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On the way to showing that LT is linear

  1. Sep 11, 2007 #1
    It seems to me that words confuse me more than mathematics does. Till the page I'm reading now, Rindler hasn't gone beyond using partial derivatives, and has used more words than mathematics. I don't have anything against it, except that I don't understand what he's trying to say most of the time. Either something is wrong with the way I read the text, or with the way he's trying to explain things. I suspect it is the former. I usually take a break and then reread the parts I don't understand. It helps sometimes, but not in this case.

    Here's something from the argument that shows the LT is linear.

    Consider a clock C moving with uniform velocity relative to a frame S. The spatial coordinates of the clock in S are [itex]x_i, i = 1,2,3[/itex]. Therefore, [itex]\frac{dx_i}{dt} = const.[/itex]. That's all fine. Here's the part that I don't understand, and I quote from the book.

    Why should [tex]\frac{dt}{d\tau}[/tex] be a constant? I assumed this to be true and read the rest of the argument, which is quite clear.

    Thanks for any help.
     
    Last edited: Sep 11, 2007
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  3. Sep 11, 2007 #2

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    If d-tau is the time interval between two “ticks” of the clock, as measured by the clock, and dt is the time between these two events as measured in the frame S, then should not this ratio be independent of where and when all this is being measured. S has been assumed to be an inertial frame of ref. That’s what is meant by homogeneity of space and time. So, d-tau/dt should be a constant, i.e., independent of the particular time and the x, y, z co-ordinates of the clock in S. Nothing more, nothing less.
     
  4. Sep 11, 2007 #3

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    It might help to think of a case where homogeneity or isotropy is violated for comparison. Suppose you try to apply the arguments to the coordinates (Rindler coordinates) of an accelerated rocketship. Then dt/dtau is not constant, but is a function of height.
     
  5. Sep 11, 2007 #4
    Thanks for the replies.

    That makes sense, sort of. But what does dt/d-tau (as a whole) represent and in which frame is it measured, if it's the ratio of some quantities measured in different frames?

    I'm not yet familiar with Rindler coordinates. I'm trying to do this from first principles, so assume I know only about the postulates of SR and the definition of an inertial frame.
     
  6. Sep 11, 2007 #5

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    OK, consider the Harvard tower experiment done on the Earth. Clocks at different heights are sometimes said to tick "at different rates". By this we mean that dt/dtau is different at different heights - dtau/dtau is always trivially equal to one. But this doesn't contradict SR, the situation isn't isotropic, there is a preferred direction, "up".

    BTW, technically, I think that this example hows that homogeneity isn't quite enough, one really needs isotropy too. Actually, the original example (based on the "gravitational field" inside an elevator) is required.
     
  7. Sep 14, 2007 #6

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    Oh yes, isotropy of course. I should have said "...independent of where and when all this is being measured and in which direction the clock is going."

    Let's not discuss the GR effects on clocks here.

    Neutrino,

    d-tau gives you the "proper time", i.e., the time between two events in the clock frame as measured by the clock, or the interval between two points on the worldline of the clock. It's an invariant. dt will depend on the relative speed between the clock and the IFR S. So, qualitatively, d-tau/dt is a measure of how fast the clock is moving wrt an IFR S. The exact relationship is given by: d-tau/dt=c*sqrt(1-v^2/c^2). Note that when v=0, d-tau=dt.
     
  8. Sep 15, 2007 #7

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    Sorry for the typo. The 'c' shouldn't be there. The correct formula is: d-tau/dt=sqrt(1-v^2/c^2).
     
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