarisation) of testing, rather than just two?).

 

[Ibix]

Once you have your assumed-isotropic two-way measurements

This is not an assumption. It has been tested. The two-way speed of light is isotropic.

you can go do two one-way tests as I described.

Given the isotropy of the two way speeds, you will get the same arrival time difference in both directions. This is true whether there is one-way anisotropy or not. That's what I showed in #143.

 

[Ibix]

Do you guys recall how the scientific mainstream simply accepted as read that there was no way to test quantum entanglement?

Until John Bell came along with a simple test .. (also involving 3 legs (polarisation) of testing, rather than just two?).

People look for violations of Lorentz covariance - it's an active topic of research. None has ever been found. And your experiment could not find it if it did, as I already explained.

 

[cmb]

This is not an assumption. It has been tested. The two-way speed of light is isotropic.

Given the isotropy of the two way speeds, you will get the same arrival time difference in both directions. This is true whether there is one-way anisotropy or not. That's what I showed in #143.

So, if the horses I referred to above took longer to run their course in one direction to the other, are you saying the interval in their arrival times at the finish line would always be the same?

I don't see why the situations are different whether horses or photons.