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Homework Help: Distance between 2 orbits of a black hole

  1. Feb 5, 2008 #1
    [SOLVED] Distance between 2 orbits of a black hole

    1. The problem statement, all variables and given/known data

    I am currently working on the black holes in the Schwarzschild metric, an exercise course asks me to calculate the physical distance between different orbits.

    [tex]d_{1}[/tex] = distance betwenn 2M and 3M
    [tex]d_{2}[/tex] = distance betwenn 3M and 4M
    [tex]d_{3}[/tex] = distance betwenn 4M and 6M

    Where (x)M are the black hole orbit's.

    How to calculate this physical distance ?

    I know that the distance between 0 and 2M is 3km M/Msun but, I think that this distance isn't realy physical because it's expressed as a function of considered mass object normalized by the mass of sun.
    I thought integrate the Schwarzschild metric on the spatial term, but isn't trivial...

    Exercise may be simple, I do not know, but I do not know the answer, so I thank you in advance for your possible assistance.

    Last edited: Feb 5, 2008
  2. jcsd
  3. Feb 5, 2008 #2


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    In general, to find a radial distance what you must do is set [tex] d\phi = \d\theta = dt = 0 [/tex] in the metric. Then ds is the physical distance between two points along a radius. You then have a relation between ds and dr (the coordinate displacement). Can you write that relationship down?

    The total physical distance between two points is the integral of both sides of this relation from the initial to the final coordinate.

    PS: You are using geometrized units to write positions in terms of masses?
  4. Feb 5, 2008 #3
    Yes it's geometrized units, I don't use the technical terme sorry.

    I understand that my distance is (for d1)

    [tex]d_{1} = \int_{2M}^{3M} (1 - \frac{2M}{r})^{-1/2} dr [/tex]

    with no intergration of [tex]dt[/tex] and [tex]d\Omega[/tex]

    But the integration is difficult not?
    I've no idea to resolve this...
  5. Feb 5, 2008 #4


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    if you have to do it by hand it's tough. But you can look it up, for example using http://integrals.wolfram.com/index.jsp. It's simply a log (or may be written as an inverse sinh) and a bunch of square roots. It's not that bad
  6. Feb 5, 2008 #5
    Ok, thank you very much, indeed I hoped a solution more elegant that the integration beast and nasty, but if it's impossible, I will do that. Thanks for all Kdv :)
  7. Feb 5, 2008 #6


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    You are welcome. if the points were very close to each other, you could Taylor expand and the integral simplifies considerably. But you can't do that in your case.
  8. Feb 12, 2008 #7
    While you had right, it was necessary to integrate it numerically, using mathematica or something like that.
    The distance between two orbits, 2M to 6M is7.2M!

    x=r/M [/tex]

    [tex]x_1= 2 [/tex] [tex]x_2= 3 [/tex]

    [tex]d_1 = M\int_2^3{(1-x/M)}^{-1/2}dx [/tex]

    [tex]d_1= 3.05M [/tex] [tex]d_2= 1.54M [/tex] [tex]d_3= 2.60M [/tex]

    [tex]d_{tot}= 7.2M[/tex]

    We can see the curvature of space-time ...

    Again, thank you for your help.

    ps: Sorry for my english, i'm french.
  9. Feb 12, 2008 #8


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    I really thought it could be done analytically. I will check on this later when I have a few minutes.

    Et ne vous en faites pas pour votre anglais, il est en fait tres bon!
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