Schwarzschild Metric

Derivation

In terms of spherical coordinate, the line element of a general spherically symmetric spacetime is $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu,$$ where \(x^\mu = t,r,\theta,\phi \). At constant \(r\) and \(t\), the surface is two-dimensional spherical surface, called a two sphere.

A line of \(\theta\)=const. and \(\phi\)=const. is orthogonal to the two-spheres. This line by definition is tangent to \(\vec{e}_r\). So, \(\vec{e}_r\cdot \vec{e}_\theta=\vec{e}_r\cdot \vec{e}_\phi=0\). This requires \(g_{r\theta}=g_{r\phi}=0\). Hence, the metric is restriced to the form $$ds^2 = g_{00}dt^2 + 2g_{0r}drdt + 2g_{0\theta}d\theta dt+2g_{0\phi}d\phi dt + g_{rr}dr^2 + r^2d\Omega^2.$$

Since not only at \(t\)=const. is spherical symmetric, a line \(r\)=const., \(\theta\)=const., \(\phi\)=const., at any time \(t\) is also orthogonal to the two-spheres. This line by definition is tangent to \(\vec{e}_t\). So, \(\vec{e}_t\cdot \vec{e}_\theta=\vec{e}_t\cdot \vec{e}_\phi=0\). This requires \(g_{t\theta}=g_{t\phi}=0\). Hence, the metric is further restriced to the form $$ds^2 = g_{00}dt^2 + 2g_{0r}drdt + g_{rr}dr^2 + r^2d\Omega^2.$$ This is the general metric of a spherically symmetric spacetime.

Define a static spactime to be a spacetime with time coordinate \(t\) that

• all metric componenets are independent of time
• geometry is unchanged if time is reversed, i.e. \(t\to -t\)

The transformation matrix of \((t,r,\theta,\phi)\to(-t,r,\theta,\phi)\) is given by \(\Lambda^{\bar{0}}_0=-1\),\(\Lambda^i_j=\delta^i_j\). So, we have \begin{align} g_{\bar{0}\bar{0}} &= (\Lambda^0_\bar{0})^2g_{00} = g_{00}\\ g_{\bar{0}\bar{r}} &= \Lambda^0_\bar{0}\Lambda^r_\bar{r}g_{0r} = -g_{0r}\\ g_{\bar{r}\bar{r}} &= (\Lambda^r_\bar{r})^2g_{rr} = g_{rr}. \end{align} To have the geometry remains unchanged, \(g_{\bar{\alpha}\bar{\beta}}=g_{\alpha\beta}\), it is required to have \(g_{0r}=0\). Hence, the metric is further restriced to the form $$ds^2 = g_{00}dt^2 + g_{rr}dr^2 + r^2d\Omega^2.$$

The Christoffel symbols of the metric are \begin{align} \Gamma^r_{rr}&=g_{rr,r}/(2g_{rr}),\, \Gamma^r_{\theta\theta}=-r/g_{rr}, \,\Gamma^r_{\phi\phi}=-r\sin^2\theta/g_{rr}, \,\Gamma^r_{00}=-g_{00,r}/(2g_{rr})\\ \Gamma^\theta_{r\theta}&=\Gamma^\theta_{\theta r}=1/r,\,\Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta\\ \Gamma^\phi_{r\phi}&=\Gamma^\phi_{\phi r}=1/r,\, \Gamma^\phi_{\theta\phi}=\Gamma^\phi_{\phi\theta}=\cot\theta\\ \Gamma^0_{0r}&=\Gamma^0_{r0} = g_{00,r}/(2g_{00}). \end{align}

