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SRT two observers will agree on their relative velocity

  1. Jun 6, 2009 #1
    Ok I am confused. As far as I understand it, in SRT two observers will agree on their relative velocity.

    How do they calculate it?

    [tex]v = \frac{\mathrm{d}x}{\mathrm{d}t}[/tex]

    In my rest frame. But in the others rest frame:

    [tex]v' = \frac{\mathrm{d}x'}{\mathrm{d}t'}[/tex]

    but now for me the length is contracted and the time delated:

    [tex]\mathrm{d}t' = \frac{\mathrm{d}t}{\gamma}[/tex]
    [tex]\mathrm{d}x' = \gamma \mathrm{d}x[/tex]
    [tex]\arrow v' = \gamma^2 \frac{\mathrm{d}x}{ \mathrm{d}t}[/tex]

    So we disagree on our differential speed, this must be wrong. What did I miss?
  2. jcsd
  3. Jun 6, 2009 #2

    George Jones

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    Re: Confusion

    Time dilation and Lorentz contraction don't tell the whole story. The worldline of the unprimed observer is the t-axis, i.e., the line [itex]x = 0[/itex]. One of the Lorentz transformation expressions is (with [itex]c = 1[/itex])

    [tex]\Delta x = \gamma \left ( \Delta x' + v \Delta t'} \right).[/tex]

    So, [itex]x = 0[/itex] gives

    [tex]v = -\frac{\Delta x'}{\Delta t'},[/tex]

    which is what one would expect, since, conventionally, the unprimed observer moves along the primed observer's negative x-axis.
  4. Jun 6, 2009 #3


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    Re: Confusion

    You need to think a little more physically about what kinds of measurements these symbols would be based on. For example, if the other guy calculated that you moved a distance of dx' in a time dt', how did he measure that? Well, he might have had a ruler of length dx' in his rest frame, and two clocks at either end which were synchronized in his frame, and observed that (time on the second clock as you passed it) - (time on the first clock as you passed it) was equal to dt'. But since he had to use two clocks at different locations to measure the time for you to cross the ruler, when considering things from the perspective of your own frame you have to consider the relativity of simultaneity--two clocks that are synchronized and a distance dx' apart in their own rest frame will be out-of-sync by v*dx'/c^2 in your frame if v is the speed of the clocks in your frame (which we don't yet assume is equal to dx'/dt'). So in your frame, if the first clock reads zero as you pass it, the second clock already reads v*dx'/c^2 at that moment. If the second clock reads dt' at the moment you pass it, then in your frame that second clock must have ticked forward by dt' - v*dx'/c^2 between the event of your passing the first clock and the event of your passing the second one. Since this clock's time is dilated, more time must have passed in your frame between these events--the time in your frame is greater than the time elapsed on the moving clock by a factor of 1/sqrt(1 - v^2/c^2), so the time in your frame between the events is (dt' - v*dx'/c^2)/sqrt(1 - v^2/c^2). And since the ruler's length is dx'*sqrt(1 - v^2/c^2) in your frame, when you were next to the first clock the second clock was dx'*sqrt(1 - v^2/c^2) away, then after a time interval of (dt' - v*dx'/c^2)/sqrt(1 - v^2/c^2) the second clock had reached your position, so the speed of the second clock in your frame must be that distance over that time, or:

    v = dx'*(1 - v^2/c^2)/(dt' - v*dx'/c^2).

    Dividing both top and bottom of the right side by dt' gives:

    v = (dx'/dt')*(1 - v^2/c^2)/(1 - v*(dx'/dt')/c^2)

    Multiply both sides by the denominator of the left side:

    v - v^2*(dx'/dt')/c^2 = (dx'/dt')*(1 - v^2/c^2)

    Add v^2*(dx'/dt')/c^2 to both sides:

    v = (dx'/dt')*[(v^2/c^2) + (1 - v^2/c^2)] = (dx'/dt')*1

    So, this is one way of showing that v, the speed of the clock in your frame, is equal to (dx'/dt'), your speed as measured in the rest frame of the clocks. It's probably easier to just use the Lorentz transformation to show this though, you could pick two events on your worldline at the same position but different time coordinates in your frame, figure out the coordinates of these events in the other frame using the Lorentz transformation, and then show that in the other frame (difference in position)/(difference in time) is equal to v (the Lorentz transformation already assumes that the other frame has a speed of v in your frame). If you're not familiar with the Lorentz transformation I could show this as well if you like.
  5. Jun 6, 2009 #4
    Re: Confusion

    Thanks a lot I have to give it some thought. I am fairly familiar with the Lorenz Transformation there are just these little problems I have with reality :)
  6. Jun 7, 2009 #5


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    Re: Confusion


    The equations you quoted,

    [tex]\mathrm{d}t' = \frac{\mathrm{d}t}{\gamma}[/tex].............(1)
    [tex]\mathrm{d}x' = \gamma \mathrm{d}x[/tex]..........(2)​

    are valid only in special cases. As you say you know about the Lorentz transform, the equations that are always valid are

    [tex]\mathrm{d}t' = \gamma (\mathrm{d}t - \frac{v}{c^2}\mathrm{d}x)[/tex]...........(3)
    [tex]\mathrm{d}x' = \gamma (\mathrm{d}x - v \mathrm{d}t)[/tex]............(4)​

    which can be inverted to give

    [tex]\mathrm{d}t = \gamma (\mathrm{d}t' + \frac{v}{c^2}\mathrm{d}x')[/tex]...........(5)
    [tex]\mathrm{d}x = \gamma (\mathrm{d}x' + v \mathrm{d}t')[/tex]............(6)​

    These equations apply to two observers O and O' each measuring the same object P.

    From (5), (1) is valid only in the special case when dx' = 0, i.e. you are measuring events that are at the same place according to O'. This includes the case P=O', when O is measuring O'.

    From (4), (2) is valid only in the special case when dt = 0, i.e. you are measuring events that are simultaneous according to O. In your scenario, this is never true.
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