Hot answers tagged

9

I think there is a misunderstanding. You are perfectly right when you write that the total micro canonical entropy of a combined system will be \begin{equation} S_\textrm{combined}(2E) = k_B\ln \sum_x \Omega(x)\Omega(2E-x) \end{equation} The micro canonical entropy ought to be a function of only the total energy, total amount of matter and total volume of ...


6

Major edit: In @gatsu's answer, it is pointed out that only the amount of energy should matter, which is correct, as there's no such thing as distinguishable microstates with only rearranged energy (think stars-and-bars-type entropy calculations). So, I've edited out that part of the first paragraph and equations (in the first draft, I dropped that part of ...


5

The entropy $S$ is extensive as long as you're consistent about what you mean by entropy. In your case you've mixed up two different definitions. One definition of the entropy has the system at fixed energy $E$ -- the other, a fixed temperature $T$. Fixed-E entropy For a system with fixed energy $E$ the entropy is defined to be $$ S = \log\Omega(E) ...


5

The energy of an electrical wave certainly undergoes energy loss through heat, so in that way is entropic. This is a result of the material's resistivity through which the electricity is conducted. The mechanism is primarily scattering of the energy through electron-phonon interactions, but electron-electron interactions do also occur. In metals, the main ...


4

If the universe is open, there's obviously more universe that you haven't included in your system. The universe, by definition, contains all energy and matter. An open system, by definition, has an outside system to exchange energy and matter with. If that outside system isn't part of the universe, then where is it?


3

At the power station electricity is generated as work from a heat engine. Work is entropy free, so we have an entropy free electron-gas at the point of generation. However, a thermodynamic gas will always equilibrate to the available degrees of freedom. In this case it is the electronic states of the conductor in the transmission wire. There will be a ...


3

It's maximizing the angular entropy in the sense that far more "available" angle vectors are inhabited. I suspect you need to be careful with the word "entropy" here. For example, the photons are not down-converted into a larger number of photons of longer wavelength (aka 'heat death of the universe').


3

We do observe spontaneous symmetry restorations in nature. This is called an emergent symmetry. See e.g. this post. A system posses an emergent symmetry if it appears symmetric at large (coarse-grained) scales although the apparent symmetry is explicitly broken by the microscopic description (typically the Hamiltonian or Lagrangian). I can give two examples ...


3

It is reversible in the first case because it satisfies the reversibility definition. A thermodynamic process is called reversible if an infinitesimal change of the external condition reverses the process. Consider a system at temperature $T$ in thermal contact with a thermal reservoir at same temperature. By an infinitesimal increase $dT$ of the reservoir's ...


2

To do it reversibly, you can heat the body from $T_1$ to $T_2$ (i.e., over a finite temperature change) using an infinite sequence of constant temperature reservoirs, in which each reservoir in turn is only dT higher in temperature than the body at any time (and also only dT higher in temperature than the reservoir before it in the sequence). Each increment ...


2

Getting cooler is the key phrase in you question. If you vigorously stir cold water you might get it to warm up if it was well insulated - duplicating Joule's clasic Mechanical equivalent of heat experiment. But if you stir a cup of hot coffee or a pot of hot water the far more significant effect effect it that the stirring accelerates cooling. Stirring ...


2

Let $\lambda_i$ be the eigenvalues of $\rho_A$. Then $$ \log \text{tr} \rho_A^n = \log \left( \sum_i \lambda_i^n \right) $$ Now, let is differentiate w.r.t. $n$. We find $$ - \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg|_{n=1} = - \frac{\sum_i \lambda_i^n \log \lambda_i }{\sum_i \lambda_i^n } \bigg|_{n=1} = - \frac{ \sum_i \lambda_i \log ...


1

It's a confusion of terms. The universe is a closed thermodynamic system whether or not it is 'open' or 'closed' in a cosmological sense. In cosmology, open and closed universes refer to the curvature of the universe, whether positive (closed, finite universe), zero (flat, open, infinite universe) or negative (curved, open, infinite universe). In ...


1

You state that: there is literally no way to squeeze more information (entropy) into a given volume than that in a black hole occupying that volume But you must keep in mind that the volume occupied by the radiation+BH system is larger than the volume of the black hole by itself. When the black hole initially forms the horizon has a radius $r$ which ...


1

Entropy is a property of the system. The change in entropy of a system as it traverses from an initial state(1) to a final state(2) is independent of the path by which the system is taken from state 1 to state 2. The path can be a reversible one, or even irreversible, the change in entropy is always the same as long as the initial and final states are the ...


1

I would like to add to what Procyon said in his(her) answer, which is right on target. The first equation in the OP should be an equality, not an inequality. For an irreversible process, the temperature of the system is typically non-uniform, and when we write $\int{\frac{dQ}{T}}$ for the Clausius inequality, what we really mean is ...


1

There is something they forgot to mention in your notes (either from ignorance, or out of omission). The temperature within the system is spatially non-uniform during an irreversible process. So what value of the temperature are you supposed to use in the integral of dq/T? The Clausius inequality calls for the use of the temperature at the boundary ...


1

There are many ways to think about the entropy of a vacuum (assuming there is no radiation and thus T=0), but all give the same result, the entropy is zero. One easy way is to notice that the walls are made of something (it doesn't matter what) that cannot change its state, so the number of microstates, $\Omega$, is equal to 1. Then $S=k\ln\Omega=0$.


1

Moving water convects heat better than static water. Take the reference frame of the water being static and the air moving. In this case, new cooler air is always sweeping in to take heat away. If the air is static, hot air remains at the interface and will not accept heat as well as cooler air. The same argument can be made for the moving water. Edit: The ...


1

Entropy is something related to thermodynamics, which is essentially a study about energy interactions. So you can use the concepts where ever you see an energy interaction. Entropy is a measure of randomness resulting from the increase in heat energy supplied to a system. When you consider an electric circuit, you have an energy source which is the emf of ...


1

All comments miss a point, or maybe I didn't get it right. I see it this way, lets start with other forms of energy (Kinetic and potential for example) of an harmonic oscillator. We know for a non damped HO the relation between position "s" and velocity "v" is given by total energy Eo = 1/2kx^2 + 1/2mv^2, k spring constant, m mass. this relationship (since ...


1

So why do we say entropy is extensive? It is a convention that is possible and useful for weakly interacting systems. Multiplicity of state of the system made of two systems of the same kind and size and energy $E$ is $$ \Omega'(2E) = \Omega(E)\Omega(E) $$ with very good accuracy, the other terms in the sum over $x$ you indicated can be neglected. ...


1

To understand the origin of black body radiation, which is what every material body with a temperature T has been measured to emit, one has to go to quantum statistical mechanics. Atoms and molecules, to start with, in any ensemble, interact with the electromagnetic radiation. At that level, the processes are quantum mechanical. In quantum mechanics the ...


1

Your definition of Holevo information is wrong. It corresponds to $C_{ea} $, the entanglement assisted capacity of the channel. See equation (5) of the paper. The Holevo information is defined for a probabilistic mixture of density matrices, or for a cq-state (cq = classical quantum state).



Only top voted, non community-wiki answers of a minimum length are eligible