In case of zero cosmological constant in vaccum, substituting the definition of Einstein tensor into the field equations, we have $$R_{\alpha\beta} = 0,$$ that is, $$\Gamma^\rho_{\beta\alpha,\rho}-\Gamma^\rho_{\rho\alpha,\beta}+\Gamma^\rho_{\rho\lambda}\Gamma^\lambda_{\beta\alpha}-\Gamma^\rho_{\beta\lambda}\Gamma^\lambda_{\rho\alpha}=0,$$ which gives us three equations \begin{align} 4g_{rr,r}g_{00}^2-2rg_{00,rr}g_{rr}g_{00} + rg_{rr,r}g_{00,r}g_{00} + rg_{00,r}^2g_{rr} &= 0 \tag{1}\\ rg_{rr,r}g_{00} + 2g_{rr}^2g_{00} - 2g_{rr}g_{00} - rg_{00,r}g_{rr} &= 0 \tag{2}\\ -2rg_{00,rr}g_{rr}g_{00} + rg_{rr,r}g_{00,r}g_{00} + rg_{00,r}^2 - 4g_{00,r}g_{rr}g_{00} &= 0 \tag{3}.\\ \end{align} Subtracting equation (1) and (3), $$g_{rr,r}g_{00} + g_{rr}g_{00,r} = 0,$$ which implies $$g_{rr}g_{00}=K,$$ where \(K\) is a constant. Substituting this into (2), $$rg_{rr,r}=g_{rr}(1-g_{rr}).$$ The general solution is $$g_{rr}=\left(1+\frac{1}{Sr}\right)^{-1},$$ where \(S\) is a non-zero real constant. So, the metric becomes $$ds^2 = K\left(1+\frac{1}{Sr}\right)dt^2 + \left(1+\frac{1}{Sr}\right)^{-1}dr^2 + r^2d\Omega^2.$$ By weak-field approximation, $$g_{00}=K\left(1+\frac{1}{Sr}\right)\approx -1 + \frac{2M}{r},$$ we have $$K=-1\, \text{and} \, \frac{1}{S}=-\frac{2M}{r}.$$ So, the metric for a spherical symmetric spacetime in vaccum is $$\bbox[5px, border: 2px solid #666]{ ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2d\Omega^2. }$$ This is known as the Schwarzschild metric. It describes the spacetime by a spherical star in vaccum.

Ref: Schutz

Example

Bob is exploring a Schwarzschild black hole of mass M. His spaceship is held stationary at \(r = R\) on the equatorial plane (\(\theta = \pi/2\)). Find the component of the 4-velocity of Bob.

Solution: Since Bob is at rest on the equatorial plane, $$u^B = \left(\frac{dt_B}{d\tau},0,0,0\right).$$ By \(g_{\alpha\beta}u^\alpha u^\beta = -1\), $$-\left(1-\frac{2M}{r}\right)\frac{dt_B}{d\tau} = -1.$$ So, we have $$\frac{dt_B}{d\tau} = \frac{1}{\sqrt{1-\frac{2M}{r}}}.$$ Hence, the 4-velocity of Bob is $$u^B = \left(\left(1-\frac{2M}{R}\right)^{-1/2},0,0,0\right).$$

Checkpoint

Alice is exploring a Schwarzschild black hole of mass M. Her spaceship is orbiting in a circular orbit at \(r = R\) on the equatorial plane (\(\theta = \pi/2\)). The angular velocity of her spaceship measured by a far away stationary observer is a constant \(\Omega = \frac{d\phi}{dt}\). Find the component of the 4-velocity of Alice.

For a far away observer, using his local coordinate $$ds^2 = dt^2 + d\phi^2,$$ we have \begin{align} dt &= dt_A\sqrt{1-\frac{2M}{r}}\\ d\phi &= d\phi_A.\\ \end{align} By \(\Omega = \frac{d\phi}{dt}\), we have $$\Omega = \frac{d\phi}{dt} = \frac{d\phi_A}{dt_A}\left(\frac{1}{\sqrt{1-\frac{2M}{r}}}\right).$$ So, the 4-velocity of Alice is $$u^A = \frac{dt_A}{d\tau}\left(1,0,0,\frac{d\phi_A}{dt_A}\right) = \frac{dt_A}{d\tau}\left(1,0,0,\sqrt{1-\frac{2M}{r}}\Omega\right).$$ By \(g_{\alpha\beta}u^\alpha u^\beta = -1\), $$-\left(1-\frac{2M}{r}\right)\left(\frac{dt_A}{d\tau}\right)^2 + r^2\sin^2\theta\left(\frac{dt_A}{d\tau}\right)^2\left(1-\frac{2M}{r}\right)\Omega^2 = -1.$$ So, we have $$\frac{dt_A}{d\tau} = \frac{1}{\sqrt{(1-R^2\Omega^2)\left(1-\frac{2M}{R}\right)}}.$$ Hence, the 4-velocity of Alice is $$u^A = \left(\left((1-R^2\Omega^2)\left(1-\frac{2M}{R}\right)\right)^{-1/2},0,0,\frac{1}{\sqrt{(1-R^2\Omega^2)}}\Omega\right).$$

