A "wormhole" is a staple of science fiction writing, but it actually describes a particular structure in general relativity that has, if not real world applications, at least academic ones. The wormhole is described by the following metric:
$ds^2 = dr^2 + (r^2 + a^2)d\Omega^2 -dt^2$
Where s a line element in this space, r is a radial coordinate, $\Omega$ is an element representing both $\theta \& \phi$, and t is time. a is a constant. Note that a normal spherical coordinate metric has $a = 0$, of course. What does this thing look like? Well, at any r, we sweep out a sphere of radius $\sqrt{r^2 + a^2}$ in constant time. Or, if you prefer, at constant t and $\theta$, it sweeps out circles of radius $\sqrt{r^2 + a^2}$. I've drawn a picture of what that looks like below
Don't worry too much about what the axis mean. What I'm really drawing is the shape of the coordinate system that I've described, for any $\theta$. Most importantly, you can see that the coordinate system allows for non-trivial solutions at $r = 0$. By comparison, the spherical coordinate system, drawn in this manner, is a cone, and vanishes exactly at $r = 0$. This is why the former is describes as a wormhole: it has no singular points, and thus may in principle allow passage from negative to positive values of r.
We can calculate the energy density and radial pressure for such an object to exist. The energy density required to maintain the wormhole is given by the zero component of the stress-energy tensor, and radial pressure by the first component:
$\rho = T_0^0$
$P_r = T_1^1$
We would like to convert this to a spherically symmetric metric, because metrics of the following form can be computed relatively easily.
$ds^2 = e^\alpha(r)dr^2 + r^2 d\Omega^2 - e^\gamma(r)dt^2$
To do so, let
$r^2 = u^2 - a^2$
$dr^2 = \frac{u^2 du^2}{u^2-a^2}$
So that
$ds^2 =\frac{u^2 du^2}{u^2-a^2} + u^2 d\Omega^2 - dt^2$
Then the functions
$e^\alpha = \frac{u^2 du^2}{u^2-a^2}$
$e^\gamma = 1$
The zero component is then given by
$8\pi u^2 T_0^0 = \frac{d}{du}\big(u e^{-\alpha} - u\big)$
This equation can be derived by calculating $R_{00}$ of the curvature tensor. I won't show that here, as it is a bit of work. It is fairly standard. This simplifies very nicely.
$8\pi u^2 T_0^0 = \frac{d}{du}\big(\frac{-a^2}{u}\big)$
Thus
$T_0^0 = \rho = -\frac{a^2}{8\pi u^4}$
The energy density is negative. Objects with negative energy density might be considered gravitationally repulsive--or have a negative mass. That is, our wormhole geometry would only exist in a dark energy or inflationary regime.
Likewise, we can calculate the radial pressure from a similar expression.
$8\pi u^2 T_1^1 = e^{-\alpha}(1 + u\gamma^\prime) - 1$
$\Gamma = 0$ for our case, so we end up with an identical expression for the pressure:
$P_r = \frac{-a^2}{8\pi u^4}$
I think it is basically coincidental that these two terms work out to be equal. They aren't actually equal in magnitude, of course, since I'm using units in which $c = G = 1$. The pressure is negative, which again, suggests that this object is inflationary--namely that there is an outward pressure that will drive objects away from the centre of the wormhole.
Now, as to the question of whether wormholes exist, well, I can't really say. We live in an expanding universe, and such a universe would have a negative pressure driving it. But beyond that, it's hard to say much. We haven't found the existence of any localized objects that display these sorts of properties. It's also unclear that you would ever be able to "pass through" the wormhole in any meaningful sense--as you approach $r=0$, the outward pressure tends to infinity, so anything approaching the wormhole would be blown out of it. As such, wormholes in the science fiction sense will probably mostly remain as just that--fiction--but maybe, possibly, somewhere in the universe, there might be an object with similar properties to this.
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