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In the derivation of Noether's theorem some authors consider not only redefinitions of the fields \begin{equation} \phi(x) \rightarrow \phi'(x) = \phi(x) +\delta\phi(x) \end{equation} but also redefinitions of the spacetime coordinates \begin{equation} x^{\mu} \rightarrow x'^{\mu} = x^{\mu} +\delta x^{\mu} \ . \qquad(*) \end{equation} It is not quite clear to me what this redefinition of spacetime coordinates means.

Until now i have thought that it is just a switch in the coordinate system. I.e. if $P$ is a point in spacetime and we assign it the coordinate $x(P)$, usually a 4-vector, then we could just as well label it with the different coordinate $x'(P)$. Therefore $x$ and $x'$ describe the same point in spacetime.

When taking this point of view, however, i have run into some trouble in the derivation of Noether's theorem. I would like to know if my conception of the meaning of $(*)$ is correct or if not what the correct interpretation is.

My question is $\textbf{not a duplicate}$ of Noether's theorem in classical field theory as that question does not address whether $x$ and $x'$ refer to the same point in spacetime.

Zarathustra
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1 Answers1

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The shift $\delta x^\mu=\epsilon^\mu$ is just a relabeling of the points of spacetime. After the relabelling the point $P$ that had coordinate $x^\mu$ now has coordinate $x^\mu- \epsilon^\mu$ and the point $P'$ that now has coordinate $x^\mu$ is point $P+ \epsilon$. The field at point $P$ is unchanged, but $\varphi(x^\mu)$ now refers to the field at $P+\epsilon$ and so $$ \varphi(x^\mu)\mapsto \varphi(x^\mu +\epsilon^\mu) =\varphi(x) + \epsilon^\mu \frac{\partial \varphi}{\partial x^\mu}+ O(\epsilon^2) $$ Similarly $$ g_{\mu\nu}(x)dx^\mu dx^\nu \mapsto g_{\mu\nu}(x+\epsilon)d(x^\mu+\epsilon^\mu)d(x^\nu+ \epsilon^\nu)\\ = (g_{\mu\nu}(x) +\nabla_\mu \epsilon_\nu+\nabla_\nu\epsilon_\mu)dx^\mu dx^\nu $$ where some covariant derivative identities are used at the lat step.

This relabeling cannot change anything physical, so, in particular, the action functional cannot be changed by it.

Now, if the equations of motion are obeyed the action is unchanged by any change in the fields. Combining these two facts, and an integration by parts leads to $$ 0=\nabla_\mu T^{\mu\nu}, $$
where $T^{\mu\nu}$ is the Hilbert energy-momentum tensor defined by $$ \delta S = \frac 12 \int d^dx \sqrt{g} T^{\mu\nu} \delta g_{\mu\nu}. $$

mike stone
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