## Why do you need to measure the speed of light in two directions?

 Note that while these experiments clearly use a one-way light path and find isotropy, they are inherently unable to rule out a large class of theories in which the one-way speed of light is anisotropic. These theories share the property that the round-trip speed of light is isotropic in any inertial frame, but the one-way speed is isotropic only in an aether frame. http://www.edu-observatory.org/physi...#one-way_tests
So let us compare light-synchronized clocks with slowly transported clocks:

a) When we find no difference, then this is in agreement with all theories in which the two-way speed is isotropic, regardless of whether the one-way speed is anisotropic or not.

b) When we find a difference, then all theories in which the two-way speed of light is isotropic are refuted

Therefore, comparing light-synchronzied clocks with slowly transported clocks says a lot about the two-way speed, though nothing about the "one-way speed" - the latter is only a useful convention helping us to simplify the arrangement of our experiments, but it's isotropy or anisotropy can never be proved nor refuted.

Regards,

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 Quote by Histspec So let us compare light-synchronized clocks with slowly transported clocks: a) When we find no difference, then this is in agreement with all theories in which the two-way speed is isotropic, regardless of whether the one-way speed is anisotropic or not. b) When we find a difference, then all theories in which the two-way speed of light is isotropic are refuted Therefore, comparing light-synchronzied clocks with slowly transported clocks says a lot about the two-way speed, though nothing about the "one-way speed" - the latter is only a useful convention helping us to simplify the arrangement of our experiments, but it's isotropy or anisotropy can never be proved nor refuted. Regards,

I have no disagreement with the above (and I don't believe anything I have said disagrees with it). However, note that once an observation of one way anisotropy refutes that two way isotropy holds in all frames, it is no longer (necessarily) true that the one way measurement is limited in the information it provides. When testing SR, you should not assume the information limitations implied by the theory you are testing.

One side note is that measurement in one particular frame that two way speed happens to be isotropic but one way speed is not, is sufficient to disprove the equivalence of inertial frames. Thus, in a hypothetical universe where the principle of relativity is false, use of one way light measurements with slow transport might establish this using only single frame, while two way measurements could require two frames to establish this.

Further, note that possible anisotropy is inherently unobservable in SR equivalent theories; while all measurements of c, two way or one way, will be c. Adopting an interpretation in which you attach significance to the unobservable anisotropy only complicates such things as interpreting Maxwell's equations. As long as one understands that you can't rule out such formulations, you are free ignore them for practical convenience.

 Quote by ghwellsjr [..] I want to put this at the time when Maxwell realized that light was a wave in the electromagnetic field his equations described and he believed it would be possible to detect the earth's motion through this field by measuring the one-way speed of light. His only problem was that technology was not available for him to perform the type of experiment that we can perform today but let's imagine that it was. So let's suppose that he took two accurate and stable atomic clocks that were synchronized at one location and slowly moved one of them some distance away and the distance was measured using a rigid calibrated ruler. (Let's stipulate that there was no error in his distance measurement.) Now let's also say that he constructs a tube or pipe that he evacuates with a perfect vacuum and he puts a light source at one end that can log the time from the atomic clock located next to it when the light is turned on and a light detector at the other end that can log the time from the other atomic clock located next to it when the light is detected. So now he does his experiment and he divides the difference between the two logged times into the measured distance. I believe he will get c as the answer and I believe there is no controversy about this, correct?
So far it looks correct.
 But let's also assume that this answer would have surprised Maxwell and so he repeats the experiment at different times of the day and at different seasons of the year.
Thus you propose two (extremely!) stable clocks that are once calibrated and synchronised, after which measurements are done at different times of the year. Such an experiment is still to be done I think (and perhaps still out of reach).
 Let's say the experiment was so easy to do that other people repeat the same experiment. They do it in every conceivable location, at the bottom of the deepest valley, at the top of the highest mountain, at the poles, at the equator, even at the bottom of the deepest ocean. They repeat the measurement with the apparatus pointed in all different directions of the compass. Everybody always gets the same value for c, correct? Everybody agrees that this is what would have happened, correct?
Sorry but no, that's certainly wrong: according to theory they must find deviations for c, due to the fact that the clocks were not re-synchronised.
 So then they put the apparatus on the longest flatbed railway car and repeat the measurement at different constant speeds. I'm assuming that their apparatus has no errors and that the accuracy is good enough that they had every reason to believe that they could measure any motion through the field for the speeds they were traveling. Everybody agrees they still always measure exactly c, correct?
No, still faulty for the exact same reason!
 So my question for you is: Is there any reason to believe that the development of science would have progressed any differently than it did as a result of MMX which was a two-way measurement instead of a one-way measurement?
Probably the development of science would have progressed quite similarly; however it could be that textbooks would present a theory that started out from positive results a little different from a theory that started out with negative results.

