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Bekenstein-Hawking formula for entropy of a black hole tells us that information content in a black hole is proportional to its area which is in fact proportional to the mass^2 of the black hole. The information content before the formation of the black hole can be different which has nothing to do with the mass. Is there any kind of information loss during the formation so that we get two black holes with an equal amount of information. Is the information due to the fields lost in the process? So we get just the bare information which comes from the mass.

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    $\begingroup$ its area which is in fact proportional to the mass of the black hole No, the area is proportional to the square of the mass. $\endgroup$ – Ben Crowell May 19 at 14:24
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Is there any kind of information loss during the formation so that we get two black holes with an equal amount of information.

If I'm understanding correctly, this sentence is describing two different sets of initial conditions leading to two black holes with the same area, and therefore the same entropy.

You're using "information" as if it were a synonym for "entropy," but they aren't the same thing. For example, information is always conserved in quantum mechanics, in the sense that the time evolution is unitary, but entropy increases. Entropy can be interpreted as a measure of the amount of information in a system, but only when we take into account the coarse-graining of the phase space.

Gravitational collapse to a black hole doesn't involve a loss of entropy, it involves a gain in entropy. It's not particularly mysterious per se for two different initial systems, with two different amounts of entropy, to evolve into final states with the same amount of entropy. For example, suppose we take two different samples of an ideal gas, in two different nonequilibrium states, but both with the same number of molecules and both confined to the same volume. They will both gain entropy, and will both end up in the same maximum-entropy macrostate.

We don't have a theory of quantum gravity, but the rough picture is probably that after the black hole has formed, there is a large amount of energy present in its microscopic degrees of freedom, and this is where the entropy resides.

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  • $\begingroup$ Still I can't understand how the entropy is only dependent on the mass. Two sets of masses can have vastly different initial entropy. After collapse into a BH it's same for both which is only dependent on the mass, that's hard for me to grasp. $\endgroup$ – Crashing_brain May 20 at 23:38
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Infomation Content in Black Holes

Your question is essentially (if i interpret it right) can stars with differing sets of initial detailed information--and thus differing information entropy-- converge to identical black holes?

The short answer is Yes --provided the no hair hypothesis is correct.

The longer answer has to take account of the measurement process applied to a Black Hole. By definition, measurements cannot be performed by an external observer to elements within the interior of a black hole, beyond the event horizon. We thus do our best to characterise the Black Hole by partitioning the event horizon of the Black Hole with elements each of minimal area (following Loop Quantum Gravity) proportional to the square of the Planck length $l_P^2$ and let N be the total finite number of partitions.

The ‘no hair’ hypothesis states that a Black Hole can be completely characterised by its classical spin, mass and total electric charge. Thus there is no preferred location on the event horizon, so that each partition element must have the same weighting. The von Neumann 'entanglement ' entropy of this partition is thus given by;

$- \sum\limits_1^N {\frac{1}{N}\log \left( {\frac{1}{N}} \right)} = \log N = \log \frac{S}{{l_p^2}} = \log \frac{{{c^3}S}}{{\rlap{--} hG}}$

with S the surface area of the black hole, which is similar to the Beckenstein-Hawking formula. This relates to classical relativity (ignoring the cosmological constant) via the macro- level Clausius definition of entropy E;

$\oint {\frac{{\delta Q}}{T}} = \Delta E$

where, integrated over a Carnot cycle, a change in the energy tensor results in a change in entropy corresponding to a change in space-time curvature as measured by the Ricci Tensor.

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  • $\begingroup$ You're making an argument that the entropy of the black hole is proportional to the log of its area, but the entropy isn't proportional to the log of the area, it's proportional to the area. $\endgroup$ – Ben Crowell May 19 at 18:46
  • $\begingroup$ quantum (von Neumann or entanglement) entropy also has a log function by analogy $\endgroup$ – Prof. J May 19 at 19:25
  • $\begingroup$ classical entropy is k logV (Boltzmann) the log function makes it additive across independent configuration spaces $\endgroup$ – Prof. J May 19 at 19:31
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    $\begingroup$ quantum (von Neumann or entanglement) entropy also has a log function by analogy Here you seem to be referring to the initial assumption of your sketch of a calculation, but I'm talking about its final result. classical entropy is k logV (Boltzmann) the log function makes it additive across independent configuration spaces This is a reasonable thing to expect, and is the result of your calculation, but it's not the behavior of the Bekenstein-Hawking entropy. $\endgroup$ – Ben Crowell May 19 at 20:13
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    $\begingroup$ There may be something very interesting going on here that is relevant to the OP's question, but I don't think the combination of this answer and the paper does a very effective job of communicating that -- certainly not at the level of the OP's question. The paper appears to be a specialized technical result, and although I'm sure you're right that it has implications for black hole entropy, that isn't evident to me as a non-specialist. The paper never uses the words "black hole." $\endgroup$ – Ben Crowell May 20 at 22:59

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