What is the volume of the sphere in hyperbolic space?

Question

I'm looking for a formula to describe surface and volume of a sphere in hyperbolic 3-space. I found some results which were generalized for any dimension, but I wasn't able to understand them.

Overall, I find there's a sort of "symmetry" between spherical and hyperbolical geometry in these formulas. The circumference of circle in hyperbolic geometry is $2\pi\sinh(r)$, in spherical geometry it's $2\pi\sin(r)$. Areas are $2\pi(\cosh(r) - 1)$ and $2\pi(1 - \cos(r))$, respectively.

I then tried to generalize it into 3D. I was trying to find surface and volume of sphere in spherical space (considered as a cap of Euclidean 4D hypersphere), and eventually obtained formulas that seem to work: $2\pi(1 - \cos(2r))$ for surface and $2\pi(r - \sin(2r)/2)$ for volume. Considering the analogy between formula for area of circle and for surface of sphere, the hyperbolic formulas should be $2\pi(\cosh(2r) - 1)$ for surface and $2\pi(\sinh(2r)/2 - r)$ for volume, but this is based mostly on intuition, so I'd appreciate if you could either confirm my line of thought or give me the real result :)

(Incidentally, what is the formula for length of an equidistant curve? I used the same analogy between goniometric and hyperbolic functions to solve it for sphere: if I have a certain length a of straight line, I can construct the perpendicular lines at the endpoints which will then intersect the equidistant curve (circle on sphere, hypercycle in hyperbolic geometry) under right angles as well. If b is the distance of the equidistant from the line, the length of such defined arc of equidistant is acos(b) on sphere, so it should be acosh(b) in hyperbolic geometry -- is this correct? Finally, what would be the formula for a part of surface equidistant from plane directly "above" a part of the plane of known volume?)

Answer 1

The volume of a sphere or a ball in hyperbolic $n$-space with sectional curvature $\kappa$ is given by $$V_\kappa(r)=\mathbf{c}_{n-1} \int_0^r \left(\frac{\sinh(\sqrt{\kappa} t)}{\sqrt{\kappa}}\right)^{n-1} \, \mathrm{d}t, $$ where $\mathbf{c}_{n-1}:=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the $n-1$-dimensional area of a unit sphere in $\mathbb{R}^n$ (see Chavel (2006), Riemannian Geometry: A Modern Introduction, equation (III.4.1) with notation from (III.3.10) and (II.5.8)).

Answer 2

To confirm the OP's conjectures: A hyperbolic circle of hyperbolic radius $r$ has circumference $2\pi R \sinh \frac rR$ (where $R$ is the absolute length). Consequently, in hyperbolic $(n + 1)$-space an $n$-sphere of radius $r$ is isometric to a Euclidean $n$-sphere of Euclidean radius $R \sinh \frac rR$. Particularly, the area of a sphere of hyperbolic radius $r$ in hyperbolic $3$-space is $$ 4\pi R^2\sinh^{2} \frac rR = 2\pi R^2\bigl(\cosh (\frac {2r} R) - 1\bigr), $$ and the volume of a hyperbolic $3$-ball of radius $r$ is $$ \int_{0}^{r} 2\pi R^2\bigl(\cosh(\frac {2t} R) - 1\bigr)\, dt = \pi R^2\bigl(R\sinh(\frac {2r} R) - 2r\bigr). $$

Answer 3

It is easy to see that the volume of a sphere of radius $s$ in ${\mathbb H}^n$ (constant curvature -1) is given by $\operatorname{vol}(S^{n-1}) \sinh^{n-1}(s)$, where $S^{n-1}$ is a unit sphere in Euclidean space of dimension $n$. To see this observe that in Minkowski space, with $x^2$ given by $x_1^2 + \cdots x_n^2 - t^2$, the path $x_1 = \sinh(s)$, $t=\cosh(s)$ is a unit speed hyperbolic geodesic. Thus a hyperbolic sphere of radius $s$ centered at $(0,\ldots,0,1)$ projects isometrically to a sphere of radius $\sinh(s)$ in $(x_1,\ldots,x_n)$ coordinates, and in this subspace the Euclidean and Minkowski metrics agree. To get the volume of a ball of radius $s$, integrate with respect to $s$.