Noether's Theorem is used to relate the invariance of the action 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?


2 Answers 2


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 $$S[\phi] = \int {\mathcal L}(\phi(x), \partial_{\mu} \phi(x), \dots) {\rm d}^4x.$$

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 $F^{\mu \nu}$.

  • 3
    $\begingroup$ Nice warning at the start ;-) (+1) $\endgroup$
    – qftme
    Jul 21, 2011 at 13:20
  • 5
    $\begingroup$ A similar discussion is given here: math.ucr.edu/home/baez/boosts.html Less mathematically inclined folks may find it more readable. $\endgroup$
    – user4552
    Jul 21, 2011 at 15:59
  • 1
    $\begingroup$ I do not understand the freedom to choose $t=0$. I thought $p$ and $E$ were independently conserved. $\endgroup$ Jan 10, 2018 at 19:20
  • 1
    $\begingroup$ The three extra components of the "angular" 4-momentum actually lead to the equivalence between momentum and energy-flux. That is, a flux of energy has momentum, and momentum is equivalent to energy flux. This is the "equivalence of mass and energy" or the fact that energy has inertia. See eg Eckart, doi.org/10.1103/PhysRev.58.919 $\endgroup$
    – pglpm
    Jun 1, 2020 at 9:19
  • $\begingroup$ How can something which is a function of t be "conserved". If t appears in the expression, doesn't that necessarily mean that quantity is not conserved (aka, time independent? $\endgroup$ May 23 at 7:21

To supplement Marek's execllent answer, I provide an alternative derivation below and provide as many intermediate steps as possible.

For an infinitesimal displacement $y^\mu=x^\mu+\xi^\mu$, a scalar field changes as

$$\phi(y)=\phi(x)+\xi^\mu \partial_\mu\phi(x)+...$$

The displacement by infinitesimal Lorentz transform $\Lambda^{\mu\nu}$ is $y^\mu=x^\mu+\Lambda^{\mu\nu}x_\nu$. Similarly the scalar field changes as: $$\phi(y)=\phi(x)+ \Lambda^{\mu\nu}x_\nu\partial_\mu\phi(x)+...$$ The variation of the field w.r.t. $\Lambda^{\mu\nu}$ is $$\frac{\delta \phi}{\delta \Lambda^{\mu\nu}}=x_\nu\partial_\mu\phi(x)-x_\mu\partial_\nu\phi(x)$$ The reason there are two terms on the right hand side is because infinitesimal Lorentz transform $\Lambda^{\mu\nu}$ is anti-symmetric, i.e. $\Lambda^{\nu\mu} = -\Lambda^{\mu\nu}$, which has only 6 independent components. (You can verify this by demanding the scalar product is unchanged after transformation, $y^\mu y_\mu = x^\mu x_\mu$)

Using Principle of Least Action, variation in Lagrangian $\mathcal{L}$ is

$$\frac{\delta \mathcal{L}}{\delta \Lambda^{\mu\nu}}=\sum_n\{\frac{\partial \mathcal{L}}{\partial\phi_n} \frac{\delta\phi_n}{\delta \Lambda^{\mu\nu}} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_n)} \frac{\delta(\partial_{\mu}\phi_n)}{\delta \Lambda^{\mu\nu}} \}$$ Applying the equation of motion $$\frac{\partial \mathcal{L}}{\partial \phi_n} -\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}=0$$ we get the conservation law: $$\frac{\delta \mathcal{L}}{\delta \Lambda^{\mu\nu}}=\sum_n\partial_\mu[\frac{\partial \mathcal{L}}{\partial_{\mu}\phi_n} \frac{\delta\phi}{\delta \Lambda^{\mu\nu}} ] $$ Substituting the expression for $\delta \phi/\delta \Lambda^{\mu\nu}$ and a similar one for $\delta \mathcal{L}/\delta \Lambda^{\mu\nu}$, we get the final conservation law $$\partial_\mu j^{\mu \lambda\sigma} = 0 $$ where the conservative current $$j^{\mu \lambda\sigma}=x^\lambda T^{\mu\sigma} - x^{\sigma}T^{\mu\lambda}$$ is the angular momentum and $$T_{\mu\nu}= \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\nu\phi-g_{\mu\nu}\mathcal{L} $$ is the momentum.


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