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I know that, on-shell, the Polyakov action $S_\text{P}$ is equivalent to the Nambu-Goto action $S_\text{NG}$.

What about off-shell? In the literature, I saw that $S_\text{P} \geq S_\text{NG}$ holds off-shell. I want to prove this but have no clue. Any comments and references are welcome.

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  • $\begingroup$ Hint: You get Nambu-Goto from Polyakov by plugging the e.o.m. for the worldsheet metric into the Polyakov action. When the e.o.m. is fulfilled, what special property does the action have at that point in field space? $\endgroup$
    – ACuriousMind
    Commented Jan 19, 2017 at 13:39
  • $\begingroup$ @ACuriousMind, I know, On the on-shell, the e.o.m of Polyakov action reduces vanishing energy-momentum tensor, and this gives some relation between intrinsic metric $\gamma_{ab}$ and induced metric on worldsheet $h_{ab}$ $i.e$, $h_{ab} = \frac{1}{2} \gamma_{ab} h_{de} \gamma^{de}$. $\endgroup$
    – phy_math
    Commented Jan 19, 2017 at 13:51
  • $\begingroup$ @ACuriousMind, what does special property means on your comment?. $\endgroup$
    – phy_math
    Commented Jan 19, 2017 at 13:52
  • $\begingroup$ @ACuriousMind: It sounds to me like you're saying that the action must be a minimum, but it could also be a maximum or a stationary point. $\endgroup$
    – Javier
    Commented Jan 19, 2017 at 13:59

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The equations of motion are the Euler-Lagrange equations of the action functional. We have that $$ S_\text{P}(X,h_c) = S_\text{NG}(X),$$ where $h_c$ is the solution to the equation of motion of the worldsheet metric, and we know that $S_\text{P}(X,h_c)$ is a stationary point of $S_\text{P}$ in the space of all possible $h$s since the E-L equations are just the equations for stationary points in field space.

If you now examine small perturbations around $h_c$ and you find that $S_\text{P}(X,h_c + \delta h) \geq S_\text{P}(X,h_c)$, then the stationary point is a minimum, and therefore $$ S_\text{P}(X,h)\geq S_\text{P}(X,h_c) = S_\text{NG}(X).$$

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