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Black hole orbital period

  1. Apr 18, 2006 #1

    pervect

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    Here's an interesting question inspired by a homework probem (not mine) - what is the orbital period (proper time) for someone in a circular orbit around a black hole as a function of the radius of their orbit (Schwarzschild r-coordinate).

    By setting the derivative of the effective potential to zero I came up with the following expression.

    [tex]\tau = \frac{2 \pi r}{c} \sqrt{\frac{2r}{r_s} -3}[/tex]

    Here [itex]r_s[/itex] = 2GM/c^2 is the Schwarzschild radius of the black hole.

    It seems reasonably sane, we get tau=0 at the photon sphere, and it varies as r^3/2 for large r.

    The circular orbits will be unstable for r < 3 r_s
     
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 18, 2006 #2
    doesn't the orbital period depens from the density, volume of the black hole, and the distance of the object orbiting?

    i don't see those metrics in your thing up here just r...
     
    Last edited by a moderator: Apr 18, 2006
  4. Apr 18, 2006 #3

    pervect

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    The orbital period certainly depends on the mass of the black hole - this is present in the formula above in a disguised form, as the Schwarzschild radius r_s of the black hole.

    r, the Schwarzschild radius of the orbit, can be losely thought of as "distance from the black hole", but it should not be confused with the distance that one would measure with a ruler. The defintion of r is that the circumference of a circle drawn around the black hole at any Schwarzschild coordinate r is [itex]2 \pi r[/itex]

    The formula doesn't depend on the density or volume of the black hole, which is not defined. The formula also does not depend on the mass of the object that is orbiting as long as the object orbiting is small. Similarly the density of the orbiting object doesn't matter, either.

    The formulas needed can be found in various textbooks, or online at

    http://www.fourmilab.ch/gravitation/orbits/

    Actually the source above has worked a fair bit of the problem, something
    I didn't realize when I posted the question. Anyway

    The differential equations that describe the orbit in Schwarzschild coordinates r and [itex]\phi[/itex], both functions of proper time [itex]\tau[/itex], are given by:

    [tex]
    (\frac{dr}{d\tau})^2 + V^2(r) = E
    [/tex]
    [tex]
    V^2(r) = (1-\frac{2M}{r})(1 + \frac{L^2}{r^2})
    [/tex]
    [tex]
    \frac{d\phi}{d\tau} = \frac{L}{r^2}
    [/tex]


    and another equation for coordinate time which we won't need. In the above expression, E and L are conserved quantites, corresponding to energy and momentum (actually energy / unit rest mass and momentum per unit rest mass). M is the mass of the black hole. Geometric units are being used at this point, i.e. G=c=1.

    These equations give us enough information to solve for [itex]r(\tau)[/itex] and [itex]\phi(\tau)[/itex].

    However, its not immediatly obvious how to solve the equations for the conditions that an orbit be circular. The tricky bit is realizing that circular orbits occur when the effective potential V(r) is at a minimum or maximum, i.e. that dV/dr = 0.

    Since dV^2/dr = 2 V dv/dr, we can alternatively set the derivative of V^2(r) to zero, as is done in the webpage.

    This then gives

    L^2 (r-3M) = M r^2

    (see the webpage on "circular orbits") or


    [tex]
    L = \frac{r}{\sqrt{\frac{r}{M}-3}}
    [/tex]

    Since [itex]d\phi / d\tau = L/r^2[/itex] and is constant, the proper time of a complete orbit is given by T, where

    [tex]\frac{2 pi} {T}= \frac{L}{r^2}[/tex]

    This yields

    [tex]T = 2 \pi r \sqrt{\frac{r}{M} - 3}[/tex]

    To get the formula quoted, we need to "un-geometrize" the above expression, as the original expressions were in geometric units which assumed that c=G=1.

    We see that the expression has units of length - to convert them into time, we divide by c (which is a factor of unity in geometric units).

    We could convert M to a length, but I chose to simply re-write the formula in terms of r_s instead. In geometric units, r_s = 2M.

    This then gives us, finally

    [tex]T = \frac{2 \pi r}{c} \sqrt{\frac{2r}{r_s} - 3} [/tex]
     
    Last edited: Apr 18, 2006
  5. Apr 18, 2006 #4
    Nice derivation.
    Did you mean [tex]r < 3/2 r_s[/tex] ?
     
