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Orbital velocities in the Schwartzschild geometry

  1. Jun 10, 2010 #1
    I'm trying to use the tensor formulation of GR to calculate the velocity of a particle in a circular orbit around a black hole.

    Here is the work I have done so far.

    What concerns me is that I end up getting zero velocity when applying the metric to the differential equations I get from the geodesic equation. I wonder if I have made a miscalculation, but I am unable to find any, so maybe there is a misunderstanding on my part.

    Any help is appreciated.
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  3. Jun 10, 2010 #2

    George Jones

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  4. Jun 10, 2010 #3


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    There seems to be a problem with signs in (13), because both terms are negative-definite.

    Using (12), and choosing t=0 to coincide with [itex]\tau=0[/itex], you can set [itex]t=\beta \tau[/itex], where [itex]\beta[/itex] is a constant. Let's also set [itex]\omega=d\phi/der t[/itex]. Then (13) gives [itex]\omega=\pm i\sqrt{2m/r^3}[/itex], where [itex]m=r_s/2[/itex]. This seems sort of right, since it coincides with Kepler's law of periods. However, it's imaginary due to the sign issue. It would also surprise me if Kepler's law of periods was relativistically exact when expressed in terms of the Schwarzschild coordinates, but maybe that's the case.

    If you can fix the sign problem, then you seem to have the right result in the nonrelativistic limit of large r. You might then want to check the result in the case of r=3m, where I believe you should obtain lightlike circular orbits.
  5. Jun 10, 2010 #4
    You are right, his equation (13) is in the "not-even-wrong" category. It is easu to see that since the correct Lagrangian, for the simplified case he's considering is:

  6. Jun 10, 2010 #5
    @George Jones , bcrowell
    Thanks for pointing that out, and thank you for the reference. I traced the sign error back to a differentiation error when calculating the Christoffel symbol.

    As for writing [itex]t=\beta\tau[/itex] I don't see how that will help since t does not appear in the other equations. I must be missing something.

    I will try to arrive at a result and do the "reality checks" you mentioned.


    Thanks for your input. I am afraid I don't know how to arrive at or what to do with the Lagrangian. From it's appearance it looks just like the metric, so L=1 here, I imagine?


    I arrived at [tex]\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{r-r_s}}[/tex]. The units match, but I doubt this is correct, since letting r=3m gives [tex]\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{\sqrt{2m}}[/tex] while my intuition tells me it should be c.
    Last edited: Jun 10, 2010
  7. Jun 10, 2010 #6
    No, the correct equations are:




    Hint: in your writeup you made [tex]dr=d\theta=0[/tex], remember? You need to think what that means.
  8. Jun 10, 2010 #7
    Yes, I arrived at these as well when I made the assumptions you mentioned. As for their meaning, I interpret it as circular (constant radial coordinate) motion around the equator ([itex]\theta=\pi/2[/itex]) with constant velocity. In addition, there was a third equation I arrived at,


    which, when I used the substitution


    which I got from the metric, gave me


    which solves to


    So now I am unsure about where I made my mistake.
  9. Jun 10, 2010 #8
    There are only two independent equations, the ones I mentioned to you.
    The third Lagrange equation, exists only if r is variable and its correct form would have been:


    But you made [tex]dr=0[/tex] (this is why I gave you the hint), so the third equation does not exist. This is the root of your errors.
    Last edited: Jun 10, 2010
  10. Jun 11, 2010 #9
    I don't understand why it shouldn't exist. I derived the general case, then assumed dr=0 and substituted that into the equations. dr doesn't even appear in that equation anymore, do I don't see how it has any influence.

    EDIT: Where does the factor of 2 come from there? I somehow didn't appear in my calculations.

    For the velocity, I obtained


    My intuition says that this is wrong by a factor of [itex]\sqrt{2}[/tex], since then it would give v=c at [itex]r=\frac{3GM}{c^2}[/itex], but I don't see how that factor dissapeared.
    Last edited: Jun 11, 2010
  11. Jun 11, 2010 #10
    You started with the metric that has [tex]dr=0[/tex]. therefore all your attempts to differentiate wrt [tex]r[/tex] should result in null terms. Yet, you clearly differentaite wrt [tex]r[/tex] in your derivation and this renders your derivation wrong.

