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B Simultaneity question

  1. Oct 25, 2016 #1
    I am reminded of Shroedinger's Cat that is only "probably alive or dead" because it is connected to a lethal device that kills if and only if a sub-atomic particle travels to a certain position or is emitted at a certain time, which quantum events have only a probability of occurring.
    I'm an English major, so let me speak plainly:
    I set things up so that two rays of light hit a balloon that explodes if and only if they hit it simultaneously--one ray before the other does not do the trick. I test the apparatus over and over. It works. No problem there.
    An observer passing by at a substantial fraction of the speed of light will not see any balloons exploding because the light rays do not reach the balloons simulaneously--or else he does see the balloons explode, since after all they do, and concludes that just one light ray is sufficient to explode a balloon in contravention to the laws of physics.
  2. jcsd
  3. Oct 25, 2016 #2


    Staff: Mentor

    That is a quantum mechanical scenario and has nothing to do with the topic of this thread.

    "Simultaneously" is an incorrect and misleading term here, because the two light rays hit the balloon at a single event--a single point in spacetime. So there is no simultaneity convention required to know whether the balloon explodes: it's purely a question of whether three curves in spacetime--the two light rays' worldlines and the worldline of the balloon--intersect at a single point. You have stipulated that they do, so the balloon explodes.

    Incorrect--you have misled yourself by using the term "simultaneously" incorrectly (see above). The two light rays still hit the balloon at the same event; that is an invariant, the same for all observers. (Since, as above, it's a matter of three curves intersecting at a point, it's just geometry, and you can't change geometry by changing coordinates.) What is different for the observer passing by is that the rays are not emitted at the same time. You didn't include the emission of the light rays in your analysis; if you do, you will see that if the light rays are emitted simultaneously in the original frame (here "simultaneously" is an appropriate term because we are talking about two different events, different points in spacetime, and asking whether they happen at the same time), they will not be emitted simultaneously in the new frame (that of the observer passing by).
  4. Oct 25, 2016 #3
    Actually I was imagining the case where the two rays of light were emitted in opposite directions at the same time at the same place where the balloon is. They are reflected off equidistant mirrors and arrive back at the same place where the balloon is at the same time, and the balloon explodes.
    Someone else observing this in translation will observe the balloon exploding, I gather from your analysis, but then must conclude that the outbound and inbound rays of light are travelling at different speeds, in contravention of that law of physics.
  5. Oct 25, 2016 #4


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    Staff: Mentor

    In this case, the moving observer will find that the two flashes of light arrive at the mirrors at different times. The distance covered by one flash of light on its outbound leg is the distance covered by the other flash on its inbound leg and vice versa. The total distance (outbound to the mirror plus inbound from the mirror) traveled by both flashes is the same, and the time between the single emission event at the center and the single explosion event at the center is equal to that distance divided by c for all observers.
  6. Oct 25, 2016 #5


    Staff: Mentor

    This is not correct. You should work through the math on this.

    What happens is that the reflection events are not simultaneous. Light still travels at c.
  7. Oct 27, 2016 #6
    Try imagining two trains passing each other in outer space. Lightning - or lets say a gama ray burst hits two points in space such that one observer on one train located between the two "hits" sees them to be simultaneous. Then the other sees them as not. If there is a second set of hits that the other observer now sees as simultaneous then the first will not see them as simultaneous. So they are simultaneous relative to one frame but not another.
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