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One-way speed of light and clock sync

  1. Feb 5, 2013 #1
    Ok, i know this makes me look like a crackpot, but suppose for an instant there is an aether or absolute space or whatever!
    Let me be on a rod that travels tranversely with respect to aether and let me be exactly on the middle point of this rod. I have a GPS device with me and send it upwards for some distance. It transmits two light signals back to the ends of the rod and I synchronize my clocks at both ends to read the same time accordingly. Light travels same distances at same speed so i am certain that my clocks are synchronized!
    Now, very slowly I turn my rod so it travels longitudinally with respect to aether. Shouldn't my clocks remain synchronized and reading the same time? If so then I measure different light speeds for the inbound and outbound legs of a two-way trip of a beam of light!
    Reichenbach and Grunbaum have stated that ε=1/2 factor Einstein uses to synchronize two
    spatially remote clocks is an arbitrary plantation.
    Is it true? Why the one-way speed of light will never be measured?
  2. jcsd
  3. Feb 5, 2013 #2


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    They perform a different motion relative to the aether. It depends on the way you want to use this aether. If you want to look at your own reference frame and assume physics is the same there: Sure, but where is the point in the aether then?

    The other way of your one-way-experiment is the GPS signal. With an aether, were you assume that light has the same speed in all directions, you can measure the one-way-speed. This is not interesting, however, as you have assumed that it is constant anyway.
  4. Feb 5, 2013 #3
    Thank you!
    In the context of the old Michelson-Morley experiment the speed of light in aether is always constant. It is the velocity of the observer which should account for the different measurements of the speed of light we should measure and we did not! (as far as the two- way speed of light is concerned)
    To clarify things here i don't care for aether! Call it absolute space if you like....
    Furthermore, this experiment never refuted the existence of an ether. It accounted for an isotropic speed of light, by introducing the γ-contraction-factor for longitudinal lengths and time. See Arthur Eddington.
    Anyway, since in my thought experiment i move tranversely wrt to absolute space it does not mind what the speed of light is! To synchronize two different clocks it suffices to know that the relative speed of light is the same for both clocks, which by symmetry holds true in my case.
    And, when i rotate the rod i do it very slowly at almost 0 speed, which does not alter the flow of time of the clocks.
  5. Feb 5, 2013 #4


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    Right. But if you have to assume that to perform your experiment at all, there is no point in measuring this speed in different directions. No theory predicts a deviation.
  6. Feb 5, 2013 #5
    I am not assuming anything...
    I was just pondering on how an ε=1/2 can lead to a theory of such inner beauty...
  7. Feb 6, 2013 #6
    On re-reading my statement, i realized i have made a mistake.
    I said that the rod is moving transversely with respect to aether. This is wrong. I should have said that the rod has a velocity directly perpendicular to its longitudinal axis of direction with respect to the static aether.
    The rest of the argument holds valid.
    Let's assume that I am an early 20th century physicist who believes that the speed of light in luminiferous motionless aether is constant.
    I have also verified by experiment that rods contract a γ-Lorentz factor √(1-v2) when they move with velocity v longitudinal to their axis and clocks delay the same amount independent of velocity direction.
    Also, forget about earth's rotational velocity, its speed around the sun, the speed of the galaxy, and expansion of the universe. All these will amount to a resultant velocity at an angle θ say to the longitudinal axis of my rod. By slowly turning the axis of my rod, I can ensure that at some point this resultant velocity will be directly perpendicular to the longitudinal axis of my rod. How can I be sure of this?
    By establishing my position exactly in the middle of the rod, and sending two wrist-watches ( both at same speed) - lol - at both ends of the rod and recording the times they read on their arrival. I shall postulate by a simple symmetry argument that if their respective reading times on arrival are identical, then the resultant relative velocity of my rod with respect to its longitudinal axis is directly perpendicular to it!
    This postulate is in accordance even with SR (were not true!) that is on a static line in the horizontal direction in Minkowskian-spacetime, two identical clocks placed in its middle will arrive on the same time at its ends, were they to move with identical speeds.
    I will synchronize, the clocks at both ends of the rod with the same identical reading of time of my moving watches, if i ever ever arrive at such a preposterous measurement and i will remain clinged to the idea that this is the only possible true synchronization!!
    Now, i very slowly rotate my rod. Length-contraction? Irrelevant! Who cares? By slow rotation, I have ensured that the rate of flow of time has not changed in any of the clocks.
    But at what cost? Now, I am having an √(1-v2) length rod with all the clocks synchronized to it at the same time in the moving frame of the rod with respect to the static aether!
    The two-way speed of light is always the same, but the one-way speed... oh no!
  8. Feb 6, 2013 #7


