- #1

Mentz114

- 5,432

- 292

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).