    Last edited: Apr 18, 2006
  6. Apr 18, 2006 #5

    George Jones

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    Yes, I agree.

    Actually, no circular orbits are possible for [itex]r < 3/2 r_s[/itex]. [itex]r = 3/2 r_s[/itex] is the photon sphere where light orbits, so, roughly, orbits inside this would have to have speed faster than light.

    For [itex]3/2 r_s < r < 2 r_s[/itex], circular orbits are wildly unstable, while for [itex]2 r_s < r < 3 r_s[/itex], circular orbits are mildly unstable.

    Wildly unstable: circular orbits spiral out to infinity when given an "infinitesimal" extra velocity.

    Mildly unstable: circular orbits spiral out to some finite value of r, then spiral back in to some minimum r, when given an "infinitesimal" extra velocity.

    These behaviours can be seen from the effective potential.

    Regards,
    George
     
    Last edited: Apr 18, 2006
  7. Apr 18, 2006 #6

    pervect

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    The last stable circular orbit is at r=6M in geometric units, i.e. r=3 r_s, where r_s = 2M.

    In the region r between 3M and 6M, circular orbits exist but are unstable. On the effective potential diagram these orbits are like a ball right at the top of a hill - the ball will tend to roll one way or the other.

    We can verify this by computing the second derivative of the effective potiential. For r=6M, the second derivative is zero. For r < 6M, the second derivative is negative.

    I imagine cl4 already knows about the photon sphere already from his question, but I'm going on a bit more to explain the physics for the benefit of readers such as GuilloB.

    Unstable circular orbits continue up to the photon sphere at r=3M, i.e. r=1.5 r_s, where the orbital velocity is equal to 'c'.

    Inside of r=3M, the "photon sphere", circular orbits do not exist, stable or otherwise.

    We can some of this graphically using the orbit simulator at

    http://www.fourmilab.ch/gravitation/orbits/

    By substiting r=5 and r=7.5 into the formula for L I derived earlier, we can see that they both have the same angular momentum, L=3.535, but different energies.

    The orbit simulator unfortunately does not label the r values, but we can type in the value L=3.535 into the simulator. By clicking at the "peak" and "valley" of the effective potential curve, we can then "launch" the orbiting particle with appropriate initial conditions. By clicking on the "peak" of the effective potential diagram, we can launch the object into an unstable circular orbit at r=5. By clicking on the "valley", we can launch it in the stable circular orbit at r=7.5. The unstable orbit tends not to last very long, of course, being unstable.

    Another interesting value of L is L = 3.464, which corresponds to the last stable orbit at R=6. There is only one stable point on the diagram, and random clicking will never quite find it - so while (marginally) stable orbits exist with this value of momentum, the intial conditions must be exactly right.
     
  8. Apr 18, 2006 #7

    Thank you. :-)
     
  9. Apr 18, 2006 #8

    George Jones

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    Here's how to see this.

    An unstable circular orbit is the top of a peak in [itex]V^{2} \left(r \right)[/itex] that occurs at

    [tex]
    r=\frac{L}{2M}\left( L-\sqrt{L^{2}-12M^{2}}\right)
    [/tex].

    Clearly, [itex]V^{2} \left(r \right) \rightarrow 1[/itex] as [itex]r \rightarrow \infty[/itex].

    Therefore, if the peak is higher than [itex]1[/itex], then the hill on the right (as [itex]r \rightarrow \infty[/itex]) is not high enough to constrain the motion when the particle "falls to the right off" the peak, and the circular orbit is wildly unstable.

    If the peak is lower than [itex]1[/itex], then the hill on the right (as [itex]r \rightarrow \infty[/itex]) is high enough to constrain the motion when the particle "falls to the right off" the peak, and the circular orbit is mildly unstable.

    To find the transition between wildly and mildly unstable, plug the above value of [itex]r[/itex] into [itex]V^{2} \left(r \right) = 1[/itex], and solve for [itex]L[/itex]. This gives [itex]L = 4M[/itex], which in turn gives [itex]r = 4M[/itex].

    Regards,
    George
     
    Last edited: Apr 18, 2006
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