Resolution 2

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\title{Velocity Distribution Function}
\author{Felipe do Nascimento Crissanto}
\date{June 2024}
\begin{document}
\maketitle
\section{Resolution of proposed problems}
\subsection{Check that the local Maxwellian (2.46) is normalized by the number density
(2.4) and that its substitution into the definitions (2.16) and (2.26) leads to
the identities.}
Firstly, we have that the Local Maxwell-Botzmann function is:
\begin{equation}
 f^{M}(t, \bs r, \bs v)=\frac{n}{(\sqrt{\pi}v_m)^3} \exp \left[\frac{({\bs v}-{\bs u} )^2}{v_m^2}\right]
\end{equation}
What is expected is that when we integrate the result it coincides with the numerical density $n$, to do so, we remember that $\bs v_m$ is the most probable speed of a fluid and $\bs u$ is the most probable speed of a particle, therefore, we consider the following:
\begin{equation}
 \bs c = \frac{(\bs v-\bs u)}{v_m}
 \label{eq:II}
\end{equation}
Isolating the variable $\bs v$, and searching for its components in three-dimensional terms, we arrive at:
\begin{equation}
 d^3\bs v = v^3_md^3\bs c
\end{equation}
Then:
\begin{equation}
 \int f^{M}(t, \bs r, \bs v)=\int \frac{n}{(\sqrt{\pi}v_m)^3} \exp \left[\frac{(\bs v-\bs u)^2}{v_m^2}\right]v_m^3d^3\bs c
\end{equation}
Simplifying, we get:
\begin{equation}
 \int f^{M}(t, \bs r, \bs v) = \frac{n}{\pi^{3/2}}\int \exp [-\bs c^2]d^3\bs c
 \label{eq:5}
\end{equation}
Writing the integral in terms of spherical coordinates, we achieve:
\begin{equation}
 4\pi\int_0^\infty \exp [-\bs c^2]d^3\bs c = (4\pi)\frac{\sqrt{\pi}}{4} = \pi^{3/2}
\end{equation}
Substituting into eq. \ref{eq:5}, finally we have:
\begin{equation}
 \int f^{M} (t, \bs r, \bs v) =\frac{n}{\pi^{3/2}}\pi^{3/2}= n
\end{equation}
With this in mind, we now consider this relationship written in terms of eq. \ref{eq:II}:
\begin{equation}
 \bs u(t, \bs r) = \frac{1}{n}<\bs v>
\end{equation}
Therefore:
\begin{equation}
 \bs u(t, \bs r) = \frac{1}{n}\int (\bs u+ v_m\bs c)f_m(\bs v) d^3\bs v
\end{equation}
Separating the terms:
\begin{equation}
 \bs u(t, \bs r) = \int\bs u f^{M}(\bs v)d^3\bs v + \int \bs c v_{m}f^{M}(\bs v)d^3\bs v
\end{equation}
Since the second integral is symmetric around the origin, its value is 0, therefore:
\begin{equation}
 \bs u(t, \bs r) = \int\bs u f^{M}(\bs v)d^3\bs v + o
\end{equation}
Solving we get:
\begin{equation}
 u(t, \bs r) = un \frac{1}{n} + 0 = u +0 = u(t, \bs r)
\end{equation}
For the temperature $T$, with $\bs V$ being the peculiar velocity, we have:
\begin{equation}
 T = \frac{m}{3nK_b}\langle \bs V^2\rangle = \frac{m}{3n\kB}\int \bs V^2(\bs v) f_m(\bs v) d^3\bs v
\end{equation}
Due to difficulties in resolution, i maintain equality as above.
\subsection{Obtain (2.33).}
Initially, we seek to understand the total energy flow, therefore, it is necessary to analyze each type of energy in the system, for example, internal kinetic energy:
\begin{equation}
 \left\langle m \frac{\bs v^2}{2} \right\rangle = \left\langle m \frac{(\bs V + \bs u)^2}{2} \right\rangle = \rho \left( e + \frac{\bs u^2}{2} \right)
\end{equation}
Where $m$ is the mass of the particle and $\bs v$ the average velocity of the particle, $\rho$ the density of the mass, $\bs V$ the peculiar velocity, and finally $e$ the internal specific energy, this portion The equation refers to the total internal kinetic energy, it is related to the fluid convection phenomenon.
 Even more, regarding the phenomenon of energy conduction, we have the following definition:
\begin{equation}
 \bs q (\bs r, t) = (\bs v\frac{m\bs V^{2}}{2})
 \label{eq:2}
\end{equation}
where $\bs q$ is the temperature flow, commonly known as heat, it is the transfer of energy in the system, finally, it is convenient to talk about the phenomenon of transport, since the forces acting in a system also influence the energy total system, for this, we define $\sf P$ as the pressure tensor:
\begin{equation}
 {\sf P} (\bs r, t) = m\bs {VV}
 \label{eq:3}
\end{equation}
Considering these definitions, we consider the flow of the tensor $\sf P$, since the analysis deals with energy variations, putting these definitions together, we find the equation that says about the energy flow vector, being:
\begin{equation}
 {\cal J}_e = \bs q + \rho\bs u(\frac{\bs u^2}{2} + e)+ \bs u \cdot \sf P
\end{equation}
We multiply the properties by $\bs u$(average particle speed), as we are looking for flows in relation to the average particle behavior.
