I have a super basic and stupid question about the Lorentz invariance of the Polyakov action (cannot skip the disclaimer..) $$S_p[X,\gamma]=-\frac{1}{4 \pi \alpha'} \int_{-\infty}^{\infty} d \tau \int_0^l d \sigma (-\gamma)^{1/2} \gamma^{ab} \partial_a X^{\mu} \partial_b X_{\mu} $$

I may write the action as (using metric tensor $\gamma^{ab}$) $$S_p[X,\gamma]=-\frac{1}{4 \pi \alpha'} \int_{-\infty}^{\infty} d \tau \int_0^l d \sigma (-\gamma)^{1/2} \partial^a X^{\mu} \partial_a X_{\mu} $$

The Lagrangian is obviously invariant under proper Lorentz transformation. $\partial^a X^{\mu} \partial_a X_{\mu}$ is a Lorentz scalar. $d\tau d\sigma$ transforms with a determinant equals 1 for proper Lorentz transformation. $(-\gamma)^{1/2} $ also transforms with a determinant equals 1.

But for the range of integration $[0,l]$, there is length contraction under a boost. Why the action is still invariant under Lorentz transformation? (or I completely mistaken something....)

  • 4
    $\begingroup$ I'm confused by your use of the term Lorentz transformation. Do you mean a transformation $X^\mu \to \Lambda^\mu{}_\nu X^\nu$. This is simply a field transformation. It has nothing to do with the transformation of $\tau$ and $\sigma$ and there is no need to invoke determinants at all. Also, the length contraction occurs in the field space $X^\mu$ not in the $\tau,\sigma$ space. $\endgroup$
    – Prahar
    Commented Aug 1, 2013 at 22:44
  • $\begingroup$ I mean a Lorentz transformation, similar to a scalar field $\phi(\mathbf{x}) \rightarrow \phi(\Lambda^{-1} \mathbf{x})$. I see your point. Thank you very much! Here $x$ corresponds to $X$, $\{\tau,\sigma\}$ embeds on $X^{\mu}$. $\endgroup$
    – user26143
    Commented Aug 1, 2013 at 23:47

1 Answer 1


You are confusing the spacetime Lorentz symmetry and the world sheet Lorentz symmetry.

The spacetime Lorentz symmetry only acts on the fields locally on the world sheet, $$ X^\mu (\sigma,\tau) \to \Lambda^\mu{}_\nu X^\nu (\sigma,\tau).$$ Note that the coordinates $\sigma,\tau$ haven't changed at all, so there is no contraction of the coordinate $\sigma$ on the world sheet. The spacetime Lorentz symmetry only acts on the Greek indices which is how we check the invariance. The physical interpretation is the usual Lorentz symmetry if string theory is used as a description of spacetime physics.

On the other hand, the world sheet theory also locally has the $SO(1,1)$ Lorentz symmetry on the world sheet which only acts on the two $\sigma,\tau$ coordinates i.e. on the Latin indices $a,b$. This symmetry is broken if the world sheet is compactified i.e. if $\sigma$ spans a periodic or compact interval. But there are still unbroken Weyl/diff diffeomorphisms on the world sheet, the conformal transformations, that must be carefully taken into account during string calculations. All these symmetries are "auxiliary" much like any gauge symmetry – all the observable states and operators in the spacetime must be invariant under them (singlets).


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