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Schwarzschild Circular Orbits Problem

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the linear velocity of a particle in a circular orbit of radius [itex]R[/itex] in the Schwarzschild geometry as would be measured as by a stationary observer stationed at one point on the orbit. (It's problem 10 in chapter 9 of Hartle, if that helps)


    2. Relevant equations
    The Scwarzschild metric, obviously,
    [itex]ds^2=- \left (1-\frac{2M}{r} \right )dt^2 + \left (1-\frac{2M}{r} \right )^{-1} dr^2 +r^2(d \theta^2 + \sin ^2 \theta d \phi ^2) [/itex]
    And in the book they derive the equation (for a circular orbit of radius R)
    [itex]\Omega \equiv \frac{d \phi}{d t}= \frac{M}{R^3}[/itex]


    3. The attempt at a solution
    OK, so in this problem, obviously [itex]\frac{dr}{d \bar{t}}=\frac{d\theta}{d \bar{t}}=0[/itex], where [itex]\bar{t}[/itex] is the time measured in the stationary observer's frame. Ok so we know the angular velocity measured from the far away observer's frame:
    [itex]\Omega \equiv \frac{d \phi}{d t}= \frac{M}{R^3}[/itex]
    since the stationary observer is, by definition not moving (w.r.t. to our far-away observer), we can easily find [itex]dt/d\bar{t}[/itex] from the metric. We then have [itex]\frac{d \phi}{d \bar{t}}.[/itex] My question is, after we get the velocity 3-vector (i.e. the velocity w.r.t. to the stationary orbit), what do we do and why?
     
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  3. Oct 23, 2011 #2

    George Jones

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    Can you write down the 4-velocity of the particle in orbit and the 4-velocity of the stationary observer?
     
  4. Oct 24, 2011 #3
    Of course, it's just:
    [tex]u_{observer} = ( dx^{\alpha}_{observer}/d \bar{t} ) [/tex]
    [tex]u_{particle} = ( dx^{\alpha}_{particle}/d \tau ) [/tex]
    Where α ranges from 0 to 3. (0 is the time component w.r.t. to the observer at ∞).
     
    Last edited by a moderator: Oct 24, 2011
  5. Oct 25, 2011 #4

    George Jones

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    Okay, now I have Hartle in front of me. I meant also the specific components of the 4-velocities with respect to Schwarzschild coordinates, i.e., (9.16) for the hovering observer and (9.47) together with (9.48) for the orbiting particle.

    What do you get when you divide the t components of the 4-velocities you wrote above? What do you get when you divide the t components of the 4-velocities from the equations I gave?
     
  6. Oct 25, 2011 #5
    ^So you want me to write out the components?

    [itex]u_{particle}=(\frac{dt}{d \tau} , 0, 0, \frac{d \phi_{particle}}{d \tau}) = ( \sqrt{1-\frac{3m}{R}}, 0, 0, \frac{M}{R^3} )[/itex] (from 9.46, 9.47, and 9.48)
    And:
    [itex]u_{observer}=(\frac{dt}{d \bar{t}} , 0, 0, 0) [/itex]
    From the metric:
    [itex]-(d \bar{t})^2=- \left (1-\frac{2M}{r} \right )dt^2 + \left (1-\frac{2M}{r} \right )^{-1} dr^2 +r^2(d \theta^2 + \sin ^2 \theta d \phi ^2) [/itex]
    since [itex] (d \bar{t})^2= - ds^2[/itex]. Of course, [itex]d \phi = d \theta = dr = 0 [/itex] since the observer is stationary, so :
    [itex](d \bar{t})^2=\left (1-\frac{2M}{r} \right )dt^2[/itex]
    Hence, rearranging,
    [itex] \left (\frac{dt}{d \bar{t}} \right ) ^2 = \left (1-\frac{2M}{r} \right ) ^ {-1} [/itex]
    i.e.,
    [itex]\frac{dt}{d \bar{t}} = \left (1-\frac{2M}{r} \right ) ^ {-1/2} [/itex]
    so that
    [itex]
    u_{observer}=(\frac{dt}{d \bar{t}} , 0, 0, 0) = \left ( \left (1-\frac{2M}{r} \right ) ^ {-1/2}, 0, 0, 0 \right ) [/itex]
     
  7. Oct 25, 2011 #6

    George Jones

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    I am again without Hartle, but shouldn't u^t above be [itex]\left( 1-3m/R \right)^{-1/2}[/itex]?

