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Relativity of Simultaneity

  1. Apr 27, 2010 #1
    As Einstein pointed out, simultaneity is relative to the reference frame one is observing from. His classic train example describes two concurrent flashes of light viewed as simultaneous to a stationary observer half way between the two sources yet not viewed as simultaneous to an observer moving towards one flash and away from the flash behind also situated at the midway point. If a buzzer was to go off when two events were simultaneous but not if not simultaneous, does this mean that the buzzer would go off in one time frame and not another (we are talking about the same two distinct yet concurrent events.) This was brought up somewhere in the past by JesseM.

    If one can observe two distinct events and make them simultaneous by adjusting the frame of reference, will the buzzer go off because we can find one frame of reference in which there is simultaneity?

    Can any two non-concurrent events be made concurrent or simultaneous by adjusting the frame of reference? (We are restricted to keeping within the light cone and staying below the speed of light.)

    Of course, if two events are too far apart that that the speed required to get from the first to the second is greater than the speed of light, they can never be simultaneous, but short of that, can any two non simultaneous events if within striking distance of the speed of light or less be made to appear simultaneous?
     
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  3. Apr 27, 2010 #2

    JesseM

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    Simultaneity is an abstract coordinate issue, a buzzer is a real physical device that is set off by local processes. All frames will agree on the outcome of these local events so however the buzzer is constructed, they'll all agree on whether it will go off (the point I brought up was just that all frames must agree on local events, if they disagreed it could lead to incompatible predictions about things like whether a bomb would explode). For example, if a buzzer is placed midway between two ends of a train, and it's designed so that it goes off if it receives light from both sides at the same moment (both light rays coinciding at the local position of the buzzer), then we can say it'll only go off it receives light from flashes on either end which were simultaneous in the train frame...but all other frames agree on whether light from two flashes reaches the buzzer at the same time, even if they disagree on whether the flashes themselves were simultaneous (in other frames the buzzer is moving towards the position of one flash and away from the position of the other, which explains how light from both flashes can strike it simultaneously even if the flashes happened at different times).
    Only if the two events have a spacelike separation can they be simultaneous in some inertial frame (i.e. neither event occurred in the past or future light cone of the other event, and the invariant interval ds^2 = dx^2 - c^2*dt^2 between them has a positive value for ds^2 rather than a negative value). But if they do have a spacelike separation, you can always find an inertial frame where they are simultaneous.
     
  4. Apr 27, 2010 #3
    Believe it or not, JesseM, I kind of understand you and I understand that timelike means that ds2 = dx2 - c2dt2 is negative (or d[tex]\tau[/tex]2 = c2dt2 - dx2 is positive.) I apologize for the position of the [tex]\tau[/tex]. I can't get it right.

    To wit, if one has a rod, one can measure both ends simultaneously and it is impossible to get from the beginning of the rod to the end of the rod in a time less than what light would take unless it were "spacelike." If one measures the end of the rod slightly later than the beginning, but still not enough time for light to make it from beginning to end, the this would still be spacelike and one could twist the frames and get the two events (measurements) at the same instant from one frame. But you are saying that if the measurement of the end of the rod occurred so far later than the measurement of the beginning of the rod that light would be able to get from the beginning to the end, then this would be timelike and one could not find a frame of reference to force these events to appear simultaneous.

    Do I have that right? Also, any clues on how to fix that tau [tex]\tau[/tex] problem? I have not inadvertently written a superscript instruction.
     
    Last edited: Apr 27, 2010
  5. Apr 27, 2010 #4
    Second question -

    I am used to, from calculus, that ds or dx or dt means a very small [tex]\Delta[/tex]s, [tex]\Delta[/tex]x, or [tex]\Delta[/tex]t but, can one use for dx, say, an actual x2 - x1 so that this [tex]\Delta[/tex] is not infinitesmal?
     
  6. Apr 27, 2010 #5
    You've got the rod analogy as far as I can see, but maybe the twin paradox is better for this, as it has a bit more to chew on than impossible rods (impossible for some of the reasons you've described).
     
