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3

The entropy of a black hole is given by the area $A$ of its event horizon according to the formula $S=\frac{kA}{4l_P}$ where $k$ is Boltzmann's constant and $l_P$ is the Planck length. For a rotating black hole with mass $M$ and a Kerr parameter $a$ the area is $A=\frac{8\pi G^2}{c^4}M(M+\sqrt{M^2-a^2})$. This is largest when $a=0$ corresponding to the ...

2

The integral carries units of $[momentum]^{3N}[space]^{3N}$ which is exactley the same as $1/h^{3N}$, so this is the factor you need for making the phase space volume dimensionless. I don't understand why you say that it has to be $[momentum]^{3N-1}$, just look directly at the integral. Furthermore, and maybe this is the main problem, the phase space VOLUME ...

1

I found this question by chance yesterday while looking for articles on Werhl entropy. I may have found a possible answer after reviewing properties of the Wigner quasiprobability distribution on http://en.wikipedia.org/wiki/Wigner_quasiprobability_distribution#The_Wigner.E2.80.93Weyl_transformation. Consider property 7 under the section "Mathematical ...

2

Because of the following. Disorder is usually equated with one's ignorance of the system - the less you know about the outcome of the random variable the more disordered it is. If the system turns out to be in a very unlikely state with low $p(x)$ you will naturally consider yourself to have been more ignorant than when it is in a state you consider very ...

5

Firstly, the logarithm needn't necessarily be to base 2. Changing the base just introduces a (scale) factor, so log10, log2 and ln are all equally useful. Log2 is convenient for people working with binary systems. Let's deconstruct the formula. I will define entropy to be $H = E[-\log(p)]$. You can see that this will reduce to a weighted average which ...

0

Anyway, consider the following two examples. (1) A random variable always has a single, definite value, i.e., it isn't really random. Then the Shannon entropy is zero. This means that you don't gain any information by being told that my puppy is cute -- puppies are always cute, with probability 1. (2) Suppose a variable is equally likely to take on $n$ ...

2

Even though the answer you chose is very good I will add my POV Take a box of gas particles. At t=0, the distribution of particles is homogeneous. There is a small probability that at t=1, all particles go to the left side of the box. In this case, entropy is decreasing. Take the statistical mechanics definition of entropy: where kB is the ...

-1

Think of entropy as a steady-state quantity related to system dynamics: Wait for a 'long time', smear out the phase space trajectories and measure the resulting volume. This means that even if all gas particles ended up on the left side of the box (unlikely, but not impossible and realized by perfectly valid microstates), entropy only would have decreased ...

4

Let me begin with the second question where you don't change the dimensionality, just the volume. The entropy never decreases when you actually compress gas. The compression means that the walls are mostly moving against the colliding molecules which means that they're recoiled backwards at higher velocities. The molecules' kinetic energy increases so they ...

0

Statistical physics doesn't tell you that entropy will increase all the time. Just that it will increase on average. The maximum-entropy state is the one with the largest number of microstates. This doesn't prevent you from observing an odd state every once in a while -- even one with very low probability -- in fact fluctuations of the state do happen and ...

7

Right, there is a small probability that the entropy will decrease. But for the decrease by $-|\Delta S|$, the probability is of the order $\exp(-|\Delta S| / k)$, exponentially small, where $k$ is (in the SI units) the tiny Boltzmann constant. So whenever $|\Delta S|$ is macroscopically large, something like one joule per kelvin, the probability of the ...

1

The entropy decreases to zero eventually. It is harder and harder to do as the number of particles is kept fixed. Said differently, the system does not like having its entropy decreased and as a consequence, the pressure increases. To answer to the edited version, if the temperature is kept fixed, a real system will crystalize and the pressure will ...

1

In the context of quantum mechanics, the entropy of a system whose initial state is given by a density matrix $\rho(0)$ is given by the so-called von Neumann entropy; $$S_\mathrm{vn}(\rho) = -k\,\mathrm{tr}(\rho\ln\rho)$$ For an isolated system, quantum mechanical time evolution is unitary; for each time $t$, there is a unitary operator $U(t)$ such that ...

0

Although summarized as an objection of macroscopic irreversibility when microscopic laws are reversible, Loschmidt's objection originally points that there has to be something breaking the time reversal symmetry in Boltzmann's derivation of the $H$-theorem. I think that Boltzmann's answer was to say that high $H$ states (in absence of external driving) are ...

1

I think most people would say the paradox is resolved - but, as the answers to this question make clear, they wouldn't necessarily agree about who resolved it or what precisely the resolution is. For my money the paradox was elegantly resolved by Edwin Jaynes in this 1965 paper. In Jaynes' argument, the symmetry is broken by the fact that we, as ...

0

How do we know the universe is expanding and the speed of light isn't slowing instead (thanks to innisfree for the idea of where to look.) From wikipedia By the 1990s and on into the twenty-first century, a number of falsifying observations have shown that "tired light" hypotheses are not viable explanations for cosmological redshifts.[2] For ...

0

Your example has a tricky issue involving angular momentum (see below), but I can address the spirit of the question using a much simpler example. Let us imagine we have a chamber containing two gases, $A$ and $B$, such that $A$-$A$ interaction terms are equal to both the $B$-$B$ interaction terms and the $A$-$B$ interaction terms. (It doesn't hurt to ...

1

"Absolute Hot" is a cute phrase, but meaningful only in a limited context, where the concept of negative temperature applies. That is, there's a finite number of energy levels in a finite energy range, and whatever particles/quanta/excitations we're studying stay in that context. For example we may be interested in spins of nuclei in a crystal, and how ...

4

Entropy isn't a force that causes things to happen. But anyway, the answer is no. Not all matter in the universe is expected to eventually collapse into black holes. See Adams and Laughlin, http://arxiv.org/abs/astro-ph/9701131 , section VD. Note also that black holes eventually evaporate, so when matter collapses into black holes, the result is that it ...

-3

At the beginning there was a single waveform expanding to the present , very high temperature /energy, and no space ie. one single wave state- low entrophy.

1

I'll explain how the entropy is calculated in fluid dynamics. For ordinary fluid flows the assumption of local equilibrium is valid. That it is assumed that the fluid can be diveded into many small volumes each of which being at equilibrium with say temperature $T$ and density $\rho$. This model also suggests that the specific entropy $s$ [J/(K kg)] is given ...

1

The connection between disorder and entropy is more complicated than that; it is made in statistical physics on the level of a microscopic model of the system, not macroscopic model. Entropy is a more general concept, applicable even if there is no microscopic model at hand, but then the connection to disorder is not applicable. Flowing fluid is not in a ...

1

I apologize "basics foundations of thermodynamics" still does not make a lot of sense to me. Steve B already provided some answer associated to one way to interprete the word "foundation" that is from statistical mechanics. I will kinda play here devil's advocate and assume that you are refering to axiomatic thermodynamics. As far as I am concerned, the ...

0

Yes, it can be proved. But the mathematical rigour only applies to ideal gas, so you might feel disillusioned after all the calculations, considering how much this can be valid in reality. If you insist you want to know, then you may read on. From the first law of thermodynamics, $$\delta Q = \frac{3}2 nRdT + \frac{nRT}V dV,$$ where internal energy ...

1

There's a group I call "thermodynamic purists" who think that thermodynamics is a self-contained system based on semi-mathematical "axioms". I disagree! I think that thermodynamics is fundamentally a consequence of statistical mechanics, and that this is the best way to think about it and understand it. I acknowledge that reasonable people can differ on ...

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