# Runge-Lenz Vector

## Main Question or Discussion Point

I'm trying to prove that the time derivative of the Runge-Lenz vector is constant. Any ideas on how I would go about doing this?

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Any help on this?

vanhees71
Gold Member
2019 Award
Simply take the time derivative! The definition of the Lenz vector is

$$\vec{A}=\vec{p} \times \vec{L}-m \alpha \frac{\vec{r}}{r}.$$

To show that this is conserved for the potential $V(r)=-\alpha/r$, we note that the angular momentum is conserved, and we have

$$\vec{L}=m \vec{r} \times \dot{\vec{r}}=m r^2 \vec{\omega}=\text{const},$$

where $\vec{\omega}$ is the momentary angular velocity.

Further we have

$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\vec{r}}{r}=\frac{\dot{\vec{r}}}{r}-\frac{\dot{r} \vec{r}}{r^2} = \vec{\omega} \times \frac{\vec{r}}{r}$$

and

$$\dot{\vec{p}}=-\vec{\nabla} V(r)=-\frac{\alpha}{r^3} \vec{r}$$

and thus

$$\dot{\vec{A}}=\dot{\vec{p}} \times \vec{L}-m \alpha \vec{\omega} \times \frac{\vec{r}}{r}=-\frac{\alpha}{r^3} \vec{r} \times m r^2 \vec{\omega}-m \alpha \vec{\omega} \times \frac{\vec{r}}{r}=0.$$