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Trying to think of a suitable ansatz for a 2 dimensional open string where both endpoints are attached to a $D0$ brane at $(0,0)$, creating a loop. The ansatz is for $F'(u)$ since dirichlet boundary conditions impose

$\left|\frac{dF'(u)}{du}\right|^2=1$ for all u.

The string $\vec{X}(t,\sigma)$ is given by

$\vec{X}(t,\sigma)=\frac{1}{2}\left(F(ct+\sigma)-F(ct-\sigma)\right)$

I know that $F'(u)$ must be of the form $[\cos(\alpha),\sin{\alpha}]$ such that the first condition is satisfied, but i'm having trouble trying to think of how to construct an ansatz.

A similar problem is in Zweibach a first course for string theory problem 7.5 & 7.6, but the open string is attached to two $D0$ branes - one at $(0,0)$ and one at $(a,o)$. The (sucessful) ansatz for that case is given by

$\vec{F}'(u)=\left(\cos\left[\gamma\cos\frac{\pi u }{\sigma_1}\right],\sin\left[\gamma\cos\frac{\pi u }{\sigma_1}\right]\right)$

Is there a way i can transform this such that the two endpoints converge at (0,0)? Or is there something easy i am missing to make a looped string?

EDIT

I have now found that for periodicity to be satisfied i also need that

$F'(u+2\sigma_1)-F'(u)=(2a,0)$

Which implies

$\int_{0}^{2\sigma_1}F'(u)du=(2a,0)$

I will try setting a to zero and see if this helps in finding my F'(u).

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    $\begingroup$ What about simply setting $a = 0,o=0$ doesn't work? Also, is this really "string theory" or are you just considering a classical string between two classical planes? Furthermore, if you want to make a "looped string", why don't you just start with a closed string? Why do you want to attach it to anything? It's not clear what your goal is here. $\endgroup$ – ACuriousMind Apr 2 '17 at 9:22
  • $\begingroup$ I am merely trying to study the properties of such a string - i know solutions for closed strings but am trying to investigate the properties of such a string for a project $\endgroup$ – byBanachTarskiIamcorrect Apr 2 '17 at 9:33
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It can be shown that the ansatz for $x$ equates to a Bessel function of the first kind (with a change of variable for $x$).

$$ \frac{a}{\sigma_1}=\frac{1}{2\pi}\int_{0}^{2\pi}\cos(\gamma\cos(x))dx=J_0(\gamma) $$

By setting $a$ to zero, we see that $\sigma_1$ is then arbitrary and we merely need to locate the zeros of the Bessel functions of the first kind. This is an intriguing example as there are an infinite number of zeros. We can set either gamma in both and $y$ to different value as long as they are a solution to $J_0(\gamma)=0$.

One of the funkier graphs I found was setting $\gamma$ in $x$ to be the 9th Bessel zero, and the $\gamma$ in $y$ to be the 8th Bessel zero. Putting $t=\sigma_1 \cdot 1.7$ we get the following graph:BesselString

Now going to try and animate this somehow.

EDIT 2

I managed to animate it!!

I present to you... Toby Functions (pending patent).

enter image description here

enter image description here

Since there are infinite Bessel zeros, there are $\infty !$ (you can change $\gamma$ for $x$ and $y$) possible animations!! If you like i can give you the code for this and you can have a play around - just message me for it or i'll post it if there is anyone out there reading this...

Note: For a correct string solution, $\gamma_1$ must equal $\gamma_2$ such that we satisfy the condition for:

$\left(\frac{dF}{du}\right)^2=1 \implies \cos^2(\theta)+\sin^2(\theta)=1$

So we have a possible $\infty !$ animations, but only $\infty$ correct string animations.

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