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 Quote by harrylin Sorry but no, that's certainly wrong: according to theory they must find deviations for c, due to the fact that the clocks were not re-synchronised.
This discussion was in a context where gwellsjr and I were hypothesizing that such effects were not quite detectable. We were also, by mutual understanding, ignoring gravity. Putting it in my words, it is proposed you had clocks good enough to distinguish that light doesn't behave like sound (noting that seasonal variations in speed and vector differences in velocity at different times of day are a thousand or more mph), but not good enough to detect slow transport differences between 1 mph and 10 mph (for example). Further, as noted, gravity is meant to be ignored.

[Edit: One other implicit assumption was that the the slow clock transport was over distances small enough that the whole experiment did not take so long as to need to worry about non-inertial motion. Say, 30 minutes or so. This whole thing was really gwellsjr 'granting for the sake of argument' a whole bunch of things he would agree are dubious, for the purpose of exploring theoretical differences with me.]

[re-edited]
 Quote by PAllen This discussion was in a context where gwellsjr and I were hypothesizing that such effects were not quite detectable.
Yes, and then he came with a "what if":

"[Maxwell's] only problem was that technology was not available for him to perform the type of experiment that we can perform today but let's imagine that it was."
And:
"Is there any reason to believe that the development of science would have progressed any differently than it did as a result of MMX which was a two-way measurement instead of a one-way measurement?"

In the case of an experiment with a purpose like that of Michelson and Morley, but with one-way light signals and clocks, the clocks have to be precise and stable enough to detect the velocity relative to the ether even if by chance the solar system is moving slowly. A null result due to inappropriate clocks would hardly, as ghwellsjr put it, "have surprised Maxwell".

In the line of the OP's question this is, I think, an interesting idea, as it highlights a significant difference between two-way measurements and one-way measurements that is rarely addressed.

ghwellsjr added "they would also conclude that the length of the pipe was changing during the course of a day". Note that a change in Lorentz contraction is a much smaller effect than a change in local synchronization.
 We were also, by mutual understanding, ignoring gravity. Putting it in my words, it is proposed you had clocks good enough to distinguish that light doesn't behave like sound (noting that seasonal variations in speed and vector differences in velocity at different times of day are a thousand or more mph), but not good enough to detect slow transport differences between 1 mph and 10 mph (for example). Further, as noted, gravity is meant to be ignored. [Edit: One other implicit assumption was that the the slow clock transport was over distances small enough that the whole experiment did not take so long as to need to worry about non-inertial motion. Say, 30 minutes or so. This whole thing was really gwellsjr 'granting for the sake of argument' a whole bunch of things he would agree are dubious, for the purpose of exploring theoretical differences with me.]
I interpreted it in the context of the topic here: there is an interesting theoretical difference between the outcome of the one-way speed-of-light measurement as suggested here by ghwellsjr to detect changes of velocity wrt the ether (if it had been technologically feasible at the time) and the two-way experiment of M-M.
 [QUOTE=ghwellsjr;--- What theory is there (that comports with reality) and claims the "one way speed of light measurement is possible"? How is this going to help the OP?[/QUOTE] Wouldn't stellar aberation suffice? Where Vt is the Earth's orbital velocity and theta is the angle of aberation: c = Vt/tan(theta)

 Quote by Tracer Wouldn't stellar aberation suffice? Where Vt is the Earth's orbital velocity and theta is the angle of aberation: c = Vt/tan(theta)
According to SR no "absolute" one-way speed of light determination is possible and it's a usual (convenient) choice of the observer to define it as half the two-way speed.

