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Homework Help: Relativity proplem

  1. Nov 5, 2009 #1
    1. The problem statement, all variables and given/known data
    An observer in reference frame S sees two events as simultaneous. Event A occurs at the point (50 m, 0, 0) at the instant 9:00:00 Universal time, 15 January 2004. Event B occurs at the point (150 m, 0, 0) at the same moment. A second observer, moving past with a velocity of 0.8c î, also observes the two events. In her reference frame S’, which event occurred first and what time interval elapsed between the events?

    2. Relevant equations
    Gama= (1-v2/c2 )^-1/2
    X’=gama *(x-v*t)
    delta time = gama * delta time proper (the moving one)
    Length=v*delta time proper = length proper/gama

    3. The attempt at a solution
    If both start at (0,0,0) then the moving observer will see the 50m event first
    X’=gama *(x-v*t)
    Gama= (1-v2/c2 )^-1/2
    .... dunno what to do rly
  2. jcsd
  3. Nov 5, 2009 #2
    What do you mean by they both start at (0,0,0)? Remember, in relativity, it is not enough to specify the spatial coordinates. You also need to specify a time.

    i.e. At what point in spacetime do the two frames S and S' coincide? (You can choose any point)

    From there, at what time coordinate in S' do the two events happen? Comparing the two time coordinates you get should let you know what to decide :-)
  4. Nov 5, 2009 #3
    Just to help a little more, for example, you can set the time the events happened in S to be 0 (or it could be any other number).
    So, if I choose the events to happen at t = 0 for S, the events coordinates in S are:
    (50, 0, 0, 0) and (150, 0, 0, 0)
  5. Nov 6, 2009 #4
    i still don't get it ,but i got
    Gama= (1-v2/c2 )-1/2
    =(1-.64) -1/2
    dunno where to use it
    only equations i got are
    X’=gama *(x-v*t)
    and delta time = gama*delta time proper
    and length = (v*delta time )/gama

    but here i don't have time ,so i can't use any of the equations ,or at least dunno how
  6. Nov 6, 2009 #5
    Okay, I see now. So I think you are missing an equation. You would need the lorentz transform for t' as well. Here are the Lorentz transforms for a frame S' moving with relative speed v in the positive x direction:
    [tex] t' = \gamma (t - \frac{v x}{c^2}) [/tex]
    [tex] x' = \gamma (x - vt) [/tex]
    [tex] y' = y [/tex]
    [tex] z' = z [/tex]

    How would you use these transforms to solve?

    I think as good advice in relativity, when in doubt, you can always use the lorentz transforms and not be wrong. Sometimes you can solve problems quicker using time dilation and lorentz contraction ([tex]\Delta t = \gamma \Delta t'[/tex] etc...) but if you don't reason the situation correctly, you could be wrong. Time dilation and lorentz contractions are derived form the lorentz transforms as well.
    Ex: If two people are on trains going the opposite direction, which one contracts?
    Answer: It depends on the frame of reference. Technically, they both do. For each observer, they see themself as stationary but the other observer moving, so each one sees the other's train contract in length. It's kinda in a way like forces too. Each object that exerts a force on another technically feels a force from the other object as well, however, in a force body diagram, you only include the force that the object feels, not the force that it's exerting on another object.

    I could give you some good problems in relativity involving lorentz contractions and time dilations with correct and incorrect answers if you wanted if you're confused on that stuff.

    Anyway, good luck,

  7. Nov 6, 2009 #6

    Gama= (1-v2/c2 )-1/2
    =(1-.64) -1/2
    t’2 =5/3(-.8(150)/c)=-.666666666*10-6

    event B occurred first
    time interval elapsed between the events =.444444444*10-6

    is that correct ?
  8. Nov 7, 2009 #7
    Sorry, thanks for messaging me again because I completely forgot about your reply!

    No, for gamma, there is a problem (it is not -1/2 it is to the power of -1/2):
    [tex]\gamma = \sqrt{\frac{1}{1-\frac{v^2}{c^2}}}[/tex]
    So if the relative velocity v is 0.8c then gamma should be:
    [tex]\gamma = \sqrt{\frac{1}{1- 0.64}} = \frac{10}{6}[/tex]

    from there, the rest was correct. Try it again, I got this number so you can compare with me (I'm doing this quick, there's always a chance I could be wrong, only you can convince yourself if what you're doing is right ;-)):
    [tex] t_2' - t_1' = [/tex]
    -4.44 10^{-7} seconds

    The important thing should be that in the lorentz transforms, x is different but t is the same, so when you subtract t1 and t2, it only depends on x.

    From this, which event happens first? Does it make sense?

    wait, I just noticed you did gamma right, sorry about that. I think this should be okay. You can also think about what happens to the time difference as objects get closer together, farther apart. Does that make sense?
    Good job on getting it ;-)
  9. Nov 8, 2009 #8
    np ,and about the bold ,i am not sure yet i get it right ,but i know that if its time is more in -ve then it came first .
  10. Nov 8, 2009 #9
    well i was just saying if you put in an arbitrary distances [tex]x_1[/tex] and [tex]x_2[/tex], you should get such a formula for the time difference:
    [tex]t2' - t_1' = -\gamma \frac{(x_1-x_2) v}{c^2}[/tex]

    When [tex]x_1[/tex] approaches [tex]x_2[/tex], the time difference become zero, which makes sense because two events happening at the same point is really just one event happening, and an event can't have a time difference between itself. As you make [tex]x_2[/tex] be much farther than [tex]x_1[/tex], the time difference starts to become infinite, which also makes sense.

    Since signals cannot go faster than the speed of light, you will receive the signal of one event before the other. This very hard to think about because I am not talking about the signal of light, I'm talking about the simultaneity of two signals. If you were in between both events (equidistant from both) and they both sent a signal of light to you, they should come at the same time, that's how one usually defines simultaneity. However, if you're moving, you're going to receive the signal from one before the other assuming the event happens when you reach the midpoint of the two events in space. So this breaks your perception of simultaneity, so your perception of space and time are different. The further apart the two events are, the more of a time delay you should notice.

    This is like the train thought experiment common in relativity books, if you make the train super long, the person on the train should see a bigger time delay between events that the non moving observer thinks are simultaneous. If the train moves faster too, of course he will see a bigger time delay too. If the train is just a point, he won't see time delay.

    I hope I've explained it well enough. I'm sure your textbook explains it. To survive your course I guess the Lorentz transformations are usually enough (and being able to distinguish which one is S and S' in the case of three reference frames sometimes etc). But reading the significance of relativity is interesting :-).
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