Q:

Given that in quantum theory information can not be lost, can information be created? That is, can (does) the total information of the universe increase, or is it steady, as in the conservation of matter/energy?

- Graydon Sharpe (age 54)

Ware, MA, USA

- Graydon Sharpe (age 54)

Ware, MA, USA

A:

The conservation of information is derived from quantum field theory via the quantum Liouville theorem. Quantum field theory works both forward and backward in time, so the conservation of entropy (or information) works both ways. If quantum field theory is correct (as it so far seems to be) then information, in the abstract, is neither created nor destroyed. Pure states remain pure states. A probabilistic combination of pure states keeps the same set of probabilities.

This may sound very strange to anyone familiar with the second law of thermodynamics, which says that entropy generally increases and never decreases. How can these claims be consistent?

The general idea is that the entropy described by the second law is the sum of the entropies of many local objects, such as the Earth, the Sun, the radiation in the nearby space, etc. These things keep getting more and more quantum-entangled. Entanglement means that only some of the possible states of one part can accompany particular states of another. Thus the number of states for the collection is not really the product of the numbers for the parts. Therefore the conserved entropy of the collection is not the sum of the entropies of the parts. The growth of the entropies of the parts is canceled by the growing negative entanglement entropy.

In practice, that entanglement entropy has no measurable consequences in ordinary circumstances. If half the possible states of the Earth, all looking very similar to the other half, can only be paired with half the possible states of Jupiter, which look just like the other half, we see no special consequences. The entropy described by the second law is the one we can monitor.

There may be some more interesting consequences of large-scale entanglement for the problem of what happens at the surface of a black hole. There has been a lot of recent discussion of that. The key words for a search would be "black hole firewall". To try to understand the overall picture may also require understanding the effects of cosmic horizons, discussed in some papers by Lenny Susskind and collaborators.

Mike W.

*(published on 09/01/2013)*

Q:

Q1) why black hole's entropy corresponds to its surface area at Schwarzschild radius? (=i.e. larger the surface area, larger the entropy.)
I read that a black hole cannot split into two because the split reduces its entropy(=surface area). However, it can still evaporate and reduce its entropy because the evaporation(radiation) process dumps the entropy into space. Is this true? How/why? the Hawking radiation has much higher entropy than its not yet evaporated black hole? How can you calculate the total entropy of a (small) black hole and the total entropy after it is completely evaporated? can you use Boltzmann formula to compute them?
Q2)I read that a spinning black hole looks asymmetric due to relativistic effect.
Does this mean that an electron also looks asymmetric due to its (intrinsic) spin?
Is relativistic effect applicable to any system/object/particle?

- Anonymous

- Anonymous

A:

For a serious discussion, you might want to look at an article by people who understand this business, e.g. .

I guess we can provide some amateur comments on some of your questions as a bridge to that. The Hawking radiation does have entropy, just like any thermal radiation. For simple equilibrium thermal radiation, one has S=(4/3)E/T, where E is energy, T is temperature, and we've used the same units for energy and temperature, i.e. used 1 for Boltzmann's constant. It's trivial to derive this from the the fact that thermal radiation energy is proportional to T^{4}, and that dS/dE=1/T by definition of T. So as the black hole loses energy dE via Hawking radiation, we have dS_{BH}=-dE/T. If the Hawking radiation were in simple thermal equilibrium, we'd gain dS_{HR}=(4/3)dE/T, for a net gain dS_{TOT}= (1/3)dE/T. I think the actual gain is less, because the Hawking radiation is not going equally in all directions, but only *out* from the horizon, so it doesn't explore the full set of states that equilibrium thermal radiation at that T would. We know the net gain isn't less than zero, because processes which lose S don't happen, and everybody who understands this business tells us that Hawking radiation does happen. If we find the exact result, we'll update this.*

On what "looks asymmetric", I think you're referring to a geometrical appearance. Although all objects obey relativity, an electron has no known intrinsic size, so there's no geometrical shape to look asymmetric. There are some related asymmetries due to spin-orbit coupling for states that have orbital angular momentum. These can be viewed as relativistic effects.

Mike W.

*Aha, see to get on the track of the exact result, which apparently is pretty much along the lines indicated here.

*(published on 09/17/2013)*

Q:

Can two entangled particles remain entangled and exchange information even if they were on opposite sides of a black hole's event horizon?

- David Cairns (age 54)

Roseburg, OR, USA

- David Cairns (age 54)

Roseburg, OR, USA

A:

I think it's not accurate to say that any entangled particles "*exchange information*". If that happened, in the ordinary sense of the words, then there would be faster-than-light communication, and that leads to paradoxes.

But, speaking of paradoxes, there's an active current discussion/debate over exactly the question you raise about entanglement across horizons. The ordinary picture of a black hole horizon, with no drama there thanks to its finite curvature, would have entanglement be maintained from the point of view of the particle crossing the horizon. From the outside point of view nobody ever quite crosses the horizon, so entanglement is also maintained. The assumption that the overall quantum state evolves via a standard unitary operation implies that this entanglement isn't lost even after Hawking evaporation. If I understand correctly, recent analysis () indicates that once most of the black hole evaporates, the outgoing radiation would then be entangled with both the leftover black hole and the earlier radiation. Such dual-entanglement is impossible, however, in anything describable by quantum mechanics. So something we have believed must be false. If you read about "black hole firewalls" (e.g. ) you'll find some of the current disputes.

Mike W.

*(published on 09/24/2013)*

Q:

From Schrödinger's cat, I understand quantum entanglement
I also now that particles have been around for billions of years.
My question is
What is the percentage of mater that is entangle starting from time zero (big bang) to present time
Seems to me that at present time, all matter is entangle.
Is that possible.
Reagards

- Eric Beausejout (age 42)

Montreal, Quebec, Canada

- Eric Beausejout (age 42)

Montreal, Quebec, Canada

A:

That's an excellent deep question, but I can only give a partial answer.

Different things are not yet close to maximally entangled. We discuss how we know that earlier in the thread in which your question has been placed. The basic argument is that the second law of thermodynamics tells us, with great reliability, that the sum of the conventional entropies of things keeps going up. The world would look radically different (and not support life) if that were not true. The only way that fact can be reconciled with the ("unitary") form of quantum field theory, which says that the net entropy of an isolated system doesn't change, is for all the parts to keep getting less isolated. In other words, they keep getting more entangled.

Since we're a long way from reaching maximum entropy, we must also be a long way from reaching maximum entanglement.

Mike W.

*(published on 11/24/2013)*