Is a finite volume with infinite extension possible with curvature of space?

Question

Let's assume a general spherically symmetric spacetime. The metric is $$\mathrm{d}s^2=-Bc^2 \mathrm{d}t^2+A \mathrm{d}r^2+r^2 \mathrm{d}\Omega.$$

and a spherical volume in that metric will be calculated with $$V = 4\pi\int_0^Rr^2\sqrt{|A|}\ \mathrm{d}r.$$

Therefore, it should be possible to construct a theoretical metric which is infinite in extension ($R$ can take every, arbitrarily large value) — but finite in volume. One only has to "tune" the curvature parameter A such that $\sqrt{|A|}$ decreases at least with $1/r^3$ for the integral to converge.

Is there any mistake in this reasoning (especially in GR) or is it therefore correct to say that it is possible to construct a metric in GR which leads to a finite volume with infinite extension?

Answer 1

Is there any mistake in this reasoning (especially in GR) or is it therefore correct to say that it is possible to construct a metric in GR which leads to a finite volume with infinite extension?

It is possible but only in the universe with negative masses.

In case of static spherically symmetric spacetime filled with matter of density $\rho$ the proper volume reads:

\begin{equation} V(r)=\int_{0}^{r}\tilde{r}^2~\Big(1-\frac{\kappa c^2}{\tilde{r}}\int_{0}^{\tilde{r}}~\rho(x)~x^2~dx\Big)^{-1/2}~d\tilde{r} \tag{1}, ~~~~~\kappa\equiv \frac{8\pi G}{c^4} \end{equation}

The derivative of $V'(r)$ goes asymptotically to zero (constant proper volume) only if the density there is negative and proportional to $r^{-(3+\epsilon)}$ ($\epsilon \geq 0$).

Answer 2

What you say is correct, but it's also not really meaningful. Whether the coordinate description is bounded or not can be changed arbitrarily, the physically meaningful distinction is whether the manifold itself has a finite volume or not.

Consider a 1D manifold with a finite volume: $x \in (-1, 1)$ with the Euclidean metric. Now, let us map this to $y \in (-\infty, \infty)$: we can do so with the inverse hyperbolic tangent, $y = \tanh^{-1}{x}$. This manifold will have a nontrivial metric $g_{yy} = (\mathrm{d}x / \mathrm{d}y)^2 = (1 - \tanh^2y)^2$, but the total volume must be unchanged: indeed, $\int_{-\infty}^{\infty} \sqrt{g_{yy}} \mathrm{d}y = 2 = \int_{-1}^{1} \mathrm{d} x$.

We now have a manifold with finite volume and two equivalent coordinate descriptions, one bounded and the other unbounded. I hope this illustrates the fact that the boundedness of coordinates is not physical, it's just an artifact of how we describe the manifold.

For a GR example (of the inverse of what you say) see Penrose diagrams: they can represent an unbounded spacetime with bounded coordinates.