\subsection{Estimate the number N and fraction $f_{N}$ of particles having the speed in the
range [$C\bs v_m, \infty $) for a gas being in equilibrium and occupying a volume of 1
$m^{3}$ at the standard conditions. Calculate this number for $(a) C = 2, (b) C = 4, (c) C = 8$.}
We have that the velocity function equation is:
\begin{equation}
 {f_{\bs v}}^{M}(\bs v) = \frac{4n}{\sqrt{\pi}{ v_m}^3}\bs v^2 \exp(-\frac{\bs v^2}{v_m^2})
\end{equation}
The numerical density number $N$ can be found via:
\begin{equation}
 N = \frac{N_L}{N} \int_{Cv_m}^{\infty} {f_{ v}}^{M}(\bs v) \, d\bs v
\end{equation}
Using its asymptotic series , the expression is simplified:
\begin{equation}
N = \frac{N_L}{\sqrt{\pi}}e^{-C^2} \left(2C + \frac{1}{C} - \frac{1}{2C^3}\right)
\end{equation}
Where, $N_L$ is the Loschmidt constant, for $C = 2$, we get closer.:
\begin{equation}
 N = \frac{N_L}{\sqrt{\pi}}e^{-2^2} \left(2 \times 2 + \frac{1}{2} - \frac{1}{2\times2^3}\right) = 1,26672 \times 10^{24} \; particles/m^3
\end{equation}
Being $f_{n_1}$:
\begin{equation}
 f_n = \frac{N}{N_L}=\frac{1,26672 \times 10^{24}}{2,68678 \times 10^{25}} = 0,044631
\end{equation}
For $C=4$:
\begin{equation}
 N = \frac{N_l}{\sqrt{\pi}}e^{-4^2} \left(2 \times 4 + \frac{1}{4} - \frac{1}{2\times4^3}\right) = 1,40601 \times 10^{19} \; particles/m^3
\end{equation}
And $f_{n_2}$:
\begin{equation}
 f_n = \frac{N}{N_L}=\frac{1,40601 \times 10^{19}}{2,68678 \times 10^{25}} = 5,23307 \times 10^{-7}
\end{equation}
And finally, with $C = 8$, we obtain:
\begin{equation}
 N = \frac{N_l}{\sqrt{\pi}}e^{-8^2} \left(2 \times 8 + \frac{1}{8} - \frac{1}{2\times8^3}\right) = 4 \times 10^{-2} \; particles/m^3
\end{equation}
Then, $f_{n_3}$:
\begin{equation}
 f_n = \frac{N}{N_L}=\frac{4 \times 10^{-2}}{2,68678 \times 10^{25}} = 1,5 \times 10^{-27}
\end{equation}
\subsection{Obtain (2.47).}
In this question, we have that the local Maxwell-Boltzmann equation is:
\begin{equation}
 f^{M} (t, \bs r, \bs v) = \frac{n}{(\sqrt{\pi} v_m)^3} \exp \left[\frac{(\bs v-\bs u)^2}{v^{2}_m}\right]
 \label{eq:14}
\end{equation}
Writing eq.\ref{eq:14} in terms of the components of the peculiar velocity $dV$ and substituting in eq.\ref{eq:2}, which is also written in terms of $\bs V$, we obtain:
\begin{equation}
 \bs q_i=\frac{mn}{2(\sqrt{\pi} v_m)^3}\int \bs V^{2}\bs V_i \exp \left(\frac{\bs V}{v_m}\right)^2 d\bs V
\end{equation}
This integral shows symmetry around the origin axis, defining this function as odd, in symmetric limits, it is equivalent to 0, therefore:
\begin{equation}
 \bs q_i = \frac{mn}{2(\sqrt{\pi} v_m)^3} \times 0 = 0
\end{equation}
Confirming the result, we can interpret that this region in the system obeys the Maxwell-Boltzmann velocity distribution function, as there is no presence of external energy entering the system, therefore, the behavior is that of a gas in equilibrium.
