Variation of action in terms of energy-momentum tensor

Question

In Di Francesco, Mathieu, and Sénéchanl, Conformal Field theory section 4.2.2 it is stated that under an arbitrary diffeomorphism $x\rightarrow x+\epsilon$ the action transforms like $$\delta S=\int d^dx T^{\mu\nu}\partial_\mu \epsilon_\nu\tag{4.34}$$ even for fields that do not satisfy the equations of motion. I don't get this. I get that via Noether's theorem $$\delta S=\int d^dx \partial_\mu(T^{\mu\nu} \epsilon_\nu)$$ for all fields. Then, if the theory has translation symmetry and the field equations are satisfied, we obtain $\partial_\mu T^{\mu\nu}=0$, which immediately leads to the first equation. However, my argument requires the use of the equations of motion. I believe I am overlooking something extremely simple. I would appreciate if anyone can point this out to me.

Answer 1

The energy momenum tensor is defined by $$ \delta S[\phi, g_{\mu\nu}]= \frac 12 \int d^dx \sqrt{g} T^{\mu\nu}\delta g_{\mu\nu}. $$ In this variation we vary the geometry but keep the fields $\phi(x)$ fixed. If we start in flat space where $g_{\mu\nu}=\delta_{\mu\nu}$ and make a diffeomorphism then $\delta g_{\mu\nu}= \partial_\mu \epsilon_\nu+ \partial_\nu\epsilon_\mu$ so $$ \delta S[\phi, g] = \frac 12 \int d^dx T^{\mu\nu}(\partial_\mu \epsilon_\nu+ \partial_\nu\epsilon_\mu). $$ No equations of motion are needed, but again the fields are to be unchanged. What does need the fields to obey their EofM is conservation $$ \partial_\mu T^{\mu\nu}=0. $$ This is because the action is unchanged a under a coordinate chage $x\to x+\epsilon$. A coordinate change requires that we change both $g_{\mu\nu}$ and $\phi(x) \to \phi(x+\epsilon)$. If the change in $\phi$ is to have no effect on $S[\phi,g]$ then we must require the field to satsify its EofM.

Answer 2

Strictly speaking, if one reads Ref. 1, then eq. (4.34) refers to eq. (2.191), which in turn refers to eq. (2.142), or better $$\delta S~=~ -\int \! d^d x~j^{\mu} \partial_{\mu}\omega_a \tag{2.140}.$$ Eq. (2.140) follows directly from Noether's first theorem, cf. e.g. this Phys.SE post, which means that the Noether current is the canonical SEM tensor. In footnotes 2 & 6 it is explained that we can improve the canonical SEM tensor into a symmetric SEM, cf. Belinfante et al, in a way so that eq. (4.34) still holds with the symmetric SEM tensor (up to possibly boundary terms).

References:

1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997; Subsection 4.2.2.