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I am very confused about coordinate invariance of actions in classical field theories on arbitrary background spacetime or even with dynamical metric. From this question, we see that if the integrated term, namely the Lagrangian density, of the action is well-defined, i.e. is a 0-tensor (a scalar), then the action is necessarily coordinate invariant.

Now, as a conformal transformation is a particular type of coordinate transformation, shouldn't any action be conformally invariant? I know that is certainly false but I am really interested in a rigorous explanation.

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Note the words "conformal transformation" can mean slightly different things in different places. But conformal transformations, in the sense that it's used in conformal field theories, are not just coordinate transformations. Instead they are a simultaneous coordinate and field transformation such that the metric is left invariant.

This often feels rather odd as conformal transformations are usually only talked about as being coordinate transformations, but keeping careful track of what's going on in sources such as the Di Francesco et at text will reveal that while they perform coordinate transformations, they always employ the flat metric. Polchinski's string book is a little more explicit about this.

But if you want the real hard proof that conformal transformations, as they appear in CFTs, are not just coordinate transformations, look no further than the transformation of the stress tensor under such a transformation. This transformation rule can be found in any source on CFTs and is usually one of the very first things written down/derived. You will note that the transformation is very simply not the transformation of a rank 2 tensor under a coordinate transformation. Instead there is an additional term (the Schwartzian) in the transformation rule which comes from the fact that we are transforming the metric directly as a field at the same time (it's simultaneous Weyl transformation).

For a couple specific references, see eq (4.31) in David Tong's string theory notes or eq (2.4.26) in Polchinski's strings book.

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  • $\begingroup$ And it's introduced everywhere as coordinate transformation... it's the recipe for confusion. Do you know any reference that discuss this topic in more details ? The fact that the stress-energy tensor doesn't transform like it would for a usual coordinate transformation is certainly very important, I didn't know that. $\endgroup$
    – xpsf
    Jun 6, 2021 at 9:11
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    $\begingroup$ @xpsf As I mentioned, I think Polchinski is more explicit about this than most other sources. The stress tensor transformation will, however, be mentioned in every source as it's quite important (though it may only be mentioned in infinitesimal form, which is a little harder to identify as being not the transformation of a rank 2 tensor...the quoted equations are also only for 2D CFTs, but something similar holds in general D). The key is to note that in all these sources they apply the "coordinate transform" but then keep using the flat metric rather than the transformed metric. $\endgroup$ Jun 6, 2021 at 19:37
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It is best to never let anyone get away with saying that all physical theories are generally covariant. Gravity is, but that's about it. If you want to consider passive co-ordinate transformations, you are just performing a substitution in the integral and \begin{align} S = \int d^dx \mathcal{L}[\phi(x)] = \int d^dx^\prime \left | \frac{dx^\prime}{dx} \right |^{-d} \mathcal{L}[\phi(x^\prime)] \end{align} are two equivalent ways of writing the action. We usually choose the former but this is just notation. There is a chance to learn something useful about the theory if we also consider active transformations. In this case, \begin{equation} S^\prime = \int d^dx^\prime \left | \frac{dx^\prime}{dx} \right |^{-d} \mathcal{L}[\phi^\prime(x^\prime)] \end{equation} is a different action because the $\phi^\prime(x^\prime)$ are different functions. How they might be expressible in terms of $\phi(x)$ again (i.e. what representation they furnish) is up to us or whomever handed us the theory. If they conspire to make $S^\prime = S$ then the particular transformation we used will be a symmetry of the theory.

For obvious reasons, fields are often defined to be in irreducible representations of the Lorentz group so that we know how to vary the action under a Lorentz transformation. And, unless we are doing something strange, the Lagrangian will also contract all indices so that we indeed get zero. For scale transformations, the situation is a bit different. Representations are again known ahead of time since the engineering dimension is easy to read off. But now the variation of an action like \begin{equation} S = \int d^dx \frac{1}{2} \partial_\mu \phi \partial_\mu \phi + \frac{1}{2} m^2 \phi^2 \end{equation} yields a term proportional to $m^2$ which is an intrinsic scale of the theory. This tells us that even though we can use centimeters instead of meters, there is no conformal symmetry. Making your ruler 100 times as long won't give you the same predictions anymore unless you also change the theory to one where the mass is 100 times as small.

As the previous answer points out, the need to consider how fields transform is paramount. However, specifics about the infinite dimensional enhancement of conformal symmetry that exists in two dimensions are tangential. The point is that, whether or not you work with fields known to transform as tensors, only very special theories are given by an action which also does.

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  • $\begingroup$ What I don't understand is that, in your first equation for example, shouldn't there be an implicit $\sqrt{-\eta}$ in the integral ? In cartesian coordinates $\sqrt{-\eta}=1$ but it's not the case in any coordinate system. This cancels the determinant in front of $\mathcal{L}[\phi(x')]$. Why isn't it he case ? $\endgroup$
    – xpsf
    Jun 9, 2021 at 8:11
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    $\begingroup$ What I wrote as $\left | \frac{dx^\prime}{dx} \right |^{-d}$ is really just the new $\sqrt{-g}$, yes. If your concern is how this affects the massive scalar, the point is that you need the metric to be dynamical to not get $S - S^\prime \propto m^2$. Then, if you also take the scalar's representation to be $\Delta = 0 \neq \frac{d - 2}{2}$, you will get a combined scalar-graviton classical theory which has a rigid scale invariance. But this is not at all enough to say you can go off and treat just the massive scalar part as a theory that's conformally invariant of a fixed background :). $\endgroup$ Jun 9, 2021 at 11:55
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    $\begingroup$ A side comment is that in this case of a scalar coupled to gravity, it's often quite natural to require not just rigid scale invariance but local scale invariance also known as Weyl transformations. In this case a mass term like $m^2 \phi^2$ will not repsect it anymore and you need to switch to the so called conformal coupling $R \phi^2$ where the "mass" now depends on the metric. $\endgroup$ Jun 9, 2021 at 11:57
  • $\begingroup$ Thank you for your explanation. In that case, what I don't understand is where is the determinant of the jacobian that comes from the transformation of $d^nx$. For me, the volume form $d^nx\sqrt{-g}$ is always invariant under coordinate change, whether $g$ is equals to $\eta$ in some coordinates, if $g$ is dynamical or if $g$ is fixed (background). That's what lead me to think that all actions are invariant under coordinate change. For me, by nature, integration on a manifold doesn't depend on the coordinates. $\endgroup$
    – xpsf
    Jun 9, 2021 at 20:09
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    $\begingroup$ Actions are invariant under passive coordinate change. But when you apply an active change, the action functional is being evaluated at different fields. If one of these fields you decide to act on is the metric itself, you can often achieve invariance under a large set of transformations (e.g. Weyl). But experimentalists will be much more interested if the action is such that you get invariance when the transformation is passive on the metric but active on the other fields. $\endgroup$ Jun 9, 2021 at 20:24

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