So, here you effectively propose stellar aberration to break the PoR. However it is expected to obey the PoR as already elaborated in Einstein's 1905 paper, in section 7:
www.fourmilab.ch/etexts/einstein/specrel/www/

 Quote by harrylin According to SR no "absolute" one-way speed of light determination is possible and it's a usual (convenient) choice of the observer to define it as half the two-way speed. So, here you effectively propose stellar aberration to break the PoR. However it is expected to obey the PoR as already elaborated in Einstein's 1905 paper, in section 7: www.fourmilab.ch/etexts/einstein/specrel/www/
Is it that there needs to be experimental proof that light speed is identical for both halves of the round trip measument of c?

If the angle of stellar aberration is the only variable measured for light passing through a measuring apparatus in each direction and the angle of aberration is the same for light travelling in both directions, then the speed of light is confirmed to be the same in both directions without the need to measure the length of an AU or the tangential velocity of the Earth's orbit. Then one clock round trip measurements of the speed of light can be truly said to be the result of equal velocities for each half of the round trip.
 There's a gravitational effect on the distance the light travels. In a two-way trip, the gravitational effect is negated.

 Quote by Tracer Is it that there needs to be experimental proof that light speed is identical for both halves of the round trip measument of c?
Quite to the contrary: there is experimental proof that we may freely choose an inertial reference system and pretend that it is "in rest", so that the speed of light becomes (or appears) identical for both halves of the round trip measurement. Since we apply the relativity principle to light, this should be true for any inertial reference system that we freely choose.

It is essential to understand that thus for each choice of reference system the speed of light becomes (or appears) according to your definition, anisotropic relative to any object or system that is moving relative to your chosen system.
 If the angle of stellar aberration is the only variable measured for light passing through a measuring apparatus in each direction and the angle of aberration is the same for light travelling in both directions, then the speed of light is confirmed to be the same in both directions without the need to measure the length of an AU or the tangential velocity of the Earth's orbit. Then one clock round trip measurements of the speed of light can be truly said to be the result of equal velocities for each half of the round trip.
There is now a thread opened on stellar aberration, so I'll only give a short answer here. If you reflect light inside an apparatus, no effect from motion is predicted by SR even if the apparatus is in inertial motion.

 Quote by harrylin Quite to the contrary: there is experimental proof that we may freely choose an inertial reference system and pretend that it is "in rest", so that the speed of light becomes (or appears) identical for both halves of the round trip measurement. Since we apply the relativity principle to light, this should be true for any inertial reference system that we freely choose.
The synchronization of clocks at A and B where tB-tA=t'A-tB stipulates that the clocks are both stationary with respect to each other...

How can this stipulation apply when one can't verify this prior to synchronization? Actually it would require multiple synchronization successes to verify points A and B are stationary with respect to one another wouldn't it?

... my point is that even if both A and B are inertial reference frames, that does not imply they are stationary relative to each other. They may be moving together or apart at a constant rate, or one or both may be free falling with a net acceleration between them.

If A and B are in either constant or free fall accelerating relative motion, both are inertial frames, but there are possible synchronization measures where tB-1A=t'A-tB will be true, yet a prior or subsequent measure will be false. So a single successful synchronization measure can give a false positive.

The reflecting mirror method stipulates stationary A and B, but what it really stipulates is that A and B happen to be a particular distance apart when the tB event occurs... and B can be moving wrt A when this happens.

The case of free falling inertial A and B may be bringing GR into this, but the case of a net constant rate distance change between A and B does not; yet the possible false positive for a successful synchronization measure still exists (meaning that the synchronization was only for a moment, the clocks were not syched before and continue to be out of synch after in spite of a momentary tB-tA=t'A-tB.

