The Schwarzschild Metric: A Newtonian Comparison

A Short Proof of Birkoff’s Theorem derived the Schwarzschild metric in units of $G = c = 1$:

\begin{equation} ds^2 = -\left(1 – \frac{2M}{r}\right)dt^2 + \left(1 – \frac{2M}{r}\right)^{-1}dr^2 + r^2d\theta^2 + r^2 \sin^2\theta d\phi^2   \label{metric} \end{equation}

and I used that metric in The Schwarzschild Metric: Part 1, GPS Satellites to show that Global Positioning System (GPS) clocks in orbit around Earth run faster than Earth-based clocks. We saw that without the appropriate general relativistic correction, GPS tracking would be off by over 11 km per day. I then used that metric in The Schwarzschild Metric: Part 2, The Photon Sphere to discuss the “photon sphere,” i.e., the radial coordinate at which photons orbit $M$. In this Insight, I will bring the Schwarzschild metric to bear on a simple Newtonian problem, i.e., throwing a ball up into the air from the surface of Earth. While this problem is trivial in Newtonian physics, it’s actually tougher than the problems solved in Parts 1 & 2 of this series using GR. While it’s certainly not advisable to use GR anytime Newtonian physics will suffice, I do so in this case to illustrate the relationship between GR and Newtonian physics.

Here are the two geodesic equations of relevance:

\begin{equation} \frac{d^2r}{d\tau^2} + \Gamma^{1}_{00}\left(\frac{dt}{d\tau}\right)^2 + \Gamma^{1}_{11}\left(\frac{dr}{d\tau}\right)^2 = 0  \label{RgdscEq} \end{equation}

and

\begin{equation} \frac{d^2t}{d\tau^2} + 2\Gamma^{0}_{10}\left(\frac{dt}{d\tau}\right) \left(\frac{dr}{d\tau}\right) = 0     \label{TgdscEq} \end{equation}

where $\tau$ is proper time along the geodesic and, as in Parts 1 & 2 of this series, $x^0 = t, x^1 = r, x^2 = \theta, x^3 = \phi$. The Christoffel symbols we need are:

\begin{equation} \Gamma^{1}_{00} = \frac{M}{r^2} \left(1 – \frac{2M}{r}\right)  \label{Christoffelrtt} \end{equation}

\begin{equation} \Gamma^{1}_{11} = -\frac{M}{r^2} \left(1 – \frac{2M}{r}\right)^{-1}   \label{Christoffelrrr} \end{equation}

\begin{equation} \Gamma^{0}_{10} = \frac{M}{r^2} \left(1 – \frac{2M}{r}\right)^{-1}   \label{Christoffelttr} \end{equation}

Putting Eq(\ref{Christoffelttr}) into Eq(\ref{TgdscEq}) gives:

\begin{equation} \frac{d}{d\tau}\left(\left(1 – \frac{2M}{r}\right)\frac{dt}{d\tau}\right) = 0  \label{TgdscEq2} \end{equation}

which means:

\begin{equation}\left(1 – \frac{2M}{r}\right)\frac{dt}{d\tau} = B \label{TgdscEq3} \end{equation}

with $B$ a constant. At $r = \infty$ and $v = 0$ the metric Eq(\ref{metric}) tells us that $d\tau = dt$, since $ds^2 = -d\tau^2$ along the geodesic, so Eq(\ref{TgdscEq3}) then tells us $B = 1$ when our object is launched with escape velocity $v = v_e$. If we get to $r = \infty$ with $v > 0$, then our metric Eq(\ref{metric}) tells us:

\begin{equation} \frac{d\tau}{\sqrt{1-\frac{v^2}{c^2}}} = dt \label{SReqn} \end{equation}

(I restored $c$ here) which tells us $\frac{dt}{d\tau} > 1$ so Eq(\ref{TgdscEq3}) says $B > 1$ when the object is launched with $v > v_e$. Therefore, we assume $B < 1$ when the object is launched with $v < v_e$. We’ll need this info on $B$ when we get to our result.

