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Derive Canonical Ensemble from Maximum Entropy Principle

A good way to start the proof is to first select some basis such that the density matrix $$\rho=\sum_{j}{p_j|\psi_j\rangle \langle\psi_j|}$$ where $p_j>0$ and $\{|\psi_j\rangle\}$ is orthonormal. ...
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5 votes

Calculating dark energy

Theories and models in physics do not have a preferred mathematical base. The physical constants and properties of substances in a model can be written out in any base - in the same way as constants ...
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1 vote

Gravitational path integral derivation of black hole temperature and entropy

You raise an excellent question in regards to how such thermodynamic quantities should be interpreted. That is, it is reasonable to ask why we attribute them directly to the black hole when we're ...
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-1 votes

What would "break" in reality if I had a perpetual motion device?

If you could have devices having 100% or even more than 100% energy efficiency , then you can easily break the first as well as second law of thermodynamics and cool your ice-cream while sitting ...
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  • 44
2 votes

Is it possible to express a density matrix of a system in a base of stationary eigenstates?

You appear to be asking if all state matrices are diagonal in the eigenbasis of an arbitrary Hamiltonian, and the answer is trivially no, that is a subset. For instance, a qubit under the Hamiltonian $...
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0 votes

Energy dissipated by friction and entropy

It would be correct to assume that the entropy increase of the universe after the block has stopped is: $$\Delta S = \frac{K}{T}$$ In response to my query, you indicated that the "ambient ...
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1 vote

Does the holographic principle imply a finite universe?

"while the actual entropy would scale cubically (with volume)(right ?)" The Bekenstien bound that gives the entropy limit to any given volume of space is specifically starting that as you go ...
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3 votes
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Proving that reduced density matrices $\rho_{\mathrm{out}}$ and $\rho_{\mathrm{in}}$ have the same eigenvalues

Here's one way to do it. Without loss of generality, I will assume $N\leq M$, and that both sets of eigenvalues are ordered from largest to smallest; i.e. $\alpha_1$ and $\beta_1$ are the largest ...
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1 vote

Unitary evolution and von Neumann entropy

Another neat method to prove this: write the von Neumann entropy as a limit of the Renyi entropies: $$ S[\rho] = \lim_{n \to 1} S^{(n)}[\rho] = \lim_{n \to 1} \frac{1}{1-n} \log \text{Tr} \rho^n $$ ...
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What is the interpretation of $dE - TdS \leq 0$?

What is the interpretation of $dE - TdS \leq 0$? If I understand correctly this is the Helmholtz free energy: $$ dF = dE - TdS $$ No. The above expression for $dF$ is wrong. The correct differential ...
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What is the interpretation of $dE - TdS \leq 0$?

I believe it's useful to remember the first law of thermodynamics, $dE = dQ + dW$, which tells us that heat Q and work W are both forms of energy (you can increase the internal energy E of you hands ...
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5 votes
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Unitary evolution and von Neumann entropy

Hint: Use the spectral decomposition to write $$\rho(0) := \sum\limits_k \lambda_k \,|k\rangle \langle k| \tag{1} ,$$ and then find an expression for $\rho(t)$ in terms of $\lambda_k$. Especially ...
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0 votes

What's the significance of a quasi-static process?

If a system initially at equilibrium in state 1 is moved to state 2, then it will not be necessarily in equilibrium in this latter state, so one will have to wait for some time that it relaxes to ...
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What's the significance of a quasi-static process?

A quasi static process is an (ideal) process where the system remains in internal equilibrium at all time, but not in equilibrium with the external environment. Internal equilibrium implies that ...
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2 votes

A question about the potential energy of ordered marbles in free space

Now imagine a huge collection of randomly positioned small massive marbles in a volume of free space, making the entropy of the collection a maximum. If the configuration is precisely defined, we are ...
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2 votes

Does an ordered abacus have a higher mass (no matter how small) than a random one?

So, if I invest energy in creating an ordered pattern on an abacus, it will have a different mass than a random one. No The energy invested will be used as work to change the potential energy of each ...
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4 votes

Does an ordered abacus have a higher mass (no matter how small) than a random one?

The answer to your question is "almost certainly not", for 3 reasons: (1) First, and probably most importantly, the paper you reference is highly speculative. We don't really understand the ...
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1 vote
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Intuition behind entropy and its differentiation

This is a development of my comment below the original question. I think that the calculation is wrong, even though the final result is correct. For a monoatomic ideal gaz: $$dU=\frac{3}{2}\,nR\,dT$$ ...
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0 votes

Energy dissipated by friction and entropy

It would be correct to assume that the entropy increase of the universe after the block has stopped is: $$\Delta S = \frac{K}{T}$$ This would be correct only provided the temperature of the "...
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1 vote

Energy dissipated by friction and entropy

If the initial temperature and the final temperature of the block are both T, then applying the complete version of the first law of thermodynamics to the block gives $Q=-K$, where Q is the amount of ...
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3 votes
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Entropy of mixing formula

Herbert B. Callen - Thermodynamics And An Introduction To Thermostatistics-Wiley (1985) has two proofs, one more theoretical and the other one is a simple thought experiment. The first start at page ...
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2 votes
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Increasing Entropy Principle Proof

You are misunderstanding the derivation of non-decrease of entropy of an "isolated system" when its subsystem undergoes a cyclic process. The wrong assumption here is that the "red"...
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1 vote

Increasing Entropy Principle Proof

Frankly, I can't make sense of the equations at the end of your post and how you reached the conclusion that $\Delta S_{universe}\le0$. Consider the following: For any combination of processes, in ...
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1 vote

Increasing Entropy Principle Proof

If I invert the processes and perform first a reversible process and then an irreversible process the result is opposite: $\Delta S_u\leq 0$. This is pretty much the point: if we could reverse the ...
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Why is the change in energy between products and reactants equal to the heat of reaction?

