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Distances between planet and observer near BH

  1. Jun 30, 2013 #1
    Imagine that we have a system that consists of a massive black hole, and the asteroid revolving around it on a stable orbit. What method can determine the distance to these asteroids observer who is still close to the event horizon?
    The first thing that came to mind is to determine the signal propagation time to the asteroid and back. This time is equal to:
    ##\Delta \tau_{observer} = \sqrt{1-\dfrac{r_g}{r_{observer}}}
    \int\dfrac{dr}{1-\frac{r_g}{r}}
    ##
    But I do not know ##r_{observer}## and ##r_{asteroid}##
    Need a different way.
     
    Last edited: Jun 30, 2013
  2. jcsd
  3. Jun 30, 2013 #2

    DrGreg

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    You'll need to tell us what you do know.
     
  4. Jun 30, 2013 #3
    Suppose, the observer knows that the orbit is circular and also knows the asteroid's rotation period determined by proper clock. A black hole is not rotating, i.e. we can use the Schwarzschild metric.
     
  5. Jun 30, 2013 #4
    NASA Disagrees?

    http://www.nustar.caltech.edu/
     
  6. Jun 30, 2013 #5

    Bill_K

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    To calculate robs and rast from the observations of the orbits you need to use Kepler's Law. A truly remarkable fact about the Schwarzschild solution is that, when expressed in terms of the Schwarzschild coordinates r and t, Kepler's Law is completely unmodified from its Newtonian form.

    In more detail, the angular velocity of a test particle in a circular orbit about a Schwarzschild mass is (dφ/dτ)2 = (GM/r3)(1 - 3rs/r)-1. Here τ is the particle's own proper time. Hopefully you recognize the leading factor (GM/r3) as Kepler's expression. Also, dφ/dτ is related to the period T of the orbit (expressed in proper time) by dφ/dτ = 2π/T.

    The proper time is related to Schwarzschild coordinate time t by dt/dτ = (1 - 3rs/r)-1/2, so (dφ/dt)2 = (dφ/dτ)2(dτ/dt)2 = GM/r3. (The relativistic factors exactly cancel!) Thus we are left with the familiar "period squared proportional to distance cubed". By measuring the period of an asteroid in Schwarzschild time we can immediately calculate from this the radius of its circular orbit.

    EDIT: Corrected some algebra!
     
    Last edited: Jun 30, 2013
  7. Jun 30, 2013 #6
    Yes but does that Explain a black hole transforming into a supermassive one then into a blazar?
     
  8. Jun 30, 2013 #7
    Bill_K, thank you. There is a simple way to get such an expression (dφ/dτ)2 = (GM/r3)(1 - 3rs/r)-1/2 from dτ2=(1-rg/r)dt2 - r22 ? Where can I get the equation in order to exclude dt2 ie dt/dτ = (1 - 3rs/r)-1/2?
     
    Last edited: Jun 30, 2013
  9. Jun 30, 2013 #8
    Use Calculus, but what I don't wanna?
     
  10. Jun 30, 2013 #9

    Bill_K

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    The key word is "circular". You have to solve the equations of motion for a test particle in a Schwarzschild field, in particular for a circular orbit.
     
  11. Jun 30, 2013 #10
    Oh, right. I will try.
     
  12. Jun 30, 2013 #11
    So, we need to do as in the case of Newton's theory, ie we need to have the solution of the field equations and the equations of motion. The expression for the metric is only a solution of the field equations.
     
  13. Jun 30, 2013 #12
    I have another question.
    To determine the distance between two points in Minkowski space, it was necessary to send a light signal forth and back. Knowing time between sending and receiving, we could determine the distance by multiplying time/2 by the speed of light. In the case of the Schwarzschild metric, it is enough for us to know the time of sending and receiving the signal to determine the distance between two points?
     
  14. Jun 30, 2013 #13

    pervect

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