# Why does a wormhole have this metric?

I asked this (or something similar) within another question and was asked to post it separately, so here goes:

The metric for flat Minkowski space is: $$ds^2 = -dt^2 +dr^2 +r^2(d\theta^2+d\phi^2\sin^2\theta),$$

and the metric for a wormhole is: $$ds^2 = -dt^2 +dr^2 +(r^2+b^2)(d\theta^2+d\phi^2\sin^2\theta).$$ Why does simply adding $$+b^2$$ into the metric change the geometry so drastically? Or more specifically, why does a wormhole emerge simply by adding $$+b^2$$ into the metric?

Many thanks.

• en.wikipedia.org/wiki/Ellis_wormhole – G. Smith Sep 10 at 18:04
• Not only do we add $b^2$ to the metric function, we also allow the $r$ coordinate to take negative values. – A.V.S. Sep 10 at 18:04
• Do you know how to calculate the geodesics for a metric? Or the curvature? – G. Smith Sep 10 at 18:05
• @G.Smith No I don't yet. I found something that talks about the space being "folliated with S^2 of radius sqroot(r^2+b^2)" - what does that mean? – Ali Chopping Sep 10 at 18:18
• I have a related answer on the Ellis geometry here – John Rennie Sep 10 at 18:22

Forget about the time part of the metric, that's irrelevant here since in technical terms our spacetimes are ultrastatic: just a 3-space extended in time, with no time dilation.

Take the flat metric

$$ds^2 = dr^2 + r^2(d\theta^2+\sin^2\theta\, d\varphi^2)$$

The two appearances of $$r$$ signify different things. The $$dr$$ part is relevant if we're measuring radial distances: the proper distance along a line from the origin to a given point at $$r=R$$ is

$$\text{radius } = \int ds = \int_0^R dr = R,$$

since $$ds=dr$$ along a curve with $$d\theta=d\varphi=0$$. On the other hand, if we want the circumference of the equator at a given radius, we have $$r=R$$ and $$\theta=\pi/2$$, implying $$ds= R\, d\varphi$$ and thus

$$\text{circumference } = \int ds = \int_0^{2\pi} R\, d\varphi = 2\pi R,$$

as expected. Note that this $$R$$ comes from the $$r^2$$ part of the metric, not the $$dr^2$$. It can also be shown that this same $$r^2$$ shows up in the $$A = 4\pi R^2$$ formula for the area.

Now what if we take the

$$ds^2 = dr^2 + (r^2+b^2)(d\theta^2+\sin^2\theta\, d\varphi^2)$$

wormhole metric? The calculation for the radius from the origin to a given point is the same, so the coordinate $$r$$ still works as radial distance. But for circumference and area we need the $$r^2+b^2$$ part, so we get that $$A = 4\pi (R^2+b^2$$).

What does this mean? In this metric, the relation between area and radius changes. This also changes the relation between area and volume, if you prefer to think that way. It should make sense that something weird is going on in a space in which we don't have $$A = 4\pi R^2$$; in fact, (I think) this can be taken as one of the definitions of curvature. For example, you could place two observers at antipodal points at $$r=R$$, and tell them to move towards the origin while measuring the distance traveled. What they will find is that after traveling a distance of $$R$$, they do not meet!

In fact, they can keep going forever, and they will never meet. That's because at $$r=0$$, the sphere doesn't have zero radius. This allows us to keep extending $$r$$ to negative values, to get a completely identical copy of the universe, which is why this is called a wormhole.