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The resolution to Maxwell's demon paradox is mostly understood to be through Landauer's principle, and it is one of the most compelling applications of information science to physics. Landauer's principle asserts that erasing information from a physical system will always require performing work, and particularly will require at least $$k_B T \ln(2)$$ of ...


5

There's really two questions here, one about the definition of distribution functions and one about the derivation of the BBGKY hierarchy. I will address them in turn. Definitions Let's define, for convenience, $\mathbf{r}^n = \mathbf{r}_1,\dots,\mathbf{r}_n$ and $\mathbf{r}^{(N-n)} = \mathbf{r}_{n+1},\dots,\mathbf{r}_N$. Next, let's denote the ...


3

Four additions to other answers and your questions: I agree with your thoughts about the door. In principle, it can be arbitrarily near to lossless. The work cost does not arise in knowing when to open the door, i.e. in measuring the state of gas particles. This was actually what Leo Szilard thought, as he discussed in 1929 in L Szilard, "Über die ...


3

Joule's law, and thermodynamics in general, is a model of the classical world. Here, classical should be interpreted as non-quantum-mechanical. Thermodynamics is the study of large collections of particles and their collective behavior. No microscopic model is assumed, and one tries to extract as many (non-trivial) features as possible based on purely ...


2

I don't know that this qualifies as a full answer, but the “intuitive” answer would be that on large enough scales, the details of the lattice won't matter: If you look from far enough away (conversely, the lattice parameter is sufficiently small), it will look like a continuum. In this picture, you would expect large-scale structures to be fully ...


2

You can also think of this in terms of information only, without invoking thermodynamics right from the start. So, you just have a system, and you don't know the exact physical state it is in. If we then consider that system including all the features needed to operate Maxwell's demon as a totally isolated system, such that even quantum decoherence is ...


2

Let us suppose the gas is confined by a harmonic potential. The bosons have, in three dimensions, energy levels $\hbar\omega(n+3/2)$ with degeneracy $n(n+1)/2$. The grand-canonical partition function of level $n$ is (without degeneracy) $$ \xi_n=\sum_{p=0}^\infty \left(\mathrm e^{-\beta \hbar\omega(n+3/2)+\beta\mu(T)}\right)^p$$ where $p$ is the number ...


2

That's a very hard question to answer with the appropriate level of detail! Very broadly speaking in an ideal metal all atoms are forming a perfectly regular crystal lattice. Conduction band electrons can move freely around these atoms, which makes it easy to pass a current trough the metal. In a (theoretical) metal with perfect crystal lattice the ...


2

It is not possible to have a state with four indistinguishable particles such that $P_{12} \psi = -\psi$ and $P_{34} \psi =\psi$, for an algebraic reason. Namely, the exchange operators have to form a representation of the permutation group $S_4$. It is rather well known that there are exactly two representations of $S_n$: the trivial representation where ...


2

Your idea is correct but the calculation is not. For $N$ moles of an ideal gas you have $U = cNRT$, look that if $T$ doesn't change, $U$ also doesn't change. So you start at the state $(T_1,V_1)$ and want to go to the state $(T_2,V_2)$. You then use two process corresponding to the following sequence of states: $$(T_1,V_1)\to (T_2,V_1)\to (T_2,V_2)$$ On ...


2

I can offer a few comments on your project, although without solving your specific problem. I believe my outlook is rather different from yours. Historically, statistical mechanics arose from macroscopic thermodynamics; you cannot divorce the two. We can take Gibbs and Boltzmann as the creators of statistical mechanics, with a hat tip to Maxwell. For ...


2

I guess you refer to the free expansion of a gas, which is an irreversible process. During free expansion, no work is done by the gas. The gas goes through states of no thermodynamic equilibrium before reaching its final state, which implies that one cannot define thermodynamic parameters as values of the gas as a whole. For example, the pressure changes ...


1

I am just trying to answer your question. I understand that you expect an answer of "yes" or "no". I am not sure if I'll succeed. I start from the formula in my comment. The variance for the initial electron energy is $$<(E - \ E_0)^2> = \Sigma_E P_{initial}(E) \ E^2 - [\Sigma_E P_{initial}(E) \ E]^2 . $$ The variance for the final electron energy ...


