Noether theorem and angular momenta

[LCSphysicist]

Homework Statement

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Relevant Equations

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Consider the invariance of a vector field by rotation, whose transformation is given by $x_{\mu'} = x_{\mu} + \epsilon^{v}_{\mu} x_{v}$ , with $\epsilon^{v}_{\mu} = - \epsilon^{\mu}_{v}$. Determine the Noether current and the conserved magnitude.

(OBS: Don't take the index positions too literal...)

Generally it is easy to deal with these type of exercises for discrete system. But since we need to evaluate it for continuous, i am a little confused on how to do it.

Goldstein/Nivaldo gives these formulas:

I am trying to understand how do we manipulate it here to get the conserved quantity.

So supposing the field is invariant (it is necessary, right?) $\psi = 0$.

I think we can write our transformation as $x_{i'} = x_{i} + \epsilon_{i j} x^{j} \theta$

So the current would be $$(\frac{\partial L}{\partial n_{p,v}} n_{p,\sigma} - L \delta ^{v}_{\sigma}) \epsilon_{\sigma j} x^{j} = (\frac{\partial L}{\partial n_{p,v}} n_{p,\sigma} \epsilon_{\sigma j} x^{j} - L \epsilon_{v j} x^{j}) $$

And the conserved charge $$ (\frac{\partial L}{\partial n_{p,0}} n_{p,\sigma} \epsilon_{\sigma j} x^{j} - L \epsilon_{0 j} x^{j}) = (\pi _{p} n_{p,\sigma} \epsilon_{\sigma j} x^{j}) $$

Is that right? I was expecting a epxression similar to angular momentum, but i was not able to achieve any!

 

[Orodruin]

So supposing the field is invariant (it is necessary, right?) $\psi = 0$.

No, it is not necessary for the field to be invariant. However, unless specified otherwise, it is probably what is assumed.

I think we can write our transformation as $x_{i'} = x_{i} + \epsilon_{i j} x^{j} \theta$

We cannot because the variation of the coordinates depend on the coordinates themselves. In order to find what the transformation actually is you must integrate the differential equation
$$
\frac{dx^i}{ds} = \epsilon^i_j x^j.
$$
Compare this to an explicit rotation in two dimensions $\delta x = -y$ and $\delta y = x$ for which
$$
x \to x \cos(s) - y\sin(s), \quad y \to x\sin(s) + y\cos(s)
$$
rather than
$$
x \to x - ys, \quad y \to y + xs.
$$

However, this is irrelevant for the issue at hand since to find the conserved current we only need $\delta x^i$.

So the current would be $$(\frac{\partial L}{\partial n_{p,v}} n_{p,\sigma} - L \delta ^{v}_{\sigma}) \epsilon_{\sigma j} x^{j} = (\frac{\partial L}{\partial n_{p,v}} n_{p,\sigma} \epsilon_{\sigma j} x^{j} - L \epsilon_{v j} x^{j}) $$

Yes, although note that your reference is using $\eta$ rather than $n$.

And the conserved charge $$ (\frac{\partial L}{\partial n_{p,0}} n_{p,\sigma} \epsilon_{\sigma j} x^{j} - L \epsilon_{0 j} x^{j}) = (\pi _{p} n_{p,\sigma} \epsilon_{\sigma j} x^{j}) $$

Assuming that you defined the canonical momenta fields
$$
\pi_p = \frac{\partial L}{\partial (\partial_t\eta_p)}
$$
then yes.

Is that right? I was expecting a epxression similar to angular momentum, but i was not able to achieve any!

Let us consider linear translations $\delta x^i = k^i$ rather than rotations. The corresponding Noether current would then be
$$
J^i = \left(\frac{\partial L}{\partial(\eta_{p,i})} \eta_{p,j} - L\delta^i_j\right) k^j = T^i_j k^j
$$
where $T^i_j$ are the components of the canonical stress-energy tensor. (Note that the canonical stress-energy tensor is not symmetric unlike the gravitational stress-energy tensor that is obtained by varying the action with respect to the metric in GR.)

Now, the interpretation of $T^i_j$ is "the $i$ component of the conserved current associated with invariance under translations in direction $j$" and therefore we can consider $p_j = T^0_j$ to be the momentum density in direction $j$. Consequently, your expression for the conserved current under rotations turns into
$$
J^i = T^i_j \epsilon^j_k x^k,
$$
with charge density
$$
J^0 = p_j \epsilon^j_k x^k
$$
which very much looks like an angular momentum density. For example, considering the $xy$ rotation mentioned earlier, we would have
$$
J^0 = p_x \delta x + p_y \delta y = p_y x - p_x y.
$$

Edit: Minor $\LaTeX$ fixes.