Motion of body in central field lies in a plane -- Lagrangian Formalism

Question

Say we have a body in 3D space inside a central field $ V(r) $. We want to show, using Lagrangian mechanics and spherical coordinates, that the body will move in a plane, which is defined by the initial values of $x, v$.

We know that it suffices to show that the angular momentum $$ \vec{l}= m \vec{r} \times \vec{v}$$ is conserved, i.e. $${ d \vec{l} \over dt} =\vec 0$$ as this means $\vec r$, $\vec v$ lie in a certain plane.

The Lagrangian in spherical coordinates is $$ L= {1 \over 2} m( \dot{r} ^2 + r^2 \dot{θ}^2 + r^2 \sin^2{\theta}\,\dot{φ}^2) - V(r) $$

I've written all 3 Euler-Lagrange equations but can't seem to find an expression that could give me the result I want.

The one that looks promising is the 3rd: $$ {d \over dt} { \partial{L} \over \partial{ \dot{φ}} }= {\partial{L} \over \partialφ }$$ or $$ {d \over dt} (mr^2 \sin^2{\theta}\,\dot{φ})=0 $$ but I think i'm not seeing something clearly. So how do I show from the Euler Lagrange equations that the angular momentum is conserved?

Answer 1

You can face the problem in various ways. One way would be to switch to the Hamiltonian picture, where it is easy to show that, with the Hamiltonian $$ H=\frac{p_x^2+p_y^2+p_z^2}{2m}+V(r) $$ all components of the angular momentum $\vec l$ are conserved through evaluation of Poisson brackets.

But the most direct way that I can think of consists in cleverly choosing your coordinates, in particular the polar axis for spherical coordinates. Suppose that you want to show that the motion is contained in the plane that contains the initial velocity vector and the center of attraction (if the vector does not point directly to the center: in this case the problem is trivially unidimensional). Now, we take as polar axis the axis that passes through the center of attraction and is orthogonal to the plane. You already know that plane, as you know initial velocity and position of the object.

If you choose coordinates in this way, you obtain that the plane is described by the equation $\theta=\frac{\pi}{2}$. The fact that the plane can be described in such a trivial way is the key here: for a point to be on the plane, it is necessary and sufficient that $\theta=\frac{\pi}{2}$. Now, take the motion equation for the coordinate $\dot\theta$: $$ \frac{d}{dt}(r^2\dot\theta)-r^2\sin\theta\cos\theta\dot\varphi^2=0. $$ The initial conditions are $\theta(0)=\frac\pi2$ and $\dot\theta(0)=0$, as the body initially lies on the $\frac\pi2$ plane. Now we use the fact that, once you obtain a solution for the motion equations (even by guessing), it is unique if the initial conditions are right.

What does that mean? Suppose that you guess a solution $$ r(t),\quad\theta(t)=\frac{\pi}{2},\quad\varphi(t), $$ where $r(t)$ and $\varphi(t)$ are solutions to the other motion equations: we can leave them unspecified. Plug $\theta(t)$ in the motion equation: the derivative is zero, as $\dot\theta(t)=0$, and the second part is zero too, as $\cos\theta(t)=0$ for all $t$. Furthermore, the initial conditions are met. So the constant solution $\theta(t)=\frac\pi2$ is the solution for the $\theta$ coordinate, and the body will always lie on the plane where it began motion.

Note that, choosing another coordinate system, you get much more complicated equations for the plane, and a constant $\theta(t)$ no longer is a solution. That's correct: a plane in spherical coordinates can be described through a complicated function $\theta(r,\varphi)$, and in general r, $\theta$ and $\varphi$ are not constant. But, if you choose your coordinates cleverly, you can eliminate $\theta$, as it is trivial.

Answer 2

We can without loss of generality rotate the coordinate system to align with

$\theta(0) = \frac{\pi}{2}$

You showed the canonical momentum of $\phi$ to be constant in time. So it follows by inserting the above:

$p_{\phi} = m r² \dot \phi \sin²(\theta(t))= m r² \dot\phi$

Therefor

$\theta(t) = const.$

Which corresponds to a planar motion.

Answer 3

Using Noether's Theorem: The transformation $q_i\rightarrow q_i+\epsilon$ is a symmetry iff $\delta L = \dfrac{dF}{dt}$. In that case, the conserved charge is given by: $$\frac{d}{dt}(\frac{\partial L}{\partial \dot q}\delta q-F)=0$$

Here when we do the transformation $\theta\rightarrow\theta+\epsilon$, $F$ comes out to be zero. (You can calculate it using Taylor Expansion) and hence the conserved charge would be: $$\frac{d}{dt}(\frac{\partial L}{\partial \dot\theta}\delta \theta)=0$$ which in this case comes out to be $\dfrac{d}{dt}(mr^2\dot{\theta})$ which implies angular momentum conservation.