Is Bekenstein entropy limit inconsistent with universal continuity?

Question

It is unknown whether the universe is discrete or continuous in its intricate quantum level structure.

See for example: Can universal continuity be experimentally falsified?

How could spacetime become discretised at the Planck scale?

Even so, all branches of modern physics rely heavily on fully continuous structures.

From the Bekenstein bound applied to a black hole, we know that the Information entropy that can be contained inside a black hole is finite and proportional to the surface area of the event horizon.

From the No hair theorem/conjecture, it is believed that the black hole is uniquely described by mass/energy, linear and angular momentum, position, and electric charge, which amounts to a total of 11 real numbers. Possibly, if magnetic monopoles exist, we can add an additional number for magnetic charge.

Most physicists will argue that these 11 numbers are continuous (i.e. not bounded rational approximations).

With an assumption of real continuity, as the black hole undergoes change, for example taking on additional mass over a period of time, the 11 numbers will change as time flows over a continuous infinitude of real numbers, with no smallest increment of time. Each of the 11 numbers then, must assume values that are rational, irrational, transcendental, non-computable and non-definable, as they continuously sweep through the real number field.

In fact, if any number is sampled at random, i.e. at a random time, it will almost surely (i.e with probability one) be non-computable and non-definable. A non-computable and non-definable number has infinite Kolmogorov complexity and carries infinite entropy, as its shortest description is its own random and infinite digit sequence. How is that consistent with the starting assumption of bounded entropy?

Does this really have anything to do with black holes? There are plenty of other things in physics that are continuous (including any wavefunction), but which are considered to have finite entropy - this argument would apply equally well to all of them. — N. Virgo, Oct 06 '15 at 07:15
True. At the time of writing, it seemed the argument is easier/clearer for black holes, but it could be made for any sphere. — Halfdan Faber, Oct 06 '15 at 15:48
I would like to just point out here that the behavior of reality is not limited in any way by our ability to define it with mathematics. I see no problem with a physical property taking any real value, even if we could never write that number down with the decimal system. — Asher, Oct 07 '15 at 02:59
Well, since continuous analysis has been so successful, physicists have come to believe that the math is the same as the physics, when perhaps the math really just is the emergent approximate limit of an underlying completely discrete and countable relational structure. — Halfdan Faber, Oct 08 '15 at 05:33
something related to that is discussed here http://www1.maths.leeds.ac.uk/~pmt6sbc/docs/davis.myth.pdf — , Jan 06 '16 at 01:47
Your assertion that "A non-computable and non-definable number has infinite Kolmogorov complexity and carries infinite entropy" is simply false. It might carry entropy in the sense of information theory and encoding, but it certainly doesn't carry entropy in the sense of thermodynamics - that e.g. a thermodynamics system with temperature $z\ \mathrm{K}$ for $z$ some non-definable number should have infinite entropy because of that temperature is patently absurd. — ACuriousMind, Jan 11 '16 at 14:25
Everything is stated in the context of information entropy. Particularly the Bekenstein bound can be stated as an upper limit on bits contained within a sphere. — Halfdan Faber, Jan 12 '16 at 19:58

Peter Shor · Answer 1 · 2016-02-14T06:39:11.517

Your argument completely fails to take quantum mechanics into account.

Consider an electron spin. If you measure the spin of an electron using a standard measurement corresponding to an observable, you can only get two answers, $|v\rangle$ and $|-v\rangle$, where $v$ is some direction in three-dimensional space. And yet, you can prepare the electron so its spin is in any of a continuum of directions (any direction in 3-dimensional space), and you can measure the spin of an electron along any axis in 3-dimensional space.

Further, the von Neumann entropy of the electron spin is one bit, despite the fact that it can be pointing any direction of space. And if you encode information in the spin of an electron, there is a theorem that you can only retrieve one bit.

Now, let's see how your arguments apply to the spin of an electron, which is a very well-understood phenomenon.

From the Bekenstein bound applied to a black hole, we know that the Information entropy that can be contained inside a black hole is finite and proportional to the surface area of the event horizon.

We know that the Information entropy that can be contained in the spin of an electron is finite and equal to one bit.

the black hole is uniquely described by mass/energy, linear and angular momentum, position, and electric charge, which amounts to a total of 11 real numbers.

The spin of an electron is uniquely described by a 3-dimensional unit vector, which can be parameterized by 3 real numbers the sum of whose squares is one – so two real parameters.

How is that consistent with the starting assumption of bounded entropy?

