The Bekenstein bound : Area versus Volume

[Demystifier]

Or is it that you are saying that the infinite cubic lattice is not even an acceptable initial condition? But then is it not too much to ask of partial differential equations (be it the EFEs) to forbid certain initial conditions based on rather global conditions?

A perfect infinite cubic lattice is an acceptable initial condition and it will not collapse. However, a perfect lattice is essentially unique; it cannot be constructed in many different ways and hence cannot contain much information. To put a lot of information into the lattice, you need some (at least small) imperfections.

But imperfect lattice will not be stable. The imperfect lattice will undergone a gravitational collapse, eventually forming black holes or other stable compact objects (stars, planets, ...). These stable compact objects will obey the Bekenstein bound.

 

[Demystifier]

Then what you say is that the all these negligible curvatures, induced by these little masses, somehow cumulate non-vanishingly in order to form a BH.

It is called Jeans instability:
https://en.wikipedia.org/wiki/Jeans_instability

 

[PeterDonis]

According the schwarzschild solution

The solution you are describing is not the Schwarzschild solution, so reasoning based on that solution does not apply to the case you are considering. If you're going to make claims about your solution, you need to first be clear about which solution it is. You can't just wave your hands; you have to actually do the math.

an isolated mass

An infinite lattice of small objects is not "an isolated mass". Such a solution is not asymptotically flat, and that is what "an isolated mass" means.

the ratio m/r is lower than the some Planck length and the metric perturbation becomes "unphysical"

Even leaving out the fact that we don't have a theory of quantum gravity and so you are speculating here in an area where there is no theory to back it up: assuming that the objects were that small, the same tentative speculations about quantum gravity that talk about perturbations smaller than the Planck length being unphysical also say that an object that small cannot store any information. So a lattice of such objects would not meet your requirements.

what you say is that the all these negligible curvatures, induced by these little masses, somehow cumulate non-vanishingly in order to form a BH.

No, I didn't say that. I said that all those small masses add up to a non-negligible global gravitational effect. Whether or not they will collapse to form a BH is a separate question that is irrelevant to refuting your claim. You are claiming that such a lattice is stable--that it can persist indefinitely in a static state, neither expanding nor contracting. When you actually do the math, you find that this claim is false.

is it that you are saying that the infinite cubic lattice is not even an acceptable initial condition?

Yes. To expand on (and somewhat correct) what I said in post #23: there are two possible cases, zero cosmological constant and nonzero cosmological constant.

If the CC is zero, then you can have an infinite lattice, but it can't be motionless, even for an instant. (In post #23 I thought it could, but I was wrong.) An infinite lattice corresponds to either the spatially flat or spatially open FRW solutions, and neither of those have even a single instant of time where they are "motionless" (neither expanding nor contracting). They either start out from a "Big Bang" singularity and expand forever into the infinite future, or (the time reverse of that) they contract forever from the infinite past to a "Big Crunch" singularity. (A finite lattice can be motionless for an instant if the CC is zero, but only for an instant.)

If the CC is nonzero, then you can have a lattice that is motionless, but it can only be finite. (This solution is basically the Einstein static universe, and it is unstable, like a pencil balanced on its point, but as an idealization it does exist.) There is no way to have an infinite lattice with the CC exactly balancing the density of matter to keep the solution static.

To see why the above is true, look at the Friedmann equations:

$$
\frac{\dot{a}^2 + k}{a^2} = \frac{8 \pi \rho + \Lambda}{3}
$$

$$
\frac{\ddot{a}}{a} = \frac{- 4 \pi \left( \rho + 3 p \right) + \Lambda}{3}
$$

"Static" (for more than a single instant) means $\dot{a} = 0$ and $\ddot{a} = 0$, which gives

$$
\frac{k}{a^2} = \frac{8 \pi \rho + \Lambda}{3}
$$

$$
\Lambda = 4 \pi \left( \rho + 3 p \right)
$$

Substitute the second into the first to obtain

$$
\frac{k}{a^2} = 4 \pi \left( \rho + p \right)
$$

Since $\rho + p$ is positive for any ordinary matter and energy, we must have $k$ positive, which means a closed, spatially finite universe.

If, instead, we assume $\Lambda = 0$ and we want our universe to be static for a single instant, then we still have to have $\dot{a} = 0$ for that instant; so at that instant, the first equation would read:

$$
\frac{k}{a^2} = \frac{8 \pi \rho}{3}
$$

If we want an infinite universe, we must have $k \le 0$; but that would require $\rho \le 0$ by the first equation, which is impossible (negative energy density). So an infinite universe can't be static even for an instant if $\Lambda = 0$. A finite universe can, since it has $k > 0$. But for a finite lattice, the second equation with $\Lambda = 0$ says:

$$
\frac{\ddot{a}}{a} = - \frac{4 \pi}{3} \left( \rho + 3 p \right)
$$

For ordinary matter and energy, $\rho + 3p$ is positive, so a finite universe with zero CC that is motionless at a single instant will start contracting; it won't stay motionless.

 

[PeterDonis]

A perfect infinite cubic lattice is an acceptable initial condition and it will not collapse.

I don't think this is correct; see my previous post. What solution do you have in mind for a perfect infinite cubic lattice that is static and stable?

 

[Demystifier]

I don't think this is correct; see my previous post. What solution do you have in mind for a perfect infinite cubic lattice that is static and stable?

You are right that it will collapse. (Actually, I liked your post for correcting my mistake). However, even though such a Universe may collapse, it will not collapse into a black hole. It will collapse into a naked singularity, similar to the Big Bang singularity.

 

[PeterDonis]

even though such a Universe may collapse, it will not collapse into a black hole. It will collapse into a naked singularity, similar to the Big Bang singularity.

Yes, agreed.