¿Qué ley de conservación corresponde a los impulsos de Lorentz?

Advertencia: esta es una derivación larga y aburrida. Si solo está interesado en el resultado, salte a la última oración.

El teorema de Noether se puede formular de muchas maneras. A los efectos de su pregunta, podemos utilizar cómodamente la formulación lagrangiana relativista especial de un campo escalar. Entonces, supongamos que se nos da una acción

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

Ahora supongamos que la acción es invariante bajo alguna transformación infinitesimal$m: x^{\mu} \mapsto x^{\mu} + \delta x^{\mu} = x^{\mu} + \epsilon a^{\mu}$(no consideraremos ninguna transformación explícita de los propios campos). Entonces obtenemos una corriente conservada

$$J^{\mu} = {\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} a_{\nu} - {\mathcal L} a^{\mu} = \left ({\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} - {\mathcal L} g^{\mu \nu} \right) a_{\nu} .$$ Obtenemos una carga conservada de él dejando $Q \equiv \int J^0 {\rm d}^3x$ya que desde $\partial_{\mu}J^{\mu} =0$tenemos eso $${\partial Q \over \partial t} = \int {\rm Div}{\mathbf J}\, {\rm d}^3 x = 0$$ que se mantiene en cualquier momento en que las corrientes decaen lo suficientemente rápido.

Si la transformación está dada por traslación$m_{\nu} \leftrightarrow \delta x^{\mu} = \epsilon \delta^{\mu}_{\nu}$obtenemos cuatro corrientes conservadas

$$J^{\mu \nu} = {\partial {\mathcal L} \over \partial \phi_{\mu}} \phi^{\nu} - {\mathcal L} g^{\mu \nu} .$$

Este objeto es más comúnmente conocido como tensor de energía de estrés.$T^{\mu \nu}$y las corrientes conservadas asociadas se conocen como momentos$p^{\nu}$. Además, en general, la corriente conservada viene dada simplemente por$J^{\mu} = T^{\mu \nu} a_{\nu}$.

Para una transformación de Lorentz tenemos

$$m_{\sigma \tau} \leftrightarrow \delta x^{\mu} = \epsilon \left(g^{\mu \sigma} x^{\tau} - g^{\mu \tau} x^{\sigma} \right)$$ (Observe que esto es antisimétrico y, por lo tanto, solo hay 6 parámetros independientes de la transformación), por lo que las corrientes conservadas son las corrientes de momento angular $$M^{\sigma \tau \mu} = x^{\tau}T^{\mu \sigma} - x^{\sigma}T^{\mu \tau}.$$ Finalmente, obtenemos el momento angular conservado como $$M^{\sigma \tau} = \int \left(x^{\tau}T^{0 \sigma} - x^{\sigma}T^{0 \tau} \right) {\rm d}^3 x .$$

Tenga en cuenta que para las partículas podemos avanzar un poco más ya que sus momentos asociados y sus momentos angulares no están dados por una integral. Por lo tanto tenemos simplemente que$p^{\mu} = T^{\mu 0}$y$M^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$. La parte de rotación de esto (escrita en la forma del pseudovector habitual) es

$${\mathbf L}_i = {1 \over 2}\epsilon_{ijk} M^{jk} = ({\mathbf x} \times {\mathbf p})_i$$ mientras que para la parte de impulso obtenemos $$M^{0 i} = \left(t {\mathbf p} - {\mathbf x} E \right)^i$$ que no es otra cosa que el centro de masa en $t=0$(Somos libres de elegir $t$ya que la cantidad se conserva) multiplicado por $\gamma$ya que tenemos las relaciones $E = \gamma m$, ${\mathbf p} = \gamma m {\mathbf v}$. Nótese la similitud con el ${\mathbf E}$, $\mathbf B$descomposición del tensor de campo electromagnético $F^{\mu \nu}$.

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:

$$\phi(y)=\phi(x)+ \Lambda^{\mu\nu}x_\nu\partial_\mu\phi(x)+...$$ La variación del campo wrt$\Lambda^{\mu\nu}$es $$\frac{\delta \phi}{\delta \Lambda^{\mu\nu}}=x_\nu\partial_\mu\phi(x)-x_\mu\partial_\nu\phi(x)$$ 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

$$\frac{\delta \mathcal{L}}{\delta \Lambda^{\mu\nu}}=\sum_n\{\frac{\partial \mathcal{L}}{\partial\phi_n} \frac{\delta\phi_n}{\delta \Lambda^{\mu\nu}} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_n)} \frac{\delta(\partial_{\mu}\phi_n)}{\delta \Lambda^{\mu\nu}} \}$$ Aplicando la ecuación de movimiento $$\frac{\partial \mathcal{L}}{\partial \phi_n} -\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}=0$$ obtenemos la ley de conservación: $$\frac{\delta \mathcal{L}}{\delta \Lambda^{\mu\nu}}=\sum_n\partial_\mu[\frac{\partial \mathcal{L}}{\partial_{\mu}\phi_n} \frac{\delta\phi}{\delta \Lambda^{\mu\nu}} ]$$ 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 $$\partial_\mu j^{\mu \lambda\sigma} = 0$$ donde la corriente conservativa $$j^{\mu \lambda\sigma}=x^\lambda T^{\mu\sigma} - x^{\sigma}T^{\mu\lambda}$$ es el momento angular y $$T_{\mu\nu}= \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\nu\phi-g_{\mu\nu}\mathcal{L}$$ es el impulso.

¿Una respuesta directa? Realmente no tiene un nombre, pero siempre se ha escrito en la literatura en forma de 3 vectores como$𝐊$, junto con el momento angular 3-vector$𝐉$.

Cualquier sistema, elemental o compuesto, relativista o no relativista, que posee un sistema de referencia en reposo (es decir, un sistema en el que la cantidad de movimiento$𝐏$es$𝟎$), y una posición del centro de masa,$𝐫$, 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 $𝐓 ≠ 𝟎$.