For a spacetime surgery wormhole, we have a manifold such that, for two connected compact sets $D_1$ and $D_2$, we remove $D_1$ and $D_2$ from the manifold and identify their boundaries. According to Friedman, solving fields on such spacetimes just requires the added conditions that, for $x_1 \in \partial D_1$ and $x_2 \in \partial D_2$, with $x_1$ the image of $x_2$ via the identification, and for the normal vectors $n_1, n_2$ orthogonal to $\partial D_1, \partial D_2$, for a field $\Phi(x)$, we have
$$\Phi(x_1) = \Phi(x_2)$$
$$n_1 \cdot \nabla \Phi(x_1) = -n_2 \cdot \nabla \Phi(x_2) $$
Which sounds reasonable enough.
With that in mind, can we still write the field expansion in terms of the original basis, or some kind of basis at least? The simplest example is just a spacetime with a closed spatial coordinate, for instance $x = 0$ identified with $x = L$, in which case we have (in 2+1 dimensions)
$$\int \frac{d^2p}{(2\pi)^2} \frac{1}{\sqrt{2\omega_p}} [f(p) e^{-i(p_y y - \omega t)} (1 - e^{-i(p_x L )}) + f^*(p) e^{i(p_y y - \omega t)}(1 - e^{i(p_x L )}) ] = 0$$
Which works out fine for $p_x L = k 2\pi$, or $f_p = \operatorname{III}_{\frac{2\pi}{L}}(p_x) f(p_y) $, giving us the classical result for the field on a cylinder.
But now let's consider a slightly more complicated example, where we have two squares identified in the manifold,
$$I^2_1 = \{ x,y \vert x \in [a, a+1], y \in [-\frac{1}{2}, \frac{1}{2}]\}$$
$$I^2_2 = \{ x,y \vert x \in [-a, -a-1], y \in [-\frac{1}{2}, \frac{1}{2}]\}$$
Two squares of side $1$ separated by a distance of $2a$, where the normals are identified such that an object entering at $x = a$ will get out at $x = -a$ (they are mirror images), so that the wormhole is orientable. I'm ignoring what happens at the corners for now.
In this case, we have
\begin{eqnarray}
\forall y \in [-\frac{1}{2}, \frac{1}{2}],\ \int &&\frac{d^2p}{(2\pi)^2} \frac{1}{\sqrt{2\omega_p}} [f(p) e^{-i(p_y y - \omega t)} (e^{-i p_x a} - e^{i p_x a})\\
&&+ f^*(p) e^{i(p_y y - \omega t)} (e^{i p_x a} - e^{-i p_x a})] = 0
\end{eqnarray}
Which corresponds to $p_x a = k \pi \to p_x = \frac{k \pi}{a}$
\begin{eqnarray}
\forall y \in [-\frac{1}{2}, \frac{1}{2}],\ \int &&\frac{d^2p}{(2\pi)^2} \frac{1}{\sqrt{2\omega_p}} [f(p) e^{-i(p_y y - \omega t)} (e^{-i p_x (a+1)} - e^{i p_x (a+1)})\\
&&+ f^*(p) e^{i(p_y y - \omega t)} (e^{i p_x (a+1)} - e^{-i p_x (a+1)})] = 0
\end{eqnarray}
Which corresponds to $p_x (a+1) = k \pi \to p_x = \frac{k \pi}{a+1}$
Let's pick $a = 1$, to have the condition that $\forall k \in \Bbb Z,\ p_x = k \pi$.
Then the other two sides, in which case we have to flip the $x$ coordinate :
\begin{eqnarray}
\int &&\frac{d^2p}{(2\pi)^2} \frac{1}{\sqrt{2\omega_p}} [f(p) e^{i\omega t} (e^{-i (p_x x + p_y/2)} - e^{-i (-p_x x + p_y/2)})\\
&&+ f^*(p) e^{-i\omega t} (e^{i (p_x x + p_y/2)} - e^{i (-p_x x + p_y/2)})] = 0
\end{eqnarray}
which works fine for $p_x = k \pi$.
Then the condition on the derivatives. We have as normals to the boundaries $\pm \partial_x$ and $\pm\partial_y$, so for $x = a$ :
\begin{eqnarray}
\forall y \in [-\frac{1}{2}, \frac{1}{2}],\ \int &&\frac{d^2p}{(2\pi)^2} \frac{1}{\sqrt{2\omega_p}} [f(p) e^{-i(p_y y - \omega t)} (p_x e^{-i p_x} + p_x e^{i p_x }) + ... ] = 0
\end{eqnarray}
Again fine.
So far all results have been identical to a cylindrical spacetime of length $L = 1$, and I have not really used the fact that those conditions only apply to specific ranges of the coordinates. I'm not quite sure how to apply those here. What would be the method to use to find the correct basis for the field?
This post imported from StackExchange Physics at 2016-06-12 18:32 (UTC), posted by SE-user Slereah
Q&A (4870)
