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Simultaneity and a thought experiment

  1. Jul 5, 2008 #1
    confused about this thought. Object A and B are d distance apart, as measured by an object exactly in the middle of them, object C. A and B are moving towards each other at c/2, again measured by C. According to C's frame of reference, all three objects will collide at the exact same moment, causing whatever effects their collision causes. But from A's frame of reference, C is moving towards A at c/2, and B is moving towards A at 0.8c. So from A's reference, A hits C first, then it hits B. How is this reconcilable? If they were, for example, pool balls, how can we figure out the velocity of C after the collision?
     
  2. jcsd
  3. Jul 5, 2008 #2

    JesseM

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    If you do a Lorentz transform you'll see that at any given moment in A's frame, the distance from A to B is not twice the distance from A to C, and all three still collide at the same moment (all reference frames must agree on local events like collisions). For example, at the moment that C is 0.5 light years away in A's frame, B is 0.8 light years away, so they both reach A 1 year later.
     
  4. Jul 5, 2008 #3

    Janus

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    From all references, A, B & C collide at the same instant. What they will not agree to is how far apart they are at any other time. For instance, When A determines that he is a distance of d from C, he will determine that B is only 0.6d from C.
     
  5. Jul 5, 2008 #4
    I'm confused by this, what's the equation being used here?

    so say that B was radiating energy and the amount of energy that hits C changes its color; the intensity of that color change depends on the amount of time of exposure and distance from C. If the average distance between B and C as calculated from A is different from the average distance between B and C as calculated from C...or is it?
     
  6. Jul 5, 2008 #5

    Dale

    Staff: Mentor

    The equation is the Lorentz transform.
     
  7. Jul 6, 2008 #6
  8. Oct 13, 2010 #7
    Can the group humour me enough to to go through a long message? I have written out the Lorentz transformations here to clarify a symmetrical sort of problem that I have with this one. Let’s simplify this one by assuming length d is 1 light minute and c is measured in light minutes. The spacetime diagram from C’s point of view looks like this:

    | 2
    / | \
    / | \
    / | \
    -1 0 1
    A C B

    The collision event is (x,t) = (0,2) in C’s reference frame.
    A is moving toward C at speed ½. The Lorentz transformation matrix for this is:
    (1 ½)
    (½ 1 )
    So (xA, tA) = (0,2)(1 ½) = (1,2)
    (½ 1)

    B is moving toward C at speed -½. The Lorentz transformation matrix for this is:
    (1 -½)
    (-½ 1 )
    So (xB, tB) = (0,2)(1 -½) = (-1,2)
    (-½ 1)
    t =2 in all three cases so simultaneity is assured.
    Is this correct?

    The puzzle I have is what if you change it so A and B start at C and leave at speed ½ in opposite directions? Do they reach a distance of 1 light-minute simultaneously? Logic and symmetry would say Yes but Lorentz seems to say No and this is because x is not zero.

    From C’s point of view the event “A reaches 1 light-minute” has co-ords (-1,2)
    The Lorentz transformation matrix for this is: (1 -½)
    (-½ 1 )
    So (xA, tA) = (-1,2)(1 ½) = (0,1½)
    (½ 1)

    From C’s point of view the event “B reaches 1 light-minute” has co-ords (1,2)
    The Lorentz transformation matrix for this is: (1 -½)
    (-½ 1 )
    So (xB, tB) = (1,2)(1 ½) = (2,2½)
    (½ 1)

    tA is not equal to tB so the events are not simultaneous. This appears to be purely because the events are not simultaneous in space.
    Is it the case that events not simultaneous in space can never be simultaneous in time, as measured by a moving observer?
     
  9. Oct 13, 2010 #8

    JesseM

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    Your numbers aren't right because you haven't included the gamma factor in the Lorentz transformation matrix, and you should also be putting -1/2 in the matrix for the transformation from C to A, not +1/2. With v=0.5c, gamma=1/sqrt(1 - 0.5^2) = 1.1547, so the matrix for transformation from C to A would be:
    [tex]\begin{bmatrix}
    \gamma&-\beta \gamma\\
    -\beta \gamma&\gamma\\
    \end{bmatrix}[/tex]
    Or in this case:
    [tex]\begin{bmatrix}
    1.1547&-0.57735\\
    -0.57735&1.1547\\
    \end{bmatrix}[/tex]

    But personally I find it easier to remember this way of writing the transformation:

    xA = gamma*(x - vt)
    tA = gamma*(t - vx/c^2)

    So with x=0, t=2, v=0.5c and gamma=1.1547 we would have:

    xA = 1.1547*(-0.5*2) = -1.1547
    tA = 1.1547*(2) = 2.3094

    And if we want to transform to the frame where B is at rest, here v=-0.5c so we have:

    xB = 1.1547*(0.5*2) = 1.1547
    tB = 1.1547*(2) = 2.3094

    In what frame are you asking if they reach a distance of 1 light-minute "simultaneously"?
    It's not really meaningful to ask if a time in one frame and a time in a different frame are "simultaneous" (after all this depends on where you arbitrarily choose to place the origin of your coordinate systems), when physicists ask if a given pair of events are simultaneous they're always asking if they both have the same time-coordinate in some particular choice of frame.
     
  10. Oct 14, 2010 #9
    Dear JesseM,

    Thanks for replying so quickly and so thoughtfully. As you said, I forgot about gamma and was never sure if my signs were the right way round. But that didn't affect my main point - that the Lorentz transformation for time depends only on x, given that v,c and t are already determined. This seems to mean that two events at separate locations, as observed from a moving reference frame can never be seen as simultaneous. Is that correct? If so this baffles me.

