Orbits of Schwarzschild Metric

From the previous post, we know that $$\tau = \int \frac{dr}{\pm\sqrt{\tilde{E}^2 - \left(1-\frac{2M}{r}\right)\left(1+\frac{\tilde{L}^2}{r^2}\right)}}.$$ Now we want to see how a testing massive particle will travel in this curved spacetime, to see if it fits well with our observed data.

Recall the 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, $$\tilde{L} = \bf{\eta}\cdot\bf{u} = r^2\sin^2\theta \frac{d\phi}{d\tau}$$

Circular Orbits

Define the effective potential as $$\tilde{V}^2 = \left(1-\frac{2M}{r}\right)\left(1+\frac{\tilde{L}^2}{r^2}\right).$$

Differentiate with respect to the proper time \(\tau\), time experienced by the particle, as energy is conserved for a circular orbit, \(\frac{dp_0}{d\tau} = 0\), we have $$2\left(\frac{dr}{d\tau}\right)\left(\frac{d^2r}{d\tau^2}\right) = -\frac{d\tilde{V}^2(r)}{dr}\frac{dr}{d\tau}.$$

Hence, $$\left(\frac{d^2r}{d\tau^2}\right) = \frac{1}{2}\frac{d\tilde{V}^2}{dr}.$$

For circular orbit, \(r\) is constant, i.e. \(\frac{dr}{d\tau}=0\), $$\frac{d}{dr}\left[\left(1-\frac{2M}{r}\right)\left(1+\frac{\tilde{L}^2}{r^2}\right)\right] = 0.$$ We can solve for \(r\), $$r = \frac{\tilde{L}^2}{2M}\left[1\pm\left(1-\frac{12M^2}{\tilde{L}^2}\right)^{1/2}\right],$$ and the specific angular momentum, $$\tilde{L}^2 = \frac{Mr}{1-3M/r}.$$ This expression of \(r\) sets a lower bound of \(r\) for circular orbits $$r_{\text{min}} = 6M.$$

As \(\frac{dr}{d\tau}=0\) for all time in a circular orbit, we know that we always have \(\tilde{E} = \tilde{V}\), $$\tilde{E} = \left(1-\frac{2M}{r}\right)^2/\left(1-\frac{3M}{r}\right).$$

The rate of change of \(\phi\) with respect to proper time is $$\frac{d\phi}{d\tau} := U^{\phi} = \frac{p^\phi}{m} = g^{\phi\phi}\frac{p_{\phi}}{m} = g^{\phi\phi}\tilde{L} = \frac{1}{r^2}\tilde{L}.$$ The rate of change of \(t\) with respect to proper time is $$\frac{dt}{d\tau}:=U^0=\frac{p^0}{m} = g^{00}\frac{p_0}{m} = g^{00}(-\tilde{E}) = \frac{\tilde{E}}{1-2M/r}.$$ Divide the first equation with the second, we obtain the rate of change of \(\phi\) with respect to coordinate time \(t\) $$\frac{dt}{d\phi} = \left(\frac{r^3}{M}\right)^{1/2}.$$ As period is the time for \(\phi\) to change by \(2\pi\), the period of the circular orbit is $$ \bbox[5px,border:2px solid #666] { P = 2\pi\left(\frac{r^3}{M}\right)^{1/2} },$$ which agrees with Newtonian mechanics.

