3

I'm reading Polchinski's Introduction to String Theory (volume I) and something got me quite puzzled in the beginning (At the top of page 19 to be precise). This part is about the open string and the transition to light-cone coordinates.

So they obtain in the middle of page 18 the following version of the Polyakov Lagrangian,

$$ L = \frac{-1}{4 \pi\alpha'} \int^{\sigma}_{0} d\sigma \Big(\gamma_{\sigma\sigma} (2 \partial_{\tau}x^{-} - \partial_{\tau} X^i \partial_{\tau} X^i) - 2 \gamma_{\sigma\tau}(\partial_\sigma Y^- -\partial_{\tau} X^i \partial_{\sigma} X^i) + \gamma_{\sigma\sigma}^{-1}(1-\gamma_{\sigma\tau}^2) \partial_{\sigma} X^i \partial_{\sigma} X^i\Big),\tag{1.3.11}$$

where a $X^{-}(\tau,\sigma)$ is separated into two pieces $x^-(\tau)$ and $Y^-(\tau,\sigma)$. These two pieces are defined as follows, \begin{align} x^-(\tau) &= \frac{1}{l} \int_{0}^{\sigma} d\sigma X^-(\tau,\sigma), \tag{1.3.12a}\\ Y^-(\tau,\sigma) &= X^-(\tau,\sigma) - x^-(\tau).\tag{1.3.12b} \end{align}

Now the argument that gets me puzzled goes as follows. The field $Y^- $ does not appear in any terms with time derivative and so is non-dynamical. It acts as a Lagrange multiplier, constraining $\partial^2_{\sigma}\gamma_{\tau \sigma}$ to vanish (extra $\partial_\sigma$ because $Y^-$ has zero mean).

I personally don't directly see how this term vanishes and what is meant with the Lagrange multiplier argument, could someone elaborate this bit?

Qmechanic
  • 201,751
Dylan_VM
  • 289

2 Answers2

4

A Lagrange multiplier works like this: consider as an example a 2D system in classical mechanics with Lagrange function $L = T - V$, but the particle is constrained to move only on the curve $y = x^2$. One way of implementing the constraint is to add a new degree of freedom $\lambda$, a Lagrange multiplier, and modifying the Lagrange function like this: $$ \tilde L(x, \dot x, y, \dot y, \lambda, \dot \lambda) = L(x, \dot x, y, \dot y) + \lambda (y - x^2) $$ The Euler-Lagrange equation for this new degree of freedom is $y(t) = x(t)^2$, thus, extremizing the modified action automatically implements the constraint.

In a nutshell we can say: If there is a d.o.f. where the derivative doesn't appear in the action, it is a Lagrange multiplier implementing the constraint that the term multiplied by it is zero.

In your case, if you integrate once by parts, you see that your action contains a term $$ \sim Y^-\, \partial_\sigma \gamma_{\sigma\tau} , $$ the constraint is $\partial_\sigma \gamma_{\sigma\tau} = 0$. The boundary terms vanish because $\gamma_{\sigma\tau}$ is zero at the boundary.

(In my version of Polchinski there is no second derivative $\partial_\sigma^2$, but if $\partial_\sigma$ is zero then also $\partial_\sigma^2$.)

Noiralef
  • 7,238
  • Thank you, If it should be just one $\sigma$ derivative then your argument indeed suffices. – Dylan_VM Oct 14 '18 at 12:31
  • 1
    It should be second derivative since $\int d\sigma Y^{-}\partial_{\sigma}\gamma_{\sigma\tau}$ is identically zero if $Y^{-}$ is constant, so the constraints only comes from non-constant $Y^{-}$ lagrange multipliers. – Nogueira Nov 18 '18 at 23:26
1

The first derivative does not necessarily have to vanish. It can also be constant since after integrating by parts, the variation of the action with respect to $Y^-$ gives $$ \delta S_{Y^-} \propto \int d \tau \int_0^\ell d\sigma \ \partial_\sigma \gamma_{\sigma \tau} \ \delta Y^-$$ which vanishes if $\partial_\sigma \gamma_{\sigma \tau} $ is constant w.r.t $\sigma$ since $\delta Y^-$ has mean $0$. Note that it has mean 0 by definition of what we mean by "variating $Y^-$" (if we allow for non-zero mean, this variation would contribute to $x^-$).

Therefore the true constraint is

$\partial_\sigma \gamma_{\sigma \tau} = \mathrm{const}(\tau) \implies \partial_\sigma^2 \gamma_{\sigma \tau} = 0$

Adrien Martina
  • 1,358