# Synchronized Clocks in Frames boosted by Acceleration

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P: 765
 Quote by ghwellsjr Now these two cables stretch from the source to the midway point where the observer is.
Yes, I agree; thought you were using cables that run all the way to the target (when taut).
 PF Gold P: 4,793 Ok, good. Now the observer takes one of these cables, along with the light detector on the far end and the electrical signal detector/light generator on the near end, and he swings the cable around so that it can pick up the reflection of the light from the target and send an electrical signal down the cable back towards him. Now he repeats the experiment. The light flashes at the source. The detector immediately generates an electrical signal that travels down the cable parallel to the flash of light outside the cable. When it gets to the near end of the cable, the electrical signal immediately produces a flash of light coincident with the flash the traveled parallel to the cable in free space and the observer sees both of these at the same time and he starts his timer. Now the flash in free space continues onward toward the target all by itself until it reaches the target. At this point, it reflects off the target and starts back toward the observer and also immediately triggers the detector in the far end of the second cable which generates an electrical signal that travels parallel to the light flash in free space coming back to the observer. After some time, the light flash and the electrical signal arrive at the observer where the detector/generator in the near end of the cable produces a flash at the same time that he also sees the light flash that got to him in free space at which point he stops his timer. Do you understand and agree with this assessment?
 P: 765 Yes George, I think this is equivalent to your original reflection test. I would like to hear why this is equivalent to Don's test.
 PF Gold P: 4,793 You previously agreed that the observer was measuring the round trip time for the light to go from his location to the target and back to him. Now with the cables in place, you have agreed that the observer is doing exactly the same thing with cables that he was doing with just light. Now replace the observer with a fast oscilloscope and remove the two detector/light generators at the receiving end and plug the two calibrated cables into the scope. You can start over and put the detector ends of both cables at the source and recalibrate as Don suggested and then move one end of one cable from the source to the target and you have exactly the test that Don devised so do you agree that his test in this case is measuring the round trip time for the light to traverse the last half of the distance from the source to the target?
P: 765
 Quote by ghwellsjr Now replace the observer with a fast oscilloscope and remove the two detector/light generators at the receiving end and plug the two calibrated cables into the scope. You can start over and put the detector ends of both cables at the source and recalibrate as Don suggested and then move one end of one cable from the source to the target and you have exactly the test that Don devised so do you agree that his test in this case is measuring the round trip time for the light to traverse the last half of the distance from the source to the target?
I will have to work through this one carefully to convince myself that all your steps from #33 onwards necessarily make your test equivalent to Don's setup, because to met it is not obvious. Thanks for your patience - I will post my conclusion when I reach it.
P: 765
 Quote by Jorrie I will have to work through this one carefully to convince myself that all your steps from #33 onwards necessarily make your test equivalent to Don's setup, because to me it is not obvious.
I have some difficulty convincing myself that Don's cable setup, with cable propagation speed less than c, measures the two-way speed of light. Here's how far I got.

His two cables are of arbitrarily, identical lengths, which I define as 1 distance unit and define time so that the 2-way speed of light comes out at $c$. I then assume that the 1-way speed of light may be $c$' (outbound) and $c$'' (inbound) and that the signal speed in the cables is equivalent to a refraction index $n$, so that

(1) $1/c' + 1/c'' = 2/c$ (vacuum) and: $n/c' + n/c'' = 2n/c$ (cables).

The time differential that Don measures on his oscilloscope:

(2) $\Delta T = 1/c' + n/c'' - n/(2c') - n/(2c'') = 1/c' - n/(2c') + n/(2c'')$

This correlates with the 2-way speed of light in vacuum only if $n$ = 1; e.g. if $n$ = 2, then $\Delta T = 1/c''$.

Or am I making a wrong assumption somewhere?
Sci Advisor
P: 2,470
 Quote by Jorrie The interesting point seems that you suggest that we can devise schemes to measure the one-way speed of light (or perhaps rather Lorentz invariance); only that we will probably always find it to equal the two-way speed?
The basic assumption usually made is that they are equal. However, on deeper level, it's irrelevant. If there is an experiment that depends on the difference, then there is an experiment to measure one-way speed of light. Even if indirectly so. Measuring difference with two-way speed of light would be an entirely acceptable measurement.

SR is built around assumption that one-way speed of light is equal to two-way speed of light. However, it should be possible to construct an equivalent theory in which they are not. Such a theory will have needless complications which simplify for any possible experiment, giving you exactly the same predictions as SR.

