My professor did the following derivation of the formula for the perihelion precession $\delta \phi = \frac{6\pi G M}{a(1-e^2)}$, but I am missing a factor of 2. I would appreciate if someone can help me spot the mistake, or tell me if the approach is totally wrong.

Consider the Schwarzschild metric, and compute the length of a geodesic as: $$S = \int_0^{\tau} d\tau' \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}} = \int_0^{\tau}d\tau' \sqrt{2T}.$$ Since the modulus of four-velocity is conserved and equal to $-1$, we have $2T=1$, and imposing the variation of the length equal to zero we get: $$\delta S = \int_0^{\tau} d\tau' \delta\sqrt{2T} = \int_0^{\tau} d\tau' \frac{\delta T}{\sqrt{2T}} = \delta \int_0^{\tau} d\tau' T.$$ Therefore, we have reduced to the lagrangian formalism where the lagrangian is given by $T$ and its functional value is conserved and equal to $1/2$: $$T = \frac{1}{2}\left[\left(1-\frac{2GM}{r}\right)\dot{t}^2 - \left(1-\frac{2GM}{r}\right)^{-1} \dot{r}^2 - r^2 \dot{\theta}^2 -r^2 \sin^2(\theta)\dot{\phi}^2 \right] = \frac{1}{2}$$ We can check using the Euler-Lagrange equation for $\theta$ that $\theta(\tau) =\frac{\pi}{2}$ is a solution, therefore the term $r^2\dot{\theta}^2$ disappears (since $\theta$ is constant) and the term $r^2 \sin^2(\theta) \dot{\phi}^2$ simplifies to $r^2\dot{\phi}^2$. At this point we can use the Euler-Lagrange equation for $t$ and $\phi$ and see that there are two conserved quantities: $$E = \frac{\partial T}{\partial \dot{t}} = \left(1-\frac{2GM}{r}\right)\dot{t}$$ which is the energy, and the angular momentum: $$L = - \frac{\partial T}{\partial \dot{\phi}} = r^2 \dot{\phi}.$$ If we invert the two above relations for $\dot{t}$ and $\dot{\phi}$ and plug them into the expression for $T$, we get a first order differential equation for $r$. After some algebraic manipulations we end up with: $$\frac{1}{2} \dot{r}^2 + \frac{L^2}{2r^2} + \frac{1}{2}\left(1-\frac{2GM}{r}\right) - \frac{GML^2}{r^3} = \frac{E^2}{2},$$ where we recognize the kinetic term $\frac{1}{2}\dot{r}^2$, so the rest of the left-hand side is an effective potential $V(r)$. Now, we can look for newtonian circular orbits by neglecting the term $\sim 1/r^3$, setting the first derivative of the potential to $0$ and solving for $r$: we get $$r_c = \frac{L^2}{GM}.$$ Now, we can expand the potential around the circular orbit, obtaining a harmonic potential, with the second derivative of the potential evaluated at $r=r_c$ playing the role of $\omega^2$, the square of the frequency of the harmonic oscillator. If we compute the second derivative at $r=r_c$ we get (I checked it with WolframAlpha): $$V''(r_c) = \omega^2 = \frac{G^4M^4}{L^6}\left(1-\frac{12G^2M^2}{L^2}\right),$$ where my professor claims that $\omega_N^2 = \frac{G^4M^4}{L^6}$ is the Newtonian orbit frequency. Now, since the GR orbit has a lower frequency than the newtonian one, it means that when the GR orbit has finished a revolution ($\phi = 2\pi$), the newtonian has travelled an additional angle given by: $$\delta \phi = \omega_N \delta T = \omega_N \left(\frac{2\pi}{\omega} - \frac{2\pi}{\omega_N}\right) = 2\pi\left(\frac{1}{\sqrt{1-\frac{12G^2M^2}{L^2}}} - 1\right) \simeq \frac{12\pi G^2M^2}{L^2},$$ which clearly differs from the famous formula (with eccentricity equal to zero) just by a factor of $2$.

I double-checked the steps multiple time, but there not seem to be calculation errors. The only thing I can think of is that there is some wrong physical assumption somewhere, but I am not sure.


1 Answer 1


Orbital precession is found from the difference between the frequency in $\phi$ and the frequency of oscillation in $r$. Both calculated in the Schwarzschild metric and both of which differ from the Newtonian result, in opposite directions.


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