How can two real numbers be characterized by one bit?

In fact, if any number is sampled at random, i.e. at a random time, it will almost surely (i.e with probability one) be non-computable and non-definable.

It's not clear whether quantum gravity lets you measure quantities with arbitrary accuracy. You certainly cannot measure the spin of an electron with arbitrary accuracy.

What your argument fails to take into account is that quantum information theory is different from classical information theory, and is extremely non-inutitive.

We don't know how to quantize space-time, but we can say that your argument that there must be a contradiction somewhere is not consistent with what we know about quantum mechanics.

score 0 · Answer 2 · answered Oct 06 '15 at 06:49

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From the Bekenstein bound applied to a black hole, we know that the Information entropy that can be contained inside a black hole is finite and proportional to the surface area of the event horizon.

And if you have astrophysical sources of collapse, then an external observer can always choose a coordinate system where the black hole has never formed and the where no event horizon has formed yet. And hence where there is no inside. So it is your choice to have them or not. And hence no physical conclusions can be drawn.

From the No hair theorem/conjecture it is believed that the black hole is uniquely described by mass/energy, linear and angular momentum, position, and electric charge, which amounts to a total of 11 real numbers.

The no hair theorem is about solutions that real solutions come closer and closer to approximating as time progresses.

With an assumption of real continuity, as the black hole undergoes change, for example taking on additional mass over a period of time,

Which is a coordinate v system artefact. You can choose one where it never forms. And as long as one person can do that, no physical conclusion can be drawn from assuming it does form.

In fact, if any number is sampled at random, i.e. at a random time,

And that is non physical as well. All you can do is interact with something based on a the material available, you don't have the ability to invoke random times. Sure if something isn't generic you shouldn't expect to be able to see it in a controlled setting. But you can't just a sum E you can do things if you can't actually do them.

How is that consistent with the starting assumption of bounded entropy?

Entropy is about knowledge, and groupings of possible states consistent with coarse graining knowledge. How is that related in the slightest?

answered Oct 06 '15 at 06:49

Timaeus

25,523

If physics gets to use all the tools from continuous analysis, then you get to use AC also, and whether you like it or not you will have non-measurable sets and non-computable numbers, and every time you sample a value it will be non-computable. Of course it is unphysical and the actual universe doesn't behave like this. But that is what you get if you insist spacetime is actually a differentiable 4-manifold, rather than just conveniently modelled as such. – Halfdan Faber Oct 08 '15 at 05:26
@HalfdanFaber I disagree that you can sample the way you claim you can sample. Even if the universe were continuous, you are just assuming you can sample in a some indeterminate way without actually saying how you would do it. And I called you on it. Physics can use continuous results to approximate discrete events in some situations and to some level of accuracy, but that's doesn't magically allow us to do things we can't do, such as make magic samples. – Timaeus Oct 08 '15 at 05:31
I complete agree. If physics is merely using continuous approximations to model discrete events or structures, then you cannot sample in any meaningful way, at least in the way I am suggesting. However, in current physics, spacetime as a differentiable 4-manifold is not considered to be an approximation. – Halfdan Faber Oct 08 '15 at 05:49
@HalfdanFaber Even if the universe were continuous, you still wouldn't be able to sample in the way you assume. A controlled experiment requires a finite specification of the experiment, and each one gets the results it gets, which is a magically kind of sampling. You have to design a method of sampling, because we don't have magic ways to sample. And the designs are limited. – Timaeus Oct 08 '15 at 09:15
Of course my argument is that the universe is entirely discrete, because an assumption of continuity leads to contradictions. – Halfdan Faber Oct 08 '15 at 17:19
However, if you insist physical spacetime is isomorphic with the continuum then you can't cherry-pick what operations/conclusions you allow. There will be continuum-many designs. Even so, there is no need to actually perform a sampling. An existence argument is sufficient. Time will flow through continuum-many non-computable reals in any arbitrarily small segment, and in those continuum-many states entropy will be infinite. You could try to resort to the uncertainty principle but that won't help (even though the UP may be a manifestation of lazy evaluation in the context of bounded precision). – Halfdan Faber Oct 08 '15 at 17:19
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@HalfdanFaber You argument is specious and fallacious. Even if the universe is continuous, you only have a discrete set of finite descriptions of experiments and that is what you can use to collect data. You don't have magic powers. And isomorphisms require morphisms. But experimentally we don't have continuum many anything, so there isn't even an invertible map, let alone a structure preserving such map. There aren't continuum many experimental designs because you don't have magic powers. There aren't continuum many descriptions. Your concept of time isn't what you measure in an experiment. – Timaeus Oct 08 '15 at 17:29
Actually, I am unsure what your argument is. In a nutshell, are you saying the universe is continuous, but we cannot measure it, nor reason about it? – Halfdan Faber Oct 08 '15 at 17:57
@HalfdanFaber In science you make models to get predictions and you make observations and experiments to get data and you compare them. When you compare them, the only things that can be compared are the corresponding parts that can be actually done with both theory and experiment. And you can't use magic to do either. You are stuck with finite descriptions of both because that is what you can manipulate. But mostly I was trying to criticize you for claiming you can do things you can't. – Timaeus Oct 08 '15 at 18:07
@HalfdanFaber Your goal should be to make models that can be compared with doable observations and experiments and I can't tell if you are even trying to do that. And when you claim you can do something you can't do that is almost the worst thing possible, worse would be lying about what you observed. You can make continuous models without using the axiom of choice. You can use continuous models and still only use them to talk about things that correspond to finite precision finite describable experiments. – Timaeus Oct 08 '15 at 18:10
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@Timeaus, OK, I see. You may find this surprising, but I agree with your line of thought. Thanks for clarifying. I think you are saying that you remain agnostic about the physical inner structure, but choose to employ a well-working continuous model and then restrict probing to what you can actually measure. I think that is a consistent approach. Thanks for your answer. – Halfdan Faber Oct 08 '15 at 18:20

