Noether's theorem for arbitrary conformal coordinate transformations

Question

I have been reading Introduction to Conformal Field Theory by Blumenhagen and Plauschinn. Equation (2.19) on page 19 states that if our theory is invariant under a general conformal transformation $x^\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu(x)$, then the conserved current is given by

$$ j^\mu = \epsilon^\nu T^\mu_{\ \nu}\tag{2.19}.$$

I have been trying to show this myself but I can't. The general procedure of Noether's theorem is to evaluate the off-shell variation and see whether it is given by the integral of a total derivative. We then equate this with the on-shell variation to yield the conserved current.

My attempt

Let the action be of the form

$$ S[\phi] = \int \mathrm{d}^3 x \mathcal{L}(\phi,\partial \phi).$$

We know that if the theory is translationally invariant under $x^\mu \rightarrow x'^\mu = x^\mu + \epsilon^\mu$, where $\epsilon^\mu$ here is a constant, the corresponding conserved currents are given by the energy-momentum tensor $T^\mu_{\ \nu}$ which takes the form

$$ T^\mu_{\ \nu} = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_{\nu} \mathcal{L}, $$

where $\partial_\mu T^\mu_{\ \nu} = 0$. Now I promote $\epsilon \rightarrow \epsilon(x)$. First, I compute the on-shell variation:

$$ \delta S_\text{on-shell} = \int \mathrm{d}^3 x \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu(\delta \phi) \\ =\int \mathrm{d}^3 x \bigg( \frac{\partial \mathcal{L}}{\partial \phi}- \partial_\mu \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \bigg)\delta \phi + \partial_\mu \bigg( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi \bigg) \\ = \int \mathrm{d}^3 x \partial_\mu \bigg(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi \bigg).$$

The translation when viewed as an active transformation on the fields means $\delta \phi = - \epsilon^\mu(x) \partial_\mu \phi(x)$. Plugging this into the on-shell variation yields

$$ \delta S_\text{on-shell}= - \int \mathrm{d}^3 x \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} (\partial_\mu \epsilon^\nu)( \partial_\nu \phi) + \epsilon^\nu \partial_\mu \bigg( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi \bigg) . $$

The first term in the integrand would vanish if $\epsilon^\mu$ was a constant. Now we evaluate the off-shell variation:

$$ \delta S_\text{off-shell} = \int \mathrm{d}^3 x \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu(\delta \phi) \\ = - \int \mathrm{d}^3 x\frac{\partial \mathcal{L}}{\partial \phi} \epsilon^\nu \partial_\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} (\partial_\mu \epsilon^\nu \partial_\nu \phi + \epsilon^\nu \partial_\nu \partial_\mu \phi) \\ = - \int \mathrm{d}^3 x \epsilon^\nu \bigg( \frac{\partial \mathcal{L}}{\partial \phi} \partial_\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\nu \partial_\mu \phi \bigg) + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu \epsilon^\nu \partial_\nu \phi \\ = -\int \mathrm{d}^3 x \epsilon^\nu \partial_\nu \mathcal{L} + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\mu \epsilon^\nu \partial_\nu \phi. $$

Equating the on and off-shell variations, we find the terms with $\partial_\mu \epsilon^\nu$ cancel, yielding

$$\int \mathcal{d}^3 x \epsilon^\nu \partial_\mu \bigg( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \partial_\nu \phi - \delta^\mu_{\nu} \mathcal{L} \bigg) = 0, $$

or

$$ \int \mathcal{d}^3 x \epsilon^\nu \partial_\mu T^\mu_{\ \nu} = 0. $$

The integrand is not of the form $\partial_\mu j^\mu$ where $j^\mu = \epsilon^\nu T^\mu_{\ \nu}$. In fact, we already know that $\partial_\mu T^\mu_{\ \nu} = 0$ from our assumption that the theory was translationally invariant already so this statement is trivially true. I am unable to pass the $\epsilon^\nu$ into the derivative too because it depends on the coordinates.

What am I missing? Have I made a mistake in my analysis?

Answer 1

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.

Answer 2

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).