# How did the big bang ever stop being a black hole?

The title says it all. With all that mass in such a small space it must have been one, but then everything would have to stay inside it. If it was still a black hole we'd have a closed universe but nobody believes that anymore. Does this mean the whole universe started out as Hawking radiation or what?

Chalnoth
The title says it all. With all that mass in such a small space it must have been one, but then everything would have to stay inside it. If it was still a black hole we'd have a closed universe but nobody believes that anymore. Does this mean the whole universe started out as Hawking radiation or what?

Well, the thing that makes it not a black hole is the distribution of mass: the mass in our universe was always distributed rather smoothly. And the equations of motion for such a universe simply aren't the equations of motion for a black hole.

So was it a black hole during the first few nanoseconds when all that mass was packed into a space the size of a golf ball?

Chalnoth
So was it a black hole during the first few nanoseconds when all that mass was packed into a space the size of a golf ball?
That's actually irrelevant. If you compare the mass of the current universe that lies within one Hubble distance ($c/H_0$), you get a Schwarzschild radius that is the same Hubble distance (and if you include more of the observable universe, the Schwarzschild radius gets larger faster). If our universe were a "black hole" then, it would still be one now.

Chalnoth, I thought the universe was a black hole interior solution, but with the clock running backwards--so a white hole.

Chalnoth
Chalnoth, I thought the universe was a black hole interior solution, but with the clock running backwards--so a white hole.
I don't think that's accurate. It's sort of similar in some respects, but the primary difficulty that I can see with it is entropy. The entropy of our universe has been increasing continuously. The entropy of a white hole decreases in time (which is also a statement that a white hole is unphysical).

Yes, well, I suppose I should think less and compare metrics more. Although the energy rather than the entropy would determine the metric, it seems.

The solution for the whole Universe is simply different from a solutions of White/Black hole.

The main difference is the momentum matter has in the expanding universe. Gravity depends not only on mass!

I think that this is a very key question.

In the standard model, the very early universe is far more dense and massive than any supermassive black hole, in fact all of them put toether and more. So how does matter overcome the gravitational pull to get beyond the Schwarzchild radius? Do we need matter to exceed the speed of light to do this? Would relativistic mass make it more difficult for matter to ever escape?

It seems that we need another type of inflation energy to make inflation happen otherwise we would be left with a permanent enormous black hole until this energy was provided in some way?

sas3
Gold Member
I heard a theory that before the forces broke apart the inflation speed exceeded the speed of light.

I remember hearing or reading that somewhere but I do not remember where or when.

I think that this is a very key question.
In the standard model, the very early universe is far more dense and massive than any supermassive black hole, in fact all of them put toether and more. So how does matter overcome the gravitational pull to get beyond the Schwarzchild radius? Do we need matter to exceed the speed of light to do this? Would relativistic mass make it more difficult for matter to ever escape?

Again, Schwarzchild radius is calculated based on the assumption that mass does not have significant momentum. In early Universe this assumption is not correct. This is why the GR solution for the whole universe is different.

Dmitry, this is probably something that I am just going to have to accept. One thing that I learned about black holes is that nothing, not even light itself (zero rest mass mass traveling at the speed of light), could ever escape from a black hole.

zsawaf
nicksauce
Homework Helper
The black hole solution is static. The expanding universe solution is not static. Comparing apples and oranges here, people.

1oldman2
Nicksauce, probably I am getting stuck trying to imagine one psuedo infinite overcoming another psuedo infinite!

nicksauce
Homework Helper
In the standard model, the very early universe is far more dense and massive than any supermassive black hole, in fact all of them put toether and more. So how does matter overcome the gravitational pull to get beyond the Schwarzchild radius? Do we need matter to exceed the speed of light to do this? Would relativistic mass make it more difficult for matter to ever escape?

By assuming the existence of a Schwarzchild radius, you have already made an error. The Schwarzchild radius is a feature of the static (or stationary) black hole solution. An expanding universe solution has no such feature.

By assuming the existence of a Schwarzchild radius, you have already made an error. The Schwarzchild radius is a feature of the static (or stationary) black hole solution. An expanding universe solution has no such feature.

