### Solution

A1  ?? Write the motion equations for both particles and show that this system is equivalent to the interaction of both particles with the effective Columb's potential with a center similar to the center of mass $CM$.

The 2nd particle acts on the 1st one with the force
$\vec{F}_{2 \to 1} = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}_1-\vec{r}_2|^3} \cdot \left( \vec{r}_1 - \vec{r}_2 \right),$
so
$m_1 \ddot{\vec{r}}_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}_1-\vec{r}_2|^3} \cdot \left( \vec{r}_1 - \vec{r}_2 \right), \quad m_2 \ddot{\vec{r}}_2=-\frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}_1-\vec{r}_2|^3} \cdot \left( \vec{r}_1 - \vec{r}_2 \right).$

Answer: $m_1 \ddot{\vec{r}}_1 = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}_1-\vec{r}_2|^3} \cdot \left( \vec{r}_1 - \vec{r}_2 \right), \quad m_2 \ddot{\vec{r}}_2=-\frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}_1-\vec{r}_2|^3} \cdot \left( \vec{r}_1 - \vec{r}_2 \right).$

The position $\vec{r}_{CM}$ of the center of mass we calculate with the definition and the position $\vec{r}$ of the 1st particle relative to $\vec{r}_{CM}$ is $\vec{r}=\vec{r}_1 - \vec{r}_{CM}$:
$\vec{r}_{CM}=\frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}, \quad \vec{r} = \vec{r}_1 - \vec{r}_{CM} = \left( \vec{r}_1 - \vec{r}_2 \right) \frac{m_2}{m_1 + m_2}.$

For $\vec{r}_1 -\vec{r}_2$ we have this differential equation:
$\ddot{\left( \vec{r}_1 - \vec{r}_2 \right)} = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}_1 - \vec{r}_2|^3} \frac{m_1 + m_2}{m_1 m_2} \left( \vec{r}_1 - \vec{r}_2 \right),$
then for $\vec{r}$ we have
$\ddot{\vec{r}} = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}|^3} \frac{m_2^2}{m_1(m_1+m_2)^2}\vec{r}.$

Finally, the 2nd Newton's law for the 1st particle in the frame of the center of mass could be written as an interaction with the Coulomb field centered at $\vec{r}_{CM}$:
$m_1\ddot{\vec{r}}= \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{|\vec{r}|^3} \frac{m_2^2}{(m_1 + m_2)^2} \vec{r}$

A2  ?? How does the angular momentum of the first particle behave?

$\begin{cases} \vec{L} = m_1 \vec{v} \times \vec{r}\\ \dot{\vec{L}} = \vec{M} = \vec{r} \times \vec{F} \propto \vec{r} \times \vec{r} = 0 \end{cases} \quad \Rightarrow \quad L = m_1 \dot{\theta} r^2 = m_1 v_1' b',$
where
$v_1' = v_1 - v_1 \frac{m_1}{m_1+m_2} = v_1 \frac{m_2}{m_1 + m_2}, \quad b' = b\frac{m_2}{m_1 + m_2}$

Answer: $L = m_1 \dot{\theta} r^2 = m_1 v_1 b \frac{m_2^2}{(m_1+m_2)^2}$

A3  ?? Write the second Newton's law in terms of $r$, $\theta$ and their time derivatives $\ddot{r}$ and $\dot{\theta}$.

Answer: $m_1(\ddot{r} - \dot{\theta}^2 r) = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{r^2} \frac{m_2^2}{(m_1 + m_2)^2}$

A4  ?? Introduce the Binet substitution $u = 1/r$ and obtain the differential equation for $u$ as a function of $\theta$ in the form of a harmonic oscillator:
$u'' + Au + B = 0,$
where $u''$ is the second derivative of $u$ with respect to the angle $\theta$.

Let's express $u''$ from $\ddot{r}$ and $u$:
$u'=\frac{du}{d\theta} = -\frac{dr}{d \theta} \cdot \frac{1}{r^2} = -\frac{dr}{dt} \frac{dt}{d \theta} \frac{1}{r^2} = -\frac{m_1\dot{r}}{L}$
$u''=\frac{du'}{d\theta} = -\frac{m_1 d(\dot{r})}{L \, d \theta} = -\frac{m_1 \ddot{r}}{L\dot{\theta}} = -\frac{m_1^2 \ddot{r}}{L^2 u^2} \quad \Rightarrow \quad \ddot{r} = -\frac{u'' u^2 L^2}{m_1^2}$

Then the 2nd Newton's law in terms of $u$ is
$-\frac{u'' u^2 L^2}{m_1} - \frac{L^2 u^3}{m_1} = \frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0} \frac{m_2^2}{(m_1 + m_2)^2} u^2$

Answer: $u''+u + \frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0 v_1^2 b^2} \frac{ (m_1+m_2)^2}{m_1 m_2^2} = 0$

A5  ?? Express the deflection angle $\chi$ in the frame of the center of mass through $b$, $v_1$, $m_{1,2}$ and $Z_{1,2}$.