 Quote by bahamagreen The synchronization of clocks at A and B where tB-tA=t'A-tB stipulates that the clocks are both stationary with respect to each other... How can this stipulation apply when one can't verify this prior to synchronization? Actually it would require multiple synchronization successes to verify points A and B are stationary with respect to one another wouldn't it?
Not at all, it's not difficult to establish if clocks have a constant distance between them and are stationary in a frame in which the laws of mechanics hold. And it's not difficult to extend this method to clocks in motion, as is nowadays done.
 ... my point is that even if both A and B are inertial reference frames,
A point is not a reference frame!
 that does not imply they are stationary relative to each other. They may be moving together or apart at a constant rate, or one or both may be free falling with a net acceleration between them. [..]
According to the original definition, free-fall is not an inertial frame; what was meant is a frame that is in uniform, straight line motion according to astronomical observations. Moreover, SR doesn't account for effects from gravitation.

 The case of free falling inertial A and B may be bringing GR into this, but the case of a net constant rate distance change between A and B does not; yet the possible false positive for a successful synchronization measure still exists (meaning that the synchronization was only for a moment, the clocks were not syched before and continue to be out of synch after in spite of a momentary tB-tA=t'A-tB.
Sure, a more complex method is required for such cases, as is done with GPS satellites. That is a completely different topic. You could start that topic if you want to discuss it.

Harald
 Harrylin, thanks, I'm figuring it out as I go... I made a model to investigate this and found the following: In the tB-tA=t'A-tB equation the "t's" are not all representing the same kind of thing. tA is the initial reading on clock A at tA but tB is not B's initial setting at tA, it is an elapsed time from B's initial clock setting at tA to the time tB t'A is also an elapsed time, the sum of tB + the time of the trip back to A tb-tA=t'A-tB only works if the initial times of both A and B clocks at time tA are set to the same time. You can't just have two clocks A and B with different time settings and run the synchronization. That is, if we call the initial clock reading of A and B as Ain and Bin at time tA, then tA=Ain=Bin. This makes it more clear that tB is really tB-Bin... This is point where I have that "Duh.." moment about what synchronized clocks really means... that at tA both clocks indicate the same value... definition of synchronous. :0 But what is interesting is that the synchronization works fine (tb-tA=t'A-tB is true) when you add a constant rate of distance change (using non-relativistic speed addition) between A and B. The path from A to B is still the same magnitude as the path from B to A at the time tB, even if the path length was different before or after tB. So the stipulation that A and B be at rest wrt each other must be to omit the possibility of relativistic effects.

 Quote by bahamagreen [..] So the stipulation that A and B be at rest wrt each other must be to omit the possibility of relativistic effects.
It wasn't a stipulation. It was just a description of a common way in those days to set up a reference system. And the following claim that "it is essential to have time defined by means of stationary clocks in the stationary system" was basically wrong: nowadays one uses mostly clocks that are moving in the stationary system. But what does that have to do with the topic?

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 Quote by bahamagreen Harrylin, thanks, I'm figuring it out as I go... I made a model to investigate this and found the following: In the tB-tA=t'A-tB equation the "t's" are not all representing the same kind of thing. tA is the initial reading on clock A at tA but tB is not B's initial setting at tA, it is an elapsed time from B's initial clock setting at tA to the time tB t'A is also an elapsed time, the sum of tB + the time of the trip back to A tb-tA=t'A-tB only works if the initial times of both A and B clocks at time tA are set to the same time. You can't just have two clocks A and B with different time settings and run the synchronization. That is, if we call the initial clock reading of A and B as Ain and Bin at time tA, then tA=Ain=Bin. This makes it more clear that tB is really tB-Bin... This is point where I have that "Duh.." moment about what synchronized clocks really means... that at tA both clocks indicate the same value... definition of synchronous. :0 But what is interesting is that the synchronization works fine (tb-tA=t'A-tB is true) when you add a constant rate of distance change (using non-relativistic speed addition) between A and B. The path from A to B is still the same magnitude as the path from B to A at the time tB, even if the path length was different before or after tB. So the stipulation that A and B be at rest wrt each other must be to omit the possibility of relativistic effects.
Einstein said at the beginning of section 1 of his 1905 paper:
 If a material point is at rest relatively to this system of co-ordinates, its position can be defined relatively thereto by the employment of rigid standards of measurement and the methods of Euclidean geometry, and can be expressed in Cartesian co-ordinates.
Points A and B are fixed with a rigid rod between them and the length has been measured. Let's say it is 1000 feet. Let's also stipulate that the speed of light is 1 foot per nanosecond to make the arithmetic easier. At A is a light source and a clock with an arbitrary time on it when the light is flashed (say 1 PM). This is tA. At B is a mirror and a second clock which also reads an arbitrary time on it when the light hits the mirror, say 2 PM. This is tB. Then when the reflected light gets back to A, the time on the clock is t'A. This time will be 2 microseconds after 1 PM or 1:00:00.000002 PM). Now we plug these number into the equation to see if they equal:

