Our universe's entropy/size vs Bekenstein's theory

[Bob Walance]

Jacob Bekenstein asserts that the entropy of a black hole is proportional to its area rather than its volume. Wow.

After watching Leonard Susskind's video 'The World as a Hologram', it seems to me that he's implying that we are all black hole stuff. Perhaps we (our galaxies and their black holes) are on the inside of a parent black hole, or maybe the totality is just one super-sized two-dimensional surface made up of whatevers (strings?).

My question to you all is this:
For the stuff that makes up the visible part of our universe - or even including the stuff that's past our 'cosmic horizon' - is there any evidence that our universe's entropy and size jibe with Bekenstein's formulas?

If you're interested, here's a link to Susskind's video. At 18:24 is a discussion about Bekenstein and his entropy-area idea.


Bob

 

[Chronos]

A finite universe is a non-starter for me. Ir creates bigger problems that any it may resolve. Embedding our universe in an endless collage of multiverses smacks of 'turtles all the way down'.

 

[PeterDonis]

For the stuff that makes up the visible part of our universe - or even including the stuff that's past our 'cosmic horizon' - is there any evidence that our universe's entropy and size jibe with Bekenstein's formulas?

Our cosmological horizon has an entropy for basically the same reason a black hole horizon has one; the formulas are basically the same (the entropy is the log of the number of Planck areas in the horizon area). Note that the formula for the cosmological horizon is derived using de Sitter spacetime; our universe is not precisely de Sitter, but now that it is dark energy dominated de Sitter is a reasonable approximation for this purpose.

The entropy we can actually "count" in our observable universe is much, much less than the Bekenstein entropy associated with our cosmological horizon. The Bekenstein entropy is an upper bound, not an exact formula.

 

[kimbyd]

Jacob Bekenstein asserts that the entropy of a black hole is proportional to its area rather than its volume. Wow.

Yup. This is accurate. The basic physical reasoning is that the area of its horizon is the only physical geometry-related quantity that a black hole can actually have. Its volume doesn't work because there's no way to measure it: the singularity at its center prevents this.

A finite universe is a non-starter for me. Ir creates bigger problems that any it may resolve. Embedding our universe in an endless collage of multiverses smacks of 'turtles all the way down'.

I don't think that's a coherent argument. Heck, the argument works even if you don't like the many worlds interpretation.

The finite universe is ideal for a number of reasons, most particularly that it offers a neat solution to the measure problem which makes it otherwise basically impossible to compute probabilities in an expanding, infinite universe. One way to understand why it is such a neat solution to the problem is this:

If you assume that our universe has a non-zero cosmological constant, then it has a cosmological horizon. There are a finite (though very, very large) possible number of degrees of freedom within that horizon. If you imagined moving far away, to another location within the universe, that location would have the same cosmological horizon, with the same exact possible number of degrees of freedom.

Furthermore, in quantum theory, if we have a physical process which allows the universe to explore all of those degrees of freedom, then that means that the physical system inside one quantum horizon will necessarily describe every possible configuration of the universe within that horizon. Since every region has the same possible configurations, they all describe the same possible configurations, such that adding a second region has added no new configuration states that haven't already been considered when examining the first.

What all of this means is that if you consider the volume inside one cosmological horizon using the many worlds interpretation, you have already considered the entire possible set of configurations of the entire universe. Adding nearby regions is superfluous.

And if you don't like the many worlds interpretation, it shouldn't be hard to see how this way of looking at the universe is absolutely identical to examining the universe as infinite with only one quantum state in any given volume, while counting each specific configuration once (that is, if any two volumes repeat, only one is counted). The states that are actually examined using this more classical interpretation are completely identical.

The construction described above, then, is going to be a valid way of examining the universe provided the following hold true:
1) The cosmological constant is a fundamental, positive, non-zero constant (note: this can be relaxed to the cosmological constant taking a finite number of positive values and the same basic result holds, though with larger numbers).
2) The laws of physics allow a complete exploration of the degrees of freedom available. This is equivalent to stating that the physical process which sets the initial conditions for the universe is homogeneous.

Nothing else is required. You don't need the many worlds interpretation at all: just consider the "single horizon" calculation to be a mathematical representation which provides a neat way of avoiding the double-counting of identical regions.

 

[PeterDonis]

If you imagined moving far away, to another location within the universe, that location would have the same cosmological horizon

I don't think this is true. In de Sitter spacetime, different "locations in space" have different cosmological horizons.

 

[kimbyd]

I don't think this is true. In de Sitter spacetime, different "locations in space" have different cosmological horizons.

If the cosmological constant is indeed a constant, then it would have the exact same shape, the exact same behavior, and result in the exact same number of degrees of freedom within the horizon.

 

[PeterDonis]

If the cosmological constant is indeed a constant, then it would have the exact same shape, the exact same behavior, and result in the exact same number of degrees of freedom within the horizon.

Yes, agreed. But the horizons centered on different points would not be "the same" horizon, as in "the same surface in 4-d spacetime". They would just have the same area and the same entropy.

 

[kimbyd]

Yes, agreed. But the horizons centered on different points would not be "the same" horizon, as in "the same surface in 4-d spacetime". They would just have the same area and the same entropy.

Sure. But if you have a full quantum universe described by the many worlds interpretation of quantum mechanics, the same area and the same entropy (combined with a homogeneous process for setting the initial conditions) result in that other place actually being completely identical when viewing the wavefunction of the universe as a whole. This is because all of the possible consequences of those homogeneous realizations are actually realized within the full wavefunction. The identical horizon area and maximal entropy actually do imply that it's the same exact system.

If you'd prefer to have an interpretation of quantum mechanics where wave function collapse is real, then the two volumes are indeed likely to be different. But if the universe is infinite, then every possible realization of those same initial conditions will happen somewhere. The many worlds picture of this system is a way of collapsing all of those possible realizations into just one volume. You can see that as a mathematical trick if you like, but the "finite universe" picture where only one volume within a cosmological horizon is examined is identical to an infinite universe with duplicate realizations removed.

 

[PeterDonis]

the "finite universe" picture where only one volume within a cosmological horizon is examined is identical to an infinite universe with duplicate realizations removed.

Ah, I see. Yes, I agree that as far as predictions go these are the same thing.