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In Zweibach's A first course in string theory, he used the least action principle to get the equations of motion for strings, wehre the variation of action(which should be zero) is : $$\delta S = \int_{\tau_i}^{\tau_f} d\tau [\delta X^\mu \mathcal{P}^\sigma_\mu]^{\sigma_1}_0 - \int_{\tau_i}^{\tau_f} d\tau \int_0^{\sigma_1} d\sigma \delta X^\mu \left(\dfrac{\partial \mathcal{P}^\tau_\mu}{\partial \tau}+ \dfrac{\partial \mathcal{P}^\sigma_\mu}{\partial \sigma} \right)$$ He then imposed the boundary conditions of the open strings endpoints (Dirichlet and Neumann) to let the first term vanish; and then he said that the second term must vanish as well. Doing so we got the equations of motion : $$\dfrac{\partial \mathcal{P}^\tau_\mu}{\partial \tau}+ \dfrac{\partial \mathcal{P}^\sigma_\mu}{\partial \sigma} =0$$ Which are for both open and closed strings.

I have three questions:

  1. How can we consider that these are the equations of motion for closed strings that don't have boundary conditions in the first place?

  2. A string's endpoint may have Dirichlet or Neumann boundary condition, not both.

  3. And the two endpoints of an open string may have different boundary conditions, so how is it logical to impose the two conditions to get the EOM?

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  1. For a closed string $\cong S^1$, there is no boundary, and hence no boundary condition (BC). Equivalently, if we think of the string as living on $\mathbb{R}$, it can be thought of as a periodic BC. The boundary term in eq. (1) therefore vanishes in any case. Hence the bulk term in eq. (1) still have be zero.

  2. For an open string $\cong I\cong[0,\sigma_1]$, there are two ways to make a product $\mathcal{P}^\sigma_{\mu}~ \delta X^\mu =0$ zero:

    • (i) The choice $X^\mu={\rm const}$ corresponds to Dirichlet BC, and
    • (ii) the choice $\mathcal{P}^\sigma_{\mu}=0$ corresponds to Neumann BC.
  3. The boundaries $\sigma=0$ and $\sigma=\sigma_1$ are independent, so one is free to pick, say, Dirichlet BC at $\sigma=0$ and Neumann BC at $\sigma=\sigma_1$.

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  • $\begingroup$ Ok it's clear now for open strings, thanks! but could you please tell me more about the closed strings, I didn't get the idea yet. $\endgroup$ – Milou Sep 12 '16 at 15:29
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Sep 12 '16 at 18:14
  • $\begingroup$ please excuse me if my question isn't so smart, but I'm self-learning those new things. How do we know that the first term is the boundary one and the second is the bulk one? $\endgroup$ – Milou Sep 17 '16 at 22:16
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    $\begingroup$ @Milou: Because the spatial worldsheet coordinate $\sigma$ can only take the values $0$ and $\sigma_1$ in the first term, while there are no such restrictions in the second term. $\endgroup$ – Qmechanic Sep 17 '16 at 22:48

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