• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,728 comments
1,470 users with positive rep
818 active unimported users
More ...

  Conversion of the Nambo-Goto action into the Polyakov action?

+ 4 like - 0 dislike

I`ve read that the Nambo-Goto action containing the induced metric $\gamma_{\alpha\beta}$

$$\tag{1} S_{NG} ~=~ -T\int_{\tau_i}^{\tau_f} d\tau \int_0^{\ell} d\sigma \sqrt{-\gamma}$$

can be converted into the Polyakov action using an intrinsic metric $h^{\alpha\beta}$

$$\tag{2} S_P ~=~ \frac{T}{2}\int d^2 \sigma \sqrt{-h}h^{\alpha\beta} \partial_{\alpha}X^{\mu} \partial_{\beta}X^{\nu} \eta_{\mu\nu}.$$

How this conversion is done is not further explained in my book so my question is if somebody can give or point me to some more explicit hints how this can be done?

(This would probably help me to see the general relationship between the two metrics which confuses me too ...)

asked Nov 22, 2011 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]

2 Answers

+ 2 like - 0 dislike

It is done as an exercise (Exercise 2.6, pg26-27) in String Theory and M-Theory by Becker,Becker,Schwarz.

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user Chris Gerig
answered Nov 22, 2011 by Chris Gerig (590 points) [ no revision ]
Care to provide the details?

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user David Z
gen.lib.rus.ec ... this is "Library Genesis", the main search engine to find all science/math textbooks to download. The text I referenced is there for you to check out.

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user Chris Gerig
Thanks @Chris Gerig, I`ll try download this textbook, it is certainly nice to have this anyway :-) ...

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user Dilaton
+ 2 like - 0 dislike

The best derivation is Polyakov's, and it is found in the long string chapter of "Gauge Fields and Strings".

The key point is that the h-field in the path integral is integrated over, but it doesn't have derivative terms, so the fluctuations in the h-field just act to replace it at each point by its stationary value. The X-parts just go along for the ride when looking for stationary points of h, so you can write the action as

$$ S = \int \sqrt{h} h^{\alpha\beta} \gamma_{\alpha\beta} $$

Where $\gamma_{\alpha\beta} = \partial_\alpha X^\mu \partial_\beta X_\mu$ is the dot product of an $\alpha$ coordinate step with a $\beta$ coordinate step, i.e. it is the induced metric. The induced metric plays the role of a source term in the h path-integral (ignoring the X path integral). The stationary point condition is found by varying h (using the important determinant variation formula $\delta h = h h^{\alpha\beta} \delta h_{\alpha\beta}$ which you learn in math class as "expansion by minors" and "the inverse-minor theorem"):

$$\sqrt{h} \gamma_{\alpha\beta} + {1\over2\sqrt{h}} h h_{\alpha\beta} h^{\kappa\delta}\gamma_{\kappa\delta} $$

If you solve for h, you find that

$$ h_{\alpha\beta} = - {\gamma_{\alpha\beta}\over {1\over 2} h^{\kappa\delta}\gamma_{\kappa\delta}}$$

This might look like an incomplete solution, but the denominator on the right is a scalar, so this is saying that the tensors h and $\gamma$ are proportional

$$ h_{\alpha\beta} = A(x) \gamma_{\alpha\beta} $$

Where the proportionality constant A(x) won't make any difference (any two A choices will give solutions, and they lead to the same action).

Substitute in the extremal value for h in the action, and remember how to take an inverse matrix: $h^{\alpha\beta} = {1\over A} \gamma^{\alpha\beta}$, and you get that the action contribution for each external source $\gamma_{\alpha\beta}$ is proportional to $\sqrt{\gamma}$ no matter what $A(x)$ happens to be, which gives the Nambu-Goto action. Then you integrate the Nambu-Goto action over the remaining path-integral variables, which are the embedding coordiantes $X^\mu$.

The Nambu-Goto path integral is hard to understand in any way other than solving it classically, defining harmonic oscillators, and quantizing these by assuming they turn into standard harmonic oscillators. This is the old approach to string theory. The Polyakov action is just used to fix a gauge for h which will turn the problem into a simple sigma-model. So the equivalence between them is more of a formal thing, which relates the harmonic oscillator expansion to the vertex operators in the h formalism. It isn't necessarily a path-integral equality, because the Nambu-Goto path integral is not clearly well defined outside of turning it into Polyakov and fixing gauge for h.

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user Ron Maimon
answered Nov 23, 2011 by Ron Maimon (7,720 points) [ no revision ]
Thanks a lot @Ron Maimon +1, this answer is already very helpful for me :-) (just started to read David McMohans String Demystified book some days ago, ha ha ...)

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user Dilaton
@Dilaton: Please, read Polyakov for this, not immitators. The current crop of string theory books is inferior to those of a generation ago.

This post imported from StackExchange Physics at 2014-04-01 13:15 (UCT), posted by SE-user Ron Maimon

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights