# Universe Hubble radius equal to Schwarzschild radius

1. Nov 4, 2015

### Gerinski

I have read that the Schwarzschild radius of a black hole with the mass-energy of the observable universe is roughly equal to the actual Hubble radius of 13.8 billion light years. And I have read that contrary to some popular esoteric interpretations such as "the universe is a black hole", "we are inside a black hole" etc, this simply means that the universe is spatially flat, or nearly so, that the equivalence of Hubble radius and Schwarzschild radius for a flat universe is derived from the Friedmann equations.

So far so good, but there are things I do not understand.

As I understand, this means that was the universe not expanding, it would collapse into a black hole, it has already the total average density of a black hole with an event horizon the size of it. But this seems highly counterintuitive, the density of the observable universe seems incredibly thin, it is by far mostly empty space. How can it have the same density as a black hole?

Alright, a big share of its energy contents is dark energy, but even so, how can we then make the equivalence to a black hole? Dark energy may contribute to the total energy density of the universe but it causes it to expand, so it goes against the tendency to collapse gravitationally. It can not be right to include dark energy in the mass-energy computation to say that the mass-energy of the observable universe is equal to that of a black hole the same size, is it?

Related question: in a black hole, the density at the singularity is infinite, whatever mass divided by zero volume. But if we express the density as the mass of the black hole divided by the volume of its Schwarzschild sphere, what would the density of a black hole be like?

Thanks

2. Nov 4, 2015

### bcrowell

Staff Emeritus
This is a non sequitur. There is no logical connection between the average density of a black hole and the density of the object that collapses to form it. The average density you're talking about (mass of the black hole divided by the volume inside its event horizon) is not especially interesting and has no special properties or logical status.

If the universe were not expanding at some moment in time, then the Friedmann equations predict collapse to a Big Crunch singularity, not a black hole singularity. This is what happens in a closed FRW cosmology with zero cosmological constant. No minimum density is required.

Black hole spacetimes have very little in common with cosmological spacetimes. Cosmological spacetimes are homogeneous. Black hole spacetimes aren't.

3. Nov 4, 2015

### Smattering

Can someone please comment on the question whether this is just a coincidence, or it results from some kind of law or principle?

Volume and mass increase with the third power of a sphere's radius, and the Schwarzschild radius is proportional to the mass. Thus, it seems to me that the sphere's Schwarzschild radius should grow faster than the radius of the sphere itself. So it seems at least possible for me that the Hubble radius and the Schwarzschild radius of the observable universe had quite different values in the past.

But then again, this over-simplistic model does not consider expansion of space and much other phenomenons that might play a role.

4. Nov 4, 2015

### bcrowell

Staff Emeritus
The Friedmann equations give $H_0\sim \sqrt{\rho}$ (where the $\sim$ means that we make simplifications such as ignoring like the contribution of pressure to the stress-energy). Since the Hubble radius is $1/H_0$, it follows that the mass inside the observable universe is on the order of the Hubble radius (in geometrical units, where $G=c=1$). This is not a coincidence.

The part about a black hole is irrelevant and misleading, however. It just happens that the radius of a black hole's event horizon is on the order of its mass in geometrical units. There is no physical analogy between the observable universe and a black hole.

5. Nov 5, 2015

### Smattering

O.k., this is very interesting. Ignoring the metric expansion of space, the observable universe's radius should grow by one lightyear per year, right? Assuming that mass is homogeneously distributed on cosmological scales, this implies that the mass should grow faster than the radius of the observable universe (volume of a sphere, etc. pp.).

So when the Friedmann equations imply that the mass stays in the same order as the radius, is this due to the metric expansion of space?

Yes, I am aware that the observable universe is not a black hole. This interpretation makes no sense when assuming that the universe continues homogeneously beyond our hubble volume. After all, what we denote as "obversable universe" or "our hubble volume" should not be different from any other hubble volume. So if our hubble volume was a black hole, then any other hubble volume would also have to be a black hole. And this just makes no sense.

6. Nov 5, 2015

### martinbn

One more thing to keep in mind is that the radius $r =2m$ is not the distance from the horizon to the centre. Nor is the black hole a ball. The spacetime inside the hole is very counter-intuitive, it is not even stationary.

