Any curved space is approximately Euclidean on small enough scales, so for a sufficiently large characteristic length (significantly larger than the radius of the smallest real stars) the answer is obviously "yes".

But what about more extreme cases? Say, if the characteristic length is 1km? That's large enough that you drop humans into such a universe and they'd survive, but small enough to seriously screw up the geometry of a planet--or a star.

A hyperbolic sphere has much more volume, thus much more mass packed closely together, than a Euclidean sphere of equal radius... but the force of gravity also decreases much more quickly with distance, so it's not obvious that results in equivalent pressures for either equal mass or equal radius; and for a given core pressure, there will be a much larger radiative surface, which will totally screw with the equilibrium temperature and gravity-radiation pressure balance.

So, yeah. Do stars work in strongly curved hyperbolic space?

  • 1
    $\begingroup$ "A hyperbolic sphere has much more volume, thus much more mass packed closely together" - did you want to say "more mass needs to be packed together to achieve the same density"? I.e. the stars will need a lot more fuel to ignite? $\endgroup$ Oct 7, 2021 at 1:53
  • $\begingroup$ Yes, but no. Yes, you need more mass within a given radius to achieve the same density. I have no idea if that means you need a lot more fuel to ignite. $\endgroup$ Oct 7, 2021 at 1:55
  • $\begingroup$ Ok, some say that we're actually living in a hyperbolic universe. Haven't read it, but it makes claims that, within this hypothesis, one doesn't need singular Big Bang or black holes, there's no need for dark matter and the Universe expansion comes natural. Intriguing, if nothing more. $\endgroup$ Oct 7, 2021 at 2:03
  • 1
    $\begingroup$ @AdrianColomitchi The expansion of the universe is a separate concept from the curvature of the universe. For instance you could have a collapsing universe that has hyperbolic curvature. $\endgroup$
    – sphennings
    Oct 7, 2021 at 2:07
  • 1
    $\begingroup$ @sphennings that was my artistic interpretation (read: reproducing it as an interpretative artist) of the claim made on PDF-page 5 in regards with the "no need of Dark Energy to explain the expansion on my particular hyperbolic universe". Based on how much I read from the book, currently know or have the necessary skills to understand it, I can't assess if its right, wrong or debatable. $\endgroup$ Oct 7, 2021 at 2:37

2 Answers 2


I think sphennings is more or less correct, but just wanted to expound on their answer with a few more quantitative comparisons. As a warning this will involve some fairly advanced math in the middle section, but I'll end with a few plots that should be parseable even if you're not well versed in the math.

The Gritty Math

First off, we have to define how exactly gravity is going to work in hyperbolic space-- I assume we want something that works exactly the same as the gravity you're used to in flat space in the limit that curvature goes to zero. Luckily, general relativity provides a pretty straightforward way to calculate this by just assuming that our metric obeys so called hyperspherical coordinates and using linearized gravity in the non-relativistic limit:

$$g_{\mu\nu} = g^{(0)}_{\mu \nu} + g^{(1)}_{\mu \nu}$$


$$ |g^{(0)}_{\mu \nu}| \gg |g^{(1)}_{\mu \nu}| $$


$$ g^{(0)}_{\mu \nu} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & S^2(r) & 0\\ 0 & 0 & 0 &S^2(r)sin^2(\theta) \\ \end{bmatrix} \quad $$


$$ S(r) = \sqrt{|k|}^{-1} sinh(\sqrt{|k|}r)$$

and our coordinates $(t,r,\theta,\phi)$ give the time, distance to origin along the shortest path, and polar and azimuthal coordinates for this path with respect to its intersection at the origin. You have to be a little careful with how you think about $\phi$ and $\theta$ because the standard euclidean definitions get messed up-- traveling north 1 mile and then west 1 mile is no longer the same as traveling west 1 mile and then north 1 mile!

Meanwhile, $k < 0$ has units of length$^{-2}$ and is a measure of how curved the space is-- as $k \to 0$, it becomes Euclidean. As $\sqrt{|k|}^{-1}$ approaches the length scale of physical processes, it will start to affect them in a way that changes them significantly from the Euclidean result. We shall see that as sphennings says, the first major impact this has is on density, not nuclear processes directly.

Now this is fine and dandy but unfortunately, I'm too lazy to deal with all the requisite Christoffel symbols and geodesic equations right now so instead I'm going to wildly wave my hands in the air and say that what we want is the following field equation, which based off a few quick calculations I made seems to be reasonable:

$$\frac{1}{\sqrt{|g^{(0)}|}}\partial_i (\sqrt{|g^{(0)}|}g_{(0)}^{ij}\partial_j \Phi) = 4\pi G \rho$$

Where $G$ is the gravitational constant, $\rho$ is mass density, and $\Phi$ is the gravitational potential that gives rise to gravitational acceleration like so:

$$\mathbf{a} = -\nabla \Phi$$

Why this equation? Well, the operator in the top equation is known as the Laplace-Beltrami operator and is basically an extension of the well known Laplacian operator to funkier geometries. And if you replace it with the Laplacian, you get $\nabla^2 \Phi = 4 \pi G \rho$-- which is exactly the formula for the Newtonian gravitational potential in Euclidean space. Now let's try to use this a bit more, by assuming our solution is spherically symmetric-- then our equations become:

$$\frac{1}{S^2(r)}\partial_r (S^2(r) \partial_r \Phi) = 4\pi G \rho$$ $$a_r = -\partial_r \Phi$$

Note that if $S(r)=r$, this is exactly the same as what we expect from Newtonian mechanics-- and indeed, if the $k\to 0$ limit is taken, this is what $S$ reduces to.

