# Poincare invariance of Dirichlet and Neumann boundary conditions

The action which describes a string propagating in a $D$ dimensional spacetime, with given metric $g_{\mu\nu}$, is given by the Polyakov action $$S_{\text{p}}=-\frac{T}{2}\int \mathrm{d}\sigma\mathrm{d}\tau\sqrt{-h}\eta^{\alpha\beta}\partial_{\alpha}X^{\mu}\partial_{\beta}X^{\nu}g_{\mu\nu}\tag{1}$$ where the symbols have their usual meaning. It is not hard to check that action is invariant under Poincare transformations $$\delta X^{\mu}(\sigma,\tau)=a^{\mu}_{~~~\nu}X^{\nu}(\sigma,\tau)+b^{\mu}\tag{2}.$$ When all the dust is settled (i.e. after gauge fixing and Weyl transformations) the Polyakov action becomes $$S_{\text{P}}=\frac{T}{2}\int \mathrm{d}\sigma\mathrm{d}\tau \left((\dot{X})^{2}-(X')^{2}\right)\tag{3}$$ where $\dot{X}=\partial_{\tau}X^{\mu}$ and $X'=\partial_{\sigma}X^{\mu}$. Variation with respect to $X^{\mu}$ yields the equation of motion $$(-\partial^{2}_{\tau}+\partial^{2}_{\sigma})X^{\mu}-T\int\mathrm{d}\tau\left[X'\delta X^{\mu}|_{\sigma=\pi}+X'\delta X^{\mu}|_{\sigma=0}\right]=0.\tag{4}$$ The $\sigma$ boundary terms tell us what type of strings we have, either closed or open strings.

• For open string equation (4) becomes $(-\partial^{2}_{\tau}+\partial^{2}_{\sigma})X^{\mu}=0$ where we assume that the end points of the string follow the Neumann boundary conditions $$\partial_{\sigma}X^{\mu}(\tau,\sigma)=\partial_{\sigma}X^{\mu}(\tau,\sigma+n).\tag{5}$$ One interesting feature is that the Neumann boundary conditions remains invariant under global Poincare transformation since \begin{eqnarray} \partial_{\sigma}X'^{\mu}|_{\sigma=0,n} & = & \partial_{\sigma}\left(a^{\mu}_{~~~\nu}X^{\nu}(\sigma,\tau)+b^{\mu}\right)|_{\sigma=0,n} \\ & = & a^{\mu}_{~~~\nu}~\partial_{\sigma}X^{\nu}|_{\sigma=0,n}\\ & = & 0 \\ \end{eqnarray}

• Whereas the Dirichlet boundary conditions $$X^{\mu}(\tau,\sigma=0)=X^{\mu}_{0}\qquad\qquad X^{\mu}(\tau,\sigma=n)=X^{\mu}_{n}$$ break the Poincare invariance, as $$X'^{\mu}|_{\sigma=0,n}=\left(a^{\mu}_{~~~\nu}X^{\nu}(\sigma,\tau)+b^{\mu}\right)|_{\sigma=0,n}\neq X^{\mu}_{0,n}$$ which simply means that under a Poincare transformation the ends of the string actually change.

Does the spectrum of string excitations keep any signature of this (non) invariance under Poincare transformations? If so, how can that result be interpreted?

• v5: How did $\pi$ become $n$ in various places? – Qmechanic Nov 19 '16 at 18:33

## 2 Answers

The Dirichlet boundary condition, in particular, breaks spacetime translation invariance. This is reflected in the string spectrum, that is, a Goldstone boson state appears in the massless spectrum of the string. This state corresponds to the collective coordinate which parametrizes small oscillations of the Dirichlet 'brane' or D-brane on which the string is constrained to move on.

For the SUPER-string, a similar effect occurs whereby a Goldstino state appears in the massless spectrum, and this indicates the breaking of some amount of spacetime supersymmetry by the D-brane.

References: String Theory, Polchinski, Volume I (equation (8.6.18) and pages 268-269), Volume II (pages 138-140).

The most obvious consequence of Poincare symmetry breaking is non-conservation of spacetime momentum. Of course, the string spectrum is also changed, and this change could be easily calculated.

• It would be better, if you elaborate your answer using maths. – AMS Nov 21 '16 at 17:14