Appendix 10 — Specific Angular Momentum

In this document, especially where we use the Schwarzschild equation, the term angular momentum is used in the form:

\[ L = m r^{2}\frac{d\phi}{dt}. \]

Since: \[ L = m v r = m r v = m r \left(r\frac{d\phi}{dt}\right) = m r^{2}\frac{d\phi}{dt}, \] this resembles classical angular momentum.

However, it is not the actual two-body angular momentum, but an approximation. Here is the explanation.

The Two-Body Problem

In the Schwarzschild formula we consider a particle moving in the gravitational field of a massive body. The reference frame is the center of that massive body. This is essentially a two-body problem.

The two bodies rotate around their common center of mass (barycenter). For circular orbits:

\[ m_{1}\frac{v_{1}^{2}}{r_{1}} = m_{2}\frac{v_{2}^{2}}{r_{2}}. \tag{1} \]

Since the periods must be equal:

\[ T = \frac{2\pi r_{1}}{v_{1}} = \frac{2\pi r_{2}}{v_{2}} \quad\Rightarrow\quad \frac{v_{1}}{v_{2}} = \frac{r_{1}}{r_{2}}. \tag{2} \]

From this it follows:

\[ v_{1} = \frac{r_{1}}{r_{2}} v_{2} = \frac{r_{1}}{r_{2}}(v - v_{1}), \tag{3} \]

so: \[ v_{1} = \frac{r_{1}}{r} v, \qquad v_{2} = \frac{r_{2}}{r} v, \tag{4–5} \] where \(r = r_{1} + r_{2}\).

The relative velocity is: \[ v = v_{1} + v_{2}. \tag{6} \]

Relation Between Masses and Distances

Substitute (3) into (1):

\[\Rightarrow m_{1} r_{1} = m_{2} r_{2}. \tag{7} \]

Thus:

\[ r_{2} = \frac{m_{1}}{m_{1} + m_{2}}\, r, \qquad r_{1} = \frac{m_{2}}{m_{1} + m_{2}}\, r. \tag{8} \]

Angular Momentum of Both Bodies

The angular momentum of \(m_{2}\) with respect to \(m_{1}\):

\[ L_{2} = m_{2} v_{2} r_{2} = m_{2}\frac{m_{1}}{m_{1}+m_{2}} r_2 v = m_2\left(\frac{m_1}{m_1+m_2}\right)^2 v r \tag{9} \]

In terms of the angular velocity \(\omega = v/r\):

\[ L_{2} = \frac{1}{m_2}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)^2\, \omega r^{2} \tag{10} \]

Similarly: \[ L_{1} = \frac{1}{m_1}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)^2\, \omega r^{2} \tag{11} \]

The total angular momentum is thus:

\[ L = L_{1} + L_{2} = \frac{m_{1} m_{2}}{m_{1}+m_{2}}\, \omega r^{2}. \]

Or in Schwarzschild form: \[ L = \frac{m_{1} m_{2}}{m_{1}+m_{2}}\, r^{2}\frac{d\phi}{d\tau}. \tag{12} \]

Reduced Mass

We define the reduced mass:

\[ m = \frac{m_{1} m_{2}}{m_{1} + m_{2}}. \tag{13} \]

The specific angular momentum is then:

\[ h = \frac{L}{m} = r^{2}\frac{d\phi}{d\tau}. \tag{14} \]

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:

\[ m = \frac{m_{2} M}{M + m_{2}} \approx m_{2}. \tag{15} \]

Thus, when \(M \gg m_{2}\), the angular momentum is determined by the particle's mass alone. This justifies the usual Schwarzschild form:

\[ L = m r^{2}\frac{d\phi}{d\tau}. \]

The specific angular momentum: \[ h = r^{2}\frac{d\phi}{d\tau} \] is then exactly the correct quantity to use in the Schwarzschild equation.