But that's exactly the thing that tells you that you can't measure one-way speed of light. At least, not within the confines of SR. If you do make a measurement, you prove relativity wrong regardless of whether you got the same or different result.
P: 3,188
 Quote by Jorrie This is the sort of statement that helps the discussion going in circles. If I correctly measure a distance and I correctly measure (in the same inertial frame) the one-way time of propagation over that distance, I define that as being a measurement of one-way speed of something. [..]
It may be just a matter of words - and yes, debates over words can help a discussion go in circles forever. According to SR (and most of physics) such measurements are reproducible. The one-way speed that you measure is the one-way speed that you (or an engineer who set up your system) first defined by free choice - that isotropy has no physical meaning. For example, if in CERN a particle is accelerated to almost the speed of light, then (according to the lab's definition and reference) light following that electron in that same direction is only slowly catching up with it, and light in the opposite direction is approaching it at almost 2c. In contrast, the definition of the electron's rest frame makes light speed isotropically c in that frame.
Note that SR does not claim that either is "right" (and both would even be contradictory); it claims that if we use that definition then the laws of physics such as that of Maxwell work wrt any inertial frame.
 I have already admitted that there is an assumption on the one-way speed of light in Lincoln's test (the 1983 definition: "The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second"). This will guarantee the outcome of his measurement to be c in every inertial frame. This may be exactly equivalent to your "two-way" argument, but IMO it's then neither here nor there.
Indeed. that definition only works if the synchronization convention was used which makes the one-way speed equal to the two way speed.
 I will rest my case.
Oops, sorry that makes my remarks like mustard after the meal... but perhaps it's still useful for you or someone else.
PF Gold
P: 4,793
 Quote by Jorrie I have some difficulty convincing myself that Don's cable setup, with cable propagation speed less than c, measures the two-way speed of light. Here's how far I got. His two cables are of arbitrarily, identical lengths, which I define as 1 distance unit and define time so that the 2-way speed of light comes out at $c$. I then assume that the 1-way speed of light may be $c$' (outbound) and $c$'' (inbound) and that the signal speed in the cables is equivalent to a refraction index $n$, so that (1) $1/c' + 1/c'' = 2/c$ (vacuum) and: $n/c' + n/c'' = 2n/c$ (cables). The time differential that Don measures on his oscilloscope: (2) $\Delta T = 1/c' + n/c'' - n/(2c') - n/(2c'') = 1/c' - n/(2c') + n/(2c'')$ This correlates with the 2-way speed of light in vacuum only if $n$ = 1; e.g. if $n$ = 2, then $\Delta T = 1/c''$. Or am I making a wrong assumption somewhere?
You're not making a "wrong" assumption but you are making assumptions that hide the fact that Don's method is measuring the two-way propagation of light for the last half of the trip and calling it the one-way propagation for the entire trip.

That is why I asked you to use cables that propagate an electrical signal the same as light in free space along a parallel path. This is something that can be measured without resorting to clock synchronization or identifying how long that propagation took or what the speed of the light or electrical signal are, just like we can measure that the propagation of light is independent of the speed of the source. See the section called "Experiments that can be done on the one-way speed of light" in the wikipedia article on the one-way speed of light that I referred you to in post #2.

That is also why I asked you to not attempt to identify the length of the cables or the distance between the source and the target. All we care about is that the observer is at the midpoint and that the two cables add up to that distance.

That is also why I asked you not to attempt to identify the speed of the round trip for light or the electrical signals. Instead, I asked you to only make a measurement of a time interval by a single timing device located at a single position with identifiable stimuli, namely when the observer sees the light and electrical signals after they propagate from the source and when the observer sees the light and electrical signals after they propagate from the target.

All I'm trying to get you to recognize, which you already agreed to, is that there is no difference between making the timing measurement with light or with cables. And you already agreed that with light, it is a two-way measurement of the last half of the distance.

Now if you want to understand the problem taking advantage of a hundred years of experience with physics then I suggest a different approach. Set up the problem in a single Inertial Reference Frame and establish the coordinate times for each event: the emission of the light at the source, the arrival of the light at the observer when he starts his timer, the arrival of the light at the target when the reflection and return starts, and the arrival of the reflected light back at the observer when he stops his timer. In this IRF, the propagation of light is defined to be c.

Now transform the scenario into another IRF moving at some high speed with respect to the first IRF. You will see that the new time coordinates for the same events do not show equal time intervals for each direction.

