Prove rotational invariance leads to conservation of ang. momentum

[xWaffle]

Homework Statement

Noether's theorem asserts a connection between invariance principles and conservation laws. In section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of L implies conservation of total angular momentum. Suppose that the Lagrangian of an N-particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the z axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from (r_α, θ_α, ϕ_α) to(r_α, θ_α, ϕ_α + ϵ) (same ϵ for all particles). Hence show that

$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}} = 0$

(b) Use Lagrange's equations to show that this implies that the total angular momentum L_z about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of L are conserved.

Homework Equations

Lagrange Equations: Translational Invariance Example

No external force.

Simply, the change in potential is:
$\delta U = U(\overline{r}_{1} + d\overline{r}_{1}, \overline{r}_{2} + d\overline{r}_{2}, ... , t) - U(\overline{r}_{1}, \overline{r}_{2}, ... , t)$
$U = \sum_{\alpha\beta} U_{\alpha\beta}(\overline{r}_{\alpha} - \overline{r}_{\beta})$

And very similarly for the Lagrangian:
$\delta\mathcal{L} = \mathcal{L}(\overline{r}_{1} + d\overline{r}_{1}, \overline{r}_{2} + d\overline{r}_{2}, ... , t) - \mathcal{L}(\overline{r}_{1}, \overline{r}_{2}, ... , t)$

Stationary requires this change in action to be zero:
$\delta\mathcal{L} = 0$

And, earlier in the semester, we proved that a conservative force can be written as the negative gradient of a scalar valued function. We will use this here:
$\delta\mathcal{L} = d\overline{r} \bullet \nabla_{1} \mathcal{L} + d\overline{r} \bullet \nabla_{2} \mathcal{L} + ... = 0$

And to redefine some of the terms in the last line:
$d \overline{r} = \hat{x} dx$
$d \overline{r} \bullet \nabla_{1} \mathcal{L} = dx \frac{ \partial \mathcal{L} }{ \partial x_{1}}$

Now, plug them into make it look a little bit more like what we want..
$dx \sum_{\alpha} \frac{ \partial \mathcal{L} }{ \partial x_{\alpha}} = 0$
$\sum_{\alpha} \frac{\partial \mathcal{L}}{\partial x_{\alpha}} = 0$

Lagrange Equations for any particle in the system:
$\frac{\partial \mathcal{L}}{\partial x_{\alpha}} = \frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}_{\alpha}} \right)$

Which is the same as saying:
$\frac{\partial \mathcal{L}}{\partial x_{\alpha}} = \frac{d}{dt} \left( p_{x\alpha} \right)$
$\frac{\partial \mathcal{L}}{\partial x_{\alpha}} = \dot{p}_{x\alpha}$

And from here, out conclusion can be shown very simply:
$\sum_{\alpha} \dot{p}_{x\alpha} = \dot{p}_{total} = 0$
$\dot{p}_{total} = 0$
$p_{total} = 0 \rightarrow consant$

The Attempt at a Solution

We've been shown an example with translational invariance leading to conservation of momentum. He used the Lagrange equations and related them to U(r) potential energy, showed that with no external forces, a change in the Lagrangian can be written as a re-written form of the difference of potentials. Which then led to the full time derivative of the Lagrangian not explicitly depending on x (only x-dot), which meant p (momentum) was constant.

I've tried to approach it in a similar way, but I'm stuck trying to wrap my head around what would be the 'rotational equivalent' of his first step. Is this the wrong way to approach it?

 

[mark726]

Should I be trying to manipulate the Lagrangian in some way instead?



Thank you for your post. It is great to see that you are actively thinking about how to approach this problem and are making connections to previous examples. Let me guide you through the steps to prove Noether's theorem for rotational invariance.

Firstly, we need to understand what rotational invariance means in terms of the Lagrangian. In the given problem, we are told that the Lagrangian is unchanged when all particles are simultaneously moved from (rα, θα, ϕα) to (rα, θα, ϕα + ϵ). This means that the Lagrangian does not depend on the angle ϕ, which is the angle of rotation about the z-axis. Therefore, we can write the Lagrangian as L(r, θ, θ-dot, ϕ-dot) = L(r, θ, θ-dot). This is the rotational invariance condition for our system.

Now, let's move on to part (a) of the problem. We are asked to show that the Lagrangian is unchanged when all particles are moved by the same angle ϵ. This means that we need to show that the change in Lagrangian, δL, is equal to zero. Using the rotational invariance condition, we can write:

δL = L(r, θ, θ-dot) - L(r, θ, θ-dot, ϕ-dot) = L(r, θ, θ-dot) - L(r, θ, θ-dot) = 0

Therefore, the Lagrangian is indeed unchanged when all particles are moved by the same angle ϵ. Now, using the same logic as in the translational invariance example, we can write:

δL = dθ \frac{\partial L}{\partial θ} + dθ-dot \frac{\partial L}{\partial θ-dot} = 0

But, as we have just shown, δL = 0. Therefore, we can write:

0 = dθ \frac{\partial L}{\partial θ} + dθ-dot \frac{\partial L}{\partial θ-dot}

Now, we can rearrange this equation to get:

dθ \frac{\partial L}{\partial θ} = -dθ-dot \