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    The way I look at it is this:

    There can be only one clock synchronization for which Newton's laws work exactly.

    Consider two objects of equal mass moving at the same velocity in opposite directions and coming to a stop. Assume that this happens with a fair, isotropic clock synchronization scheme as a starting point for the analysis.

    Now, lets look at some clock synchronization other than this "fair" one, to see what happens - to find the effect of an "unfair" or "non-isotropic" clock synchronization.

    In that case, the objects still come to a stop - the results of the experiment do not depend on how we syncrhonize our clocks. However, the measured velocities are no longer numerically equal, due to what one might call "clock synchronziation error" - the result of using something other than the isotropic clock synchronization.

    So there is a unique clock synchronization in which equal masses moving at the same velocity in opposite directions stop.

    This is theoretically sufficient to define what we want, but AFAIK it's hard to test experimentally.

    What Einstein's postulate says is that this fair, isotropic, clock synch (which is what we want) can be achieved using the principle that the speed of light is constant in any frame rather than the tedious method of colliding equal masses and making sure they come to a complete stop.
  9. Feb 6, 2013 #8


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    Ahh--the slow transport of clocks--the myth that it is better than Einstein's synchronization convention. In fact, it is no different. Let me show you why.

    First off, let me say that when you want to adopt a state of absolute aether, it is no different than selecting an Inertial Reference Frame (IRF) so you don't have to feel apologetic. But the good news is that you can use all of Einstein's Special Relativity to analyze what would happen in a state of absolute aether.

    You have stated many equivalent things in this thread, including that you can synchronize two remote clocks with a signal that originates at a location equidistant from both clocks or that you can use this technique to confirm that two such clocks are in sync, including two clocks (or watches) that were slowly transported either from the end points of a rotating rod or from the midpoint of the rod to its extremities.

    These are all different expressions of the same theme, slow transport of clocks. So in order to not have to analyze each one of them separately, I'm going to analyze just one scenario where we have three co-located clocks and then we slowly move two of them in opposite directions by the same distance and then we will use a signal sent from the stationary clock to confirm that they continue to remain in sync.

    I'm going to use a distance measure of feet and I'm going to define the speed of light to be one foot per nanosecond. We're going to move the two clocks at a speed of 0.1c for 100 nanoseconds according to their own Proper Time which will move them about 10.05 feet. Just before they get there, we're going to send signals from the stationary clock and show that they arrive at the same time according to the times on the two slowly moved clocks. Here's a spacetime diagram to illustrate this in an IRF in which the clocks started out at rest:


    The dots on each worldline represent Proper Time one-nanosecond ticks of each clock. They all started out at the origin of the frame set to zero so we see them 100 nanoseconds later where the last dot appears for each of them. We have to send the signal shown in green coming from the stationary clock at about 90.5 nanoseconds in order for them to arrive exactly when the two slowly moved clocks come to rest with readings of 100 nanoseconds on them.

    Now we look at the same thing in an IRF moving at -0.6c with respect to the first one:


    Here we see that even though the two slowly moved clocks receive the signals sent from the stationary clock at the same Proper Times for both of them, they are not in sync.

    The point of this exercise is to illustrate that it is impossible to determine by light signals or by slowly moving clocks any sense of absolute synchronization. They will both agree but still not indicate any synchronization with respect to a state of absolute aether.

    Attached Files:

    Last edited: Feb 6, 2013
  10. Feb 6, 2013 #9
    Ok, ghwellsjr, thank you!