Now, if you haven't already, you should check out this answer of mine, which details the process you can use to find the radius of a planet given it's mass by leveraging the gravitation field equation and hydrostatic equilibrium. I'll spare the gory details, but it's relatively straightforward to show that we can use pretty much the same equations as in that answer, only the $r^2$'s occurring in them should be replaced with $S^2(r)$.

The result is basically what many were speculating-- the hyperbolic curvature of space leads your gaussian spherical surfaces growing extremely quickly in surface area so that the gravitational field is less strong than you would get for the same packing in Euclidean space. In math-ier words, we can show that

$$a_r(r) = -\frac{4\pi G}{S^2(r)} \int_0^{r} S^2(r')\rho(r')dr'$$

so if $S$ is an extremely fast growing function of $r$, the evaluation of it at $r$ will outpace the average of it over the density profile and the gravitational acceleration will go down. If you really like math, this argument could probably be made more formal with some well-justified restrictions on $\rho$ and Jensen's inequality.

Fun Plots

If none of that made any sense to you, fret not-- this section should be more understandable. As I said, the net result is that for a given amount of mass, it's more difficult to pack it densely in hyperbolic space. Since nuclear processes that power fusion require very high densities, this means that a very hyperbolic universe, especially one that's expanding quickly, will have more difficulty forming stars. That being said, I'm neglecting thermal effects here so it's possible that those will play a large role for stars undergoing fusion.

These next few plots are calculated using the program from my other answer modified for hyperbolic space for a planet made of 30% iron, 70% silicates with a mass of $6 \times 10^{24} kg$-- you may recognize this as Earth. For this, I have varied $k$ a bit and you can see how more negative values lead to a smaller yet less dense planet given the same mass. Although keep in mind the planet is only small if you tunnel straight through it-- if you walk along the surface it will still seem big!

enter image description here enter image description here

I would include more extreme plots here but unfortunately my code has been running into numerical gremlins which is why I did a planet density profile instead of a star. However, there should be enough stuff here for you to play around with it on your own if you wish!

  • $\begingroup$ Could you post the modified program adapted for hyperbolic space? $\endgroup$ Oct 8, 2021 at 21:00

My back of envelop wild guess is that in a hyperbolic universe it would be harder to pack things close enough to sustain fusion.

It's not the amount of mass within a given radius that causes hydrogen to fuse it's the likelihood of colliding with another hydrogen atom. In a hyperbolic universe you can fit far more stuff within a given radius without packing it any tighter. This would reduce the likelihood of collisions, and fusion events.

Since you can pack more stuff in a given radius without packing it tighter, if you continue down this barely-scientific vein it would be easier to form a black hole.

I'll leave analyzing the interaction between harder-to-pack-tightness, easier-to-get-closeness and curvature, as an exercise for readers with far more math and physics chops than I have.

  • $\begingroup$ I imagine that a "just slightly hyperbolic" Universe with a ε-smidge departure from a flat curvature (e.g. something that requires light year distances to detect a difference of 1cm from a flat space), won't function so fundamentally different when it comes to the formation of stars. Sort of a matter of the distance scale for which the "locality of an Euclidian approximation" still holds. $\endgroup$ Oct 7, 2021 at 2:48
  • 2
    $\begingroup$ @AdrianColomitchi True. But that wouldn't be an interesting hyperbolic universe. :) $\endgroup$
    – sphennings
    Oct 7, 2021 at 2:52
  • 1
    $\begingroup$ I suspect that wouldn't really be an issue until the characteristic length becomes comparable to the mean free path of a proton in the core--with curvature still at human-survivable scales, nuclear interactions should still see effectively Euclidean geometry. $\endgroup$ Oct 7, 2021 at 3:31
  • $\begingroup$ @LoganR.Kearsley if the curvature impacts gravity, the ability of the star to accrete enough hydrogen and compress it to enough "packness" will be affected as well. $\endgroup$ Oct 7, 2021 at 4:37
  • 2
    $\begingroup$ It would be interesting whether the fact that you can pack more stuff in a smaller space balances with how the force of gravity decays faster in hyperbolic space. I'm assuming the force of gravity at distance $r$ is inverse proportional to the surface area of the sphere of radius $r$. So in Euclidean space the sphere has $r^2$ area and gravity is $1/r^2$. In Hyperbolic space the sphere has $e^{2r}$ area so gravity should be $e^{-2r}$ maybe? $\endgroup$
    – Daron
    Oct 7, 2021 at 12:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .