# Tag Info

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There is a degree to which this is just terminology, but in cosmology a distinction is somtimes made between the Heat Death and the Big Freeze. The Big Freeze is the point at which the universe reaches absolute zero, while the Heat Death is the point at which the entire universe has a constant temperature. These are not necessarily the same thing, because a ...

-1

Yes - think about it, the universe is a closed system, and will experience heat death. Heat death is the result of perpetually increasing entropy, which will exist in any closed system.

0

Some comments on the premises may be useful and answer the question: [1] Unitarity of quantum mechanics prohibits information destruction. What this really means is the von Neumann evolution equation for density matrix $D(t)$ of an isolated system prohibits change in the von Neumann entropy: $$N[D]=\sum_{k,l} -D_{kl} \ln D_{lk}.$$ Very similarly, ...

4

If I recall it correctly, the information sunk into black hole can be considered encoded in the ripples on black hole surface, much like egg impact parameters which could in principle be deciphered (at least partially; even quantum theories give us certain confidence intervals) from shattered egg fragments. Falling objects will necessarily have mass, and ...

7

If you drop an egg then it looks as though the process is irreversible and that the information about the original state of the egg is lost. However this isn't the case. The equations describing how the egg shatters are all time reversible so in principle, if not in practice, we could take the shattered egg and evolve time backwards to reconstruct it. ...

0

The information interpretation of entropy, also known as Shannon entropy, interprets entropy as an increase over time in the amount of information contained in a closed system. This interpretation is appealing to information theorists, including many computer programmers, but it does, on the surface, appear to contradict the quantum mechanics conservation ...

0

Yes, there is a subtlety there. Also notice that in quantum mechanics we have finely resolved energy eigenstates, and in a typical complicated system all degeneracies are broken so that at any given energy there is at most one state ($\Omega = 1$) but most likely there is no state at exactly that energy ($\Omega = 0$). So it would seem that $\log \Omega$ is ...

-2

Entropy argument might not work here, because amount of information is going to be influenced by number of moles, not by arrangement of individual molecules. In the same fashion as 1TB hard-drive is 1TB no matter what is written onto it. However, more precise crystal structures will tell you exact picture stored in those bits. So measuring, essentially, ...

11

Quite simply, no. Water memory doesn't appear to violate any physical laws, and the claims made about it are not well-defined or specific enough to be falsified (e.g. with an entropic argument). It's revealing that while a scientist could be convinced that he's wrong, there's nothing that could change the mind of a homeopath. The best we can do is test the ...

0

The number of microstates is usually so large that we may approximate using a derivative. After all, look at $10^{23}$ molecules in a jar. It is quite easy to think that you are looking at a continuum.

2

it is my understanding that it is a measure of the randomness and chaos of particles in as system There are many different kinds of entropy. "Measure of chaos" is very inaccurate characterization of all of them. Second law of thermodynamics has consequences for changes in macroscopic bodies that can attain thermodynamic equilibrium state and thus be ...

0

The entire problem would disappear if temperature is measured in energy units, which is quite natural. Then we can write simply S=logW, entropy is dimensionless and TS has the dimensionality of energy.

2

Where did the energy you spent go? You gained knowledge about the equation I gave. I'll try a different approach before this question is closed or migrated elsewhere. In the context of your question, the answer must be know (heh... no). Consider the case that you expend all the energy calculating incorrectly. Knowledge is, at least, justified ...

1

Consider the function $e^x$: it is monotonically increasing and yet defined for all negative $x$. Just because something increases monotonically doesn't mean it must reach infinity (or even its maximum value) in a finite amount of time. As a side note, please don't refer to entropy as disorder. It's very common but also very wrong: ...

0

In classical thermodynamics, only changes in entropy ever matter ($dS = \dfrac{dQ}{T}$ for reversible processes), so it is not meaningful (though it may be convenient) to define an absolute entropy. HOWEVER, in statistical mechanics, entropy has a probabilistic interpretation: $S = -k_B\sum_i p_i ln p_i$, where $k_B$ is Boltzmann's constant and $p_i$ is ...

