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Very simple question

  1. Oct 31, 2004 #1


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    I understand the ideas behind relativity (4d space time, different speeds see different slices of it, what not), but I've never done the math.

    If I have something travelling at v1 relative to some frame, and another thing travelling at v2, what is v2 transformed into the reference frame moving at v1?

    I'm trying to model the twin paradox in a computer program.

    Oh, and how do I calculate the slope of a reference frame relative to another reference frame? Is it just (+-)gamma?
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  3. Oct 31, 2004 #2


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    You need to use the velocity-transformations.
  4. Oct 31, 2004 #3


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    Suppose you're on a train moving with velocity u relative to the ground, and you walk forward with velocity v relative to the train, your velocity relative to the ground is not u+v but (u+v)/(1+uv). (I'm using units in which c=1. Otherwise this would be (u+v)/(1+uv/c²).)

    This is the relativistic velocity addition law.

    You seem to be asking for a slightly different version of it (if I understand you correctly). Suppose a train is moving with velocity v1 relative to the ground, and some guy is walking with speed v2 relative to the ground, then what is this guy's speed relative to the train?

    According to the velocity addition law it is (-v1+v2)/(1+(-v1)v2).

    Suppose that another observer is moving with velocity v in the coordinate system where you are stationary at the origin, and that you both measure time from an event when both of your origins were at the same place, then his t axis in a spacetime diagram that represents your coordinates is a line through the origin with slope 1/v (remember that t is in the "up" direction), and his x axis is a line through the origin with slope v. Each line that is parallell to his x axis represents a set of events that are simultaneous to him.

    Note that the when the astronaut twin has changed his direction, these lines of simultaneity ("slices" of spacetime that he thinks of as space at different times) will be very different from what they were before.
  5. Oct 31, 2004 #4


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    That's just a velocity subtractin problem. Classically the answer is v2-v1, relativistically the answer is (v2-v1)/(1+v1*v2/c^2)

    That would probably depend on how you draw the reference frame. In any event, you should be able to figure it out from the Lorentz transform.

    I prefer the following version

    [edit! Major ooops!]

    x' = gamma*(x - v t)
    t' = gamma*(t - v x)

    gamma = 1/sqrt(1-v^2)

    the inverse of which is

    x = gamma*(x' + v t')
    t = gamma*(t' + v x')

    [end edit]

    but note that the require that you use "relativistic" units, where time for example might be measured in seconds and distance in light seconds, or time in years and distance in light years.
    Last edited: Oct 31, 2004
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