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Noether's Theorem is used to related the invariance under certain continuous transformations to conserved currents. A common example is that translations in spacetime correspond to the conservation of four-momentum.

In the case of angular momentum, the tensor (in special relativity) has 3 independent components for the classical angular momentum, but 3 more independent components that, as far as I know, represent Lorentz boosts. So, what conservation law corresponds to invariance under Lorentz boosts?

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It's the center of mass. – Raskolnikov Jul 21 '11 at 10:37
Yeah, shortly after posting I found this link which led me to Landau & Lifshitz (1975), pp 41-42. I didn't find their explanation totally clear, but it seems the conserved current is the four-velocity of the centre of mass? – Warrick Jul 21 '11 at 10:38
You can answer your own question @Warrick. In fact, I think there's even a badge for that ;-) – qftme Jul 21 '11 at 10:57
To be precise, the conserved quantity for a system of free relativistic particles is the position (not the velocity) of the center of mass, which can be obtained by the application of the Noether theorem. – David Bar Moshe Jul 21 '11 at 11:15
But L&L say that the three-vector $\underline{R}=\sum_i E_i\underline{v}_i/\sum_iE_i$ doesn't represent the spatial part of any four-vector? Is this the same as the fact that the three-vector $\underline{x}\wedge\underline{v}$ also isn't part of a four-vector? – Warrick Jul 21 '11 at 11:55
up vote 48 down vote accepted

Warning: this is a long and boring derivation. If you are interested only in the result skip to the very last sentence.

Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action $$S[\phi] = \int {\mathcal L}(\phi(x), \partial \phi_{\mu}(x), \dots) {\rm d}^4x.$$

Now suppose the action is invariant under some infinitesimal transformation $m: x^{\mu} \mapsto x^{\mu} + \delta x^{\mu} = x^{\mu} + \epsilon a^{\mu}$ (we won't consider any explicit transformation of the fields themselves). Then we get a conserved current $$J^{\mu} = {\partial {\mathcal L} \over \partial \phi_{\mu}} \phi^{\nu} a_{\nu} - {\mathcal L} a^{\mu} = \left ({\partial {\mathcal L} \over \partial \phi_{\mu}} \phi^{\nu} - {\mathcal L} g^{\mu \nu} \right) a_{\nu} .$$ We obtain a conserved charge from it by letting $Q \equiv \int J^0 {\rm d}^3x$ since from $\partial_{\mu}J^{\mu} =0$ we have that $$ {\partial Q \over \partial t} = \int {\rm Div}{\mathbf J}\, {\rm d}^3 x = 0$$ which holds any time the currents decay sufficiently quickly.

If the transformation is given by translation $m_{\nu} \leftrightarrow \delta x^{\mu} = \epsilon \delta^{\mu}_{\nu}$ we get four conserved currents $$J^{\mu \nu} = {\partial {\mathcal L} \over \partial \phi_{\mu}} \phi^{\nu} - {\mathcal L} g^{\mu \nu} .$$

This object is more commonly known as stress energy tensor $T^{\mu \nu}$ and the associated conserved currents are known as momenta $p^{\nu}$. Also, in general the conserved current is simply given by $J^{\mu} = T^{\mu \nu} a_{\nu}$.

For a Lorentz transformation we have $$m_{\sigma \tau} \leftrightarrow \delta x^{\mu} = \epsilon \left(g^{\mu \sigma} x^{\tau} - g^{\mu \tau} x^{\sigma} \right)$$ (notice that this is antisymmetric and so there are just 6 independent parameters of the transformation) and so the conserved currents are the angular momentum currents $$M^{\sigma \tau \mu} = x^{\tau}T^{\mu \sigma} - x^{\sigma}T^{\mu \tau}.$$ Finally, we obtain the conserved angular momentum as $$M^{\sigma \tau} = \int \left(x^{\tau}T^{0 \sigma} - x^{\sigma}T^{0 \tau} \right) {\rm d}^3 x . $$

Note that for particles we can proceed a little further since their associated momenta and angular momenta are not given by an integral. Therefore we have simply that $p^{\mu} = T^{\mu 0}$ and $M^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$. The rotation part of this (written in the form of the usual pseudovector) is $${\mathbf L}_i = {1 \over 2}\epsilon_{ijk} M^{jk} = ({\mathbf x} \times {\mathbf p})_i$$ while for the boost part we get $$M^{0 i} = \left(t {\mathbf p} - {\mathbf x} E \right)^i $$ which is nothing else than the center of mass at $t=0$ (we are free to choose $t$ since the quantity is conserved) multiplied by $\gamma$ since we have the relations $E = \gamma m$, ${\mathbf p} = \gamma m {\mathbf v}$. Note the similarity to the ${\mathbf E}$, $\mathbf B$ decomposition of the electromagnetic field tensor $F^{\mu \nu}$. .........

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That's a very thorough answer. Thanks! – Warrick Jul 21 '11 at 12:49
Nice warning at the start ;-) (+1) – qftme Jul 21 '11 at 13:20
A similar discussion is given here: Less mathematically inclined folks may find it more readable. – Ben Crowell Jul 21 '11 at 15:59

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