I've read that the Polyakov action using an intrinsic metric $h_{\alpha\beta}$

$$\tag{1} S_P ~=~ -\frac{T}{2}\int d^2 \sigma \sqrt{-h}h^{\alpha\beta} \partial_{\alpha}X^{\mu}\partial_{\beta}X^{\nu} \eta_{\mu\nu}$$

can be converted into the Nambo-Goto action containing the induced metric $\gamma_{\alpha\beta}$

$$\tag{2} S_{NG} ~=~ -T\int_{\tau_i}^{\tau_f} d\tau \int_0^{\ell} d\sigma \sqrt{-\gamma}.$$

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?

  • $\begingroup$ It is done as an exercise (Exercise 2.6, pg26-27) in String Theory and M-Theory by Becker,Becker,Schwarz. $\endgroup$ Commented Nov 22, 2011 at 23:34
  • 1
    $\begingroup$ For the opposite question of going from the Nambu-Goto action to the Polyakov action, see physics.stackexchange.com/q/77038/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented May 29, 2015 at 12:06

1 Answer 1


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.

  • $\begingroup$ 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 ...) $\endgroup$
    – Dilaton
    Commented Nov 23, 2011 at 11:53
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    $\begingroup$ @Dilaton: Please, read Polyakov for this, not immitators. The current crop of string theory books is inferior to those of a generation ago. $\endgroup$
    – Ron Maimon
    Commented Nov 24, 2011 at 1:57

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