# Expression for distance of closest approach in Schwarzschild Geodesics

The Wikipedia article Two-body problem in General Relativity uses two length-scale variables, $a$ and $b$, to simplify the math. For some information about these, consider these statements from the article:

They are constants of the motion and depend on the initial conditions (position and velocity) of the test particle.

and

Here, b can be interpreted as the distance of closest approach.

They first occur in this equation:

$$\left( \frac{dr}{d\varphi} \right)^{2} = \frac{r^{4}}{b^{2}} - \left( 1 - \frac{r_{s}}{r} \right) \left( \frac{r^{4}}{a^{2}} + r^{2} \right)$$

Some of the prerequisite math can be found in other articles, and particularly Schwarzschild geodesics, where in the orbits of test particles section they give two equations that ostensibly define $a$ and $b$.

$$\left( 1 - \frac{r_{s}}{r} \right) \left( \frac{dt}{d\tau} \right) = \frac{a}{b}$$

$$r^{2} \left( \frac{d\varphi}{d\tau} \right) = a c$$

My first problem is that I don't understand how these define $a$ and $b$. As a previous quote implied, they should be calculable given initial conditions of a test particle, not depend on dynamic derivatives like the above equations imply. Let's return to the Two-body article. It mentions the following in-text, implying the straightforward expression of $a = h/c$.

The effective potential V can be re-written in terms of the length a = h/c

While not defined in that article, other articles on the subject give a consistent definition of $h$. Here, $L$ is angular momentum and $\mu$ is the reduced mass. Since we're only dealing with a test particle with low mass compared to the black hole, the reduced mass is its mass.

$$h = L/ \mu = L / m$$

If you don't want it, we don't even have to include the particle's mass. Let's denote angular momentum as $L = \vec{r} \times m \vec{v}$, so we can write the following. This also implies an interpretation that $a$ is the specific relative angular momentum in the geometrized unit system.

$$h = \vec{r} \times \vec{v}$$

$$a = (\vec{r} \times \vec{v}) / c$$

This is even more complicated than necessary, since most of the work on the Two-body page takes motion to be in a single plane, which would eliminate the vector notation. To complete this question, let me explain that I would just solve for $b$ from the known expression of $a$ using one of the equations I've posted, but I don't have enough information. I can see that it would be necessary to have an expression relating $d \phi$ and $d \tau$, so I've searched through the information of Schwarzschild geodesics for this, I can't find it, and I suspect that's because of the simplifications introduced to do math for a test particle.

In summary, we have an interpretation for $b$ and an explicit equation for $a$. What is an explicit equation for $b$?

-
Related: physics.stackexchange.com/q/52315/10851 (moreso the answers than the question, as they define $b$ along the way) –  Chris White Jun 21 '13 at 16:04

Wikipedia states:

Here, $b$ can be interpreted as the distance of closest approach.

Yes, but that interpretation on Wikipedia is made for massless particles with closest approach much bigger than the Schwarzschild radius $r_s$. In general that interpretation is not true.

On the other hand, OP's above equations (v1) are for massive particle, so let us focus on the massive case in what follows.

As OP writes $$ac~=~h~=~\frac{L}{m_0}$$ is the specific angular momentum $h$. In general,

$$b~=~c\frac{L}{E}~=~\frac{m_0c}{E}h~=~a\frac{m_0c^2}{E}$$

is the ratio between the (specific) angular momentum and the (specific) energy times the speed of light. The quantities $L$ and $E$ are constants of motion, which in turn reflect (some of) the Killing symmetries of the Schwarzschild metric.

The equation for the closest distance to the black hole can be deduced from the radial geodesic equation

$$\frac{1}{c^2}\left( \frac{dr}{d\tau} \right)^{2} = \left(\frac{E}{m_0c^2}\right)^{2} - \left( 1 - \frac{r_{s}}{r} \right) \left( \left(\frac{a}{r}\right)^2 + 1 \right)$$

for a massive particle in the equatorial plane by putting

$$\frac{dr}{d\tau}~=~0$$

and solve this third-order equation for $r$.

References:

1. S. Carroll, Lecture Notes on General Relativity, Chapter 7, p.172-179. The pdf file is available from his website.
-
It took me a while to process, but this is a good answer. All the constants in the resultant equation of motion are filled in. If I understand, this differential equation can give the radial position as a function of the local time. This is enough to identify everything in the two-body article. –  Alan Rominger Feb 6 '13 at 23:53

You mentioned that the constants should "not depend on dynamic derivatives like the above equations imply". Well, this particular combination of dynamic derivatives is conserved.

For example, let's take a ball attached to a string moving in a circle with constant speed. $v_x$ and $v_y$ are dynamical, i.e. they change in time, but the combination $\frac{1}{2} m (v_x^2 + v_y^2)$ doesn't. This happens to be the energy of the ball, incidentally. So a constant of motion could very well depend on the dynamic variables!

Now back to the Schwarzschild geodesic. The Schwarzschild metric has four Killing vectors. Two of the Killing vectors implies that a particle will move in a plane, so we can always choose $\theta = \pi/2$, while the other two Killing vectors are $K^\mu = (1,0,0,0)$ and $R^\mu = (0,0,0,1)$.

Now if we have a Killing vector the following is conserved along a geodesic: $K_\mu U^\mu$ where $U^\mu = dx^\mu/d\tau$. $K$, a time-like Killing vector, gives conservation of energy, \begin{align} E = -K_\mu \frac{dx^\mu}{d\tau} = (1 - r_s/r)\frac{dt}{d\tau}, \end{align} where $E$ is the energy per unit mass, but which Wikipedia has called $a/b$.

For $R$ we have conservation of angular momentum, \begin{align} L = R_\mu \frac{dx^\mu}{d\tau} = r^2 \frac{d\varphi}{d\tau}, \end{align} which Wikipedia has called $a$ (I set $c = 1$).

$E$ and $L$, or equivalently $a$ and $b$, are gotten from initial conditions, like you have rightly said. So that is where they come from. (Why define $a$ and $b$? So that there is some physical meaning to them. According to Wikipedia $b$ is the 'distance of closest approach' and $a$ is the angular momentum per unit mass, as you have pointed out)

So to answer your question about an 'explicit equation' for $b$... it is exactly as you have written. Just plug in the initial conditions and out pops $a$ and $b$.

For example, say a particle starts at rest from infinity and falls on a radial geodesic... this means that $d\varphi/d\tau|_{r = \infty} = 0$ which implies $a = 0$, and also means that $(1-r_s/r)|_{r = \infty} = 1$ and $dt/d\tau|_{r=\infty} = 1$, which implies $b = a$ so that $a/b$ is interpreted to be equal 1.

-