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Schwarzschild Four Velocity

  1. Mar 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the four velocity [itex] V^i [/itex], the four acceleration [itex] A^i [/itex] and the scalar [itex] A^i A_i [/itex] for an observer at [itex] r=r_0, \theta = \theta_0, \phi = \phi_0 [/itex] in the Schwarzschild spacetime with r>2m.


    2. Relevant equations

    The Schwarzschild Metric

    [tex] ds^2 = -\displaystyle \left(1-\frac{2m}{r} \right) dt^2 + \left( 1- \frac{2m}{r} \right)^{-1} dr^2 + r^2 \left( d\theta ^2 + sin^2(\theta) d\phi ^2 \right) [/tex]


    3. The attempt at a solution
    I'm not too sure how to even start this problem in that I don't see how the four velocity will depend on the metric, and so the only place I can see it being needed is in calculating the scalar [itex] A^i A_i [/itex]. Is the spacial part of the four velocity given by the spherical velocity [itex] ( \dot{r}, r \dot{\theta}, r \dot{\theta} sin(\theta) ) [/itex] and the time element would just be 1? (we've scaled c= 1).
     
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  3. Mar 19, 2008 #2

    kdv

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    V^i is given by [tex] \frac{dx^i}{d \tau} [/tex] Al the x^i are constant except for [tex]x^0 = t [/tex] so you will have to take the derivative of this with respect to the proper time and this is where the metric will be needed.
     
  4. Mar 19, 2008 #3
    The metric should become
    [tex] ds^2 = -\displaystyle \left(1-\frac{2m}{r} \right) dt^2[/tex]

    I figured that the space like components would be zero, but as for the time coordinate, if I want to examine [itex]\frac{dt}{d\tau}[/itex] then wouldn't I also need to consider [itex] \frac{ds}{d\tau} [/itex] or should I make some assumptions based on the fact that we assume it's a massive particle and must follow a time like geodesic?

    But if it follows a time like geodesic, it seems to me that the time will trivially be the proper time, when I have a strange inkling that it should be something with
    [tex] \displaystyle \left( 1- \frac{2m}{r} \right) [/tex].
     
  5. Mar 20, 2008 #4

    kdv

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    It's not following a geodesic since it is staying at constant r, theta and phi!
     
  6. Mar 20, 2008 #5
    It's not following a spatial geodesic, but wouldn't it still be following a geodesic in time through space time? And in that case then ds=0 and time is just proper time.
     
  7. Mar 20, 2008 #6

    George Jones

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    No.

    No.

    In special relativity, what does it mean for the interval ds to be zero?

    In special relativity, when is the time coordinate also proper time?
     
  8. Mar 20, 2008 #7
    Okay, so ds = 0 would correspond to a null geodesic. Since the object is at rest, it's following a time-like geodesic, and so I would solve

    [tex] ds^2 = -1 = -\displaystyle \left(1-\frac{2m}{r} \right) \left( \frac{dt}{d \tau}\right)^2 [/tex]

    Does this seem right?
     
  9. Mar 21, 2008 #8

    George Jones

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    No, an observer that is freely falling follows a timelike geodesic. A geodesic worldline is one on which an accelerometer would read zero. An observer at rest in Schwarzschild coordinates follows a timelike worldline, but, as kdv said, this worldline is not a geodesic. As you are supposed to show, the 4-acceleration of such an observer is non-zero.

    Not quite. On any observer's timelile worldline, [itex]ds^2 = -d\tau^2[/itex].
     
  10. Mar 21, 2008 #9
    Okay yes, I realize I wrote my above equation to be solved in a rather bad manner, but setting [itex] ds^2 = -d\tau^2 [/itex] will give me the same result so I think that's right.
     
  11. Mar 21, 2008 #10

    George Jones

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    Yes, if you get rid of everything to the left of the -1, the equation is right.
     
  12. Mar 21, 2008 #11
    So we find that [itex] V^i = \displaystyle \left( \frac{1}{\sqrt{1-\frac{2m}{r}}} , 0, 0, 0 \right) [/itex]. Now in order to find [itex] A^i [/itex] I must find [itex] \frac{D V^i}{d\tau} = \frac{\partial V^i}{\partial d\tau} + \Gamma^i_{bc} V^b V^c[/itex].

    The only non-obvious partial is [itex] \frac{\partial V^t}{\partial d\tau} [/itex]. So,
    [tex] \frac{d^2 t}{d\tau^2} = \displaystyle \frac{d}{d\tau} \left( \frac{dt}{d\tau} \right) [/tex]

    But since there's no explicit dependence on [itex] \tau [/itex]I tried using the chain rule. That is

    [tex] \displaystyle \frac{d}{d\tau} \left( \frac{dt}{d\tau} \right) = \frac{dr}{d\tau} \frac{d}{dr} \left( \frac{dt}{d\tau} \right) [/tex]
    [tex] = -\left(\frac{dr}{d\tau}\right) \left(\frac{m}{r^2 \left( 1- \frac{2m}{r} \right) ^{\frac{3}{2}} }\right) = 0[/tex]

    We conclude that [itex] \frac{\partial V^i}{\partial d\tau} =0[/itex]

    Furthermore, for non-zero [itex] \Gamma^i_{bc} V^b V^c[/itex] we are required to use [itex] \Gamma^i _{tt} [/itex]

    Now the only way [itex] \Gamma^i _{tt} [/itex] is non-zero is if we have an r derivative and so we consider [itex] \Gamma^r _{tt} =\frac{1}{2}g^{rr}(-g_{tt,r})[/itex]

    But I'm not too sure what [itex] g^{rr} [/itex] is. Our reduced metric seems to only be 1x1 and if I increase it to 2x2 then it's singular and doesn't have an inverse since dr²=0. What do I do?

    Edit: Is it really just [itex] 1-\frac{2M}{r} [/itex] ?
     
    Last edited: Mar 21, 2008
  13. Mar 21, 2008 #12
    It makes sense that [itex] g^{rr} = (1-\frac{2m}{r} ) [/itex] since then the four acceleration would be

    [tex] A^i = (0, \frac{M}{r^2}, 0, 0) [/tex]

    which makes sense since we expect to exert a force to keep the observer at rest, and [itex] \frac{M}{r^2} [/itex] happens to correspond beautifully with the Newtonian force of gravity where G=1.
     
  14. Mar 22, 2008 #13

    George Jones

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    Yes, this is [itex]g^{rr}[/itex]

    and these are components of the 4-acceleration with respect to the coordinate basis.

    To find the actual physical acceleration of a stationary observer, the 4-acceleration should be expressed with respect to an orthonormal frame, not with respect to a coordinate basis. Since the magnitude of the 4-acceleration is independent of the basis used, it suffices to calculate [itex]A^\mu A_\mu[/itex] in any basis, even a coordinate basis, to find the acceleration experienced by the observer. This expression is quite interesting.
     
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