Geodesic

The non-vanishing Christoffel symbols are \begin{align} \Gamma^t_{rt} = - \Gamma^r_{rr} &= \frac{r_s}{2r(r-r_s)}\\ \Gamma^r_{tt} &= \frac{r_s(r-r_s)}{2r^3}\\ \Gamma^r_{\phi\phi} &= (r_s-r)\sin^2\theta\\ \Gamma^r_{\theta\theta} &= r_s - r\\ \Gamma^\theta_{r\theta} = \Gamma^\phi_{r\phi} &= \frac{1}{r}\\ \Gamma^\theta_{\phi\phi} &= -\sin\theta\cos\theta\\ \Gamma^\phi_{\theta\phi} &= \cot\theta. \end{align} Define \begin{align} w(r) := 1-\frac{r_s}{r}\\ v(r) := \frac{1}{w(r)}. \end{align} The geodesic equations are \begin{align} 0 &= \frac{d^2\theta}{d\lambda^2} + \frac{2}{r}\frac{d\theta}{d\lambda}\frac{dr}{d\lambda}-\sin\theta\cos\theta\left(\frac{d\phi}{d\lambda}\right)^2 \\ 0 &= \frac{d^2\phi}{d\lambda^2} + \frac{2}{r}\frac{d\phi}{d\lambda}\frac{dr}{d\lambda} + 2\cot\theta\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}\\ 0 &= \frac{d^2t}{d\lambda^2} + \frac{1}{w}\frac{dw}{dr}\frac{dt}{d\lambda}\frac{dr}{d\lambda}\\ 0 &= \frac{d^2r}{d\lambda^2} + \frac{1}{2v}\frac{dv}{dr}\left(\frac{dr}{d\lambda}\right)^2 - \frac{r}{v}\left(\frac{d\theta}{d\lambda}\right). \end{align} By symmetry, the orbits must be planar. Let \(\theta=\frac{\pi}{2}\). Two constaints are left \begin{align} 0 &= \frac{d}{d\lambda}\left(\ln\frac{d\phi}{d\lambda} + \ln r^2\right)\\ 0 &= \frac{d}{d\lambda}\left(\ln\frac{dt}{d\lambda} + \ln w\right)\\ \end{align}

Specific Energy and Specific Angular Momentum

As the Schwarzschild metric is independent of \(\theta\) and \(\phi\), we have two Killing vectors, \(\bf{\xi} = (1,0,0,0)\) and \(\bf{\eta} = (0,0,0,1)\). So, we have two constants, the specific energy $$\tilde{E} = -\bf{\xi}\cdot\bf{u} = \left(1 - \frac{2M}{r}\right)\frac{dt}{d\tau},$$ and the specific angular momentum, $$\tilde{L} = \bf{\eta}\cdot\bf{u} = r^2\sin^2\theta \frac{d\phi}{d\tau}$$

Trajectory

As the metric is spherical symmetric, the motion is always on a single plane. We may choose \(\theta = \frac{\pi}{2}\) to make things simpler. Then, the angular momentum is contributed by \(\phi\) component of momentum. The components of the four-momentum of the particle in terms of the specific energy and specific angular momentum are \begin{align} p^0 &= g^{00}p_0 = m\left(1-\frac{2M}{r}\right)^{-1}\tilde{E}\\ p^r &= m\frac{dr}{d\tau} \\ p^\phi &= g^{\phi\phi}p_{\phi} = m\frac{1}{r^2}\tilde{L}.\\ \end{align}

By \(\vec{p}\cdot\vec{p} = -m^2\), $$-m^2\tilde{E}^2\left(1-\frac{2M}{r}\right)^{-1} + m^2\left(1-\frac{2M}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2 + \frac{m\tilde{L}^2}{r^2} = -m^2.$$

Upon solving the equation, we have $$\left(\frac{dr}{d\tau}\right)^2 = \tilde{E}^2 - \left(1-\frac{2M}{r}\right)\left(1+\frac{\tilde{L}^2}{r^2}\right).$$ Taking square root and integrate on both side, we obtain $$ \bbox[5px,border:2px solid #666] { \tau = \int \frac{dr}{\pm\sqrt{\tilde{E}^2 - \left(1-\frac{2M}{r}\right)\left(1+\frac{\tilde{L}^2}{r^2}\right)}}, }$$ which, I believe, has no analytic solution for general \(\tilde{E}\) and \(\tilde{L}\).