Conservation of Energy and Angular Momentum

Observe that in this system the total energy of the material point, which is defined by:

$$E = \frac{m \dot{\boldsymbol{r}}^2}{2} + q V,$$ (4.5)

where $\dot{\boldsymbol{r}} = \textrm{d} \boldsymbol{r}/\textrm{d} t$, is conserved. This is easy to see:

$$\begin{eqnarray*}\frac{\textrm{d}E}{\textrm{d} t} &=& \frac{m}{2} 2 \dot{\bolds... ...symbol{r}} \cdot \left(q\boldsymbol{\nabla} V\right) \\ &=& 0 \end{eqnarray*}$$

If V(r) = V(r), where r is the length of r, then another quantity that is also conserved is the angular momentum $\boldsymbol{L} = \boldsymbol{r}\times m\dot{\boldsymbol{r}}$. This is also easy to see:

$$\begin{eqnarray*}\frac{\textrm{d}\boldsymbol{L}}{\textrm{d} t} &=& \frac{\textr... ...\boldsymbol{r} \times \left( -q \boldsymbol{\nabla} V(r)\right) \end{eqnarray*}$$

Now recall that since V(r) = V(r)

$$\begin{eqnarray*}\boldsymbol{\nabla} V(r) &=& \frac{\textrm{d} V(r)}{\textrm{d... ...&=& \frac{\textrm{d} V(r)}{\textrm{d}r}\frac{\boldsymbol{r}}{r} \end{eqnarray*}$$

hence

$$\frac{\textrm{d}\boldsymbol{L}}{\textrm{d} t} = \boldsymbol... ...{d} V(r)}{\textrm{d} r}\frac{\boldsymbol{r}}{r} \right) = 0$$