Reading this article I ran into a doubt.

It derives the Polyakov action for a general $p$-brane $$S(X,g)=-\frac{T}{2}\int\text{d}^{p+1}\xi\,\sqrt{-g}\left(g^{ab}h_{ab}-(p-1)\right)\tag{7}$$ by integrating out the Lagrange multiplier from the action $$S(X,g,\Lambda)=-T\int\text{d}^{p+1}\xi\,\left(\sqrt{-g}+\Lambda^{ab}(h_{ab}-g_{ab})\right).\tag{5}$$ Indeed, the equations of motion obtained by varying $g_{ab}$ in the second action yield $$\Lambda^{ab}=\frac{1}{2}\sqrt{-g}g^{ab}.\tag{6}$$

Now, the equations of motion obtained from the first action demand that $$h_{ab}=\frac{1}{2}g_{ab}\left(g^{bc}h_{bc}+1-p\right).\tag{8}$$ Multiplying this equation by $g^{ab}$ one obtains for $p\neq 1$ that $$g^{ab}h_{ab}=p+1\tag{9}.$$ This in turn yields $$g_{ab}=h_{ab}\tag{4}$$ which was already present in the second action by varying the Lagrange multiplier. However, in $p=1$ (for a string) one can at most say that $g_{ab}\propto h_{ab}$. Thus, it seems that the second and first actions are not equivalent in this case. Why is that if we managed to prove the equivalence by varying $g_{ab}$?


This is related to that the Polyakov action$^1$ (7) is invariant under Weyl transformations $$g_{ab}\to \lambda g_{ab} \tag{15}$$ iff $p=1$. Therefore the EL eq. (8) for $g_{ab}$ must also be Weyl-covariant if $p=1$.

Now the punch-line: Since eq. (4) is not Weyl-covariant, it can not be derived from eq. (8) if $p=1$.


  1. J.A. Nieto, Remarks on Weyl invariant $p$-branes and $Dp$-branes, arXiv:hep-th/0110227; section 2.


$^1$ Although the Polyakov action (7) follows directly from the action (5), it might not be as predictive, as we eliminated the Lagrange multipliers $\Lambda^{ab}$ without using their EL eqs. And in fact it is not if $p=1$ because eq. (4) can then no longer be derived.

  • $\begingroup$ But, the action (5) doesn't have this symmetry. How is it possible that we ended up with a more symmetric action (7) by integrating out a field from (5)? In particular, the equations of motion of (5) demand that $g_{ab}=h_{ab}$, while those of (7) don't, showing the dynamics of (5) and (7) don't agree. $\endgroup$ Jul 2 '20 at 12:21
  • $\begingroup$ I agree with what you are saying. However, my point is what failed to derive the action (7) from the action (5) by plugging in the equations of motion (6)? Clearly something must have because, as you mentioned, the equation of motion (4) obtained from (5) is lost in the formulation of (7). $\endgroup$ Jul 2 '20 at 14:22
  • $\begingroup$ I guess at the end of it all my problem is conceptual. Did we really loose information? At the end of it all the induced metric is an artifact. I am just puzzled at the fact that the equation of motion changed when integrating out the Lagrange multiplier. I thought the system of equations for both Lagrangians have to be identical $\endgroup$ Jul 2 '20 at 14:42
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 2 '20 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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