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Conservation of Energy and Angular Momentum

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

\begin{displaymath}E = \frac{m \dot{\boldsymbol{r}}^2}{2} + q V,
\end{displaymath} (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

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)

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}


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

Zdzislaw Meglicki