# Particle orbits – conserved quantities

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 10; Boxes 10.1 – 10.2; Problem 10.2.

One form of the geodesic equation is

$\displaystyle \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial x^{a}}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (1)$

A geodesic is the path followed by a free particle in a given metric, so we can apply this equation to the Schwarzschild metric to discover what orbits a particle can have around a mass. The metric is

$\displaystyle ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\left(1-\frac{2GM}{r}\right)^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}\sin^{2}\theta d\phi^{2} \ \ \ \ \ (2)$

From the ${t}$ component, we get from 1:

$\displaystyle \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial t}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (3)$

Since the metric does not depend on ${t}$ the second term is zero. Also, since the metric is diagonal, the first term becomes

 $\displaystyle \frac{d}{d\tau}\left(g_{tj}\frac{dx^{j}}{d\tau}\right)$ $\displaystyle =$ $\displaystyle \frac{d}{d\tau}\left(g_{tt}\frac{dt}{d\tau}\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{d}{d\tau}\left(\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (6)$ $\displaystyle \left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}$ $\displaystyle =$ $\displaystyle e \ \ \ \ \ (7)$

where ${e}$ is a constant.

The meaning of ${e}$ can be inferred from the following argument. At ${r=\infty}$, the relation becomes

$\displaystyle \frac{dt}{d\tau}=e \ \ \ \ \ (8)$

We’ve seen that the Schwarzschild ${t}$ coordinate is the same as the object’s proper time for an object at rest, when measured by an observer at infinity. Thus for an object at rest, ${e=1}$. However, in general ${dt/d\tau=u^{t}}$, the time component of the four-velocity. In this case, ${t}$ is the time as measured by an observer at rest at infinity, and ${\tau}$ is the proper time as measured by the object, which may be moving. The four-momentum‘s time component is the energy, and the four-velocity is the four-momentum per unit mass, so ${e}$ is the energy per unit mass of the object which remains constant as the object moves in from infinity.

Now look at the ${\phi}$ component of 1. Again, since the metric does not depend on ${\phi}$ and it is diagonal, we get

 $\displaystyle \frac{d}{d\tau}\left(g_{\phi j}\frac{dx^{j}}{d\tau}\right)$ $\displaystyle =$ $\displaystyle \frac{d}{d\tau}\left(g_{\phi\phi}\frac{d\phi}{d\tau}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{d}{d\tau}\left(r^{2}\sin^{2}\theta\frac{d\phi}{d\tau}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle r^{2}\sin^{2}\theta\frac{d\phi}{d\tau}$ $\displaystyle =$ $\displaystyle l \ \ \ \ \ (12)$

where ${l}$ is another constant. If we look at motion in the equatorial plane, ${\theta=\pi/2}$ and ${d\phi/d\tau\equiv\omega}$ which is the angular speed of rotation, so we get

$\displaystyle r^{2}\omega=l \ \ \ \ \ (13)$

which is equivalent to the classical definition if ${l}$ is the angular momentum.

In fact, any initial velocity lies in some equatorial plane (merely redefine the location of the ${z}$ axis so that the velocity lies in the plane with ${\theta=\pi/2}$) so for any specific motion, we can assume ${\theta=\pi/2}$ without, as they say, any loss of generality. Also, because the metric is spherically symmetric, any motion that starts in an equatorial plane must stay in the plane, since there is no asymmetry that would push the object to one side or the other of that plane. Looking at the ${\theta}$ component of 1, we can see that an equatorial path is a geodesic:

 $\displaystyle \frac{d}{d\tau}\left(g_{\theta j}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\partial_{\theta}g_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{d}{d\tau}\left(g_{\theta\theta}\frac{d\theta}{d\tau}\right)-\frac{1}{2}\left(2r^{2}\sin\theta\cos\theta\right)\left(\frac{d\phi}{d\tau}\right)^{2}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2r\frac{dr}{d\tau}\frac{d\theta}{d\tau}+r^{2}\frac{d^{2}\theta}{d\tau^{2}}-\frac{1}{2}\left(2r^{2}\sin\theta\cos\theta\right)\left(\frac{d\phi}{d\tau}\right)^{2}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (16)$

If ${\theta=\pi/2=\mbox{constant}}$, then all three terms in the second line are zero, so the geodesic equation is satisfied. Note that this last conclusion doesn’t prove that all geodesics lie in an equatorial plane; it merely verifies that a planar path is a solution. We rely on the symmetry argument to state that all geodesics must lie in an equatorial plane. Of course, in a different metric where the mass is not spherically symmetric, non-planar paths are possible.

