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Elevator metric

  1. Aug 18, 2008 #1
    I just came across this expression

    [tex]ds^2 = (1+gz/c^2)^2(cdt)^2 - dx^2 - dy^2 - dz^2[/tex]

    in entry #19 of this thread https://www.physicsforums.com/showthread.php?t=227753 for the metric of a uniform gravitational field. Is this correct? I was wondering because it yields

    [tex]\frac{d\tau}{dt}=(1+\frac{gz}{c^2}) [/tex]

    for a stationary clock, but if

    [tex]\frac{d\tau}{dt}=\sqrt{1-\frac{v_{esc}^2}{c^2}}
    [/tex]

    between any two points of different gravitational potential, then we should get

    [tex]\frac{d\tau}{dt}=\sqrt{1-\frac{2gz}{c^2}}[/tex]

    instead. Here z=0 is the ceiling of the elevator/spaceship and z>0 as we move towards the floor, but even if we switch that around (so that z=0 is the floor and z>0 as we move towards the ceiling) it doesn't explain what happened to the "2" or the radical sign.

    Can someone point out the flaw(s) in my reasoning? Is the escape velocity/time dilation relationship not applicable here? Thanks.
     
  2. jcsd
  3. Aug 18, 2008 #2

    George Jones

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    I haven't had time to look at this, and I'm just going out the door for a walk with my wife and daughter, but I'll take a stab at it.

    Are you using the escape velocity/time dilation relationship from the Schwarzchild metric? There is no reason for this relationship to hold for a metric that isn't the Schwarzschild metric metric.

    If the relationship isn't from Schwarzschild, where is it from? How is it derived?
     
  4. Aug 18, 2008 #3
    If you take the potential difference between the edge of a rotating disk and its center and convert that into an 'escape velocity' you get the velocity of that point on the disk, which (I thought) gives you the time dilation via the SR equation (the Lorentz factor). Since it works for both this case and Schwarzschild (the relationship between escape velocity and time dilation) I thought it was universal.

    Perhaps that was the flaw right there.
     
    Last edited: Aug 18, 2008
  5. Aug 18, 2008 #4

    Mentz114

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    The metric you've quoted is a species of the Rindler spacetime, which is in constant acceleration. I'm not sure if the idea of escape velocity has meaning in this case.

    There's a singularity at g00=0, so I suppose you could calculate the velocity of a particle that falls from infinity to the singularity.
     
  6. Aug 19, 2008 #5
    By the 'escape velocity' between two points I just meant the vertical speed at which something at the lower point would have to be thrown so that the peak of its trajectory would be at or next to the higher point. So for example in a constant gravitational field with acceleration [tex]g[/tex], the 'escape velocity' between two points of height difference [tex]\Delta h[/tex] would be [tex]v_{esc}=\sqrt{2g \Delta h}[/tex]. I understand that this is not how the phrase is normally used (unless the upper height is [tex]\infty[/tex]) but I don't know what else to call it. Anyway, for a while I've believed that for any gravitational field, between any two different heights

    [tex]\frac{d\tau}{dt}=\sqrt{1-\frac{v_{esc}^2}{c^2}}[/tex]

    using the definition of [tex]v_{esc}[/tex] that I just gave, since this seems to be the case both with the Schwarzschild metric and a clock on a rotating disk. But if it's not true for the Rindler metric then I'll forget about it altogether.
     
  7. Aug 20, 2008 #6
    A follow-up - if I may - regarding constant acceleration:

    If a constant force is applied to an object of rest mass [tex]m_0[/tex], then

    [tex]
    \frac{dv}{dt}=\frac{F}{m_0}\sqrt{1-\frac{v^2}{c^2}}
    [/tex].

    Question: What function v(t) is a solution to this equation? Both arc tangent and hyperbolic tangent seem to have about the right shape, but unless there's a trig identity regarding arc tangent that I'm not aware of, neither of them work.
     
  8. Aug 20, 2008 #7

    DrGreg

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    See this thread, in particular post #19 (and the links there to another thread) and post #22.

    You will see that your equation is wrong, it should be [itex]\gamma^{-3}[/itex] rather than [itex]\gamma^{-1}[/itex]. [itex]F/m_0[/itex] is the proper acceleration, [itex]dv/dt[/itex] is the coordinate acceleration.
     
  9. Aug 20, 2008 #8
    Thanks, DrGreg. I forgot about the effective-mass-being-different-in-different-directions business.
     
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