Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is gravitating ?

  1. Sep 7, 2013 #1

    Mentz114

    User Avatar
    Gold Member

    There is a static spherically symmetric perfect fluid solution of the EFE where the energy-momentum tensor is ##diag(\rho,p,p,p)## with ##\rho=b\,\left( 2\,b\,{r}^{2}+3\,a\right) /{\left( 2\,b\,{r}^{2}+a\right) }^{2}## and ##p={b}/({2\,b\,{r}^{2}+a})##. a and b are parameters with b>0 and 0<a<1. On the surface ##r=\sqrt{(1-a)/(2b)}\equiv r_{max}## the PF metric coincides with the Schwarzschild exterior, as long as the Schwarzschild parameter m has the value ##m_s= {\sqrt{1-a}\,\left( 1-a\right) }/( {4\,\sqrt{2b}})##.

    Calculating ##M_s## the mass/energy total of the PF
    [tex]\begin{align*}
    M_s &=\ 4\pi\int_0^{r_{max}} r^2\rho\ dr = 4\pi\left[ \frac{\,b\,{r}^{3}}{2\,b\,{r}^{2}+a} \right]_0^{r_{max}}\\
    &= \frac{\sqrt{2}\,\pi\,\sqrt{1-a}\,\left( 1-a\right) }{\sqrt{b}}\\
    &= 8\pi\ m_s
    \end{align*}
    [/tex]
    This seems most satisfactory but raises the question - what happened to the pressure terms in the EMT ? It appears that the integral of the energy density accounts for all the exterior vacuum curvature. Is this an anomaly or am I right to be surprised ?

    (Actually I was very glad when the integral turned out like this - until the question of the pressure appeared).
     
  2. jcsd
  3. Sep 7, 2013 #2

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Heuristically, the positive contribution of the pressure to the mass is exactly canceled by the negative contribution of gravitational binding energy to the mass. The easiest way I know of to see how that works is to look at the Komar mass integral, in which ##\rho + 3 p## appears in the integrand, but also the "redshift factor" ##\sqrt{1 - 2m(r) / r}## appears, and the two contributions cancel each other when the integral is computed over the entire volume of the object.
     
  4. Sep 8, 2013 #3

    Mentz114

    User Avatar
    Gold Member

    Thanks. That could account for it. I'll check out the KM integral.

    This PF is more realistic than I first thought because the radius of the ball can be set to any multiple of 2m by a suitable choice of parameter a. As a-> 0 so rmax -> 4m, and as a->1 so rmax increases without bound.
     
    Last edited: Sep 8, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook