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I've been trying to understand (from several sources) how Noether's theorem for fields is derived, and reading the Wikipedia page about Noether's theorem I encountered the following:

say we have the following infinitesimal transformation of coordinates and field: $$x^{\mu} \rightarrow \xi^{\mu}=x^{\mu}+\delta x^\mu$$ $$\phi \rightarrow \alpha(\xi^{\mu})=\phi(x^{\mu})+\delta \phi(x^{\mu})$$ and the change in the action can be written as the difference between the integral of the Lagrangian over the transformed region $\Omega'$ and the integral of the Lagrangian over the original region $\Omega$: $$ \int_{\Omega'} {L(\alpha,\partial_{\nu}\alpha,\xi^\mu) d^{4}\xi} - \int_{\Omega} {L(\phi,\partial_{\nu}\phi,x^\mu) d^{4}x}$$

The article says that using divergence theorem it four dimensions and assuming the change in the region $\Omega\rightarrow\Omega'$ it can be shown that the aforementioned expression is equivalent to the following one:

$$ \int_{\Omega} {L(\alpha,\partial_{\nu}\alpha,x^\mu)+\frac {\partial} {\partial x^\sigma} [L(\phi,\partial_{\nu}\phi,x^\mu) \delta x^\sigma]-L(\phi,\partial_{\nu}\phi,x^\mu) d^{4}x}$$

I tried showing this transition is true by assuming the first integral of the original expression is a divergence of some 4-vector field but I couldn't get it right, I also trying showing the same transition based on the Jacobian of the change of variables done between the integrals and couldn't do it.

Can anyone please detail this transition so it will be clear why it is correct?

Qmechanic
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Tamir Vered
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1 Answers1

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Starting from (note that there's an error in your formula since the first lagrangian has to be the prime, transformed, lagrangian)

$$\int_{\Omega'} {L^\prime(\alpha,\partial_{\nu}\alpha,\xi^\mu) d^{4}\xi} - \int_{\Omega} {L(\phi,\partial_{\nu}\phi,x^\mu) d^{4}x}$$

if you want to change the volume of integration, we have to find the Jacobian, which, given the transformation, is simply

$$J = \frac{\partial \xi^{\sigma}}{\partial x^\sigma} = 1 + \partial_\sigma\delta x^\sigma $$

If you plug this in the integrals you find

$$\int_\Omega d^4x \left[(1+\partial_\sigma\delta x^\sigma)L^\prime-L\right] $$

which at first order leaves you with

$$\int_\Omega d^4x\,\left[(L^\prime-L)+\partial_\sigma\delta x^\sigma L\right] = \int_\Omega d^4x\left[\Delta L+\partial_\sigma\delta x^\sigma L\right]$$

where $\Delta L$ is the total variation of the lagrangian. This is given by

$$\Delta L = L^\prime(\alpha, \partial_\mu\alpha, \xi^\mu)-L(\phi, \partial_\mu\phi, x^\mu) = \frac{\partial L}{\partial \phi}\delta\phi+\frac{\partial L}{\partial\phi_{,\mu}}\delta\phi_{,\mu}+\frac{\partial L}{\partial x^\mu}\delta x^\mu+\delta L(\phi, \partial_\mu\phi, x^\mu) $$

in first order in $\delta$. Now we can do some manipulations: the integral becomes

$$\int_\Omega d^4 x\left[\frac{\partial L}{\partial \phi}\delta\phi+\frac{\partial L}{\partial\phi_{,\mu}}\delta\phi_{,\mu}+\frac{\partial L}{\partial x^\mu}\delta x^\mu+\delta L(\phi, \partial_\mu\phi, x^\mu)+\partial_\sigma\delta x^\sigma L\right] \\ = \int_\Omega d^4 x\left[\delta L + \frac{\partial L}{\partial \phi}\delta\phi+\color{red}{\frac{\partial L}{\partial (\partial_\mu \phi)}\delta\partial_\mu\phi} + \color{orange}{((\partial_\mu L)\delta x^\mu+(\partial_\mu\delta x^\mu)L)}\right]$$

where I've changed the mute index $\sigma$ to $\mu$. The red term can be rewritten as a divergence using the distribution formula for derivative

$$\frac{\partial L}{\partial (\partial_\mu \phi)}\delta\partial_\mu\phi = \partial_\mu\left(\frac{\partial L}{\partial (\partial_\mu \phi)}\delta\phi\right)-\left(\partial_\mu\frac{\partial L}{\partial (\partial_\mu \phi)}\right)\delta\phi$$

In much the same way, the orange term gives

$$ ((\partial_\mu L)\delta x^\mu+(\partial_\mu\delta x^\mu)L) = \partial_\mu(L\delta x^\mu)$$

so the integral becomes

$$\int_\Omega d^4 x \left[\delta L +\color{red}{\left(\frac{\partial L}{\partial\phi}-\partial_\mu\frac{\partial L}{\partial \phi_{,\mu}}\right)\delta\phi}+\partial_\mu\left(\frac{\partial L}{\partial \phi_{,\mu}}\delta\phi\right)+\partial_\mu(L\delta x^\mu)\right] $$

the red term is zero for the Euler-Lagrange equation. So in the end

$$\int_\Omega d^4 x \left[\delta L+\partial_\mu\left(\frac{\partial L}{\partial \phi_{,\mu}}\delta\phi+L\delta x^\mu\right)\right] $$

which is exactly your result once you write down the difference between the two lagrangians.

Just to be complete, let me end the proof by putting to zero the integral and noting that $\delta L$ can only be a total derivative $\delta L = \partial_\mu\delta \Lambda^\mu$ and getting

$$\int_\Omega d^4 x\, \partial_\mu\left(\frac{\partial L}{\partial \phi_{,\mu}}\delta\phi+L\delta x^\mu+\delta\Lambda^\mu\right) = 0 \implies \partial_\mu \left(\frac{\partial L}{\partial \phi_{,\mu}}\delta\phi+L\delta x^\mu+\delta\Lambda^\mu\right) = 0$$

The conserved current therefore is given by

$$ J^\mu = \frac{\partial L}{\partial \phi_{,\mu}}\delta\phi+L\delta x^\mu+\delta\Lambda^\mu $$

Notation:

Just occurred to me that many won't be familiar with this notation which is borrowed from general relativity, so i'll leave it here

$$\partial_\mu\phi = \phi_{,\mu}$$

Davide Morgante
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