  7. Apr 27, 2010 #6

    JesseM

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    Yes, that's almost exactly right. The "almost" is just because if there is time for a signal moving at the speed of light or slower to get from the event of one measurement to the event of the other, the separation could be timelike (if the signal could move slower than light and still make it from one measurement to the other) or it could be lightlike (if the signal would have to move at exactly the speed of light to get from one measurement to the other). "Lightlike" of course corresponds to ds^2 = 0.
    Yeah, there's a problem where sometimes if you put individual letters in Latex they appear offset from the rest of the text. I'd recomment just putting the whole equation in Latex if you want to avoid this, like [tex]d\tau^2 = dt^2 - (1/c^2)dx^2[/tex] (and the equation for tau it should be written like that, with a 1/c^2 factor next to dx^2 rather than a c^2 next to the dt^2 as with the equation for ds, since the units of proper time should be a time rather than a distance)
     
  8. Apr 27, 2010 #7
    Many thanks, I shall try it.
     
    Last edited: Apr 28, 2010
  9. Apr 28, 2010 #8
    Back to Frame-Dragger:

    If I do place a rod of certain length l0 in the moving frame S' which moves at a velocity of v to the right with respect to the stationary frame S and measure it in the stationary frame S at x1 and x2 simultaneously at t1 = t2, we get the length l0 = x2 - x1 according to the length contraction formula l0 = l/gamma. However, when we solve the simultaneous Lorentz equations for t1' and t2' in in S' which are the corresponding times for the t1 = t2 in S we note that t1' does not = t2' which means we measured the front and back of the rod at different times in S'.

    Since the rod was stationary in S' it does not make a difference for l = x2' - x1' as both x2' and x1' are fixed in S' so that l will always be the same no matter when you measure the rod even if you measure the front and back at different times in that moving frame.

    Is that logic correct? If so, this would justify the equation l0 = l/gamma

    l = length in moving frame S'
    l0 = length in resting frame S
     
    Last edited: Apr 28, 2010
  10. Apr 28, 2010 #9

    jtbell

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    Yes, exactly so. Going the other way, if we measure the ends of the rod simultaneously in S' at t2' = t1', we similarly find that in S, t2 ≠ t1. This matters because the rod is not stationary in S.
     
  11. Apr 28, 2010 #10
    To jtbell. We started this on a different thread "problems with Lorentz Transformations." In the above case where we had the rod of length l and set t1' = t2' in S' we would have t1 [tex]\neq[/tex] t2 but we would have an x1 and x2 in S in which the beginning and end of the rod were measured at different times. At this point, I don't know how I would calculate that to a t1 = t2 in S where it would be of some use. The only thing I can think of is to use the "proper length" formula s2 = (x2 - x1)2 - (t2 - t1)2/c2 and then take the square root of that. I don't know if that would work, any thoughts?
     
    Last edited: Apr 28, 2010
  12. Apr 28, 2010 #11

    JesseM

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    You have to use some physical understanding of the situation--in this case the rod is at rest in S, which naturally means that the position x2 of one end at t2 will also be the position of the same end at a different time t1.

    More generally, it may be helpful to figure out the position of each end as a function of time in each frame. If the left end of the rod is at position x=0 at time t=0 in frame S, then the left end is also at position x'=0 at t'=0 in frame S', since the origins of the two frames coincide according to the Lorentz transformation. Since the left end is at rest in S and moving in the +x' direction at speed v in S', the position as a function of time in S must be x(t)=0 and the position as a function of time in S' must be x'(t')=vt'. And since the rod is of length l0 in S and length l = l0/gamma in S', that means the position of the right end as a function of time must be x(t)=l0 in S and x'(t') = vt' + l in S'. You can use these functions to find the positions of both ends at any time in a given frame, and if you have two events on either end at the same time in one frame, you can find the coordinates of these events in the other frame and verify that they both lie on the worldlines of that end in the other frame using the equation for position as a function of time in that frame.
     
  13. Apr 28, 2010 #12

    jtbell

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    Did you see this post, in which I gave you a hint as to how to resolve this problem?

    https://www.physicsforums.com/showpost.php?p=2681947&postcount=36
     
  14. Apr 28, 2010 #13
    To jtbell

    Original post

    Your response:

    I am thinking that by "subtracting out" 9 years * 0.6c = 5.4 lt-yrs we would get... 15 - 5.4 = 9.6 lt-years - which agrees with the same calculation with [itex]t_1 = t_2 = 0[/itex]

    There must be a way to do it with proper length.
     
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