# Noether's theorem for arbitrary conformal coordinate transformations

In these types of problems, it is very useful to work on a generic spacetime with metric $$g$$. For instance, we can promote the usual free-field Lagrangian

$$S=\int\mathrm{d}^dx\,\mathcal{L}=\int\mathrm{d}^dx\,\partial_{\mu}\phi\,\partial^{\mu}\phi$$

to one on a general curved spacetime as

$$S=\int\mathrm{d}^dx\,\mathcal{L}=\int\mathrm{d}^dx\,\sqrt{g}\,g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi.$$

The general recipe here is to replace the volume form with $$\mathrm{d}^dx\,\sqrt{g}$$ and replace all derivatives with covariant derivatives.

Once this is done, the energy-momentum tensor is simply defined as

$$T_{\mu\nu}=\frac{1}{\sqrt{g}}\frac{\delta\mathcal{L}}{\delta g^{\mu\nu}}$$

(up to a possible uninteresting factor of 2).

For any coordinate (infinitesimal) transformation $$x\to x+\epsilon(x)$$, the variation of the metric tensor takes the form

$$\delta g_{\mu\nu}=\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}.$$

A vector is called a Killing vector if this variation vanishes. Plugging this in to the definition of $$T_{\mu\nu}$$, this tells us that the variation of $$\mathcal{L}$$ under this coordinate transformation is

$$\delta\mathcal{L}=\sqrt{g}\,T_{\mu\nu}\delta g^{\mu\nu}=2\sqrt{g}\,T_{\mu\nu}\nabla^{\mu}\epsilon^{\nu}=2\sqrt{g}\nabla_{\mu}\left(\epsilon^{\nu}T^{\mu}_{\,\,\nu}\right)=2\partial_{\mu}\left(\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}\right),$$

where we have used the fact that the energy-momentum tensor is conserved. Note that this expression vanishes if the field $$\epsilon(x)$$ is such that $$\delta g_{\mu\nu}=0$$ -- that is, the current

$$J^{\mu}=\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}$$

is conserved if $$\epsilon$$ is a Killing vector of $$g$$.

Now, to answer your question, a conformal transformation is one such that $$g'(x)=\Omega(x)g(x)$$. If we take an infinitesimal version $$x\to x+\epsilon(x)$$, this tells us that

$$\delta g_{\mu\nu}=\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}=\omega(x)g_{\mu\nu}.$$

A vector field $$\epsilon(x)$$ which satisfies the requirement that $$\nabla_{\mu}\epsilon_{\nu}+\nabla_{\nu}\epsilon_{\mu}=\omega(x)g_{\mu\nu}$$ for some smooth $$\omega$$ is a conformal Killing vector. If $$\epsilon$$ is a conformal Killing vector, we then have

$$\delta\mathcal{L}=\sqrt{g}\,T_{\mu\nu}\,\delta g^{\mu\nu}= \begin{cases} 2\partial_{\mu}\left(\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}\right)\\ \sqrt{g}\,\omega(x)T_{\mu\nu}g^{\mu\nu}. \end{cases}$$

Since conformal symmetry implies that the trace of $$T$$ vanishes, we see that

$$J^{\mu}=\sqrt{g}\,\epsilon^{\nu}T^{\mu}_{\,\,\,\nu}$$

is conserved. This becomes the statement you want to prove when we take the flat limit.

TL;DR: The conclusions surrounding eq. (2.19) are not true in general.

1. First of all, Ref. 1 makes it clear on top of p. 19 that $$T^{\mu\nu}$$ denotes the metric/Hilbert stress-energy-momentum (SEM) tensor, not some other SEM tensor.

2. Now it is true that if $$\epsilon=\epsilon^{\mu}\partial_{\mu}$$ is a Killing vector field (KVF), then the current (2.19) satisfies $$\nabla_{\mu} j^{\mu}~\stackrel{m}{\approx}~0,\tag{A}$$ cf. Ref. 2. [Here the $$\stackrel{m}{\approx}$$ symbol means equality modulo matter eom. The connection $$\nabla$$ is the Levi-Civita connection.]

3. Next let us for simplicity in the rest of this answer assume that the metric $$g_{\mu\nu}$$ is constant in the coordinates $$x^{\mu}$$ that we use. Ref. 1 is interested in the case where $$\epsilon$$ generates a conformal transformation, i.e. it is a conformal Killing vector field (CKVF). [This transformation is assumed to be a quasi-symmetry of the action, so that Noether's 1st theorem applies.] Concerning conformal transformations, see e.g. this Phys.SE post. In particular, the CKVF $$\epsilon$$ is not arbitrary, in contrast to what Ref. 1 incorrectly claims on the bottom of p. 19.

4. According to Refs. 3 & 4, the corresponding conserved Noether current is not eq. (2.19) but instead on the form $$j^{\mu}~=~ \epsilon_{\nu}T^{\nu\mu} + (\partial\cdot \epsilon) K^{\prime \mu} + \partial_{\nu}(\partial\cdot \epsilon)~L^{\nu\mu}, \tag{4/3.4}$$ where $$K^{\prime \mu}$$ and $$L^{\nu\mu}$$ are some $$x$$-local expressions. [It is reassuring that when the CKVF $$\epsilon$$ is a KVF, then $$\partial\cdot \epsilon=0$$ so that eq. (4/3.4) reduces to eq. (2.19).]

5. More importantly, Refs. 3 & 4 state that scale-invariance does not imply that $$T^{\mu\nu}$$ is traceless (as Ref. 1's incorrectly claims on the bottom of p.19), but rather that $$T^{\mu}{}_{\mu} ~\stackrel{m}{\approx}~-\partial_{\mu}K^{\mu}.\tag{Ca/3.6}$$ [Tracelessness of $$T^{\mu\nu}$$ follows instead from Weyl-invariance. See also e.g. this Phys.SE post.]

References:

1. R. Blumenhagen and E. Plauschinn, Intro to CFT, Lecture Notes in Physics 779, 2009; eq. (2.19).

2. S.W. Hawking and G.F.R. Ellis, The Large Scale Structure of Space-Time, Section 3.2.

3. J. Polchinski, Scale and conformal invariance in QFT, Nucl. Phys. B303 (1988) 226; eqs. (Ca) & (4).

4. K. Farnsworth, M.A. Luty & V. Prilepina, Weyl versus Conformal Invariance in QFT, JHEP (2017) 170, arXiv:1702.07079; eqs. (3.4) & (3.6).