Finally, we consider the ${r}$ component of 1. Since all four components of the metric depend on ${r}$, this takes a bit more work. We can start with the universal relation for the four-velocity ${\mathbf{u}\cdot\mathbf{u}=-1}$, which in this metric is

$\displaystyle g_{tt}\left(\frac{dt}{d\tau}\right)^{2}+g_{rr}\left(\frac{dr}{d\tau}\right)^{2}+g_{\theta\theta}\left(\frac{d\theta}{d\tau}\right)^{2}+g_{\phi\phi}\left(\frac{d\phi}{d\tau}\right)^{2}=-1 \ \ \ \ \ (17)$

From above, we know that, since ${\theta=\pi/2}$

 $\displaystyle \frac{dt}{d\tau}$ $\displaystyle =$ $\displaystyle e\left(1-\frac{2GM}{r}\right)^{-1}\ \ \ \ \ (18)$ $\displaystyle \frac{d\theta}{d\tau}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \frac{d\phi}{d\tau}$ $\displaystyle =$ $\displaystyle \frac{l}{r^{2}\sin^{2}\theta}=\frac{l}{r^{2}} \ \ \ \ \ (20)$

Therefore

 $\displaystyle -1$ $\displaystyle =$ $\displaystyle -\left(1-\frac{2GM}{r}\right)\left[e\left(1-\frac{2GM}{r}\right)^{-1}\right]^{2}+\left(1-\frac{2GM}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^{2}+r^{2}\left(\frac{l}{r^{2}}\right)^{2}\ \ \ \ \ (21)$ $\displaystyle -\left(1-\frac{2GM}{r}\right)$ $\displaystyle =$ $\displaystyle -e^{2}+\left(\frac{dr}{d\tau}\right)^{2}+\left(1-\frac{2GM}{r}\right)\frac{l^{2}}{r^{2}}\ \ \ \ \ (22)$ $\displaystyle \frac{1}{2}\left(e^{2}-1\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}+\frac{1}{2}\frac{l^{2}}{r^{2}}-GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)\ \ \ \ \ (23)$ $\displaystyle \frac{dr}{d\tau}$ $\displaystyle =$ $\displaystyle \pm\left[e^{2}-1+2GM\left(\frac{1}{r}+\frac{l^{2}}{r^{3}}\right)-\frac{l^{2}}{r^{2}}\right]^{1/2}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm\sqrt{e^{2}-\left(1-\frac{2GM}{r}\right)\left(1+\frac{\ell^{2}}{r^{2}}\right)} \ \ \ \ \ (25)$

Since the only dependent variable in this equation is ${r}$ (${l}$ and ${e}$ are constants), this is a first-order ODE for ${r}$ in terms of the proper time ${\tau}$.

As a simple example of the use of this equation, suppose we start a particle at rest at infinity and let it fall in radially towards the mass. Since the motion is entirely radial, the angular momentum is ${l=0}$. Also, since the particle starts off at rest ${e=1}$ (see above), so the equation becomes

 $\displaystyle \frac{1}{2}\left(\frac{dr}{d\tau}\right)^{2}$ $\displaystyle =$ $\displaystyle \frac{GM}{r}\ \ \ \ \ (26)$ $\displaystyle \pm\int\sqrt{r}dr$ $\displaystyle =$ $\displaystyle \sqrt{2GM}\int d\tau\ \ \ \ \ (27)$ $\displaystyle \pm\frac{2}{3}r^{3/2}$ $\displaystyle =$ $\displaystyle \sqrt{2GM}\tau+T \ \ \ \ \ (28)$

where ${T}$ is a constant of integration. If the particle is falling inwards, then we would expect ${\tau}$ to increase as ${r}$ decreases, so we need to take the minus sign on the left. The proper time interval as measured by the falling object between two radii (say, ${r_{A}=10GM}$ and ${r_{B}=2GM}$) is

 $\displaystyle \Delta\tau$ $\displaystyle =$ $\displaystyle \frac{2}{3\sqrt{2GM}}\left(\left(10GM\right)^{3/2}-\left(2GM\right)^{3/2}\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{3}\left(10^{3/2}-2^{3/2}\right)GM \ \ \ \ \ (30)$