    I have been convinced! And by joining the two end-points of the red and black lines in the second diagram, we drop back to the rest frame of the moving rod from which we emitted the slow clocks!

    Another question is: Is the Reichenbach assertion that the choice of ε=1/2 in clock synchronization is arbitrary, correct or is it a fallacy?
  11. Feb 7, 2013 #10


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    You're welcome.
    I'm not familiar with Reichenbach or his work but it sounds very much like Einstein pointing out that we can apply any synchronization we want and so he divides the roundtrip time for a light signal in half to define the time on the remote clock.
  12. Feb 7, 2013 #11
    First of all, thanks to everyone into this forum.

    The reason that i've made such a foolish mistake, to think that we can find a universally transverse moving frame in aether to synchronize our clocks within, and question the principles of SR, is because i was thinking of a rotating rod and its associated flat metric space-time.

    I am familiar with the concept of treating the helix described by any of its points as a geodesic so-to-speak, doing flat space-time tensor diff. geometry to trace how a light beam will bend to trace its course, and in the enlightment that an helix is not a true geodesic in flat spacetime, starting to think what kind of curved spacetime could create such an helix to take away all the fields creating centrifugal forces in it... hence GR.

    I think it was the rotating rod first, this and the writings of Ernst Mach that probed Einstein to GR. Anyway...

    Common knowledge has it in SR, that if we ever find ourselves in an accelerating frame ( when a rod rotates ) and suddenly the acceleration stops, then we re-synchronize our clocks to account for the discrepancies in the one-way speed of light.

    And yet, the upsetting fact remains, I was thinking overnight and could not sleep, What if there is an "aether"? What if, by accident, we hit upon a frame where a rod's velocity is perpendicular to its axis as we move into "aether" and synchronize our clocks there, we will never be sure that we have found one, as ghwellsjr very wisely demonstrated to me, but what if? And, if we rotate our rod very slowly, nothing changes in our clock-readings. Only length-contraction occurs.

    SR is a self-coherent theory, it contains no contradictions, we can't use its axioms to prove that it is wrong, but this does not prove or disprove anything....

    My foolish mistake manifests that we can't say: Ok, let's suppose ε=1/2, this leads to Lorentz-transformations, the Minkowskian pseudo-Euclidean spacetime, the metrix dτ^2=ds^2-dt^2 --- let's see if we can be led to a reduction in absurdum using its axioms... no!

    I haven't done the math yet, but another space-time with a different metric should be invented for cases other than ε=1/2. This would lead to another theory, taking none however from the validity of SR, it would be just another theory...

    I have nothing more to say upon this matter, if anyone could shed more light into this problem, you are most welcome.
  13. Feb 7, 2013 #12


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    It wouldn't even be another theory, just a new coordinate system (a non-inertial frame) in the existing theory. It could be handled the same way any non-inertial coordinate system is handled..
  14. Feb 7, 2013 #13
    Obviously, you are thinking how the rod contracts when it rotates, and you are still applying the axioms of SR with ε=1/2 when it does. Still, that's not a contradiction!

    With another ε some other contraction could be valid, as well. I do not know...
  15. Feb 7, 2013 #14


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    No, I am most defnitely not thinking that.

    Yes, the transformations between different non-inertial frames defined by ε≠1/2 would not be the same as the transforms between different inertial frames defined by ε=1/2. That would mean that the expressions for length contraction, time dilation, and relativity of simultaneity would be different also.

    Nevertheless, they are just a different set of coordinates, like any other non-inertial set of coordinates, and don't lead to any new experimental predictions.
  16. Feb 7, 2013 #15
    I am not certain as to the validity that they should be a non-inertial set of coordinates, i do not even now what kind of manifold we should be using to lead us to the same experimental predictions, namely same two-way light speed, same length-contraction and same time-dilation, take aside the one-way speed of light and simultaneity which must be different.
  17. Feb 7, 2013 #16


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    Obviously they are non inertial. The one way speed of light is c in inertial frames, by the second postulate.

    You would use a flat manifold.

