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  Derivation of the Polyakov Action

+ 6 like - 0 dislike
5138 views

As is usually done when first presenting string theory, the Nambu-Goto Action, $$ S_{\text{NG}}:=-T\int d\tau d\sigma \sqrt{-g} $$ ($g:=\det (g_{\alpha \beta})$ is the induced metric on the world-sheet and $T$ is a positive real number interpreted as the string tension), is introduced as the natural generalization of action for a relativistic point particle, which in turn is obviously a correct action as it produces the proper equations of motion (and has a nice geometric interpretation).

Not long after the introduction of the Nambu-Goto action, authors tend to introduce the Polyakov action, \begin{align*} S_{\text{P}} & :=-\frac{T}{2}\int d\tau d\sigma \, \sqrt{-h}h^{\alpha \beta}g_{\alpha \beta}=-\frac{T}{2}\int d\tau d\sigma \, \sqrt{-h}h^{\alpha \beta}\partial _\alpha X\cdot \partial _\beta X \\ & =-\frac{T}{2}\int d\tau d\sigma \, \sqrt{-h}h^{\alpha \beta}\partial _\alpha X^\kappa \partial _\beta X^\lambda G_{\kappa \lambda}(X), \end{align*} where $G_{\kappa \lambda}$ is the space-time metric, $g_{\alpha \beta}$ is the induced metric on the world-sheet, and $h_{\alpha \beta}$ is the auxiliary metric on the world-sheet ($h:=\det (h_{\alpha \beta})$). They then usually proceed to show that these two actions are equivalent, in the sense that you can deduce the equations of motion for $S_{\text{NG}}$ given the equations of motion for $S_{\text{P}}$.

Now, that's all well and dandy, but that doesn't exactly show how one would actually arrive at the Polyakov action. You can't make it as a theoretical physicist by mindlessly computing things to show you get the right answer; you have to be able to actually, you know, come up with things. Hence, instead of just pulling the Polyakov action out of a hat, it would be nice to know a way of deriving or motivating the action.

So then, imagine you are handed $S_{\text{NG}}$ and you set out to come up with an equivalent action that, at the very least, doesn't involve a square-root. How do you come up with the Polyakov action?


This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Jonathan Gleason

asked Sep 12, 2013 in Theoretical Physics by Jonathan Gleason (265 points) [ revision history ]
edited May 7, 2014 by dimension10
Note that the equations of motions of $S_P$ include the equation of motion of the metrics, which gives the Visaroso constraint : $T_{ab}=0$, and this is necessary to recover $S_{NG}$ from $S_P$. The equation of motion of the $X^\mu$ for $S_P$, alone, does not give the equation of motion of the $X^\mu$ for $S_{NG}$.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Trimok
For the opposite question of going from the Polyakov action to the Nambu-Goto action, see physics.stackexchange.com/q/17349/2451 and links therein.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Qmechanic

2 Answers

+ 6 like - 0 dislike

Quantum systems are essentially defined by their symmetries. For example, in QFT's you expect all terms not forbidden by the symmetries of the problem to appear in the Lagrangian, with irrelevant operators suppressed by large scales, etc.

So I think your first step in this approach would be to write down the most general 2D QFT respecting the 2D Diff and internal Poincare symmetries. The Diff symmetry motivates you to introduce a dynamical metric, since you've already done a similar thing for the point particle. This doesn't quite get you to the Polyakov action, since the Polyakov action has a Weyl symmetry that the NG action doesn't. You've introduced a fake degree of freedom on the worldsheet that wasn't present in the NG action, so you need some local symmetry principle to remove the redundant degrees of freedom. I don't know of a particular way to reason that this symmetry has to be Weyl invariance, maybe someone else does.

But once you believe the theory should have a local lagrangian with Diff, Poincare and Weyl symmetries, you are basically stuck with the Polyakov action. The Polyakov action (with the Euler characteristic term) is the most general 2D action with the Diff, Poincare and Weyl symmetries and the associated field content (Polchinski p 15 ).

So the guiding principle should be the symmetries of the NG action.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Dan
answered Sep 12, 2013 by Dan (220 points) [ no revision ]
+1: Would you happen to know to what extent what you describe aligns with the history? I know that Deser/Zumino and Brink/Di Vecchia wrote down the action at essentially the same time; I wonder if their process was informed primarily by symmetries.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user joshphysics
(Possible) argument for Weyl invariance: you introduce an unphysical field $h_{\mu\nu}$ which is supposed to be like a metric on the worldsheet, but you must get rid of it. $h_{\mu\nu}$ has three degrees of freedom in it, and there must be this many symmetries to eliminate them all. Diffe takes care of two, but leaves one degree of freedom ($\sim\mathrm{det}(h)$ or $\sim R$ the Ricci scalar). Weyl transformations are the only option to get rid of that. Indeed all (sufficiently smooth etc.) two dimensional manifolds are conformally flat. Bingo.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Michael Brown
I like this, but this method doesn't really convince me that putting in another auxiliary field is the way to go. Let's say in QFT we want a theory with a complex scalar field with a $U(1)$ symmetry. Of course, we could always introduce other fields into the theory, but that's not what one usually does unless you wanted the extra fields to begin with. And even if you decide that introducing a new field is the way to go, why stop at one? Surely we could add two new fields that respected all the symmetries we wanted . . .

