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A open bosonic string between two parallel branes seems to obey formulae such as

$M^2 = \big((n + {\theta_i - \theta_j \over 2 \pi}) {R' \over \alpha'}\big)^2 + {N-1 \over \alpha'} $

So that the difference $\theta_i - \theta_j$ is the distance between branes. Now I wonder, which is the formula for the superstring stretched between two parallel branes? Is it the same?

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1 Answer 1

It is very similar actually, but not exactly the same, unless you put the equation in a different form. In natural units where $\hbar =c_0=\ell_s=1$

$${m^2} = \left( {N - a} \right) + {\left( {\frac{y}{{2\pi }}} \right)^2}$$

What have I done? I wrote the "1" in the equation as $a$, the normal ordering constant. This is the important part. The normal ordering constant and number operator are what change for a superstring. The rest of the intuition/proof is the same.

Note: I also changed the notation for the separation to $y$ and I got rid of the $\alpha'=\ell_s^2$ because of the use of natural units. That makes sense, because even in your equation, you are using $\hbar=c_0=1$, just without the $\alpha'$.

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