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In Peskin equation 2.12

$$ \partial_\mu j^\mu(x) = 0, \quad {\rm for} \quad j^\mu(x) = \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi - \mathcal{J}^\mu $$

Why the current $j^\mu(x)$ is conserved when $\partial_\mu j^\mu = 0$ ?

I think the current is conserved only when $\int d^2x\,\vec{j}\cdot\vec{n} = 0$, (use divergence theorem)

so $j^\mu$ ($\mu = 1, 2, 3$) must be $0$ at spatial boundary.

This condition that $j^\mu(x)=0$ is just assumption?

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    $\begingroup$ please be more specific; can you provide equation numbers from the book? $\endgroup$
    – koldrakan
    Apr 12 '17 at 7:17
  • $\begingroup$ equation 2.12 and explanation below the equation 'This result states that the current $j^\mu(x)$ is conserved.' $\endgroup$
    – Orient
    Apr 12 '17 at 7:51
  • $\begingroup$ Yes, the boundary condition is an assumption. $\endgroup$
    – Qmechanic
    Apr 12 '17 at 8:00
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    $\begingroup$ Please do not rely on external links for your question to make sense - include all relevant equations in your question. $\endgroup$
    – ACuriousMind
    Apr 12 '17 at 10:04
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    $\begingroup$ I think conservation of the $j^\mu$ is defined to be $\partial_\mu j^\mu = 0$. $\endgroup$
    – John
    Apr 12 '17 at 12:36
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The equation implies a continuity equation. More specifically, write out $\partial_\mu j^\mu=0$ in component form gives you exactly the continuity equation (if we identify the time component as the density, and space component as the current).

A continuity equation means that the change of current (w.r.t time) within a volume equals the flux flow in/out of it. Then, it cannot be created out of nothing. This is what is meant by being conserved.

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  • $\begingroup$ Incompressible requires $\nabla \cdot \vec{j} = 0$. $\endgroup$ Apr 12 '17 at 13:03
  • $\begingroup$ @SeanE.Lake Sorry, I misuse the term "incompressible". :P $\endgroup$
    – taper
    Apr 12 '17 at 13:07
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Note that the current $j^\mu$ is a current in 4D spacetime: $j^\mu = (\rho,j^i)$, where $\rho$ is the charge density and $j^i$ is the traditional 3D current.

Charge conservation then states that for a given volume, the change in charge corresponds to the total flux into the volume. $\int\limits_V \frac{\partial\rho}{\partial t} dv = -\int\limits_S j^i da_i = -\int\limits_V \partial_i j^i dv$, where $da_i$ is an infinitesimal surface element and the last step uses the divergence theorem.

Differentiating both sides wrt volume gives: $$\frac{\partial\rho}{\partial t} = -\partial_i j^i$$ $$\frac{\partial\rho}{\partial t} + \partial_i j^i = 0 $$ $$ \partial_\mu j^\mu = 0 $$

We now derived current conservation from charge conservation. The book does it the other way around. To gain intuition, just think of the current as electrical current.

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I see two good answers but none of those really answer OP's question entirely so I'll give a third one... (I will tell some things that you appear to already know for completeness)

First of all, $\partial_\mu j^\mu = 0$ as you mentioned due to the equations of motion. Or if you want to go all fancy use Noethers theorem for translation symmetry.

You must integrate this over some 4D Volume:

$$\int dT dV \partial_\mu j^\mu = 0 \rightarrow \int dTdV \partial_T j^T = \int dTdV \partial_i j^i$$

We can now perform partial integration (=the divergence theorem) and find:

$$\int dV j^T = \int dV \rho = \int_{\text{surface at }\infty} dT dA\vec{n}\vec{J_i}$$

The last term measures the flux of the field going out trough some surface at spatial infinity. But spatial infinity is infinitly far away so there can be no flux passing trough it such that we can set this term to 0. The result is:

$$\int dV j^T = 0$$

This is what we mean with a conserved current. Note, it is not the current itself that is conserved, it is the integrated value of its temporal component. So what might have confused you is this bad "name"...


Interesting sidenote: some spacetimes have edges that are not at spatial infinity. For example, the schwarzschild metric for black holes has an event horizon that is not at spatial infinity !

In cases like this we cannot simply state that the flux trough this surface is zero and some additional care is required. But as long as you stick to Minkowski there are no problems !

I hope this helped and answered your actual question, if not let me know :)

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  • $\begingroup$ Why flux is zero at spatial infinity in Minkowski space? $\endgroup$
    – Orient
    Apr 13 '17 at 2:53
  • $\begingroup$ I'll give three reasons for this pick one you like :) 1)if the flux trough infinity is nonzero, make it such that your infinity is further away you can always put it far enough for this to be the case. 2)just be sneaky about it, the flux trough infinity is not zero but all of this happens infinity far away so why should we care about it ? You can always put $\infty$ further such that it doens't bother you... 3)mathematically you write $\phi \sim e^{ikx}$ you can always redefine $k\rightarrow k+i\epsilon, \epsilon\rightarrow0$ such that $x\rightarrow \infty, \phi\rightarrow0$ $\endgroup$
    – gertian
    Apr 13 '17 at 11:15

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