1

This is a good question. In fact, as I'll explain, it leads right to some topics of current research. First, an important fact about entropy: if you look carefully at the proof Kittel gives, it should only be applicable to a situation in which you add degrees of freedom to a small system that is exchanging energy with some heat bath. However, if you ...

0

Entropy is defined in my book as ΔS=Q/T. Already, this is not correct. In general, $$\delta S\ge\delta Q/T\;.$$ In the specific case where the system being heated is at always in thermal equilibrium, then $$\delta S=\delta Q/T$$ So, clearly, $\delta S=\delta Q/T$ can not be taken as the definition of entropy. The definition of entropy is [See ...

0

In my statistical mechanics course last year, we derived that $S \propto \frac{1}{T}$ from the following considerations: Consider two boxes (each with $N_i$ particles, $V_i$ volume, $U_i$ internal energy, and $T_i$ temperature, where $i = 1,2$) separated by a wall through which they can exchange energy (heat). Clearly $\Omega_i \propto U_i$ and $\Omega_i ... 3 You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have$\Delta S_1=\frac{\Delta Q}{T_{lower}}$and$\Delta S_2=\frac{\Delta Q}{T_{higher}}$Assume the$\Delta Q$is the same in each case. The denominator controls the "largeness" of the$\Delta S$. Therefore,$\Delta S_1 > \Delta ...

1

I find that the question here relates directly to the definition of temperature, and I'll give a short version of it. For simplicity let us consider a system generated by two sub-systems, A and B, in thermal contact (meaning they only exchange energy in the form of heat). Let me state that, for $A$ and $B$ in thermal equilibrium, the entropy $S_{A}$ and ...

2

Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature. How does this make intuitive sense? The formula defines entropy change. Since it defines new word not conceived before and thus devoid of sense, it is hard to imagine as having "intuitive sense". If your question ...

0

But when I take it through an adiabatic reversible process to the same state as I did in the irreversible process, You assume this is possible, but nobody has succeeded doing so. If someone did, it would contradict the second law of thermodynamics for the reason you mentioned.

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U is assumed to be a state function of the independent variables S,V,N. Inverting S and U, S now becomes a state function of the independent variables U,V,N.

2

I have contributed this issue with this arXiv:1411.2425 and previous works. I stress that Gibbs entropy is not "defined" but rather "constructed". It is constructed on the expression of thermodynamic forces in the microcanonical ensemble, and is constructed in such a way that it reproduces them always and exactly. The construction is actually unique. ...

2

This is a bit of a negative answer, but consider instead an expectation of some other quantity, such as the energy: $$\langle E \rangle = \sum_i p_i E_i.$$ Now, it's obvious what $E_i$ means - it's the energy of state $i$ - but what does $p_iE_i$ mean? The answer is not very much really - it's the contribution of state $i$ to the expected energy, but it's ...

2

First of all, this is a very good question. Classically black holes have no hair, and so are specified by a handful of charges (mass, angular momentum, etc). Quantum mechanically, they act like thermodynamic systems. They have temperature and a large entropy, and in fact all the laws of thermodynamics have black hole analogues. To explain away the apparent ...

1

Certain values are indeed invalid, but that usually just results in software complaining it could not find a solution. I believe the reason the software refuses to solve, is that there may be multiple solutions to a (s,x) pair. For example tracing the 50% line on this graph from wikicommons: ...

4

You are correct that the entropy of the coffee will decrease while the entropy of the cup increases. However this will not decrease the total entropy of the system. Rather, heat will continue to flow between the two objects until entropy can no longer increase. Let $\Delta S$ mean the change in total entropy as the energy of contents of the two components ...

0

Or formulated otherwise, looking backwards in time, we should expect to see an increased thermodynamic entropy. The paradoxical thing to me is that we seem to assume that in the past entropy was even smaller than today! The following assumes that the description of microscopic motion of the particles of the system is Hamiltonian (your system qualifies ...

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