## Thursday, April 29, 2010 ... /////

### The analogy between the Universe and a black hole

A Spanish translation is available!

Sean Carroll is irritated by the analogies between the black holes and our Universe - and by the claims about the depth of such comparisons. Instead, he claims that such analogies are cheaper than a cup of grande latte at Starbucks (by 4 dollars or so):

The Universe is Not a Black Hole (Cosmic Variance)
Well, while I agree with most of the detailed technical statements by Carroll, I do think that the relationship is deep and probably hides some insights - and intuition - that we haven't yet fully mastered and that may uncover some additional mysteries of gravity (and quantum gravity).

A black hole is defined as a region whose interior is separated by the event horizon from the exterior (because of the curvature caused by a sufficient mass, according to the rules of general relativity): a black hole is an object from which it's causally impossible to ever return to the exterior world which is why not even light is allowed to ever escape from the black hole again.

Such a definition may cover the whole visible Universe, too. After all, some very distant galaxies are receding away from us faster than the speed of light.

Note that such a "faster than light" behavior doesn't contradict the laws of special - and general - relativity because the speed limitation only applies locally, as long as a small enough region of space is parameterized by coordinates that look Minkowskian. To describe cosmology, however, we need to choose some arbitrary coordinates to represent the whole (curved) spacetime and it's clear that arbitrary "coordinate velocities" can't be limited by any universal speed limits.

So according to this simple argument, we may be living inside a black hole. This notion is amplified by a simple quantitative check. Let us calculate the "mass of the visible Universe". Because the spatial sections are pretty much flat, we may consider the visible Universe to be a ball of radius "R": as you learned when you were a kid, its volume is "(4/3).π.R^3".

The total density of the Universe is the critical density, "rho = 3 H^2 / 8πG" (which includes dark energy and dark matter). That's the density needed to preserve the spatial flatness which we apparently observe (at least with a surprisingly good accuracy). Multiplying the volume and the density, we obtain the mass
M = (4/3).(3/8).R^3 . H^2 / G = R^3.H^2 / (2 G).
Now, in "c=1" units, you may take the radius of the Universe to be the inverse Hubble parameter, "R = 1/H" i.e. "H = 1/R" (because at distance "R", the speed among galaxies from the expansion, "H.R", reaches the speed of light - and it's a homework for you to figure out whether this relationship remains linear for huge "R"), and you will obtain
M = R/2G   ⇒   R = 2GM
which is exactly the formula for the radius of a Schwarzschild black hole! So the analogy even works kind of quantitatively. However, the agreement in the numerical constant was somewhat coincidental (while the order-of-magnitude agreement was guaranteed by the dimensional analysis based on the correct "qualitative picture") and there are also differences that prevent you from using the analogy to achieve many goals you may have.

For example, if we live in the interior of a black hole, we should be ultimately destroyed by a singularity in the middle. However, it's very unlikely that our Universe will ever evolve to such a singularity. Instead, because of the influence of the cosmological constant, it will approach an ever more empty and ever more accurate de Sitter space; this spacetime has no singularity in the future.

The "regular future" we expect is just one reason why it turns out to be more meaningful to rebuild the analogy and claim that our visible Universe is actually the space *outside* a black hole - and the black hole interior is *behind* the cosmic horizon!

This new picture is strengthened by the fact that you may write our cosmology as a Schwarzschild-like solution where the "g_{tt}" time-time component of the metric, one that determines the redshift, goes to zero at the event horizon. In these pretty natural coordinates, "g_{tt}" is positive inside the visible Universe - much like it is positive outside a black hole (in the "+–––" convention).

So it is the inaccessible region behind the cosmic horizon that is analogous to the problematic black hole interior. In particular, if you believe black hole complementarity - i.e. the notion that the black hole interior contains just some "reshuffled" information that already exists outside the black hole (and that the operators outside and inside a black hole don't quite commute because of that) - then it is the whole infinite Universe behind the cosmic horizon (where we can't see) whose life is just a "boring" rearrangement of the quantum bits that we observe inside the visible patch.

As Carroll correctly emphasizes, there is one more crucial difference between the black hole event horizon and the cosmic horizon of our Universe: the black hole event horizon has an "objectively defined location" that all observers outside the black hole may agree upon. On the other hand, our cosmic horizon is just ours and distant observers would see their cosmic horizon elsewhere.

