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

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 $$Q=\int(a_\nu\theta^{0\nu}+K^0)d^3x$$ because $$a^\nu$$ does not drop out from the equations and Lorentz indices are contracted! What is wrong with my analysis?

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