Since we have the harmonic oscillator equation we have the solution $u + B = C\cos \theta$ where $\theta=0$ is the direction to the apsis. Thus,
$\cos \left( \frac{\pi}{2}-\frac{\chi}{2} \right) = \frac{B}{C}.$
Now we have to find the value of $C$. Let's write the equation
$\dot{\theta}r^2 = \frac{L}{m_1} \quad \Rightarrow \quad \frac{d \theta}{du} \frac{du}{dr} \frac{dr}{dt} r^2 = \frac{L}{m_1}.$
With $du/dr=-1/r^2$ l and additional $dr/dt\simeq v_1'$ at the infinity we have
$\frac{v_1'}{u'}=\frac{L}{m_1} \quad \Rightarrow \quad u'(u=0) = \frac{m_1 v_1'}{L}.$

From the oscillation equation $u' = -C \sin \theta$ and then $(u-B)^2 + (u')^2 = C^2$. Writing this expression for $u=0$ (at the infinity) we have
$C = \sqrt{\frac{m_1^2 v_1'^2}{L^2} + B^2}.$

Finally,
$\sin \frac{\chi}{2} = \frac{1}{\sqrt{1+\frac{m_1^2 v_1'^2}{L^2B^2}}} \quad \Rightarrow \quad \tan \frac{\chi}{2} = \frac{LB}{m_1v_1} = \frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0 m_1 v_1^2 b}$

Answer: $\chi = 2 \arctan \frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0 m_1 v_1^2 b}$

A6  ?? Find the $b_{90}$. Express you answer by $Z_1$, $Z_2$, $m_1$ and $v_1$.

For the $\chi = 90^\circ$ we know that $\tan( \chi/2 )= 1$. Thus,
$b_{90} = \frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0 m_1 v_1^2}$

Answer: $b_{90} = \frac{Z_1 Z_2 e^2}{4 \pi \varepsilon_0 m_1 v_1^2}$

A7  ?? In this question we work in the lab frame where the target was originally motionless.

What is the difference $\Delta p(b)$ of the momuntum of the first particle projection onto the direction of the original motion? Work with the small angle approximation $b \gg b_{90}$. Express your answer by the original momentum of the particle $p$, $m_{1,2}$, $b$ and $b_{90}$.

In the lab frame the velocity of the 1st particle was $\vec{v}_1=(v_1,0)^T$ and after the scattering it becomes
$\vec{v}_1' = \begin{pmatrix} v_1\frac{m_1}{m_1 + m_2} + v_1 \frac{m_2}{m_1 + m_2} \cos \chi \\ v_1 \frac{m_2}{m_1 + m_2} \sin\chi \end{pmatrix}$

$\Delta p = m_1(v_{1x}'-v_1) \simeq -p \frac{m_2}{m_1 + m_2} \frac{\chi^2}{2} \simeq - p\frac{m_2}{m_1 + m_2} \frac{2 b_{90}^2}{ b^2}$

Answer: $\Delta p\simeq - p\frac{m_2}{m_1 + m_2} \frac{2 b_{90}^2}{ b^2}$

B1  ?? Find the equation for $\Delta p$ in approximation $M \gg m$. Express the answer by $p$, $b_{90}$, $n_0$, $l$ and $\Lambda$.

$\Delta p = - \int\limits_{b_{90}}^{\lambda_D}p \frac{m_2}{m_1 + m_2} \frac{2b_{90}^2}{b^2} 2\pi b l \, db \simeq -p \cdot 4 \pi b_{90}^2 l n_0 \ln \Lambda$

Answer: $\Delta p = -p \cdot 4 \pi b_{90}^2 l n_0 \ln \Lambda$

B2  ?? Show, that the probability $P_\text{any}$ that the electron's velocity is any is unity, i.e.
$P_\text{any} = \int\limits_{-\infty}^{+\infty} dv_x \int\limits_{-\infty}^{+\infty} dv_y \int\limits_{-\infty}^{+\infty} dv_z f(v_x, v_y, v_z) = 1.$

Since we have infinite limits, we can shift the coordinate system to make the intergrant symmetrical:
$P_\text{any} = \int \left( \frac{m}{2\pi k_\text{B}T} \right)^{3/2} \exp \left( -\frac{mv^2}{2k_\text{B}T} \right) d^3\vec{v}.$
Accorrding to the spherical symmertry let's intergrate over the thin spheres: $d^{3}\vec{v} = 4\pi v^2 \, dv$:
$P_\text{any} =4\pi \left( \frac{m}{2\pi k_\text{B}T} \right)^{3/2} \int\limits_0^{\infty} \exp \left( -\frac{mv^2}{2k_\text{B}T} \right) v^2 dv.$
With nondementiolizing $u=v\sqrt{\dfrac{m}{2k_\text{B}T}}$ we have:
$P_\text{any} = \frac{4}{\sqrt{\pi}} \int\limits_0^{\infty} e^{-u^2}u^2 du = -\frac{2}{\sqrt{\pi}} \int\limits_0^{\infty} u \, d\left( e^{-u^2}\right) = 0+\frac{2}{\sqrt{\pi}}\int\limits_0^{\infty}e^{-u^2} du$

The integral $\int\limits_{-\infty}^{\infty} e^{-u^2} \, du$ is symmterical over $u \to -u$ then $\int\limits_{0}^{\infty} e^{-u^2} \, du = \sqrt{\pi}/2$ and $P_\text{any}=1$.