In fact, this difference is nothing else than the difference we may have already noticed: if the visible Universe "is" the exterior region of a generalized black hole, then this exterior region has a finite volume, unlike the case of the regular "localized" black holes that "float" in an infinite external space.

Let's describe this difference a bit quantitatively. Write the geometry of the visible patch of the Universe in the Schwarzschild-like way. Your coordinates will include "R", a radial one. It's small in the visible Universe which should be called the "exterior" in the black-hole analogy so it's pretty natural to do a spherical inversion and replace "R" by "Я = 1/R" (if you don't know Cyrillics, "Я" is pronounced "Ya"). The middle of the visible Universe corresponds to "R = 0" i.e. "Я = infinity".

That's great because in these "Я" coordinates, the exterior of the would-be black hole has the same large "Я" as the "R" of the exterior regions of conventional black holes. However, there's still the difference. While the conventional black holes contain an infinite volume near "R = infinity", our would-be black hole constructed out of the visible Universe has "Я = infinity" which only represents one point in space - namely us at "R = 0"!

So there are many other coordinates - obtained by spherical inversions around different origins of coordinates - where the locus "Я_2 = infinity" actually represents some extraterrestrial aliens who are as evil imperialists as Stephen Hawking determined, or "Я_3 = infinity" which denotes the home of the E.T. aliens who are compassionate conservatives instead. ;-)

Because the exterior region of the would-be black hole constructed out of the visible Universe has a finite volume, there's not enough space to "safely fix" the location of the event horizon. That's also why this event horizon turns out to be observer-dependent. It has many other negative consequences for the calculability of quantities in such a cosmic situation.

For example, it may be extremely difficult to fully determine the microstates that describe the states of the visible Universe only. On the other hand, recall that conventional black holes may be embedded in flat or anti de Sitter spacetimes which makes it possible to describe their microstates completely accurately (in string theory).

Another example of a difference between the situations is that a conventional black hole shrinks as it emits the Hawking radiation: the radiation is bringing some energy (and mass) away. The (de Sitter) Universe-as-a-black-hole doesn't shrink because the emitted thermal radiation that I will discuss below (radiation from the cosmic horizon to the black hole exterior, i.e. to the visible Universe inside) is reabsorbed by the black hole after some time. That's clearly a consequence of the finite volume of the "black hole exterior" in the case of the Universe, too: the "black hole" lies in all directions so the Hawking quanta don't have a choice and they ultimately return to such a "black hole".

However, you should also appreciate - more than Sean Carroll - that these differences do not invalidate many other conclusions that you may obtain from the analogy. For example, conventional black holes emit Hawking's thermal radiation: it's radiated from the cosmic horizon to the external region. And as I have already leaked, it's also the case that some thermal radiation is emitted by the cosmic horizon to the "external region" of the would-be "black hole" - which happens to be inside the visible patch of the Universe.

So we're immersed in the thermal sea of radiation that's coming from the cosmic horizon - it's emitted by the alien celestial bodies that are so far that we couldn't have observed them so far, not even in principle.

Well, the radiation coming from the cosmic horizon is not necessarily "quite thermal": some E.T. aliens who just became accessible yesterday could have sent us some real "signals" and not just "noise". Except that such signals were uncalculable - we didn't know anything about the aliens - so we may assume that the signals are actually thermal. Such a thermal Ansatz is just a guess but there's some sense in which it's the "best thing" we can do.

This expectation is getting increasingly accurate as the Universe behind the cosmic horizon is approaching the empty de Sitter space. (Because the cosmological constant is positive, the distance to the cosmic horizon will never exceed a certain bound: if the C.C. were zero, the radius of the visible Universe could grow indefinitely.)

That's another point where the analogy between the visible Universe and the black hole exterior becomes very useful for an overall qualitative picture. We may talk about the "generic states" of the black hole. For normal black holes, it's the black hole that has completely stabilized (by emitting the quasinormal modes) and became static. When it occurs, we may imagine that the black hole interior is completely empty and the black hole entropy is maximized.

(As the fuzzball pictures emphasize, the empty black hole interior only emerges after you average over many "generic" black hole microstates.)

Analogously, the entropy of the "Universe as a black hole" is maximized when the de Sitter space becomes empty. That's actually the point at which the surface of the cosmic horizon is maximized, too. (Any matter inside the de Sitter space - i.e. a black hole with its own event horizon - shrinks the area of the outermost de Sitter horizon.) And it's also the state in which you expect the black hole interior - i.e. the invisible Universe behind the horizon - to become empty.