7. Nov 5, 2015

### Smattering

What is the meaning of "stationary" with respect to spacetime?

8. Nov 5, 2015

### Gerinski

Thanks, but it must have some value. It has mass X and event horizon's sphere volume (as measured from outside) Y, so it must be possible to say what X/Y is like for typical black holes. I'm just intrigued if such a value would be counterintuitively small for laymen like me. I know this does not reflect any physical density. The space inside of the black hole's event horizon does not contain any stuff, all of its mass resides at the hypothetical singularity, not in the space enclosed within the event horizon.
But still we can divide X/Y and see what turns out. Perhaps the density of a black hole measured in that way would be (say for example) not much denser than lead? (just guessing).

9. Nov 5, 2015

### martinbn

10. Nov 5, 2015

### Smattering

O.k., I think now I understand what you were referring to by "not stationary".

11. Nov 5, 2015

### bcrowell

Staff Emeritus
That's an interesting paper. I find it easy to believe that we can't meaningfully assign a volume of $(4/3)\pi r^3$ to the interior, since obviously that's a Euclidean formula, and nothing here is Euclidean. What is less obvious to me is why they think the interior volume can and should be characterized as the maximal volume of a spherically symmetric spacelike surface. Even in the Minkowski case, this surprises me. I would have expected that one could have made the interior volume arbitrarily large or arbitrarily small, based on intuition from the fact that a spacelike geodesic neither maximizes nor minimizes length. I guess there is something subtle and (to me) non-obvious that happens when you increase the number of dimensions and require symmetry.

12. Nov 5, 2015

### bcrowell

Staff Emeritus
You can divide the charge of an electron by the S&P 500 index and add Barack Obama's year of birth, and it will have some value. That doesn't mean it's a meaningful thing to consider.

Why don't you just go ahead and calculate it? It's going to depend on the mass of the black hole, which can take on any value.

13. Nov 5, 2015

### Smattering

When they say "long spacelike 3d cylinder" - I guess they are referring to a cylinder with a 3d surface, right?

But if there is there is really so much space inside and the singularity is located at the other end of the cylinder - how long will it take for an infalling object to reach the singularity?

14. Nov 5, 2015

### Gerinski

Of course, and the mass defines its Schwarzschild radius as well. So I will try go ahead and calculate it, it does not seem too difficult a task. Thanks for the big help.

15. Nov 5, 2015

### bcrowell

Staff Emeritus
We can't say that there is "really" that much space inside. The paper makes a somewhat arbitrary definition of the volume. It's not the only possible definition.

The proper time to reach the singularity is quite short -- on the order of r/c, where r is the radius of the event horizon.

16. Nov 6, 2015

### martinbn

A space-like geodesic doesn't minimize or maximize the space-time interval, but if you restrict to a space-like hypersurface, which will have a definite metric, there should be a maximal one i.e. one with maximal length where by length one means the length from the induced metric. (Of course I am vague here. If you consider curves on a non-compact subset, say missing points, there need not be a maximal length.)

17. Nov 6, 2015

### Smattering

O.k. I understand that. But do you have an idea what the authors mean by the following quote:

Why does matter have "newer and newer space" to fall into? And how can it be that the time to reach the singularity stays in the order of r/c if matter has newer and newer space to fall into?

18. Nov 6, 2015

### martinbn

For the first one because space-time is dynamic so space changes with time. For the second, because the singularity is like a moment of time, no matter how much new space you get Sunday is coming in two days.

19. Nov 6, 2015

### Smattering

I can understand that in the sense that mass bends space-time and the curvature of space-time changes the movement of masses which results in yet a different curvature of space-time and so on ...

But on the other hand, I would have thought that a singularity has already maximum density so that there cannot be any further movement of masses--at least not in the Schwarzschild case where the singularity is located in a vacuum. And if the distribution of mass (or energy) stays constant, why does the curvature of space-time keep changing?

But isn't the singularity also like a location in space?

Best regards,
Robert

20. Nov 6, 2015

### Staff: Mentor

No. The singularity is spacelike, not timelike. A "location in space" would be described by a timelike curve in spacetime.