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Special Relativity, light down a rocket

  1. Mar 30, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A rocket of length ##L_o## flies with constant velocity ##v## (in frame S' relative to a frame S in the x direction). At time t=t'=0, the capsule on the top of the rocket passes the point ##P_o## in S. At this moment, a light signal is sent from the top of the rocket to the bottom.
    1)In the rest frame of the rocket, how long does it take the light signal to reach the end of the rocket?
    2)In the rest frame of the observer, S, at which time does the signal reach the end of the rocket?

    2. Relevant equations
    Lorentz transformations, proper length and proper time.

    3. The attempt at a solution
    1)The wording of the question at the beginning is confusing, but I take it that S is the rest frame of the observer, S' is the rest frame of the rocket and S' moves wrt S at velocity v. I also took ##L_o## to be the proper length of the rocket (i.e the value measured in S', the rest frame of the rocket). Are these correct interpretations?

    At ##t_o' = 0## the signal leaves the top of the rocket. At ##t_1'## it arrives at the bottom in frame S'. Speed of light same in all inertial frames (S' inertial since it moves with constant velocity v wrt another inertial frame) and ##L_o## the length of the rocket in S'. So ##t_1' = L_o/c## is the time taken in S'.

    2)In S, as the rocket passes, the observer would see the rocket contract so expect time for signal to reach bottom in S to be smaller than that in S'. Why is ##t_1 = \gamma t_1'## not valid here?
    Thanks.
     
  2. jcsd
  3. Mar 30, 2014 #2

    BruceW

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    yeah, for part 1) I agree. The wording is a bit unclear. But I think you have interpreted it the right way. and yes, that would mean ##L_0/c## is the time taken according to the rocket rest frame. For part 2), using the time dilation formula would not work (as you've guessed). To understand why, think about the spacetime coordinate difference in the two frames of reference. And in what (special) case does the time dilation formula apply? (it is not a general case). So, to answer the question, what is the equation you must use for the general case?
     
  4. Mar 30, 2014 #3

    CAF123

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    Hi BruceW,
    Consider S' first. Then there are two events of interest - the light leaving the top of the rocket (denote by ##(x_o', t_o')## and the light reaching the bottom (denote by ##(x_1', t_1')##)
    I think ##(x_o', t_o') = (P_{o,x}, 0)## and ##(x_1', t_1') = (P_{o,x} + L_o, L_o/c)##, where ##P_{o,x}## is the x component of ##P_o##. Is that right? I see your point and it is obvious that the rocket observer will notice that the light hitting the bottom and leaving the top are at two different locations within the rocket. I think it is correct to say that the time dilation formula is applicable when the two events of interest happen at the same spatial coordinate in the frame of reference.

    In S, ##(x_o, t_o) = (P_{o,x}, 0)## as before and ##(x_1, t_1) = (P_{o,x} + l, t_1)## where ##l## is the length of the rocket in S. Can this be found? I think I can use the Lorentz transformation ##\Delta x \equiv l = \gamma(\Delta x' + v \Delta t') = \gamma(L_o + v L_o/c)##?

    So $$t_1 - t_o = t_1 = \Delta t = \gamma ( \Delta t' + v\Delta x'/c^2) = \gamma (L_o/c + vL_o/c^2)$$
     
    Last edited: Mar 30, 2014
  5. Mar 30, 2014 #4
    The rear of the rocket in the S' frame is located at x' = -L0, and, as you said, the signal reaches this location at t' = L0/c. What does the inverse Lorentz Transform tell you about the coordinates of the event as reckoned from the S frame of reference?

    Chet
     
  6. Mar 30, 2014 #5

    CAF123

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    Hi Chestermiller,
    Do you mean to say the two events of interest are located at (Po,x,0) and (Po,x-Lo, Lo/c)? I think I made a sign error in my previous post.


    See my previous post but instead, correcting the sign error, I have ##\Delta x = \gamma (\Delta x' + v \Delta t') = \gamma(-L_o + v L_o/c)##. So is this the length of the rocket in the frame S?

    Similarly, the time interval in S for the two events is $$\Delta t = \gamma(\Delta t + v \Delta x'/c^2) = \gamma(L_o/c - v L_o/c^2).$$
     
  7. Mar 30, 2014 #6
    Yes. But, please factor out the L0/c, and do some cancellation of terms with those γ's.

    Chet
     
  8. Mar 30, 2014 #7

    CAF123

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    Yes, so $$\Delta t = \frac{L_o}{c} \sqrt{\frac{1-v/c}{1+v/c}}$$ which is always less than ##L_o/c## since v<c (I think this was to be expected, in my first post I noted that in S the rocket would be contracted and so in that frame the time taken for the light signal to reach the other end would be shorter, so I think the result makes sense.)

    I also mentioned in my first post that we had a proper length in S'. But the space time coordinates in each of the two frames are:

    S': light leaving top (Po,x, 0) ; light hitting bottom (Po,x-Lo, Lo/c)

    S: light leaving top (Po,x, 0) ; light hitting bottom (Po,x-l, t1), where ##l = \gamma(-L_o + vL_o/c)## and ##t_1 = \Delta t## given above.

    Neither events have a common spatial or temporal coordinate in common so all distances/ time measured in this process are said to not be proper. Is that correct? When it is said ##L_o## is the length of the rocket, is that meant to be it's rest length, assumed to be measured by an observer when the rocket is grounded?
     