score -3 · Answer 3 · answered Jan 11 '16 at 14:20

Is Bekenstein entropy limit inconsistent with universal continuity?

Yes, but that doesn't mean the Bekenstein bound is correct and everything is fine. Entropy can be considered as "sameness", related to available energy. And the expression $S \leq \frac{2 \pi k R E}{\hbar c}$ has a c in it, but the coordinate speed of light at the event horizon is zero.

It is unknown whether the universe is discrete or continuous in its intricate quantum level structure.

A photon has its E=hf quantum nature, but that doesn't mean it approaches you in steps. It's quantum field theory, and the wave nature of matter. IMHO there's no evidence at all for a universe that has some "discrete intricate quantum level structure".

Even so, all branches of modern physics rely heavily on fully continuous structures.

I'd say the universe appears to be continuous.

From the Bekenstein bound applied to a black hole, we know that the information entropy that can be contained inside a black hole is finite and proportional to the surface area of the event horizon.

We don't actually know that, it's hypothetical.

From the No hair theorem/conjecture

Which is also hypothetical.

it is believed that the black hole is uniquely described by mass/energy, linear and angular momentum, position, and electric charge, which amounts to a total of 11 real numbers. Possibly, if magnetic monopoles exist, we can add an additional number for magnetic charge.

There's potential issues with the angular momentum and charge and magnetic monopoles. But I'll go with the flow.

Most physicists will argue that these 11 numbers are continuous (i.e. not bounded rational approximations).

Note that a particle such as an electron has unit charge. But again, I'll go with the flow.

With an assumption of real continuity, as the black hole undergoes change, for example taking on additional mass over a period of time, the 11 numbers will change as time flows over a continuous infinitude of real numbers, with no smallest increment of time.

Fair enough.

The 11 numbers must then each assume values that are rational, irrational, transcendental, non-computable and non-definable, as they continuously sweep through the real number field.

OK.

In fact, if any number is sampled at random, i.e. at a random time, it will almost surely (i.e with probability one) be non-computable and non-definable. A non-computable and non-definable number has infinite Kolmogorov complexity and carries infinite entropy, as its shortest description is its own random and infinite digit sequence. How is that consistent with the starting assumption of bounded entropy?

The number does not actually exist. A bl;ack hole exists, and a photon exists. This photon can fall into the black hole increasing its mass/energy by E=hf, and E can take any value. But we can't say that a black hole consists of n photons or is anything to do with statistical mechanics or information theory. Note the black hole information paradox: "The black hole information paradox[1] is an observational phenomenon that results from the combination of quantum mechanics and general relativity which suggests that physical information could permanently disappear in a black hole, allowing many physical states to devolve into the same state". For all you know the photon totally loses its identity, and the black hole is like one big boson, where everything is the same. For all you know the black hole could be something like a BEC, and subject to a Bosenova.

Sorry this doesn't give you anything definite, but so much of this stuff is hypothetical.

Is Bekenstein entropy limit inconsistent with universal continuity?

3 Answers3