Why can't Schwarzschild wormholes exist? So, I've recently been reading up on Schwarzschild wormholes and I've learned that they cannot exist becuase they violate the 2nd Law of Thermodynamics.
What I'm asking is: Why do they violate the Law?
I probably sound like an idiot, but I just can't understand why they violate Thermodynamics.
 A: I don't know why they should violate thermodynamics either, but they don't exist because they're static. They cannot be created at any finite time - they must have existed since the beginning of time and will exist forever. The physically realistic Schwarzschild solution is created from collaps and does not have the second asymptotic region. 
A: I am now also studying wormhole. based on so far I learned, First the Schwarzschild wormhole is unstable under the Kruskal time that is {T,X}. It could be reflected by the Kruskal spacetime diagram and its coordinates relationships with original Schwarzschild wormhole coordinates.  Now recall the global coordinate transformation between {T,X} and {t,r}:$T^{2}-X^{2}=\left(1-\frac{r}{2  M}\right) e^{r / 2  M}$, which is defined the whole space except the black region. it is easy to verify that the r=const is hyperbole (as shown in fig the blue dot line) in Kruskal space time. In region A and A', we have r>2M, while in Region B and W, 0<r<2M. So as to illustrate the dynamic of wormhole, we could select three slices (T>0, T=0, T<0) each of them represent a whole spatial geometry at that time. In each slice, the r is determine by X only :$T_{0}^{2}-X^{2}=\left(1-\frac{r}{2 G M}\right) e^{r / 2 G M}$. therefore, we could find the $r_{min}$(the throat of a wormhole ) at some value of X. Obviously,$r_{min}$ is obtained when $X =0$.   We found that$r_{min}^{T=\sqrt(0.4)}=r_{min}^{T=-\sqrt(0.4)}=1.649286999,r_{min}^{T=0}=2$, it shows with T increasing, the throat firstly open up to the maximum radius at T=0, then It shrinks again. It is found the time is so small that can not travel through. (see this paper). Second, the Schwarzschild wormhole is unstable, which is studied by RedMount by using'cutting-slide' model, which shows the blue sheet apeares at horizon, where the energy momentum is approaching infinite.
the key is there is Dirac function with respect with coorditate v. here I post the resut of whithe hole while as for the worm hole , the mechnisim is similar by cutiing both horizon(you can see ).The final non-zero component of energy-momentum tensor is  $T^{vv}=-{\frac {{\rm e} \left( \cos \left( \theta \right)  \left( {\frac {
\partial }{\partial \theta}}a \left( \theta,\phi \right)  \right) \sin
 \left( \theta \right) + \left( \cos \left( \theta \right)  \right) ^{
2}a \left( \theta,\phi \right) - \left( \cos \left( \theta \right) 
 \right) ^{2}{\frac {\partial ^{2}}{\partial {\theta}^{2}}}a \left( 
\theta,\phi \right) -a \left( \theta,\phi \right) +{\frac {\partial ^{
2}}{\partial {\theta}^{2}}}a \left( \theta,\phi \right) +{\frac {
\partial ^{2}}{\partial {\phi}^{2}}}a \left( \theta,\phi \right) 
 \right) {\it \delta} \left( v \right) }{32\,{M}^{4} \left( \sin
 \left( \theta \right)  \right) ^{2}}}$
The detailed calculation could be done by software, here is my code. hope it helps:
with(DifferentialGeometry);with(Tensor);with(Tools);
DGsetup([u, v, theta, phi], M1, verbose)
r := solve((r/(2*M) - 1)*exp(r/(2*M)) = -(u + Heaviside(v)*a(theta, phi))*v, r)
g := evalDG(-32*M^3*exp(-r/(2*M))*(1/2*(du &t dv) + 1/2*(dv &t du) + 1/2*Heaviside(v)*diff(a(theta, phi), theta)*(dtheta &t dv) + 1/2*Heaviside(v)*diff(a(theta, phi), theta)*(dv &t dtheta) + 1/2*Heaviside(v)*diff(a(theta, phi), phi)*(dphi &t dv) + 1/2*Heaviside(v)*diff(a(theta, phi), phi)*(dv &t dphi))/r + r^2*((dtheta &t dtheta) + sin(theta)^2*(dphi &t dphi)))  %%% the coefficient before the cross term dudv should be careful which should make up the dvdu part which is half of original metric form in paper given that the metric is symmetric.
Ricci := RicciTensor(g)
DGinfo(EinsteinTensor(g), "CoefficientSet")

