I have seen written (for example in Peskin's and Schroeder's QFT book) that $\partial_\mu j^\mu=0$ means that the current $j$ is conserved. I do not understand how this comes to be: I do not understand what "conserved" means in this context: usually I would say that $j$ is conserved if $\frac{dj}{dt}=0$, and this is clearly not what is in stake here.
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3$\begingroup$ Look up the continuity equation (hint: that's the equation you're asking about). The charge density is the zero component of $j$. A spacial integral over this density gives you a thing (charge) that's conserved (in the sense of the word you seem to prefer), as implied by that equation. In field theory it is standard terminology for "conserved" to just mean that equation is satisfied. $\endgroup$– Adomas BaliukaCommented Jul 26, 2017 at 23:37
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2 Answers
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Let's rewrite that formula a little clearer. I'll (suggestively) write $j=(\rho,\vec{j})$, where $\rho=j^0$ is the 0th component of the 4-vector, and $\vec{j}$ is the spacial component of the 4-vector. I'll call $\rho$ the density, and $\vec j$ the current. Then the statement $\partial_\mu j^\mu=0$ becomes
$$ \frac{\partial \rho}{\partial t}=-\vec\nabla\cdot\vec{j} $$
This says that the divergence of the current $\vec{j}$ is equal and opposite to the change in the density. Put another way, if you consider an infinitesimal volume, the current flux out of the boundary of the volume exactly equals the decrease in the density within the volume. This is usually what we mean when we say something (say a fluid) is conserved: If the fluid is flowing out of an area, the density (concentration of the fluid) decreases proportionately.
Note that if you assume that all currents go to zero at infinity, you can also make a global claim that you might recognize more as a conservation law:
$$ \frac{\partial}{\partial t}\int \rho = -\int\vec\nabla\cdot\vec j= 0 $$ In other words, $\int\rho$ is globally conserved. The first statement is usually referred to as local conservation, since it specifies that any decrease in $\rho$ has to be caused by a current flux, and doesn't allow $\rho$ to spontaneously decrease in one region and increase in another region far away.
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$\begingroup$ But is conservation of charge equivalent to $\partial_\mu j^\mu=0$? How can you go from $\frac{d}{dt}\int\rho=0$ to $\partial_\mu j^\mu=0$? i.e. does the global conservation of charge imply what apparently one calls conservation of current (which is local in some sense)? Peskin and Schroeder seem to suggest this. $\endgroup$– soapCommented Jul 27, 2017 at 13:08
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1$\begingroup$ @Soap No, local conservation of charge is strictly stronger than global conservation of charge. It's possible to go from local conservation to global conservation, but if you only know global conservation you can't derive local conservation. $\endgroup$ Commented Jul 27, 2017 at 18:07
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$\begingroup$ @Soap Intuitively, global conservation says that total charge is conserved, but doesn't forbid charge from spontaneously teleporting across the universe. Local conservation says that to decrease the charge, you need to have a finite current density out of a volume. So things can't teleport instantaneously. $\endgroup$ Commented Jul 27, 2017 at 18:08
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The charge is $Q=\int d\mathbf{x} j^0$ so $\dot{Q} = \int d\mathbf{x}\partial_0 j^0=-\int d\mathbf{x}\partial_i j^i$. Integrating by parts, $\dot{Q}$ is a boundary term that vanishes if $\mathbf{j}$ vanishes fast as $r\to\infty$. Then $Q$ is conserved.