At every point in the quasistatic chemical reaction the system is in equilibrium with its environment, that is the Gibbs free energy takes minimum value, i.e., $$ dG = \sum_i\mu_idN_i=0 $$ see ...
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What is the meaning of an entropy current?

we can define the energy and entropy currents What you defined are the rates of change ($\dot U$, $\dot S$). In the case of energy, which is conserved, the rate of change must correspond to the net ...
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2 votes
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Natural log introduced in microstates derivative with respect to energy in equilibrium equation

Recall the chain rule: $$ \left(f\circ g \right)' =\left(f'\circ g\right) \cdot g',\tag{1} $$ where $f\circ g\equiv f\left(g\left(x\right)\right)$. In your case, $f(\cdots)\leftarrow\ln(\cdots)$ and $...
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3 votes

What is an intuitive explanation for $T = \mathrm{const}$ when $\Omega(E) = e^E$?

The basic intuition here is that temperature is not about number of microstates as such. Rather, it is about how the number of microstates varies with the energy---the standard definition of ...
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3 votes
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What is an intuitive explanation for $T = \mathrm{const}$ when $\Omega(E) = e^E$?

I'd say the relevant quantity here is $\omega (E):=\ln\,(\Omega (E)) = - \ln \,(1/\Omega(E))$. You can view $\omega$ as a measure of information or rather of missing information, in the sense of how ...
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2 votes
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Confusion regarding the total number of microstates of a $N$ particle system

Note: it would be better to indicate the sum of the total microstates as $\Omega=\sum_{\{n_i\}}^* W{\{n_i\}}$, where $^*$ indicates that the sum is performed over the sets $\{n_i\}$ that satisfy the ...
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Why can't $pV$ work generate entropy in a reservoir?

Why can't $pV$ work generate entropy in a reservoir? The short answer is it can, if the $pV$ work is irreversible. Reversible work neither generates nor transfers entropy. So what it boils down to is ...
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Question about my understanding of Clausius inequality and changes in entropy

For a cyclic process on a closed system, $\Delta S=\oint{\frac{\delta Q_{rev}}{T_{syst,rev}}}=0$. So, for a cyclic process on a closed system, the Clausius inequality tells us that $$\Delta S=0\ge \...
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1 vote

Landauer's Extended Principle

You can erase information without any heat emission if you randomize degrees of freedom in e.g. a computer memory or a set of spins initially in the same direction. You certainly need some mass-energy ...
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Question about my understanding of Clausius inequality and changes in entropy

You mention an expression of the second law that holds only for systems going through a cyclic transformation (due to the $\oint$ notation). Since you also mention the universe, which has no reason to ...
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0 votes

Does a book get lighter if you rearrange the letters?

Well if you compress the more ordered book into code and write the code in a new book. Yes, the new book will be lighter. The original book will not be lighter because although you have ordered the ...
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1 vote
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Using quasistatic processes to calculate quantities

The piston is now released and after some oscillations, comes to rest in a final equilibrium situation corresponding to a larger volume of the gas. Does the entropy increase? Yes. Because the process ...
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Extracting Work From Mixing Of Ideal Gases?

Mixing of ideal gases is just two expansion processes into the same final volume. Replace the partition with one permeable to gas B but not gas A, and using a piston permeable to gas A but not gas B, ...
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-1 votes

Bekenstein entropy black hole v.s Hawking entropy black hole

In informational terms, the relation between thermodynamic entropy $S$ and Shannon entropy $H$ is given by relation between $S$ & $H$: $$ S=kH\ln(2)$$ whence $$ H \le 2πRE/\hbar c\ln(2) $$ where $...
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Can a quantum measurement breach the third law of thermodynamics?

Third law of thermodynamics is meant as limitation on macroscopic processes using heat transfers and work to do refrigeration on macroscopic bodies. I make a quantum measurement of energy with ...
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Can a quantum measurement breach the third law of thermodynamics?

The uncertainty principle states $\Delta E$ $\geq$ $\dfrac{\hbar}{(2\Delta t)}$. A temperature of exactly $0 \ \text{K}$ implies $\Delta E = 0$. But this would mean $\Delta t = \infty$ or an infinite ...
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1 vote
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Proof of the expression for entanglement entropy from correlation matrix

It should, of course, be cautioned that the quantum state corresponding to the correlation matrix $C$ is a Gaussian state -- for a pure state, this simply means it's a Slater determinant, but more ...
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1 vote

Proof of the expression for entanglement entropy from correlation matrix

The general strategy for non-interacting bosons or fermions in this case is as follows. First, note that the quantity you are interested (entropy) is basis independent. Thus, you can move into a basis ...
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How is the 'lost' work related to the entropy change in free expansion? Why is it not just defined by energy 'lost' but energy lost per kelvin?

As Denbigh has shown, for a closed system in contact with a single constant temperature reservoir held at the same temperature as the initial temperature of the system T (i.e., all heat transfer takes ...
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How is the 'lost' work related to the entropy change in free expansion? Why is it not just defined by energy 'lost' but energy lost per kelvin?

The change in entropy would be, however I don't really understand this formula: $$ \Delta S = nR \ln \frac{V}{V_0}= nR \ln 2$$ The formula assumes (1) an insulated container ($Q=0$), (2) an ideal gas,...
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