Conservation of the Laplace-Runge-Lenz vector in a Central Field

[stephenklein]

Homework Statement

I'm tasked with showing the LRL vector is conserved in time when the potential is Newtonian. I.e. $U(r) = -\frac{\alpha}{r},$ with $\alpha$ constant.

Relevant Equations

$\vec{L} = \vec{v} \times \vec{M} - \frac{\alpha \vec{r}}{r},$ where $\vec{L}, \vec{M}$ are the LRL vector and angular momentum, respectively

I actually have worked through the solution just fine by taking the derivative of \vec{L}:

\frac{d \vec{L}}{dt} = \dot{\vec{v}} \times \vec{M} - \alpha \left(\frac{\vec{v}}{r} - \frac{\left(\vec{v} \cdot \vec{r}\right)\vec{r}}{r^{3}}\right)
I permuted the double cross product:
\dot{\vec{v}} \times \vec{M} = \dot{\vec{v}} \times \left(m\vec{r} \times \vec{v}\right) = m\vec{r}\left(\dot{\vec{v}} \cdot \vec{v}\right) - m\vec{v}\left(\dot{\vec{v}} \cdot \vec{r}\right)
Here's where I'm running into trouble. In both Landau and my lecture notes, the next step is apparently to invoke
m\dot{\vec{v}} = \frac{\alpha \vec{r}}{r^{3}},
which comes from the fact that the force is Newtonian. But this implies the acceleration is only in the radial direction, which is only true for motions in circular orbit for which the angular velocity is constant. Obviously, for elliptical orbits, the condition that angular momentum is conserved implies that the angular velocity is at a maximum when r is at a minimum, and vice versa. Therefore, \dot{\phi} is not constant and m\dot{\vec{v}} will have an angular component as well. This can be seen as well from the equations of motion derived from the Lagrangian in polar coordinates.

Am I missing something? The derivation in Landau came under the section for Kepler's problem, so it doesn't seem obvious that I should assume the motion is perfectly circular, with no angular acceleration.

 

[TSny]

Therefore, \dot{\phi} is not constant and m\dot{\vec{v}} will have an angular component as well.

In polar coordinates, the angular component of acceleration is given by $a_{\phi} = 2\dot r \dot\phi + r \ddot \phi$. For reference see this document.

So, nonzero $\ddot \phi$ does not necessarily imply nonzero $a_{\phi}$

 

[stephenklein]

So, a nonzero $\ddot \phi$ does not necessarily imply nonzero $a_{\phi}$

Wow, I feel like a dummy for not just looking up acceleration components in polar coordinates. That was immensely helpful, thank you.

So to flip the argument around, can we say that because $m\dot{\vec{v}}$ has only a radial component, that $a_{\phi} = 0$, meaning $2\dot{r} \dot{\phi} = -r\ddot{\phi}$? And physically, what would that refer to?

 

[TSny]

So to flip the argument around, can we say that because $m\dot{\vec{v}}$ has only a radial component, that $a_{\phi} = 0$, meaning $2\dot{r} \dot{\phi} = -r\ddot{\phi}$?

Yes

And physically, what would that refer to?

I'm sorry, but I'm not sure what you are asking here.

You could say that it is a consequence of the conservation of angular momentum. The angular momentum is $\vec M = mr^2\dot\phi \; \hat k$ where $\hat k$ is a constant unit vector perpendicular to the plane of the orbit.

Since $\dot {\vec M} = 0$, you have $\frac {d }{dt}(mr^2 \dot \phi) = 0$. This implies $2\dot{r} \dot{\phi} + r\ddot{\phi} = 0$.