How to Prove d<L>/dt = <N> for a Particle in Potential V(r)?

[Tuneman]

ya that is the x component, of $$r \ X\ (-\nabla V)$$ and then I can simply argue if a add the three components, $$L_x, L_y, L_z$$ knowing $$r = x,y,z$$, I can say $$\frac{dL}{dt} = r \X\ (- \nabla V)$$

Wow...that makes total sense... Ty very much for helping me solve the problem, I really appreciate it.

 

[Galileo]

Yep, since similar results hold for $L_y$ and $L_z$. You can just permutate the indices cyclicly (is that a word?)

$\vec L$ is just a operator with 3 components: $\vec L=L_x \vec i+L_y\vec j +L_z \vec k$, so you can just equate the components of both sides.

 

[Tuneman]

awesome, thanks for the help!

and to show that to show that for a sphyrically symmetric potential $$\frac{d<L>}{dt} = 0$$, someone mentioned to me to try this

$$\frac{d<L>}{dt} = [H,L] + \frac{dh}{dt}$$

where the right term on the right hand side equals zero because the hamiltonian isn't time dependent, and then I guess I am left off where i started in the previous problem. Should I be thinking of this equation, or should I be trying a different approach.

 

[Galileo]

The equation you derived is general, ofcourse.

$$\frac{d}{dt}\langle \vec L \rangle = \langle \vec r \times (-\vec \nabla V) \rangle$$

So try to evaluate the right hand side for the case where $V=V(r)$, $r=|\vec r|$.

Hint: Use $\vec \nabla$ in spherical coordinates.

 

[Tuneman]

Ah I see!
$$r \times -\nabla V = r(r \times r)(\frac{-dV}{dr})+ r(r\times \theta)\frac{1}{r}(\frac{-dV}{d\theta})+r(r\times\phi)\frac{1}{rsin\theta}(\frac{-dV}{d\theta})$$
Where the things getting cross multiplied are the unit vecotrs (i don't know how to put the hats on them) and of course those are partial deriviatives.
The first term has r x r, which is of course 0, and the other two terms have me differentiating V(r) with respect to the two angular coords, so of course they are 0! so the whole thing equals zero!

 

[Galileo]

Yep. The quick way of seeing this intuitively is by recalling that $\vec \nabla V$ is the vector which, at eacht point, points in the direction of maximum increase of V. If V is spherically symmetric, what other direction can this be but radially outward? In the tangential direction it must be zero.
So $\vec \nabla V(r)$ points in the direction of $\vec r$ everywhere, so its cross product with $\vec r$ is zero everywhere.

 

[harsh]

Actually, you could have used the equation is suggested. The commutator [H,L] is zero, and <dl/dt> is obviously zero. Its a nifty little formula, and its very useful.

- harsh