The most general form of the Noether's current (see here and here) is given by $$j^\mu(x)=\sum\limits_a\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\delta\phi_a -\theta^{\mu\nu}\delta x_\nu-K^\mu\tag{1}$$ where $$\theta^{\mu\nu}=\frac{\partial\mathscr{L}}{\partial(\partial_\mu\phi)}\partial^\nu\phi-\eta^{\mu\nu}\mathscr{L}.\tag{2}$$Using $(1)$, let us determine the conserved current due to spacetime translations for which $\delta x^\mu=a^\mu$ (a spacetime independent constant) and $\delta\phi_a=0$. Therefore, $$j^\mu(x)=-\theta^{\mu\nu}a_\nu-K^\mu.\tag{3}$$ Now conserved current implies $$\partial_\mu(\theta^{\mu\nu}a_\nu+K^\mu)=a_\nu\partial_\mu\theta^{\mu\nu}+\partial_\mu K^\mu=0.\tag{4}$$ If $K^\mu$ were zero (as is assumed in Ryder's book, for example), since $a_\nu$ is arbitrary, we would immediately get the usual conservation laws $$\partial_\mu\theta^{\mu\nu}=0\tag{5}$$ from which we obtained four conserved quantities: $P^\nu$.

In the general case, when $K^\mu\neq 0$, assuming that both $\theta^{\mu\nu}$ and $K^\mu$ vanish sufficiently rapidly at spatial infinity, I obtain, $$\frac{d}{dt}(a_\nu\theta^{0\nu}+K^0)d^3x=0.\tag{6}$$ The disturbing thing about this is that now I get only one conserved quantity $$Q=\int j^0d^3x=-\int(a_\nu\theta^{0\nu}+K^0)d^3x\tag{7}$$ because $a^\nu$ does not drop out from the equations and Lorentz indices are contracted! What is wrong with my analysis?

Addendum The analogue of Eq.$(7)$ is like obtaining $$p_x a_x+p_ya_y+p_za_z={\rm constant}$$ under a general spatial translation ${\bf r}\to {\bf r}+{\bf a}$ where ${\bf a}=a_x\hat{x}+a_y\hat{y}+a_z\hat{z}$. What would please me is that if I could show $p_{x,y,z}$ are intividually constants. Please let me know if I am making a conceptual/notational mistake.

Question How to apply general expression for Noether's current to get the energy-momentum conservation law?


1 Answer 1


Your $Q$ still is four independent quantities - one for each independent choice of the translation direction $a^\nu$.

Noether's theorem states that for a one-parameter continuous quasi-symmetry \begin{align} x^\mu & \mapsto x^\mu + \epsilon \delta x^\mu \\ \phi^a & \mapsto \phi^a + \epsilon \delta \phi^a \end{align} where the theory is quasi-symmetric under this transformation in a neighbourhood of $\epsilon = 0$ for constant $\delta x^\mu, \delta \phi^a$, you get the conserved current expression in your question. The transformation \begin{align} x^\mu & \mapsto x^\mu + a^\nu \\ \phi^a & \mapsto \phi^a \end{align} is not of this form, it is really a family of one-parameter symmetries, additionally parametrized by the four-vector $a^\mu$ - it is a generic infinitesimal translation, and the translation group is $\mathbb{R}^4$. There are four independent one-parameter continuous transformations here: \begin{align} x^\mu \mapsto x^\mu + \epsilon e^0 \\ x^\mu \mapsto x^\mu + \epsilon e^1 \\ x^\mu \mapsto x^\mu + \epsilon e^2 \\ x^\mu \mapsto x^\mu + \epsilon e^3 \end{align} for $e^\mu$ the unit vector in the $\mu$-direction, and to each of these transformations Noether's theorem associates a conserved quantity. The usual derivation you cite where the $a^\nu$ is kept generic is just an efficient way to derive the conserved quantities for each of these transformations at once, where we can then choose e.g. $a = e^2$ at the end if we want the conserved quantity associated to the translation in the 2-direction.


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