12
$\begingroup$

I often see in physics articles on gravity such notion as conformal gauge and Weyl transformation.

They use Conformal gauge to change coordinates to transform metrics from arbitrary $$ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}$$ to $$ds^2=-e^{2\omega}dx^{+}dx^{-}. $$Equations of motion change dramatically (from my point of view) to a simpler form.

Also one can use Weyl transformations to convert the action in a simpler form, so the equations of motion change too.

I understand how to do technical details (I mean how to find new curvature, Christoffel symbols, how to write new action and equations of motion), but it seems to me that I DO NOT understand the meaning of these transformations.

Why do they call it "gauge"? I mean I know that in gauge theories gauge transformations do not change equations of motion and action, so they keep "physics" untouched. But it is not so in case of conformal gauge, isnt' it? And why are we satisfied when we obtain the solutions for metrics and fields in conformal gauge for action, which we get after weyl transformation etc? What does it tell us about initial system?

$\endgroup$
2
  • $\begingroup$ a conformal mapping is a mapping which respects angles (in projective geometry meaning) and not lengths. In the case of Weyl it is a mapping that makes the scale (or length) a gauge or a parameter. So the metric is defined up to a multiplicative constant which fixes the (unit) length (in contrast to GR metric which is absolute in this sense) $\endgroup$
    – Nikos M.
    Oct 26, 2014 at 20:38
  • $\begingroup$ For the existence of conformal gauge, see this Phys.SE post. $\endgroup$
    – Qmechanic
    Jun 10, 2018 at 19:30

2 Answers 2

9
$\begingroup$

The two-dimensional Polyakov action for a string with worldsheet $\Sigma$ and worldsheet metric $h_{ab}$

$$ \frac{T}{2}\int_\Sigma \sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu$$

has full conformal symmetry under the Virasoro algebra and under Weyl transformations1 , which can be seen as gauge degrees of freedom. It follows that we can always treat the worldsheet metric as being flat up to a Weyl factor (or even without it). Consequently,

$$h_{ab} = \mathrm{e}^{2\phi} \eta_{ab}$$

is known as the conformal gauge. It should be noted that Weyl symmetry is special to two worldsheet dimensions, since the action is not invariant under Weyl transformations for other dimensions due to $\sqrt{-h}\mapsto(\mathrm{e}^{2\phi})^{D/2}\sqrt{-h}$ in $D$ dimensions under a (local or global) Weyl transformation $h_{ab} \mapsto \mathrm{e}^{2\phi}h_{ab}$.


1For the difference between the two, see this question and answer.

$\endgroup$
5
  • $\begingroup$ It seems to me that you haven't answer my question. I asked about the role and meaning of these transformations in gravity. The connection is not obvious. $\endgroup$
    – xxxxx
    Oct 26, 2014 at 18:20
  • $\begingroup$ @xxxxx: You asked what conformal gauge and Weyl transformation mean - this is what they are. The connection of string theory to quantum gravity is indeed non-obvious - but if you are reading articles where string theoretic methods are applied to gravity, surely they explain why they are treating gravity as stringy? $\endgroup$
    – ACuriousMind
    Oct 26, 2014 at 18:26
  • $\begingroup$ I mean that they do them while analyzing CGHS model and similar model. I do not understand the following thing: "why are we satisfied when we obtain the solutions for metrics and fields in conformal gauge for action, which we get after weyl transformation etc?" Because action and equations in, for example, CGHS model has no invariance under weyl transformation and conformal gauge, hasn't it? $\endgroup$
    – xxxxx
    Oct 26, 2014 at 18:29
  • 1
    $\begingroup$ @xxxxx: Ah, you should have said that we are looking at CGHS here. CGHS, as a 2D theory that is also used in non-critical string theory, also possesses conformal and Weyl symmetry - if you look closely, the CGHS action is nothing else than the Polyakov action with a two dimensional Einstein-Hilbert term and a dilaton added. For a detailed discussion of the CGHS action and its generalizations, see for example arxiv.org/abs/hep-th/9606097. $\endgroup$
    – ACuriousMind
    Oct 26, 2014 at 18:39
  • $\begingroup$ Thanks for the reference, I will chack it a bit later ;) $\endgroup$
    – xxxxx
    Oct 26, 2014 at 18:42
6
$\begingroup$

For clarity let's work with a Lorentzian signature. Our $g$ is a metric for a 2 dimensional Lorentzian manifold $M$.

It is well-known that any two dimensional pseudo-Riemannian manifold is conformally flat, that is

$$g = e^{2\omega}\eta$$

Where $\eta$ is the flat 2D Minkovski metric.

Your Lightcone gauge example

You didn't define your $x^\pm$s but I presume you mean:

$$x^\pm = x^0 \pm x^1$$

Where $(x^1, x^0)$ are some local coordinates, then:

$$- e^{2\omega}dx^+ dx^- = - e^{2\omega}[(dx^0)^2 - (dx^1)^2] \\= e^{2\omega}[\eta_{\mu\nu}dx^\mu dx^\nu] = (e^{2\omega}\eta_{\mu\nu})dx^\mu dx^\nu \\= g_{\mu\nu}dx^\mu dx^\nu $$

Where the last line follows from the equation at the top. This is answers the first part of your question.

Why is it called a gauge

One way to answer this is to consider the tangent bundle $TM$, then indeed choosing a coordinate map can be viewed as choosing a local trivialization in the tangent bundle - choosing a local trivialization in a bundle is what's usually called a gauge fixing.

I personally have not worked out whether this is a gauge fixing if you view your field theory as things in an associated bundle of a principal G-bundle where G is the conformal group - would be nice if someone else sheds light on this.

And if you're looking for a physicist explanation, the story is usually as follows:

  • There is some symmetry (in this case a conformal transformation) that leaves our theory invariant, as a result you obtain extra degrees of freedom when trying to write down a solution to the theory - because any other solution related to this solution by that symmetry transformation is also a solution.
  • a gauge fixing is a procedure where fix these extra degrees of freedom by hand, leaving you with a single solution to work with.

As you see, this is exactly what is happening here.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.