    What is [itex]d \bar{t} / d \tau[/itex] in Schwarzschild coordinates? What is [itex]d \bar{t} / d \tau[/itex] in special relativity?
     
    Last edited: Oct 25, 2011
  8. Oct 27, 2011 #7
    ^Probably. Typo.

    Also:
    [itex]d \bar{t} / d \tau = \frac{dt}{d \tau} \frac{d \bar{t}}{dt} = \left ( 1 - \frac{3M}{R} \right ) ^ {-1/2} \left (1-\frac{2M}{R} \right ) ^ {1/2} [/itex]

    And in SR, isn't it just the gamma-factor?
     
  9. Oct 27, 2011 #8

    George Jones

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    Yes. What happens when set these expressions equal?
     
  10. Oct 28, 2011 #9
    ^So you just solve for v?

    Why can you do that? Is it b/c the observer at R is in a local inertial frame?

    And if so, can't you just use [itex]v^2 = \eta_{ij} v^{i} v^{j} [/itex] where [itex]\eta_{ij}[/itex] is the flat space metric (in spherical coordinates) and you sum over 1 to 3 and [itex]v^{i}=\left(\frac{dr}{d \bar{t}}, \frac{d \theta}{d \bar{t}}, \frac{d \phi}{d \bar{t}}\right)[/itex]?

    ...Am I being an idiot here? xD
     
    Last edited by a moderator: Oct 29, 2011
  11. Oct 30, 2011 #10

    George Jones

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    Which observer at R? Both observers have constant [itex]r = R[/itex]. The hovering observer H has non-zero 4-acceleration and the orbiting observer O has zero 4-acceleration, so O is inertial and H is not. But, as you suspect, a local inertial coordinate (LIC) can be used to demonstrate the result.

    Let [itex]p[/itex] be an event of coincidence for O and H. Let [itex]\left\{ x^{\mu '} \right\}[/itex] be an LIC that has the following properties:

    1) [itex]p[/itex] is at the origin;

    2) H is at rest at [itex]p[/itex];

    3) at [itex]p[/itex], O moves in the positive [itex]x^{1 '}[/itex] direction.

    More about 2). Even though H has non-zero 4-acceleration, 2) is possible, but, because of the non-zero 4-acceleration, H will not be at rest away from [itex]p[/itex]. I used H instead of O for 2), because the final is not dependent on one of the observers having zero 4-acceleration.

    Let [itex]\mathbf{u}[/itex] and [itex]\mathbf{w}[/itex] be the 4-velocities of H and O respectively. In the LIC, [itex]\mathbf{u}_p = \left( 1, 0, 0, 0 \right)[/itex] and [itex]\mathbf{w}_p = \left( \gamma, \gamma v, 0, 0 \right)[/itex]. Consequently, in the LIC,
    [tex]
    \gamma = - g_{\mu \nu} u_p^\mu w_p^\nu .
    [/tex]
    Because [itex]g_{\mu \nu} u_p^\mu w_p^\nu[/itex] is a coordinate-invariant scalar, [itex]\gamma = - g_{\mu \nu} u_p^\mu w_p^\nu[/itex] in any coordinate system in both special and general relativity! This useful result gives the physical (not coordinate) between any two coincident observers. Use Schwarzschild spherical coordinates and this expression to do the problem. In coordinate-free notation,
    [tex]\gamma = - g \left( \mathbf{u}_p , \mathbf{w}_p \right)[/tex]
    This works in Minkowski spacetime, but I don't see how to use this here, and it doesn't work for general curvilinear coordinates in Minkowski spacetime. In Hartle's problem, we can constuct a LIC around [itex]p[/itex], and then use this LIC to construct spherical coordinates locally around [itex]p[/itex], but there won't be an obvious relationship between the local spherical coordinates and Schwarzschild spherical coordinates.
    No.
     
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