# The Schwarzschild Metric: Part 3, 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.

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