We analyze the pressure tensor in this region, as follows, replacing eq.\ref{eq:14} in eq.\ref{eq:3}, still in terms of the peculiar velocity$V$, we obtain:
\begin{equation}
 {\sf P_{ij}} = \frac{mn}{(\sqrt{\pi} v_m)^3} \int \bs V_{i}\bs V_{j} \exp \left(\frac{\bs V^2}{ v^{2}_m}\right) d\bs V
\end{equation}
Now we analyze for $i \neq j$, in this case:
\begin{equation}
 \int \bs V_{i}\bs V_{j} \exp \left(\frac{\bs V^2}{ v^{2}_m}\right) d\bs V = 0
\end{equation}
From a physical point of view, this means that the region's shear stress is equivalent to 0, coinciding with what is expected in a gas at equilibrium, with no influences from external forces that interfere in the system and no difference in speed between the particles in this region, we can confirm the validity of the Maxwell-Boltzmann equation by checking when the dimensional component $ i = j$, that is, if there are normal forces and consequently a total resultant force in the system, then:
\begin{equation}
 \int \bs V_{i}^2 \exp \left(\frac{\bs V^2}{ v^{2}_m}\right) d\bs V = \frac{{\pi}^{3/2}}{2} v^{5}_m
\end{equation}
Showing what was expected, the behavior is that of an ideal gas:
\begin{equation}
 {P_{ij}} = \frac{mn}{(\sqrt{\pi} v_m)^3} \frac{{\pi}^{3/2}}{2}v^{5}_m \neq 0
\end{equation}
Manipulating that equation, we achieve:
\begin{equation}
 {P_{ij}} =\frac{ mn v^2_{m}}{2}
 \label{eq:34}
\end{equation}
Understanding the existence of the following relationship:
\begin{equation}
 v_m = \sqrt{\frac{2\kB T}{m}}
\end{equation}
Then:
\begin{equation}
 {v_m}^2 = \frac{2\kB T}{m}
\end{equation}
Substituting into eq. \ref{eq:34}, we get:
\begin{equation}
 {P_{ij}} =\frac{ mn\frac{2\kB T}{m}}{2} = \frac{n2\kB T}{2} = n\kB T
 \label{eq:37}
\end{equation}
We know that the pressure $p$ per number density derived from the equation of state is:
\begin{equation}
 p = n\kB T
\end{equation}
Therefore, when substituting in eq.\ref{eq:37}, we achieve:
\begin{equation}
 {P_{ij}} = n\kB T = p
\end{equation}
\subsection{Obtain (2.40).}
We have that the distribution function is:
\begin{equation}
 {f_{\bs v}}^{M}(\bs v) = {\bs v^2} \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi} f_m(\bs v) \sin \theta \, d\theta \, d\varphi = 4\pi n \left( \frac{m}{2\pi \kB T} \right)^{3/2} \bs v^2 \exp\left( - \frac{m \bs v^2}{2 \kB T} \right)
 \label{eq:21}
\end{equation}
Deriving and simplifying, we achieved:
\begin{equation}
 \frac{df^{M}_v}{dv} = \frac{4\kB mnT\bs v-2m^{2}n\bs v^3}{{\kB}^{2}T^{2} \exp \left[\frac{m\bs v^{2}}{2{\kB}T}\right]}
\end{equation}
Equating to 0 we see that the numerator can adopt a null value, we obtain:
\begin{equation}
 4{\kB}mnT\bs v = 2m^{2}n\bs v^3
\end{equation}
Isolating $\bs v_{m}$, we get:
\begin{equation}
 {v_{m}}= \sqrt{{\frac{2{\kB
 }T}{m}}}
\end{equation}
Where $\bs v_{m}$ is the average velocity of the fluid, we can achieve this same result by considering only a significant portion of the equation, in this case:
\begin{equation}
 \bs v^2 \exp{-\frac{m\bs v^2}{2\kB T}}
\end{equation}
Deriving and simplifying, we get:
\begin{equation}
 \frac{2v}{\exp{\frac{m\bs v^2}{2\kB T}}} - \frac{m\bs v^3}{\kB T \exp{\frac{m\bs v^2}{2\kB T}}}
\end{equation}
equating to zero:
\begin{equation}
 \frac{2v}{\exp{\frac{m\bs v^2}{2\kB T}}} = \frac{m\bs v^3}{\kB T \exp{\frac{m\bs v^2}{2\kB T}}}
\end{equation}
Simplifying, we have:
\begin{equation}
 2 = \frac{m\bs v^2}{\kB T}
\end{equation}
Finally, isolating the variable speed $\bs v$ we got its final form:
\begin{equation}
 \bs v = \sqrt{\frac{2\kB T}{m}}
\end{equation}
\subsection{Calculate the mean speed $\langle \bs v\rangle$ for helium and xenon at $T = 300 \; K$}
We are aware of the following relationship:
\begin{equation}
\langle v\rangle =\frac{1}{n}\int_{0}^{\infty}f^{M}_{\bs v}(\bs v) d\bs v = \sqrt{\frac{8\kB T}{\pi m}}
\end{equation}
Where $\kB$ is the Boltzmann constant and $T$ is the temperature , substituting the mass of helium under standard conditions, we obtain:
\begin{equation}
 {\bs v_1} = \sqrt{\frac{8(1,38\times 10^{-23} )300}{\pi \times 6,64 \times 10^{-27}}} = 1259,47 \; m/s
\end{equation}
For xenon, we have:
\begin{equation}
 {\bs v_2} = \sqrt{\frac{8(1,38\times 10^{-23} )300}{\pi \times 2,189 \times 10^{-25}}} = 219,45 \; m/s
\end{equation}
With this, we conclude that the mass of the particle influences its speed inversely, that is, the lighter it is, the faster it is.
\end{document}