# Trick for deriving the stress tensor in any theory

In D. Tong's notes on string theory (pdf) section 4.1.1 he explains a trick for deriving the stress-energy tensor which arises from translations in the base manifold of the field theory (in this case the worldsheet). The problem is that I don't understand exactly how the procedure works. I need to look at some worked examples.

EDIT: Perhaps I should explain a little bit more where I'm standing.

Usually, to derive the energy momentum tensor we make a translation in the base manifold, say $x^\mu$ in the usual QFT notation. $$x^\mu\to x'^\mu=x^\mu+\epsilon^\mu$$ without changing the field in a direct way: $$\phi(x)\to \phi'(x')=\phi(x)$$ $$\Rightarrow \delta\phi(x)=-\epsilon^\mu\partial_\mu\phi(x)$$ So in the variation of the action is $$\delta S=\int_R d^4 x [\frac{\partial \mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial \mathcal{L}}{\partial\partial_\mu\phi}\delta\partial_\mu\phi]+\int_{\partial R}d\sigma_\mu \mathcal{L}\epsilon^\mu$$ where the second integral comes from the change of variables $x\to x'$. Thus after integrating by parts the first integral we get the Euler-Lagrange equations which give zero and we are left with $$\int_{\partial R}d\sigma_\mu [\mathcal{L}\epsilon^\mu-\frac{\mathcal{L}}{\partial\partial_\mu \phi}\epsilon^\nu\partial_\nu \phi(x)]=\int_{\partial R}d\sigma_\mu J^\mu$$ where $J^\mu=\mathcal{L}\epsilon^\mu-\frac{\mathcal{L}}{\partial\partial_\mu \phi}\epsilon^\nu\partial_\nu \phi(x)$ has to be conserved by imposing $\delta S=0$. From $J$ we extract the energy-stress tensor: $$\Theta^{\mu\nu}=\frac{\partial\mathcal{L}}{\partial\partial_\mu \phi}\partial^\nu \phi-\mathcal{L}\eta^{\mu\nu}$$ So, still in QFT notation, what Tong says is to promote $\epsilon$ to a function of $x$ so that the surface integral (using Stokes theorem): $$\delta S=\int_Rd^4x \partial_\mu J^\mu=\int_Rd^4x \partial_\mu( \Theta^{\mu\nu}\epsilon_\nu)=\int_Rd^4x [\partial_\mu( \Theta^{\mu\nu})\epsilon_\nu+ \Theta^{\mu\nu}\partial_\mu\epsilon_\nu]$$ but this is not quite the same as eq. 4.3.

The trick is given in equation 4.4 of the attached article:

First couple the theory to gravity, (by introducing a metric tensor in the integration measure and for each index raising) obtaining the action:

$S = \int_M d^4x \sqrt{-g} \mathcal{L}$

Then vary the action with respect to the metric tensor:

$T_{\alpha\beta} = \frac{1}{\sqrt{-g}} \frac{\delta S}{\delta g^{\alpha\beta}}$

Then replace the metric tensor by a flat one to return to the original theory. The resulting stress energy tensor is called Belinfante - Rosenfeld stress energy tensor.

The Belinfante - Rosenfeld stress-energy tensor is not equal to the canonical stress-energy tensor corresponding the translations (derived from Noether's theorem) , but differs by the divergence of an anti-symmetric 3-tensor, thus will be conserved whenever the canonical one is conserved.

The 3-tensor is the canonical conserved current corresponding the Lorentz symmetry. Thus for spinless fields both tensors will be equal.

The Belinfante - Rosenfeld stress-energy tensor is usually considered to be better, since it is always symmetric and gauge invariant.

The reasoning why both methods give the same stress- energy tensors (“up to a total divergence) is as follows:

When one covariantizes the theory, it becomes invariant under diffeomorphisms. Thus the variation of the action with respect to the diffeomorphisms vanishes. Now, this variation is composed from the variation due to the (matter) fields and the variation due to the metric. In flat space the variation due to the fields is the canonical Noether current due to translations, thus the variation with respect the metric should cancel this contribution. Therefore if we vary with respect to the metric, keeping the fields fixed, we obtain the same variation with an opposite sign.

The subtlety of not obtaining the same stress-energy tensor is explained in the following article by: Gotay and Marsden. This article contains a few examples of the derivation of the symmetric (Belinfante - Rosenfeld ) stress-energy tensor. Please, see also the following article by Forger and Romer for further explanations and examples.

• – Qmechanic Jan 3 '15 at 16:16