PHYS6562 W4 Daily Question

PHYS6562 F4 Daily Quaetion

## Ergodicity

The system of classical particles bouncing back and forth is not ergodic since each particles can only occupy certain close volume in phase space, with fixed *x*, *y* position, *P*_{x}, *P*_{y} momenta and ±*p*_{z}. Therefore, the given volume in phase space of the ensemble will not evolve with time.

## Heat engine

(a) True.

(b) True, since the work in done on the gas.

(c) False, it’s a refrigerator.

(d) True.

\begin{equation}
W = (P_0)(3V_0) + \int_{4V_0}^{V_0} \frac{4P_0V_0}{V} dV = 3P_0V_0 - 4P_0V_0\log4
\end{equation} (e) True, a part does no work so *Δ**U* = *Q*. \begin{equation}
\Delta U = \frac{3}{2} nRT_c - \frac{3}{2} nRT_h = -\frac{3}{2} (3P_0V_0) = -\frac{9}{2}P_0V_0
\end{equation} where n is the number of particles in mole and *R* = 8.317 is the ideal gas constant. (f) True. The whole cycle *Δ**U* = 0, the heat input and the net work done *onto* the gas must equal the heat output. \begin{equation}
Q_h + W = Q_c
\end{equation}

PHYS6562 M5 Daily Question

## Undistinguished particles and the Gibbs factor

(a) \begin{equation} \Omega = \frac{V^N}{N!} \end{equation}

\begin{equation}
S = k_B \space \log(\Omega) = k_B(N\log(V) - B\log(N) + N) = Nk_B\left(1+\log\left(\frac{V}{N}\right) \right)
\end{equation} The average volume of a region for single particle is *v* = *V*/*N* with the entropy of *s* = *k*_{B}log(*v*). Therefore, the configuration entropy is related to N times the entropy of a particle in its region with average volume.

(b) \begin{equation}
S = Nk_B\left(\frac{5}{2}-\log(\rho\lambda^3)\right) = Nk_B\left(\frac{5}{2}+\log\left(\frac{V}{N}\lambda^3\right)\right)
\end{equation} The configuration entropy for an ideal gas particle is also related to the small average volume *v*. \begin{equation}
S_c = Nk_B\log(v)
\end{equation}

But the ideal gas particles also occupy the momentum space and each particle has the average length of *λ* in each dimension. So

\begin{equation} S_p = 3Nk_B\log(\lambda) = Nk_B\log(\lambda^3) \end{equation}

Combine the configuration and momentum entropy for one particle, \begin{equation} S = S_c+S_p = Nk_B\log(v*\lambda^3) = Nk_B\log\left(\frac{V}{N}*\lambda^3\right) \end{equation} The total entropy is related to the N times the space occupied both in configuration and momentum space.

PHYS6562 W5 Daily Question

## Entropy of socks

For a child’s room containing 100 things in the room, we only need to consider the configuration space since those things are stationary in the room. The room before tightening up, the number of configurations is

\begin{equation} \Omega = V^N = 5^{100} \end{equation}

Let’s consider the center-of-mass location accuracy of 1 cm. First of all, the center-of-mass can be located itself at any point of the room. Therefore,

\begin{equation} \Omega_{CM} = 5 \end{equation}

Then we can assume that the center of mass in *x* direction is at 0. For each randomly located thing *A*1 at *x*, there would be the another thing *A*1′ located at the −*x*. We now have only 50 things we can throw at will. For the *A*50′, since the deviation of 1cm is acceptable, *A*50′ can have roughly 1m of freedom in the *x* direction. Therefore,

\begin{equation} \Omega_{item} = (L^{50}*1)^3 = 5^{50} \end{equation}

where *L* = 5^{1/3}.

\begin{equation} \Omega' = \Omega_{CM}\times\Omega_{item} = 5^{51} \end{equation}

Entropy decrease can be found to be

\begin{equation} \Delta S = k_B\log{\Omega'} - k_B\log{\Omega} = -49k_B\log{5} \end{equation} \begin{equation} \Delta S_{total} = 10^9\times\Delta S = 1.09\times10^{-12} \end{equation}

For 1 liter baloon contracting by 1% in volume, \begin{equation}
\Delta S = Nk_B\log{0.99V} - Nk_B\log{V} = \frac{PV}{T}\log(0.99)
\end{equation} where *P* = 1*a**t**m*, *V* = 1*L* and *T* = 300*K*. \begin{equation}
\Delta S = -0.0034 (J/K)
\end{equation}

PHYS6562 F5 Daily Quaetion

## Loaded dice, and sticky spheres

(a)

For discrete system, the entropy \begin{equation}
S_{discrete} = -k_B \sum p_i\log p_i
\end{equation} \begin{equation}
S_{discrete} = \frac{3}{2}k_B \log 2
\end{equation}

(b)

Entropy for being hit by a comet with $\rho_c(\theta,\phi) = \frac{1}{4\pi}$: \begin{equation}
S_c = -k_B \iint \rho_s \log (\rho_s) d\Omega = k_B \log (4\pi)
\end{equation} Entropy for being hit by an asteroid with $\rho_a(\theta,\phi) = \frac{cos(\theta)}{\pi^2}$: \begin{equation}
S_a = -k_B \iint \rho_a \log (\rho_a) d\Omega = -k_B \int \left( \frac{2\cos^2\theta}{\pi}\log(\cos\theta) - \frac{4\cos^2\theta}{\pi}\log\pi \right)d\theta
\end{equation} \begin{equation}
S_a = -k_B \left(\frac{1}{2}(1-\log 4) - 2\log(\pi) \right) = k_B \log \left(\frac{2\pi^2}{\sqrt{e}} \right)
\end{equation}

The entropy difference \begin{equation} \Delta S = S_c - S_a = k_B\log\left(\frac{2\sqrt{e}}{\pi} \right) \end{equation}

AEP 4830 HW3 Root Finding and Special Functions

AEP 4830 HW2 Numerical Integration

VLASS Pilot Survey Design

and 5 collaborators

This document describes the design for the VLA Sky Survey (VLASS) Pilot Survey which will be carried out in the 27 May – 5 Sep 2016 time-frame.