Gravitating Perfect Fluid Solution & Mass/Energy Total

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SUMMARY

The discussion centers on a static spherically symmetric perfect fluid solution of the Einstein Field Equations (EFE) characterized by an energy-momentum tensor of the form ##diag(\rho,p,p,p)##. The energy density ##\rho## and pressure ##p## are defined with parameters a and b, where b>0 and 0 PREREQUISITES

  • Understanding of Einstein Field Equations (EFE)
  • Familiarity with perfect fluid solutions in general relativity
  • Knowledge of the Komar mass integral
  • Basic concepts of energy-momentum tensors
NEXT STEPS
  • Study the derivation and implications of the Komar mass integral
  • Explore the properties of static spherically symmetric solutions in general relativity
  • Investigate the role of energy-momentum tensors in gravitational theories
  • Examine the relationship between pressure, energy density, and gravitational binding energy
USEFUL FOR

Researchers in theoretical physics, particularly those focused on general relativity, cosmology, and gravitational theories, will benefit from this discussion. It is also relevant for students studying advanced concepts in gravitational physics.

Mentz114
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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 &amp;=\ 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 /> &amp;= \frac{\sqrt{2}\,\pi\,\sqrt{1-a}\,\left( 1-a\right) }{\sqrt{b}}\\<br /> &amp;= 8\pi\ m_s<br /> \end{align*}<br />
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).
 
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Mentz114 said:
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.
 
PeterDonis said:
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.
 
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