B Measuring the One Way Speed of Light

[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.