Using this result and putting our other two Christoffel symbols Eq(\ref{Christoffelrtt}) and Eq(\ref{Christoffelrrr}) into our first geodesic equation Eq(\ref{RgdscEq}) gives:

\begin{equation} \left(1 – \frac{2M}{r}\right)\frac{d^2r}{d\tau^2} + \frac{M}{r^2}\left(B^2 – \left(\frac{dr}{d\tau}\right)^2 \right) = 0  \label{RgdscEq2} \end{equation}

Now I will start making approximations to show this leads to the Newtonian conservation of energy equation:

\begin{equation} \frac{1}{2}v^2 – \frac{GM}{r} = \frac{E}{m}  \label{Newton} \end{equation}

where $\frac{E}{m}$ is the (conserved) total energy per unit mass of the launched object (I restored $G$). Start by assuming $\frac{d^2r}{d\tau^2} = -g = -\frac{GM}{R^2}$ where $R$ is the radius of Earth and M is the mass of Earth, i.e., $g = 9.8 m/s^2$ per usual. Next let $r = R+y$ so that $\frac{dr}{d\tau} = \frac{dy}{d\tau} = v$, i.e., $y$ is the height above Earth’s surface of our projectile; we will assume $y$ is small compared to $R$. Now Eq(\ref{RgdscEq2}) gives us:

\begin{equation} -\left(1 – \frac{2GMR}{c^2R^2\left(1+y/R\right)}\right)g + \frac{GM}{R^2\left(1+y/R\right)^2}\left(B^2 – \frac{v^2}{c^2} \right) = 0  \label{RgdscEq3} \end{equation}

where I have restored $G$ and $c$. Expanding:

\begin{equation}\left(1 + \frac{y}{R} \right)^{-1} \approx 1 – \frac{y}{R} \label{approx1} \end{equation}

and

\begin{equation} \left(1 + \frac{y}{R} \right)^{-2} \approx 1 – 2\frac{y}{R} \label{approx2} \end{equation}

and putting these into Eq(\ref{RgdscEq3}) and rearranging we obtain:

\begin{equation}\left(\frac{1}{2} – \frac{y}{R}\right)v^2 + gy = \left(\frac{1}{2} – \frac{y}{R}\right)B^2c^2 – \frac{c^2}{2} + gR \label{RgdscEq4} \end{equation}

Since $\frac{y}{R} \ll 1$, we have $\left(\frac{1}{2} – \frac{y}{R}\right) \approx \frac{1}{2}$, so Eq(\ref{RgdscEq4}) is now:

\begin{equation}\frac{1}{2}v^2 + gy = \frac{\left(B^2 – 1\right)c^2}{2} + gR \label{RgdscEq5} \end{equation}

The $gy$ on the LHS of Eq(\ref{RgdscEq5}) and $gR$ on the RHS come from $-\frac{GM}{R\left(1+y/R\right)}$ expanded on the LHS of the Newtonian conservation of energy equation Eq(\ref{Newton}) then using $g = \frac{GM}{R^2}$. That means $\frac{\left(B^2 – 1\right)c^2}{2}$ in Eq(\ref{RgdscEq5}) is our total conserved energy per unit mass $\frac{E}{m}$ from the RHS of Eq(\ref{Newton}). Notice that for $v = v_e$ at launch we have $B = 1$ and the RHS of Eq(\ref{RgdscEq5}) is just $gR$ as expected (total energy equals zero in Eq(\ref{Newton})). If $v > v_e$ at launch, $B > 1$ and the RHS of Eq(\ref{RgdscEq5}) is a little larger than $gR$ (total energy is positive in Eq(\ref{Newton})). If $v < v_e$ at launch, $B < 1$ and the RHS of Eq(\ref{RgdscEq5}) is a little smaller than $gR$ (total energy is negative in Eq(\ref{Newton})).

That concludes my outline of how the geodesic equations for the Schwarzschild metric give rise to standard Newtonian physics for an object launched slowly upwards near the surface of Earth.