So some calculations are impossible to "export" from one context to the other because of various qualitative differences (which are quite universally rooted in the "compactness" of the exterior region of the "black hole" in the cosmic case).

But there are also many arguments that are unaffected by the differences. In these contexts, it becomes very useful to keep the relatively close analogy in mind. At the end, the analogy may boil down to the shared "horizons" as coordinate singularities allowed by general relativity, or even to the common description by "general relativity" (at this point, the analogy becomes pretty vague), but it's just wrong to deny that there are many deep links between the two pictures.

And that's the memo.

#### snail feedback (8) :

Wow. So someone - you, as it happens - answered a question I had archived for 30 years now, when I pondered about the Hubble constant and concluded (extrapolating...) that stuff might be going over the speed of light at the limit; but that being there, the limit would be elsewhere... and so on? I asked that question to a physics prof (thermodynamics, I'm afraid) and he said I might have a point, but he couldn't know.

A second unanswered question I still have is, are we to expect relativistic effects in an ultracentrifuge? These things produce about a half a million Gs centrifugal force. The answer I got was - no, there's no referential change (?)

Dear Baco, good to be solving some problems.

Concerning the ultracentrifuges, what matters for special relativistic effects is the speed, not the acceleration. Ultracentrifuges don't have more than 100 rotations per second so it's just hundreds of meters per second, or millionths of the speed of light.

So the relativistic effects go like v^2/c^2 which is "parts per trillion" and are too small to be measured. But they're still real. Rotating reference frames are not inertial.

So for example, the proper circumference of the rotating ultracentrifuge will no longer be 2.pi.R where R is the proper radius. It will be shorter because of Lorentz contraction that only acts on the "longitudinal" directions along the motion (circumference) but not the "transverse" directions that are orthogonal to motion (the radial direction).

So the circumference will genuinely be by 10^{-12} of the radius shorter than you would expect from Circ=2.pi.R.

Best wishes
Lubos

"Rotating reference frames are not inertial."

Ah! That's that! Thanks a lot! I suppose then (allow me...) that the relativistic acceleration stuff only works with gravitational fields?

Dear Baco,

in special relativity, only inertial frames are allowed to formulate the laws of physics in the same simple way: rotating frames are not.

In general relativity, you may use any frame or system of coordinate - curved, rotating, accelerating - but you must allow arbitrary gravitational fields that may partly "undo" the acceleration etc.

So you may look at the centrifuge from the rotating viewpoint but you can no longer assume that the space is flat in that system. Instead, it will have the same intrinsic geometry - and e.g. the modified circumference/radius ratio - that you may derive from the inertial frame, too. ;-)

Cheers
LM

"but you must allow arbitrary gravitational fields ....

"it will have the same intrinsic geometry you may derive from the inertial frame...

Ah! Lubos, thanks a lot! I knew they hadn't told me the full story :-)

Oh, just a note - it doesn't change a thing - but ultracentrifuges do 70.000 - 80.000 rpm (they did 15 years ago, that is).

Regards.

If there is a singularity in our past then we must still be inside it's Schwarzschild radius, no? The Big Bang did not blast us out of the black hole around the primordial singularity, that is impossible isn't it? If so doesn't that mean the Big Bang just created the space we see now and we are seeing it all from the inside?

Dear pat, unfortunately, I find your questions a bit ill-defined.

What you mean by "its [not it's] Schwarzschild radius"? The mass of the Universe is not conserved during the cosmological evolution - only the mass of dust is conserved.

The mass stored in radiation is decreasing (as 1/R) while the mass stored in the dark energy is increasing (as R^3), and so on.

But even if they were not changing, I don't know how you would make the conclusions about the black hole. The initial mass, after the Big Bang singularity, was given a lot of outgoing "momentum".

More generally, your reasoning rooted in "Schwarzschild radius" etc. seem to be designed for a static geometry. But the Universe, in its evolution, was very far from static: this time dependence is really what the word "cosmology" in modern times, so the naive computations that are good for static solutions can't be immediately imported to the evolving Universe.

Best wishes
Lubos

Black holes are the result of believing that bodies move by gravitational geodesic. The first and second array element of the Schwarzschild metric are the mathematical inverse of each other. But if they were equal, then, even independently of the possible zeros and infinite, light cones never be closed. There would be no event horizons or black holes.

Connected theory, the only alternative to Einstein's general relativity, is the solution to the problems of theoretical physics today. See Bubok.com: