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B Special relativity; mixed up time and space due to motion

  1. Jun 19, 2017 at 7:55 PM #1
    I just want to be sure I understand this correctly:

    Suppose there's a long spaceship moving in uniform motion relative to me (an inertial frame). Now someone in the front* of the spaceship calls someone (on the phone) in the back of the spaceship.
    *(this long length of the ship is oriented along the relative motion; front and back are then as you would think.)
    So then special relativity says that if the spaceship is long enough and moving fast enough (rather, if the product is large enough) then I will see the person at the back answer the phone call before I see the person at the front make the call?
     
  2. jcsd
  3. Jun 19, 2017 at 8:32 PM #2
    I can not figure how you could see that happen.. If the signal is traveling through the telephone wire at lightspeed, then that signal will reach the back of the ship at the same time as the light from the person, traveling toward you (assume the ship is moving away). Then that light along with the answerer will travel to you. If the person could answer instantly, then you would "see" both events happen at the same time.
     
  4. Jun 19, 2017 at 8:33 PM #3
    Do you really want to deal with the finite amount of time that it takes for the phone signal to travel from the front to the back of the spaceship (in the rocket frame of reference), or would you be willing to settle for simultaneous lightning strikes at the front and rear of the rocket, in the rocket frame of reference?
     
  5. Jun 19, 2017 at 9:07 PM #4
    Ok let us ignore the time between calling and answering and just say that two events at the front and back occur simultaneously in the frame of the spaceship. Then I will see the event in the back occur first right?
     
  6. Jun 19, 2017 at 10:42 PM #5
    It you're standing half-way between the locations where the two strikes occur, then yes.
     
  7. Jun 19, 2017 at 11:01 PM #6
    And if I'm not? Sorry I am new to relativity and trying to make my understanding sound.
     
  8. Jun 20, 2017 at 12:24 AM #7

    Ibix

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    It depends what you mean by "see". If you mean it literally then what you see depends on where you stand.

    However, in special relativity we typically correct for the travel time of light. So if I receive light from my clock right next to me at 12.00 and eight minutes later I receive light from a solar flare starting on the Sun, I would regard those things as simultaneous. You do hear that referred to (sloppily) as "seeing them as simultaneous".

    It's the latter that usually gets used in relativity, and I'm guessing that's what you actually mean. So - I proceed on that assumption.

    Think of yourself at rest in a frame S, which uses coordinates x and t. The ship is at rest in a frame S', which uses coordinates x' and t' and is moving at speed v in the +x direction.

    If two things that happen simultaneously in S', the rocket frame, are the t' coordinates the same or different? Does the front of the rocket have a higher or lower x' coordinste than the back? What do the Lorentz transforms tell you about the t coordinates of the two events?
     
  9. Jun 20, 2017 at 12:53 AM #8

    Nugatory

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    Yes, and there will be other observers who would find that it's the other way around and the event at the front of the spaceship is the one that happened first. There is no onserver-independent way of saying that one came before the other. However...
    That time actually makes a very big difference. The telephone signal is traveling at a speed no greater than the speed of light, and when you do allow for the signal's travel time you will find that all observers everywhere will agree that the calling event happened before the answering event.
     
  10. Jun 20, 2017 at 3:50 AM #9
    Define "see."


    All observers who correct for the travel time of light will agree upon that, correct? But there could conceivably be observers who see (as in with their eyes, without correcting for the travel time of light) the phone answered first, could there not? EDIT- oh wait never mind. At best it would be simultaneous because as you just said the signal from one phone to the other travels at best at light speed too. Besides, there'd be a delay as the person answered the phone.

    EDIT-2: Wait. What if the observer and the two phones form a triangle, with the hypotenuse the leg from which the light from the calling phone would travel. In that case, could we not have a scenario where the time it takes the CALL signal to go to the other phone PLUS the time it takes the light signal from the answering phone to reach the observer be smaller than the time it takes for the light from the calling phone to travel along the hypotenuse of this triangle? All it would take is an extremely large hypotenuse of this triangle.
     
  11. Jun 20, 2017 at 4:12 AM #10
    t' is the same, x' is larger at the front, these with the transforms imply the event at the back happened sooner in 'my' frame (the event at the back occurs at smaller t than the front).

    Is this in accordance with your descriptions?

    I see. If we take Ibix's description of the coordinates then the time of each event in 'my' frame is given by t=ϒ(t'+(v/c^2)x') so then the time between calling and answering will be Δt=ϒ(Δt'+(v/c^2)Δx' where Δx' is the rest length of the ship and -Δt'≥Δx'/c is the signal time. Then Δt≥ϒΔx'(-1/c + v/c^2) but if v<c, then Δt<0 i.e. the front event (calling) happens before the back event (answering) in my frame as well, by virtue of this limitation on signal speed.
     
  12. Jun 20, 2017 at 4:53 AM #11

    Nugatory

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    No. If you try drawing a spacetime diagram that includes the light cones of the calling and answering events you'll see why. The answering event has to be in the future light cone of the calling events and all observer's worldlines must be timelike; that's enough to ensure that light from the answering event cannot reach any observer before light from the calling event.
    The distance along two sides of a triangle is always longer than the third side, so there can be no such triangle.
     
  13. Jun 20, 2017 at 8:27 AM #12
    There is an easier way to understand this. Rather than you being the sole observer in your frame of reference, imagine that there is a team of observers in your frame strung out along the direction of the rocket. Each member of your team carries a clock, and these clocks are all synchronized to your clock. One observer from your team is exactly at the location (in your frame) of the front of the rocket when the lightning strike occurs there, and another observer from your team is exactly at the location (in your frame) of the rear of the rocket when the lightning strikes there. Each of these two observers notes down on a piece of paper the time on his clock that the corresponding lightning strike occurred. Then they get together and compare notes. They find that the time (in your frame) that the lightning strike occurred at the rear of the rocket (as noted by the observer to the rear) was earlier than the time (in your frame) that the lightning strike occurred at the nose of the rocket (as noted by the observer toward the front). So, while the strikes occurred simultaneously in the frame of reference of the rocket, they did not occur simultaneously in your frame of reference.
     
  14. Jun 22, 2017 at 8:00 AM #13
    Triangle inequality?


    What if it isn't flat spacetime? Could we have a geometry where that law doesn't hold? Is this conceivable?
     
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