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I'm not sure if I understand the concept correctly. Given an infinitesimal transformation

$$\phi \rightarrow \phi + \alpha \Delta\phi$$

the change in the Lagrangian density $\mathcal{L}(\phi,\partial_\mu \phi)$ is

$$\mathcal{L} \rightarrow \mathcal{L} + \alpha \Delta\mathcal{L}$$

For the transformation to be a symmetry, the new Lagrangian can differ only by a four-divergence so that

$$\Delta\mathcal{L} = \partial_\mu J^\mu$$

for some four-vector $J^\mu$.

Now, we have, using E.L. equations, the identity

$$\partial_\mu J^\mu = \partial_\mu\left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\right)$$

From which finally the "conserved current" is:

$$j^\mu \equiv J^\mu-\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi$$

Anyway, I'm trying to do a calculation for a concrete example of the Lagrangian $\mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi$ and transformation $\phi \rightarrow \phi + \alpha$ for constant $\alpha$.

For this, $\Delta\mathcal{L} = 0 = \partial_\mu J^\mu$. Also $\Delta\phi = 1$. So

$$\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi = \partial^\mu \phi$$

And

$$j^\mu = J^\mu - \partial^\mu \phi$$

Peskin & Schroeder say that the conserved current is just $j^\mu = \partial^\mu \phi$. I suppose this is because it's defined up to a 4-divergence. So in this case $J^\mu$ can be omitted and also the minus sign doesn't matter since it's defined up to a multiplicative constant as well.

Please correct my understanding of this. What I'm having most trouble understanding here is how the different objects are defined 'up to' something.

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    $\begingroup$ As $\Delta\mathcal L=0$, you have $J^\mu=0$. Therefore, $j^\mu\propto\partial^\mu\phi$ (you set to $-1$ the constant of proportionality, P&S to $+1$: the actual value is irrelevant. If $j^\mu$ is conserved, so is $\tilde j^\mu=A j^\mu$ for any $A\in\mathbb C$). You found the correct answer, but the thing is, the answer is not unique! that's why yours and P&S's differ. (note that $\partial_\mu J^\mu=0$ doesnt imply that $J^\mu=0$, but as $j^\mu$ is defined modulo a closed field, you can set $J^\mu=0$ WLOG). $\endgroup$ – AccidentalFourierTransform Dec 25 '15 at 12:41
  • $\begingroup$ Please elaborate what you mean by 'defined modulo a closed field'. I suspect this is the crux of the thing. $\endgroup$ – Spine Feast Dec 25 '15 at 12:47
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    $\begingroup$ I would recommend to first derive the Noether theorem without the $J^{\mu}$ and only then realise than if one adds an additional current nothing really changes very much. $\endgroup$ – gented Dec 25 '15 at 12:54
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    $\begingroup$ @DepeHb A closed field $f^\mu$ is any field such that $\partial_\mu f^\mu=0$. For example, $j^\mu$ is a closed field. If you add two closed fields, the result is also closed ($\partial_\mu(f^\mu+j^\mu)=\partial_\mu f^\mu+\partial_\mu j^\mu=0$). This means that the Noether current $j^\mu$ is not unique: if you add any closed field to $j^\mu$, the resulting current is also conserved, and so equally valid as the original current. In your example, you dont know what $J^\mu$ is, but you are certain that it is closed. Thus, you can set $J^\mu=0$ WLOG. $\endgroup$ – AccidentalFourierTransform Dec 25 '15 at 12:55
  • $\begingroup$ What if there's a complex scalar field? Is there two conserved currents? $\endgroup$ – Spine Feast Dec 25 '15 at 17:35
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For the given case since $\partial _\mu J^{\mu} = 0$, there is no need of adding this boundary term to the conserved current $j^{\mu}$ as the addition of it is meaningless as it won't play any part except adding up to a meaningless factor. We already have for the equation of the conserved current $\partial_{\mu} j^{\mu} = 0$ and we can choose to add any term $a^{\mu}$ to $j^{\mu}$ as long as it satisfies $\partial _{\mu} a^{\mu} = 0$. Hence it is really irrelevant to have the $J^{\mu}$ part.

As for the second part of the question, you are right in saying that the minus sign can be omitted as it is defined upto a multiplicative constant.

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