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In the context of string theory, and world sheets the Dirichlet boundary conditions can be written as: $$\frac{\partial X^\mu(\tau,\sigma_1)}{\partial \tau}=0$$ where $\sigma_1$ is the value of the parameter $\sigma$ at the end of the 'string'. This however, seems to imply that $$\delta X^\mu(\tau,\sigma_1)=0$$ But I cannot see why, so please can you explain?

Here are my thoughts (which are wrong since I get the wrong outcome):

It is my assumption that $\delta X^\mu \equiv dX^\mu$ in this context although I could be wrong. This therefore means that: $$\delta X^\mu(\tau,\sigma_1) =\frac{\partial X^\mu(\tau,\sigma_1)}{\partial \tau}d\tau + \frac{\partial X^\mu(\tau,\sigma_1)}{\partial \sigma}d\sigma$$ So subbing in my first equation we get: $$\delta X^\mu (\tau,\sigma_1)=\frac{\partial X^\mu(\tau,\sigma_1)}{\partial \sigma}d\sigma$$ which is generally not equal to $0$. Thus my first equation does not necessary imply my second, as it should.

References:

  1. A first course in string theory by Barton Zwiebach (2 e.d.) pg 114
  2. http://www.damtp.cam.ac.uk/user/tong/string/three.pdf
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This seems to be a simple matter of confusion regarding which variables are held constant. His notation $$\frac{\partial X}{\partial\tau}(\tau,\sigma_*)=0$$ is misleading. What he really means is $$\frac{\partial}{\partial\tau}\left(X(\sigma_*)\right)(\tau)=0$$ In other words, we fix $\sigma$ to be one of the end points and look at how it changes with respect to $\tau$. Your expansion above is correct, but you must take $\delta \sigma=0$ because we focus attention on one value of $\sigma$, namely an end point.

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Your assumption that $\delta X^{\mu} = dX^{\mu}$ as well as the expansion is not correct. $\delta X^{\mu}$ is a variation and you can not rwright variation of a parameter (the functional derivative relates a change in a functional to a change in a function on which the functional depends).

\begin{equation} \delta X^{\mu} = X^{'\mu}(\tau,\sigma^{*}) - X^{\mu}(\tau,\sigma^{*}) \end{equation} whereas $dX^{\mu}$ is an exterior derivative. \begin{equation} dX^{\mu}(\tau, \sigma) = \frac{\partial X^{\mu}}{\partial\tau}d\tau + \frac{\partial X^{\mu}}{\partial\sigma}d\sigma \end{equation} This means that the condition(1) that $ \frac{\partial X^{\mu}(\tau, \sigma^{*})}{\partial\tau} = 0 $ is only true when $ X^{\mu}(\tau,\sigma^{*}) = constant$ that is for static D branes, So this is a sufficient condition not necessary. However, $ X^{\mu}(\tau,\sigma^{*}) = f(\tau)$, which means that the D brane can be dynamical then only that $\delta X^{\mu} = 0$ is satisfied.

Dirichlet Boundary Condition is $\delta X^{\mu}(\tau,\sigma^{*}) = 0$ but for a special case when the brane is static, we can write it as $ \frac{\partial X^{\mu}(\tau, \sigma^{*})}{\partial\tau} = 0 $ .

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