For example, let's say that the source and start of the experiment are at the origin of the IRF's and the observer is located at x=0.5 and the target is at x=1 in the first IRF. Here are the coordinates of the events for the first IRF:

Start of light: x=0 t=0
Light reaches observer: x=0.5 t=0.5
Light reaches target: x=1 t=1
Reflected light reaches observer: x=0.5 t=1.5

Now transform the time coordinates of these events into an IRF traveling at 0.6c with respect to the first IRF:

Start of light: t=0
Light reaches observer: t=0.25
Light reaches target: t=0.5
Reflected light reaches observer: t=1.5

The difference between the time coordinates of the light reaching the observer and the reflected light reaching the observer is 1.25. This is partitioned into 0.25 for the light to get from the observer to the target and 1.0 for the reflected light to get from the target back to the observer.

Now when traveling at 0.6c, gamma is 1.25, so the observer's clock is time dilated by that factor so that when he measures the time for the light to make its trip from him to the target and back, his clock will advance by 1 unit, just like it did in the first IRF. He cannot tell that the light got to the target in one quarter of the time that the light took to get back from the target to himself.

Now if you understand this explanation, you can go ahead and repeat it for cables that propagate signals identically to light or at some reduced rate.

[George: prepare yourself for the barrage of criticism that you are mixing coordinates from two different IRF's.] Ok, I'm prepared.
P: 765
 Quote by ghwellsjr Now if you understand this explanation, you can go ahead and repeat it for cables that propagate signals identically to light or at some reduced rate. [George: prepare yourself for the barrage of criticism that you are mixing coordinates from two different IRF's.] Ok, I'm prepared.
I understand what you said and it is very clear when using only light, because then Don Lincoln's setup becomes exactly equivalent to your setup. It is not so clear in Don's actual setup, because I have shown in #42 that his result could depend on any anisotropy of light propagation:
 The time differential that Don measures on his oscilloscope: (2) $ΔT=1/c′+n/c′′−n/(2c′)−n/(2c′′)=1/c′−n/(2c′)+n/(2c′′)$
I agree that Don cannot claim to have measured the one-way speed of light, because components of light propagation in both directions (c' and c'') are present in this result. Which is essentially what this thread has cleared up now.

Thanks for your patience.
PF Gold
P: 4,793
 Quote by Jorrie Thanks for your patience.
You're welcome--and thanks for the feedback.

Now, are you going to have another discussion with Don Lincoln? If you do, I would hold out the possibility that he is already well aware that his test is not really measuring the one-way speed of light independent of a synchronization convention but it is an excellent and practical way to follow Einstein's convention to measure the propagation of light in the lab, something that he and his co-workers are probably doing all the time at Fermi-Lab.
P: 765
 Quote by Jorrie The time differential that Don measures on his oscilloscope: (2) $\Delta T = 1/c' + n/c'' - n/(2c') - n/(2c'') = 1/c' - n/(2c') + n/(2c'')$
Don Lincoln has made one remark that I still need to consider: if the $\Delta T$ that he measured correlates with $c$ for the specific length of the cables, then for his lab he has established that $c' = c'' =c$, within experimental error. How else?

This shows that it is equivalent to having a synchronized clock at each end of the cable; but still, he needed only one clock, the oscilloscope. One clock, two clocks? Probably a moot point.
PF Gold
P: 4,793
 Quote by Jorrie Don Lincoln has made one remark that I still need to consider: if the $\Delta T$ that he measured correlates with $c$ for the specific length of the cables, then for his lab he has established that $c' = c'' =c$, within experimental error. How else?
"Established"--now that's a very interesting word. It can mean to make or bring into existence or it can mean to determine the truth of something. So which do you mean? I hope you mean the first meanings because that is what he's doing.

And now I see that Don Lincoln used that same word in his description of the test setup.
 Quote by Jorrie This shows that it is equivalent to having a synchronized clock at each end of the cable; but still, he needed only one clock, the oscilloscope. One clock, two clocks? Probably a moot point.
How 'bout three clocks? Two synchronized cables and the scope.
P: 765
 Quote by ghwellsjr "Established"--now that's a very interesting word. It can mean to make or bring into existence or it can mean to determine the truth of something. So which do you mean? I hope you mean the first meanings because that is what he's doing.
Yes, he first establishes simultaneity and then determines the one-way speed of light for the lab - which is then a given...

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