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Gravitational Binding Energy in GR

  1. Mar 19, 2014 #1
    What is the gravitational binding energy in GR in the spherically symmetric case?

    I calculate ##E=mc^2(1-\frac{1}{\sqrt{1-\frac{r_s}{R}}})##

    where ##m## is the mass of the body, ##r_s## is the Schwarzschild radius, and ##R## is the area radius as in the Birkhoff theorem.
     
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  3. Mar 19, 2014 #2

    WannabeNewton

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    For a spherically symmetric static star it is ##E = M_p - M## where ##M_p## is the total proper mass of the star and ##M## is the total mass of the star.
     
  4. Mar 19, 2014 #3

    pervect

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    We need a bit more information, were you assuming a constant density interior?
     
  5. Mar 20, 2014 #4
    We can assume the mass of a star is concentrated at a single point and a body of mass ##m## is a distance ##R## from the point. The binding energy is the energy required to move the body of mass ##m## from ##R## to infinity.

    Essentially I am asking what is the GR counterpart to the Newtonian formula ##-\frac{GMm}{r}##.

    In my calculations I derived ##E=mc^2(1-\frac{1}{k})## where ##k=\sqrt{1-\frac{r_s}{R}}## and ##r_s## is the Schwarzschild radius.
     
  6. Mar 20, 2014 #5

    pervect

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    MTW's gravitation has a formula for binding energy on pg 604, but their formula won't converge for the case which you describe. Certainly if you plug 0 into the Newtonian formula you won't get a finite binding energy.

    Note: MTW's expression uses geometric units, where G=1.

    I believe a taylor series expansion of MTW's formula will show that it gives an answer which is always greater than the Newtonian one, so I don't see how you can get any value other than infinity when r=0.

    MTW's formula:
    [tex]-\int_0^r \rho \left[ \left( 1-2m/r \right)^{-\frac{1}{2}} -1 \right] \,4 \pi r^2 dr [/tex]

    ##\rho## is the density, which in general a function of r that depends on the equation of state of whatever is composing the mass. m is not the total mass, but the mass inside radius r, so it's also a function of r. ##\rho## is the proper density, i.e. the density in a locally Minkowskii frame.

    [add]
    ## m(r) = \int 4 \pi r^2 \rho dr## see pg 603. This equation for m(r) is only valid in Schwarzschild coordinates.

    Taylor expansion for m/r < 1
    [tex]-\int_0^r \rho \left[ \frac{1}{2} \frac{2m}{r} + \frac{3}{8} \left( \frac{2m}{r} \right) ^2 + \frac{5}{16} \left( \frac{2m}{r} \right)^3 + ... \right] 4 \pi r^2 dr
    [/tex]

    If you take only the first term of the taylor series expansion, you get the Newtonian result.
     
    Last edited: Mar 20, 2014
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