Elliptic orbit

The rate of change of \(\phi\) with respect to proper time is $$\frac{d\phi}{d\tau} := U^{\phi} = \frac{p^\phi}{m} = g^{\phi\phi}\frac{p_{\phi}}{m} = g^{\phi\phi}\tilde{L} = \frac{1}{r^2}\tilde{L}.$$ Hence, $$\left(\frac{dr}{d\phi}\right)^2 = \frac{\tilde{E}^2-(1-2M/r)(1+\tilde{L}^2/r^2)}{\tilde{L}^2/r^4}.$$

To get rid of the \(r^4\), define $$u := 1/r.$$ Substitute into the previous equation to obtain the rate of change of \(u\) with respect to \(\phi\) $$\left(\frac{du}{d\phi}\right)^2 = \frac{\tilde{E}^2}{\tilde{L}^2} - (1-2Mu)\left(\frac{1}{\tilde{L}^2} + u^2\right).$$ When the particle is far away from the source, we may neglect the \(u^3\) terms, $$\left(\frac{du}{d\phi}\right)^2 = \frac{\tilde{E}^2}{\tilde{L}^2} - \frac{1}{\tilde{L}^2}(1-2Mu) - u^2.$$ As \(u = M/\tilde{L}^2\) for circular orbits, define $$y = u - \frac{M}{\tilde{L}^2},$$ to indicate how much the orbit is different from a circular orbit. Then, we have $$\left(\frac{dy}{d\phi}\right)^2 = \frac{\tilde{E}^2 - 1}{\tilde{L}^2} + \frac{M^2}{\tilde{L}^4} - y^2.$$ The solution is $$y = \left[\frac{\tilde{E}^2 + M^2/\tilde{L}^2 - 1}{\tilde{L}^2}\right] \cos(\phi + B),$$ for some arbitrary constant \(B\). Solve for \(r\), $$ \bbox[5px,border:2px solid #666] { \frac{1}{r} = \frac{M}{\tilde{L}^2} + \left[\frac{\tilde{E}^2 + M^2/\tilde{L}^2 - 1}{\tilde{L}^2}\right]\cos(\phi + B), }$$ the equation of an ellipse.

Precession of Mercury

Assume the orbit is nearly circular, such that \(y\) is small and \(y^3\) is negligible, $$\left(\frac{dy}{d\phi}\right)^2 = \frac{\tilde{E}^2 + M^2/\tilde{L}^2 - 1}{\tilde{L}^2} + \frac{2M^4}{\tilde{L}^6} + \frac{6M^3}{\tilde{L}^2}y - \left(1-\frac{6M^2}{\tilde{L}^2}\right)y^2.$$ The solution is $$y = y_0 + A\cos (k\phi + B),$$ where \(B\) is an arbitrary constant and \begin{align} k &= \left(1-\frac{6M^2}{\tilde{L}^2}\right)^{1/2}\\ y_0 &= 3M^3/k^2\tilde{L}^2\\ A &= \frac{1}{k}\left[\frac{\tilde{E}^2 + M^2/\tilde{L}^2 - 1}{\tilde{L}^2} + \frac{2M^4}{\tilde{L}^6} - y_0^2\right]^{1/2}. \end{align} The particle returns to the same \(r\) when \(k\phi = 2\pi\), so $$\phi = \frac{2\pi}{k} = 2\pi\left(1- \frac{6M^2}{\tilde{L}^2}\right)^{-1/2}.$$ For nearly Newtonian orbits, $$\phi \approx 2\pi \left(1 + \frac{3M^2}{\tilde{L}^2}\right).$$ The shift from one orbit to the next is then given by $$\Delta \phi = 6\pi M^2/\tilde{L}^2 \text{ radians per orbit}.$$ Approximate the specific angular momentum $$\tilde{L}^2 = \frac{Mr}{1-3M/r} \approx Mr.$$ Then the shift is about $$\Delta \phi \approx 6\pi\frac{M}{r}.$$ Now, substitute the data of orbit of Mercury. \(r = 5.55 \times 10^7 \text{km}\) and \(M = 1.47 \text{km}\), we obtain $$\Delta \phi_{Mercury} = 4.99 \times 10^{-7} \text{ radians per orbit}.$$ Since each orbit takes 0.24 year, the shift per century is $$\Delta \phi_{Mercury} = 43''/\text{century},$$ which agrees very well with the observed data. Yeah!

Reference: A first course in general relativity by Bernard F. Schutz