(Cf. Di Francesco et al, Conformal Field Theory, pp. 40-41) I am trying to derive eqn. (2.142) or

$$\delta S = \int d^d x ~\omega_a~\partial_{\mu}j^{\mu}_a \tag{2.142}$$

in the book CFT by Di Francesco et al. I have obtained the final expression

$$\delta S = \int d^d x\,\partial_{\mu} \omega_a \left[\frac{\delta F}{\delta \omega_a} \frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \frac{\delta x^{\nu}}{\delta \omega_a}\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)} + \frac{\delta x^{\mu}}{\delta \omega_a}L\right] +$$ $$ \omega_a\left[ \frac{\delta F}{\delta \omega_a}\frac{\partial L}{\partial \Phi} + (\partial_{\mu}\frac{\delta F}{\delta \omega_a})\frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu} (\frac{\delta x^{\nu}}{\delta \omega_a})\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)} + \partial_{\mu} (\frac{\delta x^{\mu}}{\delta \omega_a})L\right]\tag{A}$$

and indeed the terms multiplying $\partial_{\mu}\omega_a$ there are exactly $j^{\mu}$ as obtained in eqn. (2.141). The problem I am having is that the terms multiplying $\omega_a$ don't appear to vanish. (The first two do as a consequence of the classical equations of motion but the last two do not)

The method Di Francesco employs is to assume a position dependent parameter $\omega = \omega(x)$, then make it constant at the very end.

So, if we make $\omega$ independent of position at the end (i.e impose the rigid transformation), then $$\partial_{\mu}\omega_a = 0\tag{B}$$ identically. In which case, we are left with $$\omega_a\int d^dx [..] = 0,\tag{C}$$ by further considering a symmetry transformation, where [..] are the terms multiplying $\omega$ in the expression above.

So I am not sure how Di Francesco is left with

$$ \delta S~=~-\int d^d x~ j^{\mu}_a ~\partial_{\mu} \omega_a \tag{2.140}.$$

The paragraph preceding eqn. (2.140) seems to be contradictory to me (in particular the first and last sentence) and if indeed he is imposing a rigid transformation then shouldn't $\partial_{\mu}\omega_a = 0$ be zero in (2.140)?


3 Answers 3


From $$\delta S = \int_{\mathcal D} \text{d}^d x \,\partial_{\mu} \omega_a [\dots]_1 + \int_{\mathcal D} \text{d}^d x \,\omega_a [\dots]_2 \overset{!}{=} 0,$$ we deduce that for a global variation of the symmetry transformation parameters $\omega$'s $$\omega_a \int_{\mathcal D} \text{d}^d x \,[\dots]_2 = 0.$$ By arbitrariness of the domain of integration $\mathcal D$, it follows $[\dots]_2 = 0$ identically.

Now consider a local variation $\omega_a = \omega_a(x)$. Then from the demand that $\delta S=0$, $$\int_{\mathcal D} \text{d}^d x\, (\partial_{\mu} \omega_a [\dots]_1 + \omega_a [\dots]_2) = 0.$$

The important point is that as $[\dots]_2$ is independent of the $\omega$'s and we inferred the vanishing of this coefficient by restricting to the global symmetry, we deduce now too that $$\int_{\mathcal D} \text{d}^d x\, \partial_{\mu} \omega_a [\dots]_1 \overset{!}{=} 0 \overset{!}{=} \int_{\mathcal D} \text{d}^d x\, \omega_a \partial_{\mu} j^{\mu}, $$ where in the latter equality we integrated by parts, simplified the result on the physical assumption that the fields $\Phi$ and their derivatives vanish at spacetime infinity and made the identification $[\dots]_1 \equiv j^{\mu}$. The $\omega$ independence of the coefficient is only clear after following Di Francesco's notation, naively it seems otherwise.

As $\mathcal D$ and the $\omega$'s are without further restriction, we obtain $$\partial_{\mu} j^{\mu} = 0.$$

  • $\begingroup$ I don't think it is valid to assume that $\delta S=0$ for any domain $\mathcal D$. For example consider a free scalar field theory, which is translation invariant. Take $\mathcal D$ to be some compact domain and consider a field configuration $\phi(x)$ localized in this domain. Then $S[\phi]\neq 0$, but if we translate the field $\phi\rightarrow\phi'$ so that the support of $\phi'$ is now completely disjoint from $\mathcal D$ then $S[\phi']=0$, and hence $\delta S\neq 0$. $\endgroup$ Nov 21, 2020 at 1:23
  1. OP writes:

    The problem I am having is that the terms [in eq. (A)] multiplying $\omega_a$ don't appear to vanish.

    Let's call term in eq. (A) that multiplies $\omega_a$ for $k^a$. The term $k^a$ vanishes$^1$ off-shell $$ k^a~=~0 \tag{K}$$ because of the main assumption in Noether's theorem: The action has a global (i.e. $x$-independent) symmetry.$^2$

  2. Eq. (K), OP's eq. (A), and the definition (2.141) of $j^{\mu}_a$ then together imply that the eq. (2.140) holds off-shell for any $x$-dependent $\omega^a(x)$.

  3. Finally integrate eq. (2.140) by parts to deduce the sought-for eq. (2.142).


$^1$ If we allow boundary terms, the term $k^a$ could contain total divergence terms. See also this related Phys.SE post.

$^2$ By a symmetry is meant an off-shell symmetry. An on-shell symmetry is a vacuous notion.


I came up with this idea: Suppose $\omega_a$ is independent of $x_\mu$, then

\begin{align} 0 = \delta S &= \int d^d x (\omega_a \text{ term}) + (\partial_\mu\omega_a \text{ term}) \\ &= \int d^d x (\omega_a \text{ term}) \end{align}

So $$ \int d^d x (\omega_a \text{ term} ) = 0$$

Then we can safely say $$ \delta S = \int d^d x (\partial_\mu \omega_a\text{ term}) $$ For rigid transformation ($\partial_\mu \omega_a =0$).

(Actually I am unsure of this. Just a thought ...)


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