    Length contraction and time dilation are frame variant coordinate artifacts, they are not experimental predictions. Experimental predictions are all invariants.
  18. Feb 7, 2013 #17
    Sorry, I was refering only in the sense of the Michelson-Morley experiment and the other experiment with the longer arm, i always forget its name, that established time-dilation.

    I am trying to forget everything about SR, to see what happens with other ε's.

    But, i don't have to argue about anything, about SR or its postulates.
  19. Feb 7, 2013 #18


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    It isn't that difficult. Just write down the transform between an inertial frame and a non-inertial frame, and use that to derive the metric in the non-inertial frame. Once you have the metric, then everything else follows the normal rules.
  20. Feb 7, 2013 #19
    But, how can i do this since i haven't defined any flat space-time yet?

    So far, all experiments have convinced me of the following 3 facts:

    1. The invariance of the 2-way speed of light.

    2. The existence of a Lorentz factor γ, for lengths and times in moving frames.

    3. SR is an ε-dependent theory ( with ε=1/2 ) that makes accurate predictions.

    If some-one has proven that ε=1/2 is the only possible value, because there can be no-other flat spacetime with properties 1 and 2, other than this with ε=1/2 ( the space-time of SR ), I would be very glad to accept it.
  21. Feb 7, 2013 #20


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    But you don't need any of that to work out the transform. Just look at how ε is defined and you should be able to relate a coordinate system (t,x) in which ε=½ to another coordinate system (T,X) with arbitrary ε and get
    cT &= ct + (2\epsilon - 1) x \\
    X &= x
    \end{align}[/tex]You can then work out what [itex]c^2 dt^2 - dx^2 [/itex] is in terms of dT and dX.
  22. Feb 7, 2013 #21


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    So define one. All it takes is a statement like "in the absence of gravity" or "neglecting gravity" or "far from any significant gravitational source".

    I don't see what your remaining concern is. Take the transform posted by DrGreg, calculate the metric, and there you have a complete description of the physics.
  23. Feb 7, 2013 #22
    Hi I have a question.
    Could you explain what you meant when you said the constancy of c wrt inertial frames was natural and not simply a convention?
  24. Feb 8, 2013 #23


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    Hi Austin0, I am not sure who you are talking to, but I think that you must be referring to some other thread. Nobody has said anything like that in this one.
  25. Mar 19, 2013 #24
    Ok, I'll go for DrGreg's transformations to find the null rays in the deformed frame (X,T):

    This can be done both ways: either by setting x=t and working out: X=something*T, or by setting the new interval ds=0 in (X,T) coordinates and thus arriving at: dX=something*dT, both expressions being equivalent.

    I have one last stupid question before putting this thread to rest, and I must ask it:

    What do I take the new speed of light to be? If i remember well from my calculations, there is an 4ε(ε-1) coefficient in the dX^2 factor in the new ds. Do I measure the new speed of light as: dX/dT, or do I take it: √(4ε(ε-1))dX/dT ?

    I am almost certain that this must be: √(4ε(ε-1))dX/dT, but am I right?

    Again, many thanks to DrGreg he was most helpful.
  26. Mar 20, 2013 #25
    I think any question without LATEX is being frowned upon. I'll ask properly this time.

    I will go for DrGreg's tranforms with c=1 and evaluate the new speed of light in the deformed frame.

    On setting x=t one gets: X = [itex]\frac{1}{2ε}[/itex]T or upon evaluating the new ds=[itex]\sqrt{4ε(1-ε)dX^{2}-dT^{2}+2(2ε-1)dXdT}[/itex] = 0 [itex]\Rightarrow[/itex] dX = [itex]\frac{1}{2ε}[/itex]dT, both expressions being equivalent.

    How do I measure the new speed of light?

    Is it [itex]\frac{dX}{dT}[/itex] = [itex]\frac{1}{2ε}[/itex], or is it: [itex]\frac{\sqrt{4ε(1-ε)}dX}{dT}[/itex] = [itex]\sqrt{\frac{1-ε}{ε}}[/itex] ?
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