What is gravitating ?

  • Thread starter Mentz114
  • Start date
  • #1
Mentz114
5,432
292
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).
 

Answers and Replies

  • #2
38,396
16,173
what happened to the pressure terms in the EMT ?

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.
 
  • #3
Mentz114
5,432
292
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.
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:

Suggested for: What is gravitating ?

Replies
10
Views
10K
Replies
3
Views
3K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
9
Views
2K
Replies
12
Views
858
Replies
11
Views
3K
Replies
28
Views
4K
Replies
15
Views
1K
Top