1

The specific heat of a molecule depends on the number of degrees of freedom the molecule has. There are several degrees of freedom available: translation (3), rotation (3), vibration (depends on the number of bonds in a molecule) and electronic modes. Now, for something that is monatomic, you have 3 translational modes (x,y,z directions), zero rotation ...


1

You might be best off creating a bunch of trial systems and evolving each one independently using the (much easier) Hamilton's equations, which are good old ordinary differential equations. After all this is how the Liouville equation was constructed in the first place by people like Gibbs---in the limit of an infinite number of trial systems, you get the ...


1

It's a sum over states. The partition function can also be written as \begin{align} Z = \sum_\alpha g_\alpha e^{-\epsilon_\alpha/kT} \end{align} where $\alpha$ is an index which labels levels, $\epsilon_\alpha$ is the energy of level $\alpha$, and $g_\alpha$ is the degeneracy of level $\alpha$. This follows from the sum over states by noting that in that ...


1

Suppose the spins on the sites of the square lattice are coupled nearest neighbor so that the Hamiltonian is defined as $$H=-J\sum_{<ij>}S_iS_j$$ where $<ij>$ denotes nearest neighbors $i$ and $j$ (each site has 4 nearest neighbors). The ground state of the Hamiltonian will be given by finding a configuration of spins $\{S_i\}$ for which the ...


1

It means that every state (wavefunction) that has the same total energy has also the same probability, that is, if you have the system prepared in the same way many times, and measure the state, the probability of measuring a specific state is not a function of the energy. Is it saying all the wave functions are 'equal' to one another in the sense ...


1

The energy needed to measure the spins can be essentially nought, or at least, in a thought-experimental point of view, it can be made arbitrarily small. However, if you think that the second law is salvaged by a nonzero work needed for measurement you are, even though ultimately wrong, in the most excellent company. None less than the great Leo Szilard ...


1

The first law of thermodynamics, is as the name suggest a law. It states that if you consider some process a thermodynamic system undergoes then $$\Delta U = W + Q$$ The point is that $W$ and $U$ are things that depend on what you are studying. Pick the infinitesimal version just for simplicity $$dU = \delta W + \delta Q$$ Then for a gas it trully makes ...


1

Maxwell's thermodynamics and E&M were both atomistic. For example, in his Treatise on Electricity and Magnetism §255, he found it extremely natural to suppose that the currents of the ions are convection currents of electricity, and, in particular, that every molecule of the cation is charged with a certain fixed quantity of positive electricity, ...


1

I'm not sure if there is much to physically understand in the equation itself, the derivation is where all the physical insight takes place (Kirkwood's superposition approximation). What of the equation, then? Why is it important and why should anyone care? It's because the BBGKY hierarchy, while exact, cannot be solved: It is a relation between the ...


1

The final distribution is a function of the initial distributions in this sense: the initial distributions determine thermodynamic quantities - energy, temperature - which determine the thermodynamic quantity $T_C$ after the partition is removed, which finally determines the new statistical distribution. The best way to solve this problem is to find $T_C$ ...


1

Not sure if this answers your quesitons, or if this is more of a lenghty comment, but here goes. So for the grand canonical the particle operator $\hat{N}$ is defined through $\hat{N}|i, N\rangle = N|i, N\rangle$, and so $\hat{\rho} = \frac{1}{Z} \sum_i \sum_N e^{-\beta(E_i-\mu N)}|i, N\rangle\langle i, N| = ...


1

Work is force acting over a distance. That means there can only be work done on the gas if the piston moves, i.e. if the gas changes volume. In that case, $W = \int F \text{d} x = \int P\, \text{d}V$, where P is the pressure difference between the inside and outside of the calorimeter. Note, however, that in your example the gas is doing work on the ...


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A wavefunction is a normalised vector (or a ray) in the Hilbert space of vector states (I will assume finite degrees of freedom, so the C*-algebra of the system can be taken to be $B(H)$, with $H$ an $L^2(\mathbb R^n)$ space). Spin gives a superselection rule, and therefore there must be superselection sectors. It follows from the general theory that vector ...



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