tB - tA = t'A - tB
2:00:00.000000 - 1:00:00.000000 = 1:00:00.000002 - 2:00:00.000000
1:00:00.000000 ≠ -0:59:59.999998

Whoops--they're not equal. The clocks are not synchronized according to Einstein's definition.

Let's subtract the time on clock B by one hour and repeat the experiment the next day (now tB = 1:00:00.000000):

1:00:00.000000 - 1:00:00.000000 = 1:00:00.000002 - 1:00:00.000000
0:00:00.000000 ≠ 0:00:00.000002

Still not synchronized. Now let's advance the time on Clock B by 1 microsecond and repeat the next day (now tB = 1:00:00.000001):

1:00:00.000001 - 1:00:00.000000 = 1:00:00.000002 - 1:00:00.000001
0:00:00.000001 = 0:00:00.000001

Hooray, now they're synchronized.

Your statement about the clocks having initial times on them that are the same is meaningless. That's the whole point of defining a synchronization process--we can't tell when or if the times on remotely separated clocks have the same time on them. Saying "tB is really tB-Bin" means you have missed the whole point of what Einstein is saying. You need another "Duh..." moment. You should not think that there is any reality to the times on remote clocks apart from us putting meaning into those times. It's not that we are figuring out what nature is trying to tell us--we can't--instead, we are arbitrarily putting meaning into nature, at our own whim.
 Alright, I see that tB is not elapsed time, but I'm still seeing synchronized clocks as indicating identical times in what Einstein called common "time" of A and B, or "the time of the stationary system.” Let tA=10 and t'A=20 tB-tA=t'A-tB tB-10=20-tB tB=20+10-tB tB=30-tB 2(tB)=30 tB=15 tB-tA=t'A-tB 15-10=20-15 5=5 So A and B are synchronized if tB=15 If A assumes travel time for AB to equal BA then A may calculate what his A clock indicated when tB occurred. The elapsed round trip t'A-ta=20-10=10, so AB=BA=5 A may conclude that he sees tB 5 seconds after it happens, so may calculate that tB happened when his A clock was reading t'A-5=20-5=15 So, with regard to what you wrote, "You should not think that there is any reality to the times on remote clocks apart from us putting meaning into those times.", are you saying that this calculation of tB's occurrence with respect to A's clock time is meaningless? When two clocks side by side show the same time we say that those clocks are synchronous. When two clocks are distant but synchronous, they no longer show the same time, but knowing t'A and assuming AB=BA one can calculate what was the A observer's local time that corresponded to the B time of a distant event. Is this not correct? "But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an “A time” and a “B time.” We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A." My interpretation of Einstein's quote is that the common "time" for A and B IS defined when assuming AB=BA... that what is synchronous is the time readings of synchronous clocks in this common time, which takes into account the propagation delay of light. When A calculates that tB=15 and then calculates that A's clock was at 15 when tB occured, is this not the synchronization of the common time of A and B? Is this common time not the same as the "time of the stationary system"? "It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it “the time of the stationary system.”
 Recognitions: Gold Member You seem to now understand Einstein's synchronization process but it appears that you are thinking that Einstein was describing a way to "discover" or "determine" the common "time" by that process, rather than the process "creating" the common "time". You don't seem to appreciate that fact that Einstein's synchronization is merely one of many different ways to "create" the common "time".