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?


3 Answers 3


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 $F^{\mu \nu}$.

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    $\begingroup$ A similar discussion is given here: math.ucr.edu/home/baez/boosts.html Less mathematically inclined folks may find it more readable. $\endgroup$
    – user4552
    Commented Jul 21, 2011 at 15:59
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    $\begingroup$ I do not understand the freedom to choose $t=0$. I thought $p$ and $E$ were independently conserved. $\endgroup$ Commented Jan 10, 2018 at 19:20
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    $\begingroup$ 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
    Commented Jun 1, 2020 at 9:19
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    $\begingroup$ I think a better interpretation of $M^{0i} = (t\mathbf p - \mathbf xE)^i$ is as the center of mass/moment of mass in the rest frame of the particle at all times, rather than that in all frames at $t=0$ (in the set of frames with the same origin at $t=0$). This makes no assumptions about the history of the particle's motion (in particular, having a constant velocity) and is also is not trivially a constant of motion by virtue of being a literal constant, i.e. an initial condition. $\endgroup$
    – Er Jio
    Commented Mar 22 at 14:32

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.


A direct answer? It doesn't really have a name but has always been written in the literature in 3-vector form as $𝐊$, alongside the angular momentum 3-vector $𝐉$.

Any system, elementary or composite, relativistic or non-relativistic, that possesses a rest frame (i.e. a frame in which the momentum $𝐏$ is $𝟎$), and a center of mass position, $𝐫$, it may be called a "Tardyon" (or "Bradyon"). For such systems, the angular momentum decomposes into $𝐉 = 𝐫×𝐏 + 𝐒$, where $𝐒$ is its internal angular momentum; i.e. its angular momentum taken with respect to its center of mass position. When the system is elementary, and not composite, then this angular momentum component is referred to as its "spin". The other part, $𝐋 = 𝐫×𝐏$ is the "orbital" angular momentum of the system - which is the angular momentum it has by virtue of its motion around the origin $𝐫 = 𝟎$.

The other quantity, for such systems, decomposes as $𝐊 = M𝐫 - 𝐏t + 𝐓$. This has no official name and (as you can see) is explicitly time-dependent, by virtue of the appearance of $t$ for time. For the lack of a better name, you may call it the "moving mass moment", because of the dependence on the time and on the momentum $𝐏$.

The mass here, $M$, is itself a "moving" mass, as well. In non-relativistic theory it is $M = m$, equal to the mass $m$ of the system in its own rest frame - its rest mass. In Relativity, it has a dependence on the momentum given as $$M = m \sqrt{1 + \frac{1}{c^2}\frac{P^2}{m^2}}.$$ Nowadays, the "total energy" $E = Mc^2$ is usually used in the relativistic literature, in place of the moving mass $M$, but this obscures the discussion of what $𝐊$ is; which is just that it's the mass moment that the system would have had at $t = 0$, if its position is projected back to $t = 0$ by treating it as having moved between then and the present at the velocity $𝐯 = 𝐏/M$.

The additional quantity $𝐓$ - an analogue of internal angular momentum - is dependent on $𝐒$. For non-relativistic Tardyons, $𝐓 = 𝟎$, while in Relativity $$𝐓 = \frac{1}{c^2}\frac{𝐏×𝐒}{m + M} = \frac{𝐏×𝐒}{mc^2 + E}.$$ So the fact that it is non-zero and spin-dependent is a purely Relativistic effect.

In both the relativistic and non-relativistic cases, this can be solved for $𝐫$ in terms of the canonical quantities $𝐉$, $𝐊$, $𝐏$ and $M$; with the result being the classical version of the Newton-Wigner operator for the center of mass position vector.

For all systems - Tardyons, Luxons, Tachyons - if there is no internal angular momentum - more precisely: if $𝐖 = 𝟎$, where $𝐖 ≡ M𝐉 + 𝐏×𝐊$ is the Pauli-Lubanski 3-vector, then solely by virtue of this fact, there is a decomposition into $𝐉 = 𝐫×𝐏$ and $𝐊 = M𝐫 - 𝐏t$. So, they're called "spin 0", which is an abuse of terminology, if the system is not a Tardyon.

Tachyons belong to a class which possess frames of reference in which $M = 0$, but $𝐏 ≠ 𝟎$ - an infinite speed frame. In that frame, $Π^2 = P^2$ is the square of an impulse, its value being $Π = P \sqrt{1 - M^2c^2/P^2}$ or $Π = P \sqrt{1 - E^2/(Pc)^2}$. The non-relativistic equivalent of this type of system has no name, so I've called it a "Synchron". Correspondingly, the infinite speed frame for a tachyon could be called a "synchron" frame: in it, it does not appear as an entity moving in time, but just as a spatial object existing at an instant. The non-relativistic version, the "synchron" is essentially the instantaneous transfer of non-zero impulse across space: the "-on" for action-at-a-distance dynamics.

The issue is muddy on how $𝐉$ and $𝐊$ decompose, except in the "spin 0" case.

Luxons have neither an infinite speed "Synchron" frame nor a rest frame. This can only happen if $P = Mc = E/c$.

Likewise, here, the issue is muddy on how $𝐉$ and $𝐊$ should decompose, except in the "spin 0" case. But, this time, there is also a marginal exception for the case where $𝐖$ and $𝐏$ align, with a fixed ratio $𝐖 = η𝐏$. This sub-class has no official name, either. So, I've referred to it as the "helical" case - or "Helion". A similar subclass exists for the Synchron class; so helical synchrons can be considered as the non-relativistic limit of helical luxons.

Photons belong to the helical sub-class.

The proportion $η$ is a fixed property of the system and is directly related to the component of the angular momentum $𝐉$ parallel to $𝐏$, which is called "helicity", with its value being $ηc$.

Photons don't have spin. They have helicity.

It's possible to write $𝐉 = 𝐫×𝐏 + η𝐏/M$, and $𝐊 = M𝐫 - 𝐏t$ but, this time, $𝐫$ is not quite a bona fide "center of mass position". It's possible to fix it to make it "canonical", but only at the cost of making $𝐫$ a singular function of the other canonical quantities $𝐉$, $𝐊$, $𝐏$ and $M$ (or $E$).

In the spin-0 case, for a non-interacting system, all the canonical quantities are constant, in time - as an expression of the conservation laws for them each. But since $𝐊$ is explicitly time-dependent, its constancy makes $𝐫$ a function of time: $𝐫 = 𝐫_0 + 𝐯t$, expressing a motion in a constant direction at a constant speed, located at $𝐫_0 = 𝐊/M$ at time $t = 0$, moving with constant velocity $𝐯 = 𝐏/M$. So, the conservation law for $𝐊$ is actually just the law of inertia, itself.

You'll have to work out what $𝐫$ looks like, as a function of time $t$ for the case of Tardyons with spin, where $𝐓 ≠ 𝟎$.

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    $\begingroup$ Could you give a source for this? $\endgroup$
    – Riemann
    Commented Mar 3 at 18:52
  • $\begingroup$ Also, why is this a purely relativistic effect? The term $\bf{P}\times \bf{S}$ can be nonzero in nonrelativistic situations. $\endgroup$
    – Riemann
    Commented Mar 3 at 19:23
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    $\begingroup$ That's actually the Newton-Wigner operator - in classical form. You'll notice the 1/c² factor up front. That becomes 0 in the non-relativistic version of the Newton-Wigner operator. $\endgroup$
    – NinjaDarth
    Commented Mar 4 at 1:56
  • $\begingroup$ Thank you, that makes sence. $\endgroup$
    – Riemann
    Commented Mar 4 at 9:39

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