Gravitating Perfect Fluid Solution & Mass/Energy Total

[Mentz114]

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
$$\begin{align*}<br /> M_s &=\ 4\pi\int_0^{r_{max}} r^2\rho\ dr = 4\pi\left[ \frac{\,b\,{r}^{3}}{2\,b\,{r}^{2}+a} \right]_0^{r_{max}}\\<br /> &= \frac{\sqrt{2}\,\pi\,\sqrt{1-a}\,\left( 1-a\right) }{\sqrt{b}}\\<br /> &= 8\pi\ m_s<br /> \end{align*}$$
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).

 

[PeterDonis]

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.

 

[Mentz114]

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 r_max -> 4m, and as a->1 so r_max increases without bound.