# Lorentz covariance of the Noether charge

The invariance under translation leads to the conserved energy-momentum tensor $\Theta_{\mu\nu}$ satisfying $\partial^\mu\Theta_{\mu\nu}=0$, from which we get the conserved quantity$$P^\nu=\int d^3\mathbf x\Theta^{0\nu}(x)$$But I cannot see explicitly how this quantity is a four-vector covariant under Lorentz transformation, since $d^3\mathbf x$ is part of the invariant $d^4x$, $\Theta^{0\nu}(x)$ is part of the covariant tensor $\Theta^{\mu\nu}(x)$, neither of which transforms covariantly.

So can someone show me how this becomes correct?

And generally, how to show that a Noether charge $Q$ corresponding to the Noether current $j^\mu$, $$Q=\int d^3\mathbf x j^0(x)$$ , is a Lorentz scalar?

• Weinberg, gravitation and cosmology, p40-41 – user26143 Nov 21 '13 at 6:50
• Thank you very much @user26143, that really solved my problem! – LYg Nov 21 '13 at 22:03

You may use the following notation for hypersurfaces in four dimensions :

$d\sigma_\mu = \epsilon_{\mu\alpha\beta\gamma}dx^\alpha dx^\beta dx^\gamma$

For instance $d\sigma_0= d^3x$

The expression of the momentum-energy is then :

$P_\nu = \int d\sigma^\mu \Theta_{\mu\nu}$

The same kind of expression could be used with the charge :

$Q = \int d\sigma^\mu j_{\mu}$

[EDIT]

How make the connection with the OP formulae ?

One may adopt the following point of view, take for instance the formula for the charge $\tilde Q = \int d\sigma^\mu j_{\mu}$, this means :

$\tilde Q = \int d\sigma^0 j_{0} + \int d\sigma^1 j_{1} + \int d\sigma^2 j_{2}+ \int d\sigma^3 j_{3} \\ =\int dx~ dy~ dz ~j_{0}+\int dy~ dz~ dt ~j_{1}+\int dz~ dt~ dx ~j_{2} + \int dt~ dx~ dy ~j_{3} \\=Q + \int dy~ dz~ dt ~j_{1}+\int dz~ dt~ dx ~j_{2} + \int dt~ dx~ dy ~j_{3}$

Now, take one of the residual integrals, for instance $I_1=\int dy~ dz~ dt ~j_{1}$, it is an integral at $x$ constant, and one may choose $x=\pm\infty$. At infinity, we may suppose that the current is zero : $j_1(\pm \infty)=0$. So, assuming a zero current $j_1$ at spatial $x$ infinity, we get $I_1=0$, and one may have the same demonstration for the other 2 integrals.

So, finally , with the hypothesis of taking spatial slicing of the residual integrals at spatial infinity, and vanishing currents at spatial infinity, we have $\tilde Q = Q$

• But I couldn't see how your definition $P_\nu=\int d\sigma^\mu\Theta_{\mu\nu}$ coincides with $P_\nu=\int d^3\mathbf x\Theta_{0\nu}=\int d\sigma^0\Theta_{0\nu}$, respects – LYg Nov 22 '13 at 1:00
• @LYg : I edited the answer. – Trimok Nov 22 '13 at 10:50
• However, what if you haven't chosen $x=\pm\infty$, do you mean $I_1=\int dy~ dz~ dt ~j_{1}$ is independent of $x$ ? but can you show this, seeing $j_i$ only vanishes at spatial infinity, not at time infinity, so that you cannot say $I_1$ is independent of $x$ due to vanishing space-time surface by Gauss's law. – LYg Nov 22 '13 at 13:46