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I have a question the variation of the Polyakov action, related to this Phys.SE post.

For Polyakov action $$ S_p[X,\gamma]=-\frac{1}{4 \pi \alpha'} \int_{-\infty}^{\infty} d \tau \int_0^l d \sigma (-\gamma)^{1/2} \gamma^{ab} \partial_a X^{\mu} \partial_b X_{\mu}.$$

Consider the variation (p. 14 in Polchinski's string theory book)

$$ \delta S_p~=~ \frac{1}{2\pi \alpha'} \int_{-\infty}^{\infty} d \tau \int_0^l d \sigma (-\gamma)^{1/2} \delta X_{\mu} \nabla^2X^{\mu}$$ $$ - \frac{1}{2\pi \alpha'} \int_{-\infty}^{\infty} d \tau (-\gamma)^{1/2} \delta X_{\mu} \partial^{\sigma} X^{\mu} |^{\sigma=l}_{\sigma=0} \tag{1.2.27} $$

The open-string and closed-string boundary conditions

$$\partial^{\sigma} X^{\mu}(\tau,0)=\partial^{\sigma} X^{\mu}(\tau,l)=0 \tag{1.2.28} $$ $$ X^{\mu}(\tau,l) = X^{\mu}(\tau,0), \partial^{\sigma} X^{\mu}(\tau,0)=\partial^{\sigma} X^{\mu}(\tau,l), \gamma_{ab}(\tau,l)= \gamma_{ab}(\tau,0) \tag{1.2.30} $$

will make the "surface term" (the second term in the RHS of Eq. (1.2.27)) vanishes. It is said (p. 14-15 of Polchinski)

The open-string boundary condition (1.2.28) and closed string boundary condition (1.2.30) are the only possibilities consistent with $D-$dimensional Poincare invariance and the equation of motion.

The boundary condition (1.2.28)/(1.2.30) will make the surface term vanish. The resulting equation of motion is tensor equation.
$$(-\gamma)^{1/2} \nabla^2 X^{\mu}=0$$ My question is, why non-vanishing surface term will break the Poincare invariance? E.g. a Lorentz transformation $X_{\mu} \rightarrow \Lambda_{\mu}^{\nu} X_{\nu}$ will not affect $\sigma,\tau$ space

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In fact, to respect Poincaré invariance we need, for open strings, Neumann boundaries conditions (the end of the open string moves freely in space).

For instance, If we have one or more Dirichlet conditions for $\sigma = 0$ and/or $\sigma = l$, this means that one or two of the extremities of the open string is ending on D-branes, so the Poincaré invariance is broken (for instance, the translation invariance is broken, because the D-brane has specific position in target space-time).

See exercices 1.6, 1.7 page 31, and solution of problem 1.7 page 7 of corrected exercices

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