# Integrating the geodesic equation to obtain Newtonian energy of test particle

I am studying General Relativity, and have come across a question that I am finding rather intractable:

In Newtonian Theory, the energy equation for a test particle in orbit around a point mass is: $$\frac{v^{2}}{2} + \frac{\ell^{2}}{2r^{2}} - \frac{GM}{r} = \mathcal{E}$$ Where $r$ is the radius, $v$ is the radial velocity, $\ell$ is the angular momentum per unit mass, $\mathcal{E}$ is the constant energy per unit mass and $-GM/r$ is the gravitational potential. For the Schwarzchild solution show that the integrated geodesic equation may also be written in the form: $$\frac{v_{s}^{2}}{2} + \frac{\ell_{s}^{2}}{2r^{2}} + \Phi_{s}(r) = \mathcal{E}_{s}$$ Where $v_{s} = \mathrm{d}r/\mathrm{d}\tau$, and $\ell_{s}$ and $\mathcal{E}_{s}$ are constants.

I am not really sure what it means by integrating the geodesic equation. My understading is that the geodesic equation is given by:

$$\frac{\mathrm{d}^{2} x^{\lambda}}{\mathrm{d} \tau^{2}} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} = 0$$

Where $\Gamma^{\lambda}_{\mu\nu}$ is the affine connection. However, this leads to a system of equations, none of which, when integrated, yield the form given in the question.

I'm sure that I am missing something fundamental and simple, but I just cannot find it in any of the references or materials that I have been given.

I thought that perhaps integrating the geodesic equation directly would be the way to go, but I do not end up with the correct solution:

$$\int\left(\frac{\mathrm{d}^{2} x^{\lambda}}{\mathrm{d} \tau^{2}} + \Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\right)\:\mathrm{d}\tau = \mathcal{E}_{s}$$

We note that as integration is distributive over addition, we find:

$$\frac{\mathrm{d}x^{\lambda}}{\mathrm{d}\tau} + \int\left(\Gamma^{\lambda}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\right)\:\mathrm{d}\tau = \mathcal{E}_{s}$$

We note that if we let $\lambda = r$, then this becomes proportional to $v_{s}$ and thus I am missing a factor of $v_{s}$.

Edit 2: After the very helpful comments below, I went back and did the question again, this time getting stuck at a different point.

So we find that in the $\lambda = t$ case, the only Christoffel symbols $\Gamma_{\mu\nu}^{t}$ that are non-zero are when $\mu = t$, $\nu = r$ or vice versa. Thus we find:

$$\frac{\mathrm{d}t}{\mathrm{d}\tau} + GM\int\frac{\mathrm{d}r}{\mathrm{d}\tau}\frac{\mathrm{d}t}{\mathrm{d}\tau}\frac{\mathrm{d}\tau}{r(r - 2GM)} = \mathcal{E}_{s}$$

We can rewrite the integral in terms of $r$, to give us:

$$\frac{\mathrm{d}t}{\mathrm{d}\tau}\left(1 - \frac{\ln(r)}{2} + \frac{\ln(r - 2GM)}{2}\right) = \mathcal{E}_{s}$$

Which is nowhere near the form that is asked for?

Let $g_{\mu\nu}$ be the metric given by $$\tag{1}\mathrm{d}s^2=-A(r)\mathrm{d}t^2+B(r)\mathrm{d}r^2+r^2\mathrm{d}\Omega^2.$$ The geodesic equation corresponds to the Euler-Lagrange equation of the action $$S[x]=\int\sqrt{-g_{\mu\nu}\dot x^\mu\dot x^\nu}\mathrm{d}\tau$$ Now either (i) vary this functional directly or (ii) calculate the Christoffel symbols of (1) and write down the geodesic equation. Note that $\delta S/\delta x^\mu=0$ is entirely equivalent to $\ddot x^\mu+\Gamma^\mu{}_{\nu\rho}\dot x^\nu\dot x^\rho=0$. Some calculations give $$\tag{2}\frac{\delta S}{\delta t}=0\implies\frac{\mathrm{d}}{\mathrm{d}\tau}[A(r)\dot t]=0$$ $$\tag{3}\frac{\delta S}{\delta \phi}=0\implies \frac{\mathrm{d}}{\mathrm{d}\tau}[r^2\sin^2\theta\dot\phi]=0$$ $$\tag{4}\frac{\delta S}{\delta\theta}=0\implies \frac{\mathrm{d}}{\mathrm{d}\tau}[r^2\dot\theta]-r^2\sin\theta\cos\theta\dot\phi^2=0$$ We can solve (2) and (3) instantly by integration: $$\tag{5}\dot t=\frac{\epsilon}{A},\quad\dot\phi=\frac{\ell}{r^2\sin^2\theta}$$ where $\epsilon$ and $\ell$ are constants of integration. Additionally, (4) is easily solved by fixing the equatorial plane at $\theta=\pi/2$. We also have the affine parameter equation $$g_{\mu\nu}\dot x^\mu\dot x^\nu=-1=-A(r)\dot t^2+B(r)\dot r^2+r^2\underbrace{\dot\theta^2}_0-r^2\underbrace{\sin^2\theta}_1\dot\phi^2$$ Plugging everything into this equation, we get $$\tag{6}\frac{\epsilon^2}{A}-B\dot r^2-\frac{\ell^2}{r^2}=-1$$ Then, use $$A(r)=1-\frac{2GM}{r}=B(r)^{-1}$$ (the Schwarzschild solution) and the desired equation comes out with a little modification of the symbols to fit those in the OP.
We would like to note that there is a better method for finding first integrals to the geodesic problem. Let $\xi^\mu$ be a Killing vector of the metric $g_{\mu\nu}$ and $\dot x^\mu$ the tangent vector to a geodesic $x^\mu(\tau)$. Then $$g_{\mu\nu}\dot x^\mu\xi^\nu=\text{const.}$$ i.e. it is independent of $\tau$. This is exactly a first integral and may be used to solve the geodesic problem, but there are no actual integrals involved.