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Simultaneity: disagreement, in the same reference frame

  1. Oct 2, 2015 #1

    DAC

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    Hello PF.
    There are two lightning flashes , and two stationary observers. Observer 1 who is equidistant from the flashes, sees them as simultaneous
    Observer 2 stands one step to the side of observer 1, and is not equidistant. He sees the flashes as separate, the light having unequal distances to travel.
    I thought the relativity of simultaneity resulted from different reference frames.
    Thanks.
     
  2. jcsd
  3. Oct 2, 2015 #2

    Dale

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    Both account for the finite speed of light, so both agree that the flashes were simultaneous
     
  4. Oct 2, 2015 #3

    DAC

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    If there was a moving train observer would he account for the finite speed of light and conclude simultaneity?
    Thanks.
     
  5. Oct 3, 2015 #4

    Orodruin

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    No. This is the entire point of relativity of simultaneity. It is not related to when different observers actually see the flashes, it is about when the events actually occured in their reference frames.
     
  6. Oct 3, 2015 #5

    DAC

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    Could you give me an example? Thanks.
     
  7. Oct 3, 2015 #6

    Nugatory

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    Suppose on January 1 2016 you look up in the sky and see a star explode. The star was three light-years distant. Then on January 1 2018 you're looking up at the sky and you see another star explode. This one was five light-years distant.

    Did both stars explode at the same time, and if so, when was it? The answers are "yes" and "January 1 2013".

    Einstein's train thought experiment shows that an observer who is moving relative to me will answer this question "no, they did not both explode at the same time". That's relativity of simultaneity at work.
     
  8. Oct 3, 2015 #7

    DAC

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    But the observers are not moving, but still disagree on simultaneity
     
  9. Oct 3, 2015 #8

    Orodruin

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    No. If they are not moving relative to each other they agree on simultaneity. Consider an observer two lightyears closer to the star we saw exploding first. This observer would see the stars explode in the inverse order but, taking the travel time of light into account, still conclude that they exploded simultaneously. This changes when the observers are movingrelative to each other.
     
  10. Oct 3, 2015 #9

    Nugatory

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    They will not disagree if they are at rest relative to one another.

    You on the earth were looking up at the sky on January 1 2016 and saw the first star explode; and then you saw the second star explode on January 1 2018. You allowed for the light travel time and realized that both stars exploded at the same time, January 1 2013.

    Now, suppose that I am on a spaceship one light-year away from earth and closer to both stars, and at rest relative to the earth. I'm watching the first star through a telescope and I see it explode on January 1 2015. Then I see the second star explode on January 1 2017. Now, bearing in mind that I am two light-years away from the first star and four light-years away from the second, what do I conclude about whether the two stars exploded at the same time?

    I conclude that both stars exploded at the same time, January 1 2013. That agrees with you back on earth.
     
  11. Oct 3, 2015 #10

    Janus

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    Here's an example dealing with two clocks. Here we will just 1 light flash to start two clocks. The flash originates at the midpoint between the clocks and starts each clock when the light strikes it.
    This is what happens according to someone at rest with respect to the clocks.
    synch1.gif

    Now it doesn't matter whether where you are with respect to these clocks.The light travels outward as a circular front from the point of emission and reaches both clocks at the same moment. As long as you are at rest with respect to them, you will agree that the clocks start at the same time and then keep the same time afterward.

    Now here is the same thing as according to someone moving with respect to the Clocks ( or to someone which considers the clocks as moving with respect to themselves, it makes no difference. Keep in mind we are talking about the same clocks and the same flash of light, the only difference is the relative motion between the clocks and the observer.
    synch2.gif

    The light is emitted from the midpoint between the clocks, and due to the invariant nature of the speed of light (it must have a constant value with respect to the observer.) it spreads out at at the same speed in all directions forming a circular front. However, the clocks are seen as moving from left to right and the left clock meets the light before the right clock does. Thus the left clock start running first and then the right clock. After which they run at the same speed, but out of sync. Anyone who has the same relative motion with respect to the clocks will conclude the same thing.

    So what you end up with is two frames of reference in motion with respect to each other, one the says the clocks read the same time, and one that says that they do not.
     
  12. Oct 4, 2015 #11
    And the important thing to remember (because it is counterintuitive) is that even though the clocks cannot be Einstein synchronized in both frames, no observer is entitled to claim which frame's clocks are actually correctly synchronized.
     
  13. Oct 4, 2015 #12

    http://www.popsci.com/science/article/2010-09/superaccurate-clocks-prove-your-head-older-your-feet
     
    Last edited: Oct 4, 2015
  14. Oct 5, 2015 #13

    DAC

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    Thanks PF for the replies. But, if the non equidistant second embankment observer, sees the flashes as not simultaneous, but calculates they are, why can't the train observer do the same.
    Thanks.
     
  15. Oct 5, 2015 #14

    Orodruin

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    Because the events are not simultaneous in his frame. If he accounts for travel time, he will conclude that the events were not simultaneous. This is the very essence of the relativity of simultaneity.
     
  16. Oct 5, 2015 #15
    Maybe you mean that that observer receives the flashes at different times, and based on that information he calculates that the flashes were emitted at the same time.
    It may be much clearer if you make a calculation example with numbers, and tell us what you find.
    For that you only need to use the following formula: travel time = distance / speed.
     
  17. Oct 5, 2015 #16

    Janus

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    Because the train observer is also equidistant by his measure. The embankment observer determines that the strikes occurred simultaneously because he is at the midpoint between the points of the embankment where the strikes occurred and he sees them at the same moment. The train observer determines that they did not occur simultaneously because he is midway between the points on the train where the strikes hit and he sees them at different moments.

    Other observers on the train would agree with this. Take for example an observer on the train that is passing the embankment observer at the moment the light from the strikes reaches that point. He, like the embankment observer, will see the light flashes at the same time. He is, at that moment halfway between the points on the embankment where the strikes hit. From the embankment, it looks like this:

    train1.gif
    The strikes occur simultaneously and the light arrives at both observers at the same place and moment.

    If we switch to the train car's frame, the light still arrives at both observers at the same moment, and both observers are halfway between the points on the embankment where the lightning strikes hit. But this means that the train observer was closer to one strike than he was to the other when they occurred. Since the light from both strikes has to travel at the same speed relative to him, one strike has to occur before the other in order for the light from the strikes to reach him at the same time.

    train2.gif
     
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