    Imagine someone on a platform observing a train carriage passing at fixed speed, left to right. There are two people standing in the middle of the carriage, and simultaneously they walk at the same speed in opposite directions - one towards the front of the carriage and one towards the back. Logic and calculation of relative speeds tell me they should reach the ends of the carriage at the same time, as observed by someone in the carriage or by someone on the platform. But the Lorentz transformation will give different values for t and t'. Of course, if it was two light rays sent out from the center of the carriage one towards the front and one towards the back then they would not touch the ends simultaneously, but that's because c is constant for all observers. This is not the case for the walking speed of the people on the train.

    The puzzling thing is that the Lorentz transformation does not distinguish between events involving light beams moving at fixed speed c, and events involving ordinary speeds which add to the speed of the train, v. Yet surely the constancy of the speed of light is the whole reason for the failure of simultaneity.

    This is something I've been chewing over for a while. I'm glad to be able to share it with some-one.
     
  11. Oct 14, 2010 #10

    JesseM

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    There is one exception in the sense that if the axis between the two events in the first frame is orthogonal to the axis of motion of the second frame as seen in the first frame, then the events will still be simultaneous in the second frame as well. But otherwise, yes, if two events are simultaneous in the first frame they are non-simultaneous in the other frame. This is a necessary consequence of the fact that both frames must measure the speed of light to be the same in both directions--just imagine that the two events occurred at either end of a ruler at rest in the first frame, and that earlier a flash had been set off at the exact midpoint of the ruler, at just the right time so that the light from the flash would reach the ends of the ruler at the moment the two events occurred at either end. It's not hard to show that in the second frame where the ruler is moving, if the light from the flash still moves at c, it must reach the ends of the ruler at different times, since in this frame the ruler is moving, and one end of the ruler is moving towards the position where the flash occurred and the other is moving away from it. Likewise if we two events occur at either end of the moving ruler simultaneously in the second frame where the ruler is moving, then if we imagine each event is a flash that sends light towards the center of the ruler, we can see that an observer at rest relative to the ruler and sitting at its center will receive the light at different times, and thus is forced to conclude the events happened non-simultaneously if he wants to believe both light beams moved at c. See here and here if this isn't clear to you (also sections 8 and 9 of Einstein's book where he introduced the famous train/lightning thought experiment).
    Keep in mind that "relative speed" doesn't work the same way in relativity as in Newtonian physics. If both walk at speed v1 in the frame of the carriage, and the carriage is moving at speed v2 in the frame of the platform, in Newtonian physics the person walking toward the back of the carriage would have a speed of v2 - v1 in the platform frame while the person walking towards the front would have a speed of v2 + v1. But in relativity you would instead have to use the relativistic velocity addition formula, which would tell you that in the frame of the platform the person walking towards the back would have a speed of (v2 - v1)/(1 - v2*v1/c^2) while the person walking towards the front would have a speed of (v2 + v1)/(1 + v2*v1/c^2). Remember, each frame is defining "speed" in terms of distance/time as measured by a system of clocks and rulers at rest in that frame, but a ruler at rest in one frame will appear shrunk in a second frame, and a pair of synchronized clocks at rest in one frame will appear slowed-down and out-of-sync in another frame. So, conceptually it shouldn't be too surprising that the velocity addition formula (which tells you the velocity of an object in one frame if you know its velocity in a second frame and you know the velocity of the second frame in the first frame) would work differently in relativity than it does in Newtonian physics. Also note that the Newtonian formula would allow v2 + v1 to exceed c even if both v2 and v1 were smaller than c individually, while the relativistic formula ensures that as long as v2 is smaller than c and v1 is smaller than or equal to c, the velocity in the second frame will also be smaller than or equal to c.
    No, but the timing of the people reaching each end of the train is determined by looking at clocks at rest in the observer's frame that are right next to each end of the train at the moment the person reaches that end, and that pair of clocks is assumed to have been "synchronized" in the past using light-signals (the Einstein synchronization convention), with the assumption that light travels at c in both directions in the frame of the observer where the clocks are at rest. So, for reasons I gave above, it's natural that a pair of clocks that are synchronized in one frame (like two clocks at rest relative to the track, one positioned so it's next to the back end of the train at the moment the person reaches the back, and the other positioned so it's next to the front end of the train at the moment the person reaches the front) must be out-of-sync in a different frame (like the rest frame of the train).
    Yes, exactly. Again the key is that the time coordinate of arbitrary events is assumed to be assigned in a local way by synchronized clocks at rest in whatever frame you're using--if you have clocks at rest at different positions in your frame, you assign a time-coordinate to any event you see by looking at what time showed on the clock of yours that was right next to the event at the moment it occurred, that way you don't have to worry about signal delays between when the event occurred and when the light from it actually reaches your eyes. And all these clocks have been "synchronized" in your frame using the assumption that light travels at c in your frame, so you could make sure two clocks are synchronized by setting off a flash at the midpoint between them, and making sure both clocks read the same time at the moment the light hits them. In other frames this will result in all your clocks being out-of-sync for reasons I mentioned above (unless the axis between a pair of clocks is orthogonal to the direction the other frame is moving relative to yours).
     
  12. Oct 15, 2010 #11
    Thank you JesseM, I see that now. The problem with simultaneity in the reception of light signals is clear because of the constancy of the speed of light. The problem with simultaneity of slower-than-light events is not so clear at first but becomes evident when you realise the only way to be if events are simultaneous is to measure them with synchronised clocks, which you can't do outside an inertial reference frame. Thanks for taking the time to clear that up for me.
     
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