8.1 Areal Quantities
The signed areal velocity along a curve \(\gamma \) is
\[ \frac{1}{2m} \left\langle \frac{L(0)}{\| L(0)\| }, q(t) \times p(t) \right\rangle , \]
where \(L(0) = q(0) \times p(0)\). The swept area on a time interval \([t_1, t_2]\) is:
\[ \mathcal{A}(t_1, t_2) = \int _{t_1}^{t_2} \text{(signed areal velocity)}\, dt. \]
If \(L \neq 0\), the signed areal velocity is constant and equal to \(\| L\| /(2m)\).
Proof
By definition,
\[ \frac{d\mathcal{A}}{dt} = \frac{1}{2m}\hat L_0 \cdot (q(t) \times p(t)), \]
where \(\hat L_0 = L(0)/\| L(0)\| \). By conservation of angular momentum, \(q(t) \times p(t) = L(0)\). Therefore
\[ \frac{d\mathcal{A}}{dt} = \frac{1}{2m}\hat L_0 \cdot L(0) = \frac{\| L\| }{2m}. \]
Two time intervals of the same length sweep the same area.
Proof
The swept area on \([t_1, t_1 + \Delta t]\) equals \((\| L\| /2m) \cdot \Delta t\), which depends only on the length \(\Delta t\). So any two intervals of length \(\Delta t\) sweep equal areas.