B3  ?? Find the mean value of the velocity $\langle v \rangle$. Ignore any corrections associated with the presence of $v_d$.

$\langle v \rangle = \int \left( \frac{m}{2\pi k_\text{B}T} \right)^{3/2} v \exp \left( -\frac{mv^2}{2k_\text{B}T} \right) d^3\vec{v} \simeq 4\pi \left( \frac{m}{2\pi k_\text{B}T} \right)^{3/2} \int\limits_0^{\infty} \exp \left( -\frac{mv^2}{2k_\text{B}T} \right) v^3 dv$
Now it's more convinent to use the substitution $u=\dfrac{mv^2}{2k_\text{B}T}$
$\langle v \rangle = \sqrt{\frac{8k_\text{B}T}{\pi m}} \int\limits_0^{\infty} e^{-u} u \, du = -\sqrt{\frac{8k_\text{B}T}{\pi m}} \int\limits_0^{\infty} u \, d\left( e^{-u} \right) =\sqrt{\frac{8k_\text{B}T}{\pi m}} \int\limits_0^{\infty}e^{-u} du = \sqrt{\frac{8k_\text{B}T}{\pi m}}$

Answer: $\langle v \rangle = \sqrt{\frac{8 k_\text{B}T}{\pi m}}$

B4  ?? Estimate the mean force $\langle \vec{F} \rangle$ that acting on the volume $\Delta V$ from the ions. For simplicity, assume that $\Lambda$ is independent of the electron velocity $v$ and use the value of $\Lambda$ for $v=\langle v \rangle$. Also, ignore that $|\vec{v}|$ (not a vector, just it's absolute value) depends on the $\vec{v}_d$. Finally, you understand approximations properly if
$\langle \vec{F} \rangle \propto \vec{v}_d \int u^3 e^{-u^2} du.$

Note, that all of these approximations don't change the crux of the phenomena, the only thing that might be affected is a different numerical coefficient before.

On the electron moving with the velocity $\vec{v}$ acts the force
$\vec{F}(\vec{v}) =-4\pi mn_0\vec{v}_1 v_1 b_{90}^2 \ln \Lambda$

Then the mean (over electrons) force acting on the one electron from the volume $\Delta V$ is
$\langle \vec{F} \rangle = -4\pi mn_0 b_{90}^2 \ln \Lambda \int (\vec{v}_d + \vec{u}) ue^{-\frac{mu^2}{2k_\text{B}T}}\, d^3\vec{u}.$

The second term is equal to zero according the symmetry. The first term could be found with methods which we already developed:
$\vec{v}_d\int\limits_0^{\infty} 4 \pi u^3 e^{-\frac{mu^2}{2k_\text{B}T}} du = \vec{v}_d 8\pi \left( \frac{2k_\text{B}T}{m} \right)^2 \int\limits_0^{\infty} x e^{-x} dx = \vec{v}_d 8\pi \left( \frac{2k_\text{B}T}{m} \right)^2$

Finally,
$\langle \vec{F} \rangle = -32 \vec{v}_d \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda$

Answer: $\langle \vec{F} \rangle = -32 \vec{v}_d \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda$

B5  ?? Write the second Newton's law for $\Delta V$. Consider, that the interaction with neighboring volumes of the electron gas is canceled out.

Answer: $n\Delta V m \ddot{\vec{r}} = -32\dot{\vec{r}} n \Delta V \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda - en\Delta V \tilde{E} e^{-i \omega t}$

B6  ?? Find the current density $\vec{j}_t$ in the model under the external field $\tilde{E} e^{-i \omega t}$.

Let's substitute the periodic motion $\vec{v} = \vec{v}_0 e^{-i\omega t}$:
$-i\omega m \vec{v}_0 e^{- i \omega t} = -32 \vec{v}_0 e^{- i \omega t} \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda - e\tilde{E} e^{- i \omega t}$
then
$\vec{v}_0 = \frac{e\tilde{E}}{i \omega m - 32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda} \quad \Rightarrow \quad \vec{j}_t =\frac{ne^2}{32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda - i \omega m} \tilde{E} e^{-\omega t}$

Answer: $\vec{j}_t =\frac{ne^2}{32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda - i \omega m} \tilde{E} e^{-\omega t}$

B7  ?? With $\sigma = \mathfrak{Re} \left\lbrace \dfrac{j}{\tilde{E}e^{-i\omega t}}\right\rbrace$ find the conductivity $\sigma$ of the plasma.

$\sigma = \operatorname{\mathfrak{Re}}\left\lbrace \frac{ne^2}{32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda - i \omega m} \right\rbrace = ne^2\frac{32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda }{ \omega^2 m^2 + \left(32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda \right)^2}$

Answer: $\sigma = ne^2\frac{32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda }{ \omega^2 m^2 + \left(32 \sqrt{\frac{2\pi k_\text{B}T}{m}} mn_0 b_{90}^2 \ln \Lambda \right)^2}$