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[General Relativity] Find the acceleration of an object

  1. Jul 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the metric
    [tex]ds^2=-e^{2\phi}dt^2+\frac{1}{1-\frac{b(r)}{r}}dr^2[/tex]
    find the time component of the 4-acceleration of an object moving with velocity v in the r direction.


    3. The attempt at a solution
    The four-velocity of the object is [tex]u^a=(t', v)[/tex]
    where the prime stands for the derivative with respect to the proper time of the object.
    I know that the four velocity satisfies [tex]g_{ab}u^au^b=-1[/tex] so I find
    [tex]t'=\left(1+\frac{v}{1-\frac{b}{r}}\right)\exp(-2\phi)[/tex]
    Now I use the formula
    [tex]a^0=u^a\nabla_au^0=u^0(\partial_0\dot t+\Gamma_{0c}^0u^c)
    +u^1(\partial_1\dot t+\Gamma_{1c}^0u^c)[/tex]
    and I find
    [tex]a^0=t'\dot\phi v+v\left[\dot\phi t'+\frac{v}{2\exp(2\phi)}
    \frac{b/r^2}{\left(1-b/r\right)^2}\right][/tex]
    where the dot stands for the derivation with respect to r and I've supposed that [itex]\partial_0 t'=\partial_1 t=0[/itex].
    Could this be right?
     
    Last edited: Jul 17, 2012
  2. jcsd
  3. Jul 17, 2012 #2

    George Jones

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    Velocity v with respect to whom?
     
  4. Jul 17, 2012 #3

    gabbagabbahey

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    I get something a little different... perhaps you should show your calculation
     
  5. Jul 17, 2012 #4
    With respect to an observer at rest who sees the metric [itex]ds^2[/itex]

    I'm a little tired...I did some stupid mistakes :blushing:
    The metric is
    [tex]ds^2=-e^{2\phi}dt^2+\frac{1}{1-\frac{b(r)}{r}}dr^2[/tex]
    and I get
    [tex]
    t'=\pm\sqrt{1+\frac{v^2}{1-\frac{b}{r}}}\:\:e^{\phi}
    [/tex]
    Now I'm going to correct all the others formula (I've also found an error in a Christoffel symbol :blushing: )....
     
  6. Jul 17, 2012 #5

    George Jones

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    Sorry, but I don't know what this means. Do you mean "At rest with respect to an observer who has contant [itex]r[/itex]"?
     
  7. Jul 17, 2012 #6

    George Jones

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    I think that this must be the case. Another qestion. Is [itex]v[/itex] a coordinate velocity, or is it an actual physical velocity?

    Here is a (slick) way to calculate relative physical (not coordinate) speed between two observers who are coincident at an event. Suppose the two 4-velocities are [itex]u[/itex] and [itex]u'[/itex]. Then, [itex]-\gamma = -\left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta.[/itex] This is an invariant quantity, and, consequently, can be calculated using any coordinate system/basis.

    This works in all coordinate systems in both special and general relativity.

    In order to use this method, you first need to (easily) find the 4-velocity of an observer who sits at constant [itex]r[/itex].
     
  8. Jul 18, 2012 #7
    Yes, I mean that. And v is the physical velocity.

    I don't understand this. If I am at constant r, then my four velocity is
    [itex]u^a=(e^{-\phi}, 0)[/itex] while the 4-velocity of the moving object is
    [itex] u'^b=(k, v)[/itex] with k chosen so that [itex]u'^bu'_b=-1[/itex]. Where am I wrong?
     
  9. Jul 18, 2012 #8

    George Jones

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    This isn't correct. Is v the spatial component of 4-velocity in special relativity?
     
  10. Jul 18, 2012 #9

    George Jones

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    There two ways that you can do this:

    1) in a coordinate basis;

    2) in an orthonormal basis (like in the other thread).

    1) In a coordinate basis, the 4-velocity of the stationary observer has components [itex]\left( e^{-\phi} , 0\right)[/itex] and call the components of the other 4-velocity [itex]\left( u'^0 , u'^{1}\right)[/itex]. Use the method I give above to find [itex]u'^0[/itex].

    2) In an orthonormal basis, what are the component of the 4-velocity of the stationary observer? The components of the other 4-velocity are the same as the components of 4-velocity in special relativity. What are these?

    Notes: you don't have to do 2) to do 1); don't confuse the two methods.
     
  11. Jul 18, 2012 #10
    I got it! I used the coordinate basis method ad I've found the right answer
    [tex]u'=(\gamma e^{-\phi}, \pm v\gamma\sqrt{1-\frac{b}{r}})[/tex].
    But I don't understand why
    [tex]-\gamma = -\left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta[/tex]
    In flat spacetime where [itex]u=(\gamma, \gamma v)[/itex] this is obvious, but how can you generalize it to a generic curved spacetime?
     
  12. Jul 18, 2012 #11

    George Jones

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    Great!

    In the orthonormal basis, the components of the two 4-velocities are [itex]\left( 1 , 0 \right)[/itex] and [itex]\left( \gamma , \gamma v \right)[/itex], just like in special relativity. Consequently, [itex]u^\mu u'_\mu = - \gamma[/itex] in the the orthonormal basis. But, [itex]u^\mu u'_\mu[/itex] is a scalar, and thus its value is independent of basis. Consequently, [itex]u^\mu u'_\mu = - \gamma[/itex] in every basis, including a coordinate basis.
     
  13. Jul 18, 2012 #12
    First of all, thank you very much for your help!
    As you can see my knowledge of general relativity is very limited...
    Why do you say that the components of the 4-velocities in an orthonormal basis is just like in special relativity? I'm studying GR on "Wald" and it doesn't explain these things
     
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