When we deal with the Klein Gordon Lagrangian, we can find some conserved quantities.

For example, we remark that under a space time translation $a^\mu$ we can find that the quantity: $a_\rho T^{\mu \rho}$ is a conserved current where $T^{\mu \rho}$ is the energy impulsion tensor.

That means that $\partial_\mu (a_\rho T^{\mu \rho})$.

My questions:

First question:

In fact we can show that $\partial_\mu T^{\mu \rho}=0$. Could we have guess it without explicitely calculating this quantity? I mean is it obvious that if the 4-current conserved is $a_\rho T^{\mu \rho}$ then $\partial_\mu T^{\mu \rho}=0$?

[edit] By asking the question I just thought it could be because $a_\rho \partial_\mu T^{\mu \rho}=0$ is true for any, $a$ thus $\partial_\mu T^{\mu \rho}=0$. Am I right?

Second question:

We also can prove that there are charged conserved associated with this symmetry.

For example we have : $P_\mu = \int_{\mathbb{R}^3} d^3x T_{0 \mu}$ that is a conserved charged (it means that $\frac{d P_\mu}{dt}=0$).

It is almost the same question : could we have immediately guess from the conserved current that we would have this conserved charge ?

Is it because as the fields go to zero at infinity, $ \int div(\vec{j}) d^3x=0$ and as $\frac{\partial j^0 }{\partial t}=div(\vec{j})$, we have $\int \frac{\partial j^0}{\partial t} d^3x$

So because the field go to zero at infinity we immediately deduce a conserved charge when we have a conserved 4-current?

[edit]: In fact I don't think it is that obvious as we have the fields $\phi$ that goes to zero at infinity but not necessarily the spatial part of the $4$-current conserved in a general case. Am I right?

  • 2
    $\begingroup$ Yes, you are absolutely right. Please mark the end of your questions with a single question mark instead of three. $\endgroup$ – Prof. Legolasov Apr 19 '17 at 0:07
  • $\begingroup$ Thank you for your comment. So just to be sure : having a conserved 4-current doesn't ensure me to have a conserved charge. It is very often the case but we could imagine case where it is not (for example if $\vec{j}$ contains other terms than $\phi(x)$ or $\pi(x)$). $\endgroup$ – StarBucK Apr 19 '17 at 0:17
  • $\begingroup$ Generally, for each region $\mathcal{R}$ bounded by the closed surface $\partial\mathcal{R}=\Sigma$ you have $\frac{d}{dt} Q_{\mathcal{R}} = - \intop_{\Sigma} \vec{j} \cdot \vec{d\sigma}$. This is the general form of the charge conservation law. $Q$ is conserved globally for the whole space only if the appropriate boundary conditions are satisfied. $\endgroup$ – Prof. Legolasov Apr 19 '17 at 0:24

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