  9. Mar 30, 2014 #8
    No. In an object's rest frame, the coordinates of its two ends don't change with time, so they don't have to be measured at the same time. L0 is the proper length of the rocket.

    Chet
     
  10. Mar 30, 2014 #9

    CAF123

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    Ok, so while the rocket is flying, in the rest frame put a light emitter in the middle of the rocket and send light pulses towards the top and bottom of the rocket. If there are clocks at the ends, they will both read a common time T, say and from that deduce the length.

    Are the space time coordinates of the emission of light at the top and the receiving of light at the bottom in S and S' given in my previous post correct? Thanks.
     
  11. Mar 30, 2014 #10
    I don't know why you would do it this way, instead of just using a tape measure. If you were measuring the width of your kitchen, would you use light signals and clocks? The rocket is at rest within its own rest frame.
    Yes. They are perfect.

    Chet
     
  12. Mar 31, 2014 #11

    CAF123

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    That would make sense, but in what circumstance would you use light signals and clocks?

    Is it also correct to say that ##\Delta t' = L_o/c## is a proper time since it was measured in the rest frame of the rocket? It seems intuitive to be the case, but ##\Delta x' \neq 0## when I look at the space time coordinates.

    Thanks.
     
  13. Mar 31, 2014 #12

    BruceW

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    You've got to think about what spacetime events you are talking about. I'm guessing that here, you are talking about (light leaves front of spaceship, time t1) and (light reaches back of spaceship, time t2). Clearly, this is not a proper time or proper length, as you say, because the spacetime difference is non-zero for both the time and space coordinates. To get a proper length or time, it is not enough to say that something was measured in the rest frame. It must be measured in the rest frame, and have one of either space or time difference be equal to zero.
     
  14. Mar 31, 2014 #13
    I respectfully beg to differ. In the rest frame of an object, you don't need to measure the coordinates of both ends of the object at the same time. This is because, in the object's rest frame, the coordinates of its ends don't change with time. So the proper length of the rocket is L0, since it is always at rest within its own rest frame.

    Chet
     
  15. Mar 31, 2014 #14
    If you were trying to calibrate the clocks
    ##\Delta t' = L_o/c## is the amount of proper time that elapses at each end of the rocket during the time interval that the signal travels from one end of the rocket to the other.
     
  16. Mar 31, 2014 #15

    BruceW

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    yeah, that's true. but if the spaceship is stretching or something, then we have the more general case where it's ends do change with time. Also, I feel it is more intuitive to think about proper length (between two spacetime events) as the spatial coordinate difference, in a frame where the time coordinate difference is zero.
     
  17. Mar 31, 2014 #16

    CAF123

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    Intuitively, it seems correct to say that ##\Delta t' = L_0/c## is a proper time (and you have suggested above that this indeed is the case), but more generally, is the proper time not defined as the time in a frame in which two events happen at the same spatial location? In which case, neither in S or S' are the spatial components the same for the events of interest.
    Thanks.
     
  18. Mar 31, 2014 #17
    I'm going to ask one of the relativity mavens in Homework Helpers to put their two cents in on this.

    Chet
     
    Last edited: Mar 31, 2014
  19. Mar 31, 2014 #18

    BruceW

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    ##\Delta t' = L_0/c## is the proper time that has elapsed for a clock, which is attached to the rocket (i.e. stationary in the rocket rest frame). for example, we could organise things so that the clock at the rear of the rocket starts ticking when the light is emitted at the front of the rocket (according to the rocket frame). And, we know that the light takes time ##\Delta t' = L_0/c## to reach the clock (according to the rocket frame). So for the clock, ##\Delta t' = L_0/c## proper time has passed, since the clock is also in the same reference frame as the rocket.

    edit: I think the most important thing to keep in mind is which worldline are you talking about. If we are talking about the worldline of the light beam, then the spacetime interval along that worldline is zero. But, if we are talking about a clock attached to the back of the spaceship, then it's spacetime interval will be nonzero. So it depends on which worldline you are talking about.

    edit again: I think I've used slightly incorrect terminology. I think worldline is only for paths through spacetime that are timelike. So the path of the beam of light is not a worldline, so what I really meant is just the more general 'path through spacetime'.
     
    Last edited: Mar 31, 2014
  20. Mar 31, 2014 #19

    CAF123

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    Thanks.
    This helps me see it intuitively but, in the reference frame of the rocket S', the light leaves the rocket at the top when x1' = Po,x and hits the bottom when x2' = Po,x-Lo with a corresponding time interval Lo/c. The proper time is the time in some frame between two events which occur at the same spatial coordinate. Since x1' is not equal to x2' , according to the definition, the time between the two events in this frame is not proper. So I get intuitively why it is a proper time, but I don't see how it agrees formally with the definition. Thanks.

    Edit: I now see your edit, but this would suggest to me that now the proper time is in the frame of the light beam, not the rocket.
     
    Last edited: Mar 31, 2014
  21. Mar 31, 2014 #20

    BruceW

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    between any two spacetime events you can calculate a spacetime interval. And if ##| \Delta t | < | \Delta x |## then that spacetime interval is a proper length. So clearly, you can't just choose any two spacetime events and say the length of the spaceship in its rest frame is equal to any proper length. A similar thing goes for proper time. The time which passes according to a clock is equal to the proper time along the worldline of the object. You can't just choose any worldline and associate it to the proper time of the clock.
     
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