    From the Euler-Lagrange equations.

    Yes, it is very wrong.
    From the correct equation [tex]\frac{d^2\phi}{ds^2}=0[/tex] you should obtain (no surprise):


    The trajectory is completed by the other obvious equation


    You get one more interesting equation, that gives u the time dilation. Start with:

    [tex]ds^2=(1-r_s/R)dt^2-(Rd\phi)^2[/tex] and you get:




    The last equation gives you the hint that:


    The last expression is what you were looking for.
  12. Jun 12, 2010 #11
    The only thing I said is that I have constant r. If you have a function, say f(x)=x2, you can say that x is fixed at three, but you can still get the slope at x=3 by differentiating wrt x.

    As for the last equation you posted. I don't doubt its validity, but I don't see how it will get me anywhere, since neither v nor [tex]\omega[/tex] are known. In fact, since [tex]v=\omega R[/tex], v cancels on both sides.
  13. Jun 12, 2010 #12
    Umm, no. If you did things correctly, then you'd have realised that [tex]dr=d\theta=0[/tex] reduces the metric to :


    So, your Christoffel symbols need to reflect that. They don't.

    Yet, the result is extremely important since it tells you that the orbiting object has constant angukar speed and the trajectory is a circle.

    No, [tex]v[/tex] is not [tex]\omega R[/tex].
    Last edited: Jun 12, 2010
  14. Jun 12, 2010 #13
    How do you define v?

    Of course the trajectory is a circle. I imposed that restriction by setting r=constant after deriving the general case geodesic equations. The equations saying d2t/ds2=0 and d2φ/ds2=0 are neccesary consequences.

    What I am seeking is an expression which gives the orbital velocity as a function of r. I define [tex]v=\frac{\text{d}\phi}{\text{d}\tau}r[/tex], and except for the factor [tex]\sqrt{2}[/tex] my result reduces to the Newtonian formula at large r, which makes me beleive my derivation is valid, the erronous factor [tex]\sqrt{2}[/tex] notwithstanding.
    Last edited: Jun 12, 2010
  15. Jun 12, 2010 #14
    You don't get to "define", you need to "derive" :


    The last equation gives you the hint that:


    since, in GR:

  16. Jun 12, 2010 #15
    And is v, in your case, measured by an observer from infinity? You have to give a definition. [tex]v=\frac{\text{d}\phi}{\text{d}\tau}r[/tex] and [tex]\frac{\text{d}\phi}{\text{d}t}r[/tex], for example, aren't the same, so you have to specify.

    Aside from that, how can your equation be used to calculate the orbital period as a function of r only?
  17. Jun 12, 2010 #16
    You can do it all by yourself by remembering that [tex]\frac{d\phi}{d\tau}=\omega[/tex] (see the derivation from the Euler-Lagrange equation)

    [tex]\phi=\omega \tau[/tex]. Make [tex]\phi=2\pi[/tex]
    The orbital period is not a function of r.
  18. Jun 12, 2010 #17
    For circular motion, it is easy to obtain the relationship [tex]v=\omega r[/tex]. You get it directly from the definition of the radian. Do you claim [tex]v=\frac{\text{d}\phi}{\text{d}\tau} r[/tex] is not a valid definition of v? If so, please explain.

    Please link to the Euler-Lagrange derivation, and I'll do my best to understand it.

    How do you arrive at that conclusion? It is obvious that the orbital period is a function of r. That's why Mercury's orbital period is shorter that Earth's.
  19. Jun 12, 2010 #18
    Not in GR. You are fixated on galilean physics. I am sorry, untill you get off your fixations, I can't help you.
  20. Jun 12, 2010 #19
    Then please explain the situation in GR.
  21. Jun 12, 2010 #20
    What do u think I've been doing for you starting with post 4?
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