# Dr Phill’s Science Made Simple

## The Schwarzschild Interior Metric

Having discovered the Schwarzschild Outer Metric is an exact solution to Einstein’s Field Equations for the vacuum of space around a static and symmetric body in 1915, Karl Schwarzschild went on to discover the Schwarzschild Interior Metric in 1916. The metric is based on the assumption that the body is a spherical non-rotating incompressible fluid.

### Define the metric

The metric can defined in the same way as for the outer metric.

$ds^2 = A(r)dt^2 - B(r)dr^2 - r^2d\theta^2 - r^2 \sin^2\theta d\phi^2$

Where $$A$$ and $$B$$ are unknown functions of $$r$$. The interior metric must agree with the outer metric at $$r = r_g$$.

The metric and its inverse are now defined to be:

$g_{\alpha\beta} = \begin{bmatrix} A & 0 & 0 & 0\\ 0 & -B & 0 & 0\\ 0 & 0 & -r^2 & 0\\ 0 & 0 & 0 & -r^2 \sin^2 \theta \end{bmatrix}$ $g^{\alpha\beta} = \begin{bmatrix} \frac{1}{A} & 0 & 0 & 0\\ 0 & -\frac{1}{B} & 0 & 0\\ 0 & 0 & -\frac{1}{r^2} & 0\\ 0 & 0 & 0 & -\frac{1}{r^2 \sin^2 \theta} \end{bmatrix}$

### The Stress-Energy Tensor

The stress-energy tensor needs to be defined in terms of the density $$\rho$$ and the pressure $$p$$ inside the body.

Given that the mass of the body in $$M$$ and the radius of the body is $$R$$, then the density is a constant.

$\rho = \frac{3M}{4\pi R^3}$

The mass function defines the mass of the body inside a given radius.

$m = \frac{4}{3}\pi r^3\rho$

The stress-energy tensor can be constructed from the density, pressure, and the metric.

$\begin{gather} T_{\mu\nu} = c^2 \begin{bmatrix} \rho A & 0 & 0 & 0\\ 0 & p B & 0 & 0\\ 0 & 0 & pr^2 & 0\\ 0 & 0 & 0 & pr^2\sin^2\theta \end{bmatrix} \end{gather}$

### Einstein Tensor

The Einstein tensor is calculated in the same way as for the outer metric.

\begin{align*} G_{tt} &= -\frac{A}{r^2B}(1 - r\frac{B'}{B} - B)\\ G_{rr} &= \frac{1}{r^2}(1 + r\frac{A'}{A} - B)\\ G_{\theta\theta} &= \frac{r^2}{2B}\bigg(\frac{A''}{A} + \frac{A'}{rA} - \frac{B'}{rB} + \frac{A'(AB)'}{2A^2B}\bigg)\\ G_{\phi\phi} &= G_{\theta\theta}\sin^2\theta \end{align*}

### Field Equations

The field equations are:

\begin{align*} -\frac{A}{r^2B}\bigg(1 - r\frac{B'}{B} - B\bigg) &= c^2\kappa\rho A\\ \frac{1}{r^2}\bigg(1 + r\frac{A'}{A} - B\bigg) &= c^2\kappa pB\\ \frac{r^2}{2B}\bigg(\frac{A''}{A} + \frac{A'}{rA} - \frac{B'}{rB} - \frac{A'(AB)'}{2A^2B}\bigg) &= c^2\kappa pr^2 \end{align*}

Rearrange terms.

$\frac{1}{B} - \frac{rB'}{B^2} = 1 - c^2r^2\kappa\rho\qquad{(1)}$ $\frac{1}{r^2 B} + \frac{A'}{rAB} - \frac{1}{r^2} = c^2\kappa p\qquad{(2)}$ $\frac{A''}{2AB} + \frac{A'}{2rAB} - \frac{B'}{2rB^2} - \frac{A'}{4AB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = c^2\kappa p\qquad{(3)}$

### Solve the First Equation

Equation (1) can be written as:

$\frac{1}{B} - \frac{rB'}{B^2} = \frac{d}{dr}\bigg(\frac{r}{B}\bigg) = 1 - c^2r^2\kappa\rho$

Integrate from $$r = 0$$.

$\frac{r}{B} = r - \frac{1}{3}c^2r^3\kappa\rho$

Divide by $$r$$.

$\frac{1}{B} = 1 - \frac{1}{3}c^2r^2\kappa\rho\qquad{(4)}$

Define the mass function $$m = m(r)$$ as density multiplied by volume.

$m = \frac{4}{3}\pi\rho r^3\qquad{(5)}$

Now $$\kappa = \frac{8\pi G}{c^4}$$ and let $$\mu = \frac{G}{c^2}$$.

$\frac{1}{B} = 1 - \frac{8\pi G\rho r^2}{3c^2} = 1 - \frac{2\mu m}{r}\qquad{(6)}$

This must agree with the outer solution when $$r = R$$ and $$m = M$$.

$B = \bigg(1 - \frac{2GMr^2}{c^2R^3}\bigg)^{-1}\qquad{(7)}$

### Solve the Second Equation

This is more complex as the pressure is dependent on the metric and both are unknown.

Multiply (2) by $$r^2$$ and rearrange.

$\frac{1}{B}\bigg(\frac{rA'}{A} + 1\bigg) = c^2\kappa pr^2 + 1\qquad{(8)}$

Differentiate with respect to $$r$$.

