Having discovered the Schwarzschild Outer Metric is an exact solution to the 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.
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 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.
\[\begin{gather} T_{\mu\nu} = c^2 \begin{bmatrix} \rho A & 0 & 0 & 0\\ 0 & p B & 0 & 0\\ 0 & 0 & pr^2 & 0\\ 0 & 0 & 0 & pr^2\sin^2\theta \end{bmatrix} \end{gather}\]
The Einstein tensor is calculated in the same way as for the outer metric.
\[\begin{align*} G_{tt} &= -\frac{A}{r^2B}(1 - r\frac{B'}{B} - B)\\ G_{rr} &= \frac{1}{r^2}(1 + r\frac{A'}{A} - B)\\ G_{\theta\theta} &= \frac{r^2}{2B}\bigg(\frac{A''}{A} + \frac{A'}{rA} - \frac{B'}{rB} + \frac{A'(AB)'}{2A^2B}\bigg)\\ G_{\phi\phi} &= G_{\theta\theta}\sin^2\theta \end{align*}\]
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.
\[\frac{1}{B} - \frac{rB'}{B^2} = 1 - c^2r^2\kappa\rho\](1) \[\frac{1}{r^2 B} + \frac{A'}{rAB} - \frac{1}{r^2} = c^2\kappa p\](2) \[\frac{A''}{2AB} + \frac{A'}{2rAB} - \frac{B'}{2rB^2} - \frac{A'}{4AB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = c^2\kappa p\](3)
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 \(r\).
\[\frac{1}{B} = 1 - \frac{1}{3}c^2r^2\kappa\rho\](4)
Define the mass function \(m = m(r)\) as density multiplied by volume.
\[m = \frac{4}{3}\pi\rho r^3\](5)
Now \(\kappa = \frac{8\pi G}{c^4}\) and let \(\mu = \frac{G}{c^2}\).
\[\frac{1}{B} = 1 - \frac{8\pi G\rho r^2}{3c^2} = 1 - \frac{2\mu m}{r}\](6)
This must agree with the outer solution when \(r = R\) and \(m = M\).
\[B = \bigg(1 - \frac{2GMr^2}{c^2R^3}\bigg)^{-1}\](7)
This is more complex as the pressure is dependent on the metric and both are unknown.
Multiply ([2]) by \(r^2\) and rearrange.
\[\frac{1}{B}\bigg(\frac{rA'}{A} + 1\bigg) = c^2\kappa pr^2 + 1\](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.
\[\frac{A''}{2AB} + \frac{A'}{2rAB} - \frac{B'}{2rB^2} - \frac{A'}{2AB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = \frac{1}{2}c^2\kappa p'r + c^2\kappa\](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.
\[-\frac{A'}{2A}\frac{1}{rB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = c^2\kappa p'\](10)
Divide ([1]) by \(r^2\), rearrange, and add ([2]).
\[\frac{1}{rB}\bigg(\frac{A'}{A} + \frac{B'}{B}\bigg) = c^2\kappa(\rho + p)\](11)
Substitute ([11]) into ([10]) and divide by \(c^2\kappa\).
\[p' = -\frac{A'}{2A}(\rho + p)\](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.
\[\frac{A'}{2A} = \frac{c^2\kappa pr^3 + 2\mu m}{2r(r - 2\mu m)}\](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.
\[\frac{\rho + p}{\rho + 3p} = \sqrt\frac{1 - \lambda\rho R^2}{1 - \lambda\rho r^2}\](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.
\[\rho + p = \frac{2\rho Y}{3Y - y}\](15)
Differentiate with respect to \(y\).
\[\frac{dp}{dy} = \frac{2\rho Y}{(3Y - y)^2}\](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.
\[ds^2 = c^2d\tau^2 = \frac{1}{4}\bigg(3\sqrt{1 - \frac{2GM}{c^2R}} - \sqrt{1 - \frac{2GMr^2}{c^2R^3}}\bigg)^2dt^2 - \bigg(1 - \frac{2GMr^2}{c^2R^3}\bigg)^{-1}dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2\]