Consider the metric describing the interior of a star,

$$d s^{2}=-e^{2 \alpha(r)} d t^{2}+e^{2 \beta(r)} d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)$$

defined for $0 \leqslant r \leqslant r_{0}$ by

$$e^{\alpha(r)}=\frac{3}{2} e^{-\beta_{0}}-\frac{1}{2} e^{-\beta(r)},$$

with

$$e^{-2 \beta(r)}=1-A r^{2}$$

Here $A=2 M / r_{0}^{3}$, where $M$ is the mass of the star, $\beta_{0}=\beta\left(r_{0}\right)$, and we have taken units in which we have set $G=c=1$.

(i) The star is made of a perfect fluid with energy-momentum tensor

$$T_{a b}=(p+\rho) u_{a} u_{b}+p g_{a b}$$

Here $u^{a}$ is the 4-velocity of the fluid which is at rest, the density $\rho$ is constant throughout the star $\left(0 \leqslant r \leqslant r_{0}\right)$ and the pressure $p=p(r)$ depends only on the radial coordinate. Write down the Einstein field equations and show that they may be written as

$$R_{a b}=8 \pi(p+\rho) u_{a} u_{b}+4 \pi(\rho-p) g_{a b}$$

(ii) Using the formulae given below, or otherwise, show that for $0 \leqslant r \leqslant r_{0}$, one has

$$\begin{aligned} &4 \pi(\rho+p)=\frac{\left(\alpha^{\prime}+\beta^{\prime}\right)}{r} e^{-2 \beta(r)} \\ &4 \pi(\rho-p)=\left(\frac{\beta^{\prime}-\alpha^{\prime}}{r}-\frac{1}{r^{2}}\right) e^{-2 \beta(r)}+\frac{1}{r^{2}} \end{aligned}$$

where primes denote differentiation with respect to $r$. Hence show that

$$\rho=\frac{3 A}{8 \pi} \quad, \quad p(r)=\frac{3 A}{8 \pi}\left(\frac{e^{-\beta(r)}-e^{-\beta_{0}}}{3 e^{-\beta_{0}}-e^{-\beta(r)}}\right)$$

[The non-zero components of the Ricci tensor are

$$\begin{aligned} &R_{00}=e^{2 \alpha-2 \beta}\left(\alpha^{\prime \prime}-\alpha^{\prime} \beta^{\prime}+\alpha^{\prime 2}+\frac{2 \alpha^{\prime}}{r}\right) \\ &R_{11}=-\alpha^{\prime \prime}+\alpha^{\prime} \beta^{\prime}-\alpha^{\prime 2}+\frac{2 \beta^{\prime}}{r} \\ &R_{22}=1+e^{-2 \beta}\left[\left(\beta^{\prime}-\alpha^{\prime}\right) r-1\right] \\ &R_{33}=\sin ^{2} \theta R_{22} \end{aligned}$$

Note that

$$\left.\alpha^{\prime}=\frac{1}{2} A r e^{\beta-\alpha} \quad, \quad \beta^{\prime}=A r e^{2 \beta} .\right]$$