0
$\begingroup$

I'm trying to prove via Noether's Theorem that the angular momentum of a massive particle in a gravitational field is conserved. The attempt follows:

OBS: I'm working in the euclidean space so I'll make no distinction between upper and lower indices

The lagrangian of this particle is given by: $L = \frac{1}{2} m \dot{x}^i \dot{x}_i - m \phi(x)$ where $\phi(x)$ is the gravitational potential.

For an infinitesimal rotation by an infinitesimal angle vector $\vec{\theta}$ with components $\theta ^k$ such that $x^i \rightarrow x^i + \epsilon_{ijk}x^j \theta ^k$, we have

\begin{equation} \delta L = - m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k + m\dot{x}^i \epsilon_{ijk} \dot{x}^j \theta ^k. \end{equation}

The second term in RHS is zero because $\dot{x}^i \dot{x}^j$ is symmetric and $\epsilon_{ijk}\theta ^k$ is antisymmetric.

My problem comes with the first term in RHS, because I think that it should be zero, such that I would get the conserved quantity $Q = \epsilon_{ijk}m\dot{x}^i x^j \theta^k = \vec{l} \cdot \vec{\theta}$, with $\vec{l}$ being the angular momentum vector, and $\vec{l} \cdot{\vec\theta}$, the angular momentum in the direction of $\theta$. It is necessary that $- m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k = 0$ to get conservation of angular momentum or did I make some mistake?

$\endgroup$
  • $\begingroup$ You are mixing polar and Cartesian coordinates. Work in either one but do not mix. $\endgroup$ – my2cts Mar 15 at 7:32
  • $\begingroup$ Hint: You need to use rotational symmetry of potential. $\endgroup$ – Qmechanic Mar 15 at 7:55
  • $\begingroup$ In $- m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k$ we have $\partial ^i \phi(x) = \frac{\partial \phi}{\partial x} \partial ^i x$, where $x = \sqrt{x^a x^a}$, a= 1,2,3, hence $ - m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k = - m \frac{\partial \phi}{\partial x} \frac{x^i}{x} \epsilon_{ijk} x^j \theta^k = 0 $, because $\epsilon_{ijk} x^i x^j = 0$. Is this correct? $\endgroup$ – Lil'Gravity Mar 15 at 19:12
0
$\begingroup$

You have an extra time derivative in your expression for $Q$, there should be just one. (Your present expression for $Q$ vanishes, for the same reason you have described in the question). Having made that alteration, if we demand that $\frac{dQ}{dt}=0$, we will need to substitute the equation of motion. Upon substitution, this turns out to be precisely the first term.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.