Why is this? The matter is still momentarily at least the size of a singularity with pseudo infinite density. Sorry I am not being argumentative, I just dont understand.

nicksauce
Homework Helper
Okay, this isn't the correct way to think about it, but it might help you out anyway.

The event horizon isn't dependent on density, but rather the mass to radius ratio. We need
$$M/R = \frac{c^2}{2G}$$

Now let's see what we get for M/R for our universe. Note that $$M/R = \rho\,R^2$$ We can estimate R as the Hubble radius
$$R \sim \frac{c}{H}$$. From the Friedmann equations, in a radiation dominated universe, we have $$H=H_0(\sqrt{\Omega_R}a^{-2})$$, so $$R = \frac{ca^2}{H_0\sqrt{\Omega_R}}$$. Meanwhile, $$\rho=\rho_c\Omega_ra^{-4}$$, so we get $$\rho\,R^2 = \frac{\rho_cc^2}{H_0^2}$$. Putting in the value of $$\rho_c$$, we get $$M/R = \frac{3c^2}{8\pi\,G}$$

And since 3/8pi < 1/2, we never get the condition, for the Schwarzchild radius.

Because in GR it is not the mass that is creating gravity. You always think just about 'how much mass you have in some volume'.

PhilKravitz
It seems to me there are two seperate questions
1) black hole inside the universe
2) the whole universe as one big black hole

On the first the thing is the matter/energy is so uniformly distributed that even though a sphere may contain enough stuff to make the escape velocity at the surface be greater than c the stuff outside the sphere contributes an equal and opposite force and the net force is near zero (no black hole).

On the second well I guess we have two cases the universe is open and the universe is closed.

If open then a photon can travel arbitrarily far from some reference point so it does not seem like a black hole. It seems to me for that to work there would need to be matter/energy beyond the observable universe. This gets confusing to me.

If closed then a photon can only go around in circles. Do we want to call this a black hole interior? Darned if I know.

Hi nicksauce,

I know what m, r, c, G, rho (and even pi) are, but what are H, H0, Omega, a and rho-c?

Chalnoth
Hi nicksauce,

I know what m, r, c, G, rho (and even pi) are, but what are H, H0, Omega, a and rho-c?

First, $a$ is the scale factor of the universe. By convention, typically $a=1$ is defined as now. So $a=0.5$ would be when galaxies in the universe were, on average, half as far apart as they are now. The expansion rate $H$ is then defined as:

$$H(t) = {1 \over a}{da \over dt}$$

For nearby galaxies, where we can neglect the fact that $H$ changes over time, this definition of $H$ gives it the property that the recession velocity of a galaxy is simply given by $v = Hd$, with $d$ being the distance to the galaxy. The current Hubble expansion rate is then defined as $H_0$.

$\rho_c$ is the amount of matter/energy density, for a given expansion rate, that is required to give a universe with flat space. It is defined as:

$$\rho_c = {3 H^2 \over 8 \pi G}$$

Lastly, $\Omega$ is a given matter or energy density of the universe divided by $\rho_c$.

People are saying I should include other energies besides mass. Fair enough, but that would only seem to make the problem worse.

I also don't understand why momentum means we can't talk about a Schwarzchild radius. The stress-energy tensor is gonna be bigger, so does this mean its other elements are actually opposing the effect of the 0,0 element?

I believe pressure contributes to black hole collapse, so we'd be relying on the off-diagonal elements to save us. Is it something to do with a determinant? I never understood those field equations anyway, but it seems to me that if they were right we wouldn't need dark matter.

The reason the early universe wasn't a black hole and didn't become one was because of the distribution of matter/energy/pressure. It was uniform throuought all of space and therefore there was no concentration to create an interior event horizon. It's very much like the way there is no gravity at the center of the earth.

The early universe may have packed everything into a golf ball but there's nothing outside the golf ball. No concentration, no black hole.

Hi Antiphon,

That makes sense actually. I suppose we're talking about a closed universe in the sense that you'd end up where you started if you kept going in the same direction, except that by the time you got back more space would have appeared in between, and you'd need to exceed the speed of light to catch up. Right?

This argument sounds quite different from either of nicksauce's though.