Current and charges conserved in Klein Gordon equation. What are the links between them?

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 $\partial_\mu T^{\mu \rho}=0$?

 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?

: 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?

• Yes, you are absolutely right. Please mark the end of your questions with a single question mark instead of three. – Prof. Legolasov Apr 19 '17 at 0:07
• 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 $\pi(x)$). – StarBucK Apr 19 '17 at 0:17
• 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. – Prof. Legolasov Apr 19 '17 at 0:24