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Jonathan Gleason
. . . And hell, if we're going to add in yet another field, we could probably find get another symmetry along with it.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Jonathan Gleason
. . . Of course, I guess you might just say that the simplest one that works is the way to go (in this case, an action with just one extra field). Still, I have to admit, I'm not completely satisfied with this answer.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Jonathan Gleason
I think it is possible to motivate introduction of the metric using your criterion in the question. You would like to get rid of the square root to arrive at a local lagrangian for the physical fields. Since you are planning to quantize, you should also actually want your physical fields X to have the canonical kinetic term with two derivatives. We cannot simply use the ordinary kinetic term, it will not be Diff invariant (if it were, gauge-fixing with conformal gauge wouldn't be gauge fixing). So we should actually couple these scalars minimally to gravity. So we trade sqrt(dXdx) for sqrt(h)

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Dan
Regarding introducing more fields and finding more symmetries, this is of course what we have to do in order to obtain realistic models with fermionic degrees of freedom. Add in worldsheet fermions, find worldsheet superconformal invariance, etc. In this sense the NG action is just a jumping off point.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Dan
+ 3 like - 0 dislike

I) The closest cosmetic resemblance between the Nambu-Goto action and the Polyakov action is achieved if we write them as

$$\tag{1} S_{NG}~=~ -\frac{T_0}{c} \int d^2{\rm vol} ~\det(M)^{\frac{1}{2}} , $$

and

$$\tag{2} S_{P}~=~ -\frac{T_0}{c}\int d^2{\rm vol}~ \frac{{\rm tr}(M)}{2} , $$

respectively. Here $h_{ab}$ is an auxiliary world-sheet (WS) metric of Lorentzian signature $(-,+)$, i.e. minus in the temporal WS direction;

$$\tag{3} d^2{\rm vol}~:=~\sqrt{-h}~d\tau \wedge d\sigma$$

is a diffeomorphism-invariant WS volume-form (an area actually);

$$\tag{4} M^{a}{}_{c}~:=~(h^{-1})^{ab}\gamma_{bc} $$

is a mixed tensor; and

$$\tag{5} \gamma_{ab}~:=~(X^{\ast}G)_{ab}~:=~\partial_a X^{\mu} ~\partial_b X^{\nu}~ G_{\mu\nu}(X) $$

is the induced WS metric via pull-back of the target space (TS) metric $G_{\mu\nu}$ with Lorentzian signature $(-,+, \ldots, +)$.

Note that the Nambu-Goto action (1) does actually not depend on the auxiliary WS metric $h_{ab}$ at all, while the Polyakov action (2) does.

II) As is well-known, varying the Polyakov action (2) wrt. the WS metric $h_{ab}$ leads to that the $2\times 2$ matrix

$$\tag{6} M^{a}{}_{b}~\approx~\frac{{\rm tr}(M)}{2} \delta^a_b~\propto~\delta^a_b $$

must be proportional to the $2\times 2$ unit matrix on-shell. This implies that

$$\tag{7} \det(M)^{\frac{1}{2}} ~\approx~ \frac{{\rm tr}(M)}{2},$$

so that the two actions (1) and (2) coincide on-shell, see e.g. the Wikipedia page. (Here the $\approx$ symbol means equality modulo eom.)

III) Now, let us imagine that we only know the Nambu-Goto action (1) and not the Polyakov action (2). The the only diffeomorphism-invariant combinations of the matrix $M^{a}{}_{b}$ are the determinant $\det(M)$, the trace ${\rm tr}(M)$, and functions thereof.

If furthermore the TS metric $G_{\mu\nu}$ is dimensionful, and we demand that the action is linear in that dimension, this leads us to consider action terms of the form

$$\tag{8} S~=~ -\frac{T_0}{c}\int d^2{\rm vol}~ \det(M)^{\frac{p}{2}} \left(\frac{{\rm tr}(M)}{2}\right)^{1-p} , $$

where $p\in \mathbb{R}$ is a real power. Alternatively, Weyl invariance leads us to consider the action (8). Obviously, the Polyakov action (2) (corresponding to $p=0$) is not far away if we would like simple integer powers in our action.

This post imported from StackExchange Physics at 2014-03-12 15:52 (UCT), posted by SE-user Qmechanic
answered Sep 12, 2013 by Qmechanic (3,120 points) [ no revision ]

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