Appendix 10 — Specific Angular Momentum
In this document, and especially where we use the Schwarzschild equation,
the term angular momentum is used in the form:
\begin{equation}
L = m r^{2}\frac{d\phi}{dt}.
\end{equation}
Since:
\begin{equation}
L = m v r = m r v = m r \left(r\frac{d\phi}{dt}\right)
= m r^{2}\frac{d\phi}{dt},
\end{equation}
this resembles classical angular momentum.
However, this is not the true two-body angular momentum, but an approximation.
The explanation follows below.
The two-body problem
In the Schwarzschild formulation, we consider a particle moving in the gravitational field
of a large massive body.
The reference frame is the center of that large body.
This is, in fact, a two-body problem.
The two bodies orbit around their common center of mass (barycenter).
For circular orbits:
\begin{equation}
m_{1}\frac{v_{1}^{2}}{r_{1}}
=
m_{2}\frac{v_{2}^{2}}{r_{2}}.
\label{eq:R03}
\end{equation}
Since the periods must be equal:
\begin{equation}
T = \frac{2\pi r_{1}}{v_{1}}
= \frac{2\pi r_{2}}{v_{2}}
\quad\Rightarrow\quad
\frac{v_{1}}{v_{2}} = \frac{r_{1}}{r_{2}}.
\label{eq:R04}
\end{equation}
It follows that:
\begin{equation}
v_{1} = \frac{r_{1}}{r_{2}} v_{2}
= \frac{r_{1}}{r_{2}}(v - v_{1}),
\label{eq:R05}
\end{equation}
thus:
\begin{equation}
v_{1}
= \frac{r_{1}}{r} v,
\qquad
v_{2}
= \frac{r_{2}}{r} v,
\label{eq:R06}
\end{equation}
where \(r = r_{1} + r_{2}\).
The relative velocity is:
\begin{equation}
v = v_{1} + v_{2}.
\label{eq:R07}
\end{equation}
Relation between masses and distances
Substitute (\ref{eq:R05}) into (\ref{eq:R03}) :
\begin{equation}
\frac{m_1v^2_1}{r_1}=\frac{m_1v^2_2}{r_1}\left(\frac{r_1}{r_2}\right)^2=\frac{m_2v^2_2}{r_2}
\Rightarrow \frac{m_1}{r_1}\left(\frac{r_1}{r_2}\right)^2=\frac{m_2}{r_2}
\end{equation}
\begin{equation}\Rightarrow
m_{1} r_{1}
= m_{2} r_{2}.
\label{eq:R09}
\end{equation}
From this it follows:
\begin{equation}
r_{2}
= \frac{m_{1}}{m_{1} + m_{2}}\, r,
\qquad
r_{1}
= \frac{m_{2}}{m_{1} + m_{2}}\, r.
\label{eq:R10}
\end{equation}
Angular momentum of both bodies
The angular momentum of \(m_{2}\) relative to \(m_{1}\):
\begin{equation}
\begin{aligned}
L_{2}
= m_{2} v_{2} r_{2}
&=m^2\frac{r_2}{r}v\frac{m_1}{m_1+m_2}r_2v
= m_{2}\frac{m_{1}}{m_{1}+m_{2}} r_2v
\\ &\quad
=m_2\left(\frac{m_1}{m_1+m_2}\right)^2vr
\label{eq:R11}
\end{aligned}
\end{equation}
In terms of the angular velocity \(\omega = v/r\):
\begin{equation}
L_{2}
= \frac{1}{m_2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)^2\, \omega r^{2}
\label{eq:R12}
\end{equation}
Similarly:
\begin{equation}
L_{1}
= \frac{1}{m_1}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)^2\, \omega r^{2}
\label{eq:R13}
\end{equation}
The total angular momentum is therefore:
\begin{equation}
L = L_{1} + L_{2}
= \frac{m_{1} m_{2}}{m_{1}+m_{2}}\, \omega r^{2}.
\end{equation}
Or in Schwarzschild form:
\begin{equation}
L
= \frac{m_{1} m_{2}}{m_{1}+m_{2}}\, r^{2}\frac{d\phi}{d\tau}.
\label{eq:R15}
\end{equation}
Reduced mass
We define the reduced mass:
\begin{equation}
m = \frac{m_{1} m_{2}}{m_{1} + m_{2}}.
\label{eq:R16}
\end{equation}
The specific angular momentum is then:
\begin{equation}
h = \frac{L}{m}
= r^{2}\frac{d\phi}{d\tau}.
\label{eq:R17}
\end{equation}
Limit of a very large central mass
If \(m_{1} = M\) is a very large mass (e.g., a star or black hole)
and \(m_{2}\) is a small particle:
\begin{equation}
m = \frac{m_{2} M}{M + m_{2}}
\approx m_{2}.
\label{eq:R18}
\end{equation}
Thus, when \(M \gg m_{2}\), the angular momentum is determined by the mass of the particle alone.
This justifies the commonly used Schwarzschild form:
\begin{equation}
L = m r^{2}\frac{d\phi}{d\tau}.
\end{equation}
The specific angular momentum:
\begin{equation}
h = r^{2}\frac{d\phi}{d\tau}
\end{equation}
is then exactly the correct quantity to use in the Schwarzschild equation.
