# Tag Info

5

A footnote on Wikipedia actually explains this. The $h$ appearing in the partitioning of the phase space is not Planck's constant, it is simply the size of a phase space cell in which we do not distinguish single states anymore. It's initially arbitrary. Since Planck's constant provides a natural scale below which we should not apply classical thinking, it ...

4

Momentum, unlike energy, is a vector quantity. This means that if you have a large number of particle moving at random their momenta tend to cancel out, rather than adding the way a scalar quantity like energy would. If the total momentum of the component parts of your system does not add to 0, then this simply means that your entire system is moving in ...

3

This derivation does not make direct use of the Hamiltonian of the system and I am supposed to find out at what part it is faulty. I.e. the derivation uses an assumption that does not hold in general. This part seems suspicious: Since the velocities in each room direction are independent, we have $p(v_x,v_y,v_z)=f(v_x^2)f(v_y^2)f(v_z^2)$ In ...

2

It is true that classical thermodynamic equations emerge from statistical mechanics. And that the increase in entropy depends on the increase in the number of microstates. Decays also increase the number of microstates. They are irreversible because decay releases energy and the thermodynamic system cannot deliver enough energy and combination of ...

2

G.J.M. Hagelaar and L.C. Pitchford give an elegant derivation of fluid equations in the scope of two-term formulation of the Boltzmann equation. Yours equation above appears in (39) (see "Solving the Boltzmann equation to obtain electron transport coefﬁcients and rate coefﬁcients for ﬂuid models", http://dx.doi.org/10.1088/0963-0252/14/4/011)

2

Indeed they are conserved quantities so there are canonical ensembles with linear momentum exchange and angular momentum exchange with a reservoir. In fact Gibbs used angular momentum as an example to demonstrate the canonical ensemble's use beyond simply energy (Gibbs 1902, see around eqn 98). Considering linear and angular momentum together, the canonical ...

2

If you want the gas particles to have extended size then the first approximation would be to use the hard sphere gas model. Assume you have $N$ non-interacting hard spheres each of volume $v_0$ in a box of volume $V$. Then the entropy can be calculated as $S = Nk_B \ln [\frac{e}{N}(V - \frac{Nv_0}{2})(\frac{4\pi mEe}{3Nh^2})^{3/2}]$. C.f. Kardar ...

2

The problem with the Boltzmann definition is, as you have neatly shown, that its usefulness depends on the assumption that your system is in equilibrium with its surroundings. Without first assuming equilibrium and subsequently setting the temperatures as equal, one cannot show that the Boltzmann entropy satisfies the First Law and hence meaningfully define ...

2

Boltzmann's version of entropy require a finite number of states, and Planck had asserted that probability has no meaning without a “finite number of equally likely configurations.”{1} That is, in order to be able to use Boltzmann's equation and obtain finite results, he needed to use a discrete number of states, but light was supposed to be continuous. ...

1

The condition you wrote, namely that the partial derivative of the phase density with respect to time vanishes, is a standard one placed on phase densities that describe equilibrium systems. See, for example, page 29 of Eric D'Hoker's statistical mechanics lecture notes which can be found here: http://www.pa.ucla.edu/content/eric-dhoker-lecture-notes On ...

1

It seems to me you are conflating micro and macrostates. Entropy is not a property defined for a specific microstate but for an ensemble. When we describe a system in terms of pressure, entropy, etc instead of the momentum and position of each particle we are giving up the possibility of discussing specific microstates. The 'states of high order' you are ...

1

An ideal gas should consist of pointlike particles that are non-interacting, except if they collide, in which case they should do so elastically, without losing kinetic energy. I do not think there is any distinction here between a collision and a repulsive force. Any short-range repulsive force between particles (short-range compared with the average ...

1

The reasoning in the question is correct. If you have a box with gas particles placed in half of a box but otherwise uniformly random and with random velocities then it is overwhelmingly likely that it entropy will increase with time, but if reverse the velocities, you will still have randomly distributed velocities and the same argument will apply. By time ...

1

There are two issues in this question, one is how to compute the "reduced" phase space density (i.e. the marginal distribution on the reduced phase space), and the other is how to deal with continuum phase spaces and distributions on it. Let's face the first problem, in a simplified situation. Suppose you have two sets $A = \{a_{1},a_{2}\}$ and ...

1

The rationale is this: Assuming that $x$ and $x'$ are random vriables Their variances are given as $\left<\Delta x^2 \right>$ and $\left< \Delta x'^2 \right>$ Now the covariance of both $x$ and $x'$ is different if $x$ and $x'$ are correlated as random variables or not. If they are not correlated the covariance (joint correlation) is just the ...

1

Let me simplify the problem: assume a container of unit LENGTH along the X axis, assume that we have only 3 molecules instead of 5, and that the molecules move only along the X axis, i.e. they don't have a transversal velocity. Now, what you need to find (in my understanding) is the probability to find the molecules 1 and 2 separated by a distance R. ...

1

I am not sure what the exact system is from the description. However, if you are considering the Hamiltonian of the form $\mathcal{H} \sim J_{ex}\sum \sigma_i \sigma_j$ or something similar, then the corresponding partition function will be $Z \sim \sum e^{-\frac{J_{ex}}{k_B T}\sum \sigma_i \sigma_j}$. It is pretty clear that the physics only depends on the ...

1

For a system in the grand-canonical ensemble, the symbols $E$ and $N$ are not fixed values of the energy and particle number respectively. They are specified values for the ensemble averages of these quantities. Explicitly: \begin{align} E = \langle \hat H\rangle, \qquad N = \langle \hat N\rangle, \end{align} where angled brackets denote ensemble ...

1

There was a recent publication addressing exactly this question. From the abstract: Here, we show that the overall foaming-over process can be divided into three stages where different physical phenomena take place in different time scales: namely, the bubble-collapse (or cavitation) stage, the diffusion-driven stage, and the buoyancy-driven stage.

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