Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Schwarzschild solution and velocity of stationary observer

  1. Oct 11, 2009 #1
    Hey all,

    I suddenly find myself very confused about velocity and coordinate systems. I have a feeling this is very simple, but sometimes the mind just curls up, you know? ;)

    When you ask what an observer observe, you need to see things from his point of view - his reference frame. And his reference frame must be a frame moving with him. But this reference frame can look very different depending on what coordinates he choose, right? So this frame can e.g. be an inertial frame, but it can also be some weird twisted frame, right?

    Consider the Schwarzschild solution and a stationary observer at a distance r from the center (beyond 2GM). He must of course be using some sort of thrust to stay still.

    Now, my book says: "Work in inertial coordinates such that the observer is in the rest frame. Then the velocity of the observer is U = (1,0,0,0)." So the observer, using inertial coordinates, sees himself as having velocity U=(1,0,0,0)?

    Another place in my book, the velocity of this stationary observer is described as

    [tex]U=((1-\frac{2GM}{r})^{-1/2},0,0,0)[/tex]

    This is obviously in a coordinate system where the metric is the Schwarzschild metric itself. So this is another kind of reference frame. But is this what the observer himself measures? I think I have a tendency to consider a omnipresent observer looking down at the whole S.T., and I think I have considered

    [tex]U=((1-\frac{2GM}{r})^{-1/2},0,0,0)[/tex]

    as the velocity that this omnipresent observer sees the stationary observer having. But this is wrong, right?

    So basically, when you want to know your velocity, you choose a coordinate system, find the corresponding metric and then calculate the velocity from the relation

    [tex]U^\mu U^\nu g_{\mu \nu} = -1. [/tex]

    And what you find is the velocity that you see yourself having. Right?
     
  2. jcsd
  3. Oct 11, 2009 #2

    Jonathan Scott

    User Avatar
    Gold Member

    An observer's own four-velocity is always (1,0,0,0), isn't it?
     
  4. Oct 11, 2009 #3

    atyy

    User Avatar
    Science Advisor

    Yes.

    Edit: Ooops, I meant just the LHS, not sure if the RHS is right - does it work out in Schwarzschild coordinates?
     
    Last edited: Oct 11, 2009
  5. Oct 11, 2009 #4

    Mentz114

    User Avatar
    Gold Member

  6. Oct 11, 2009 #5
    The equation holds in any coordinate system, so it works in Schwarzschild coordinates as well. This is the neat thing about tensors equations and scalars - they don't care much about coordinates. :)
     
  7. Oct 11, 2009 #6
  8. Oct 11, 2009 #7
    No, only in special relativity.
     
  9. Oct 11, 2009 #8

    Jonathan Scott

    User Avatar
    Gold Member

    OK - I guess the way I learned GR it had never occurred to me to use "own velocity" in a different way from SR.
     
  10. Oct 11, 2009 #9

    atyy

    User Avatar
    Science Advisor

    OK, good, that's what I thought from Fermi-Walker transport, but sometimes I need to do the computation in another coordinate system for a sanity check :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Schwarzschild solution and velocity of stationary observer
  1. Schwarzschild solution (Replies: 6)

Loading...