PHYS6562 W4 Daily Question

## No attractors in Hamiltonian system

It’s not true that neighboring initial conditions will converge to the fixed state after long time. As we saw from the Jupiter problem, there are some regions where the initial conditions for chaotic and periodic trajectories lies infinitesimally close but lead to different results. The chaotic initial conditions will not make the system converge to certain state. Therefore, there is no attractor in this 3-body problem.

Stable Hamiltonian systems can withstand a minor perturbation *in the process* and the system will still end up where it’s supposed to.

Unstable Hamiltonian system will diverge if perturbed *in the process* and the system will not come to the original equilibrium regardless of whether the system becomes chaotic or not.

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

AEP 4830 HW3 Root Finding and Special Functions

# Plotting of Bessel Functions

The Bessel Functions of the first kind *J*_{ν}(*x*) and the second kind *Y*_{ν}(*x*) are two functions of interest in this program. The program incorporates two header files **“nr3.h”** and **“bessel.h”** and outputs the function values between *x* = 0 to *x* = 20. The first two Bessel functions *J*_{0}, *J*_{1}, *Y*_{0} and *Y*_{1} are built in within **“bessel.h”** and we can use the recurrence formula to construct higher order Bessel functions. \begin{equation}
J_{\nu+1}(x) = \frac{2\nu}{x}J_\nu(x) - J_{\nu-1}(x)
\end{equation} \begin{equation}
Y_{\nu+1}(x) = \frac{2\nu}{x}Y_\nu(x) - Y_{\nu-1}(x)
\end{equation} The first three *J*_{ν} and *Y*_{ν} are plotted by MATLAB as shown in Fig. 1 and Fig. 2.

AEP 4830 HW2 Numerical Integration

# Calculation of *I*/*I*_{0} with varying *n*

The analytical Fresnel Diffraction Theory gives the light intensity at *x*_{0} on a viewing plane with distance *z* behind the edge the following form.

\begin{equation}
I(x_0) = \frac{I_0}{\lambda z} \left| \int^{x_0}_{-\infty} \exp \left( \frac{i\pi x^2}{\lambda z} \right) dx \right| ^2
\end{equation}

where *λ* and *I*_{0} are the light wavelength and intensity respectively\cite{Saunders1980}. It can be transformed into the following form by substituting $u_0 = x_0 \sqrt{\frac{2}{\lambda z}}$: \begin{equation}
\frac{I(u_0)}{I_0} = \frac{1}{2}\left[ (C(u_0)-C(-\infty))^2 + (S(u_0)-S(-\infty))^2 \right]
\end{equation} where \begin{equation}
C(u_0) = \int^{u_0}_0 \cos \left( \frac{\pi u^2}{2} \right) du\\
S(u_0) = \int^{u_0}_0 \sin \left( \frac{\pi u^2}{2} \right) du\\
\end{equation}

As a result, instead of calculating the complex integral in Eq. (1), we simplify the task by evaluating *C*(*u*_{0}) and *S*(*u*_{0}).

First, I’ll start with calculating both *C*(*u*_{0}) and *S*(*u*_{0}) by the trapezoid rule\cite{Recipe}. The inputs of the program are the number of iterations *n* and the integral upper limit *u*_{0}. For *m*^{th} iteration, the interval will be cut into 2^{m} equally spaced trapezoids and the intensity ratio in Eq. (2) as well as the errors in *C*(*u*_{0}) and *S*(*u*_{0}) will be printed out. The output is shown in the following tables. As *n* increases, the errors in both functions are reduced dramatically, giving the convergence of the intensity ratio, shown as bold face numbers.

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.