I How certain are we about black hole mass density?

[PeterDonis]

OK, you've made up quite a new definition of vacuum (we both now its not, there is at least a singularity, and there is mass anyway)

The singularity is not part of the manifold.

The complete spacetime of a realistic black hole does include a non-vacuum region (occupied by the object that originally collapsed to form the hole). But that does not mean that non-vacuum region is "still inside" the hole forever. I am not changing the definition of "vacuum" at all.

we surely can measure the change of the horizon by shooting a laser and seeing if it hit a detector somewhere or elsewhere

This is not a local measurement. It's a global one.

We just can compute that all the particle of the universe seems to be at the center of a singularity 13.7 by old.

You're missing the point. The spacetime geometry of the universe as a whole is not the Schwarzschild geometry. It's not even close to that geometry. The Schwarzschild geometry is static outside the horizon and asymptotically flat--i.e., the metric goes to Minkowski at infinity. The spacetime geometry of the universe as a whole is static nowhere and has no "infinity" at all, let alone an asymptotically flat one. The observable universe is a portion of the universe as a whole, but also is static nowhere and has no "infinity". So any numerical similarity between some computation you make about the universe and some computation you make about the Schwarzschild geometry is physically meaningless. It's like saying the point on Earth where the prime meridian meets the equator is "somehow the same" as the origin of a Euclidean plane, because they both happen to have coordinates $(0, 0)$.

 

[PeterDonis]

I think I heard once Susskind say something like that (kind of: that's like if we are inside a BH)

Yet another reason for me not to like Susskind.

 

[Boing3000]

This is not a local measurement. It's a global one.

OK. The question become, does this global measurement will detect a change inside that shell ? The answer seems to be no.

You're missing the point. The spacetime geometry of the universe as a whole is not the Schwarzschild geometry. It's not even close to that geometry.

I didn't miss that point because your explanations are perfectly clear.

It's like saying the point on Earth where the prime meridian meets the equator is "somehow the same" as the origin of a Euclidean plane, because they both happen to have coordinates $(0, 0)$.

More like having a formula to compute the surface of the oceans, and obtaining, by coincidence, the age of the ocean. My only point is that is is bizarre.

Actually it is more like having a formula to compute the surface of the oceans in some metric, and obtaining the volume of the ocean in another one. Totally unrelated, not even the same units. But not that bizarre, given the obvious (but missing) link (average depth).

 

[Ibix]

Actually it is more like having a formula to compute the surface of the oceans in some metric, and obtaining the volume of the ocean in another one.

In geometric units, the Schwarzschild radius is 2M. Taking your word for the fact that this is the same as the age of the universe, T, you are saying 2M=T. Either this is just a coincidence (we just happen to be around at the time when T=2M, which happens at some point for a very large range of possible relationships between M and T), or this holds for all times (i.e., the mass of the observable universe is directly proportional to its age). I'm not sure, but I do not believe the latter is correct.

 

[Boing3000]

In geometric units, the Schwarzschild radius is 2M. Taking your word for the fact that this is the same as the age of the universe, T, you are saying 2M=T.

I may very well have made a mistake I used this formula $r_s= \frac{2GM}{c^2}$. That gave me this code that you could check in your browser javascript console (F12)

(function(c,G,yearS,solarMass){ console.log("Mass:"+(((Math.pow(c,2) * ((13.7e+9)*yearS*c)) / (2*G))/1/*solarMass*/).toExponential()+"kg");})(299792458, 6.674e-11, 365 * 24 * 60 * 60, 1.98847e+30)

Either this is just a coincidence (we just happen to be around at the time when T=2M, which happens at some point for a very large range of possible relationships between M and T), or this holds for all times (i.e., the mass of the observable universe is directly proportional to its age). I'm not sure, but I do not believe the latter is correct.

I don't either, that's why I am surprised.
I have heard that we are a a somewhat special time in the universe that make the expansion rate very detectable (it won't be anymore in the far future). Maybe that's the reason.

 

[PeterDonis]

does this global measurement will detect a change inside that shell ?

For the case you described, no, by the shell theorem.