$-\frac{B'}{B^2}\bigg(\frac{rA'}{A} + 1\bigg) + \frac{A'}{AB} + \frac{rA''}{AB} - \frac{rA'A'}{A^2B} = c^2\kappa p'r^2 + 2c^2\kappa pr$

Divide by $$2r$$ and rearrange.

$\frac{A''}{2AB} + \frac{A'}{2rAB} - \frac{B'}{2rB^2} - \frac{A'}{2AB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = \frac{1}{2}c^2\kappa p'r + c^2\kappa\qquad{(9)}$

Subtract (3) from (9).

$-\frac{A'}{4AB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = \frac{1}{2}c^2\kappa p'r$

Multiple by 2, divide by $$r$$, and rerrange.

$-\frac{A'}{2A}\frac{1}{rB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = c^2\kappa p'\qquad{(10)}$

Divide (1) by $$r^2$$, rearrange, and add (2).

$\frac{1}{rB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = c^2\kappa(\rho + p)\qquad{(11)}$

Substitute (11) into (10) and divide by $$c^2\kappa$$.

$p' = -\frac{A'}{2A}(\rho + p)\qquad{(12)}$

Substitute (6) into (8) and multiply by $$r$$.

$(r - 2\mu m)\bigg(\frac{rA'}{A} + 1\bigg) = r + c^2\kappa pr^3$

Rearrange and divide by 2.

$\frac{A'}{2A} = \frac{c^2\kappa pr^3 + 2\mu m}{2r(r - 2\mu m)}\qquad{(13)}$

Eliminate the $$A$$ terms from (12) and (13).

$\frac{p'}{\rho + p} = -\frac{c^2\kappa pr^3 + 2\mu m}{2r(r - 2\mu m)}$

Let $$\lambda = \frac{c^2\kappa}{3} = \frac{8\pi G}{3c^2}$$, and hence $$2\mu m = \lambda\rho r^3$$.

$\frac{p'}{\rho + p} = \frac{3\lambda p r^3 + \lambda\rho r^3}{2r(r - \lambda\rho r^3)}$

Multiply by $$2\rho$$ and reorganise.

$\frac{2\rho p'}{(\rho + p)(\rho + 3p)} = \frac{\lambda\rho r}{1 - \lambda\rho r^2}$

Write in integral form.

$\int^0_p{\bigg(\frac{3}{\rho + 3p} - \frac{1}{\rho + p}\bigg)dp} = \int^R_r\frac{\lambda\rho r}{1 - \lambda\rho r^2}dr$

Integrate.

$-\ln(\rho + 3p) + \ln(\rho + p) = \frac{1}{2}\ln(1 - \lambda\rho R^2) - \frac{1}{2}\ln(1 - \lambda\rho r^2)$

Take exponentials.

$\frac{\rho + p}{\rho + 3p} = \sqrt\frac{1 - \lambda\rho R^2}{1 - \lambda\rho r^2}\qquad{(14)}$

Let $$y^2 = 1 - \lambda\rho r^2$$ and $$Y^2 = 1 - \lambda\rho R^2$$.

$\frac{\rho + p}{\rho + 3p} = \frac{Y}{y}$

Solve for $$p$$.

$p = \frac{\rho(y - Y)}{3Y - y}$

Add $$\rho$$ to both sides and simplify.

$\rho + p = \frac{2\rho Y}{3Y - y}\qquad{(15)}$

Differentiate with respect to $$y$$.

$\frac{dp}{dy} = \frac{2\rho Y}{(3Y - y)^2}\qquad{(16)}$

Rewrite (12) in integral form and change variable from $$p$$ to $$y$$.

$\int^r_0\frac{1}{\rho + p}dp = \int^y_1\frac{1}{\rho + p}\frac{dp}{dy}dy = -\int^r_0\frac{A'}{2A}dr$

Substitute (15) and (16) to eliminate $$p$$ and multiply by $$2$$.

$2\int^y_1\frac{3Y - y}{2\rho Y}\frac{2\rho Y}{(3Y - y)^2}dy = 2\int^y_1\frac{dy}{3Y - y} = -\int^r_0\frac{A'}{A}dr$

Integrate and negate.

$2\ln(3Y - y) - 2\ln(3Y - 1) = \ln A - \ln A_0$

Take exponentials.

$A = A_0\bigg(\frac{3Y - y}{3Y - 1}\bigg)^2$

We need to find the constant of integration $$A_0$$. At the surface of the body $$r = R$$ and $$y = Y$$. The value of $$A$$ must agree with the outer solution. Substitute the values.

$1 - \frac{2GM}{c^2R} = Y^2 = A_0\bigg(\frac{2Y}{3Y - 1}\bigg)^2$

Hence.

$A = \frac{1}{4}(3Y - y)^2$

Substitute for $$y$$ and $$Y$$.

$A = \frac{1}{4}\bigg(3\sqrt{1 - \frac{2GM}{c^2R}} - \sqrt{1 - \frac{2GMr^2}{c^2R^3}}\bigg)^2$

The Schwarzschild interior metric is now known.

$ds^2 = c^2d\tau^2 = \frac{1}{4}\bigg(3\sqrt{1 - \frac{2GM}{c^2R}} - \sqrt{1 - \frac{2GMr^2}{c^2R^3}}\bigg)^2dt^2 - \bigg(1 - \frac{2GMr^2}{c^2R^3}\bigg)^{-1}dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2$