2
$\begingroup$

I'm a bit confused about Noether's theorem (or about calculus of variations in general) when it comes to the translational symmetry $x^\mu\mapsto {x'}^\mu=x^\mu-a^\mu$. My professor just wrote that if the Lagrangian density depends on $\phi$ and $\partial_\mu\phi$ only, then it transforms like

$$ \tag{1}\mathcal L\mapsto\mathcal L'=\mathcal L+a^\mu\partial_\mu\mathcal L=\mathcal L+a^\mu\partial_\nu({\delta^\nu}_\mu\mathcal L) $$

I don't understand why this is the case and how it is derived.

This is what I have tried so far to derive it:

Let's make the "infinitesimal" translation $x^\mu\mapsto x^\mu-a^\mu$ a bit more concrete by introducing a parameter $\alpha$ that is meant in the limit $\alpha\to 0$. The translation is then $x^\mu\mapsto x^\mu-\alpha\,a^\mu$, so we have

\begin{align} \frac{d\phi}{d\alpha}=\frac{\partial\phi}{\partial {x'}^\mu}\frac{\partial{x'}^\mu}{\partial\alpha} = -a^\mu\partial_\mu\phi \end{align}

For $\mathcal L$, I got:

\begin{align} \frac{d\mathcal L}{d\alpha} & =\frac{\partial\mathcal L}{\partial\phi}\frac{\partial\phi}{\partial\alpha} + \frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\frac{\partial(\partial_\nu\phi)}{\partial\alpha} \\ & = \frac{\partial\mathcal L}{\partial\phi}\frac{\partial\phi}{\partial\alpha} + \frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\partial_\nu\frac{\partial\phi}{\partial\alpha} \\ & = -a^\mu\left(\frac{\partial\mathcal L}{\partial\phi}\partial_\mu\phi+\frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\partial_\nu\partial_\mu\phi\right)\\ & = -a^\mu\left(\frac{\partial\mathcal L}{\partial\phi}\partial_\mu\phi+\partial_\nu\left(\frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\partial_\mu\phi\right)-\left(\partial_\nu\frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\right)\partial_\mu\phi\right)\\ & = -a^\mu\left(\partial_\nu\left(\frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\partial_\mu\phi\right)+\underbrace{\left(\frac{\partial\mathcal L}{\partial\phi}-\partial_\nu\frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\right)}_{=0}\partial_\mu\phi\right)\\ & = -a^\mu\partial_\nu\left(\frac{\partial\mathcal L}{\partial(\partial_\nu\phi)}\partial_\mu\phi\right) \end{align}

The last term in the brackets is something else than the last term in the brackets in $(1)$. It looks more like the first part of the Noether current $j^\mu=\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\Delta\phi-\mathcal J^\mu$. However, it should be the second part, the $\mathcal J^\mu$, which is defined as the variation of $\mathcal L$ under the symmetry: $\mathcal L\mapsto\mathcal L+\epsilon\partial_\mu\mathcal J^\mu$.

What's the problem here?

$\endgroup$
  • 1
    $\begingroup$ This question (v1) is essentially a field theoretic version of this Phys.SE question for point mechanics. $\endgroup$ – Qmechanic Dec 21 '15 at 16:57
  • 1
    $\begingroup$ The Lagrangian $\mathcal{L}$ can be viewed as a function of $x$ as $\mathcal{L}_{\phi}(x)=\mathcal{L}(\phi(x),(\partial_{\mu}\phi)(x))$ for a given field $\phi$. $\endgroup$ – Dominik Dec 21 '15 at 17:33
  • 1
    $\begingroup$ Try this. I recomend understanding really really good this aspect since most of modern physics is based on QFT. $\endgroup$ – Mikey Mike Dec 21 '15 at 17:35
  • 1
    $\begingroup$ @bass its just a Taylor expansion... $\endgroup$ – AccidentalFourierTransform Dec 21 '15 at 20:06
  • 2
    $\begingroup$ $\mathcal{L}_{\phi}(x+\alpha a)=\mathcal{L}_{\phi}(x)+\alpha a^{\mu}(\partial_{\mu}\mathcal{L}_{\phi})(x)+\mathcal{O}(\alpha^{2})$. $\endgroup$ – Dominik Dec 21 '15 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.