Problem set 18

1. The Lagrangian of a particle of charge $e$ and mass $m$ in applied electric and magnetic fields is given by $L = \frac{1}{2}mv^2+ e\vec A\cdot \vec v - e\phi$, where $A$ and $\phi$ are the vector and scalar potentials corresponding to the magnetic and electric fields, respectively. Which of the following statements is correct?
1. The canonically conjugate momentum of the particle is given by $\vec p= m\vec v$
2. The Hamiltonian of the particle is given by $H = \frac{p^2}{2m}+ \frac{e}{m}\vec A\cdot \vec p + e\phi$
3. $L$ remains unchanged under a gauge transformation ofthe potentials
4. Under a gauge transformation of the potentials, $L$ changes by the total time derivative of a function of $r$ and $t$
The canonical momentum is given by $$p_i=\frac{\partial L}{\partial \dot x_i}=mv_i+eA_i$$ The Hamiltonian is given by \begin{align*} H&=\sum\limits_ip_ix_i-L\\ &=\frac{1}{2}mv^2+e\phi\\ &=\frac{1}{2m}\left|\vec p-e\vec A\right|^2+e\phi \end{align*} The equations of gauge transformation of the potentials are given by $${\scriptstyle\vec A(\vec x,t)\rightarrow \vec{A'}(\vec x,t)=\vec A(\vec x,t)+\vec\nabla\Lambda(\vec x,t)}$$ $${\scriptstyle\phi(\vec x,t)\rightarrow \phi'(\vec x,t)=\phi(\vec x,t)-\frac{\partial\Lambda(\vec x,t)}{\partial t}}$$ $${\scriptstyle L\rightarrow L'(\vec x,t)=L+e\left(\nabla\Lambda\cdot\vec v+\frac{\partial\Lambda}{\partial t}\right)}$$ $$L'(\vec x,t)=L+e\frac{d\Lambda}{dt}=L$$ Because, total time derivative of$\Lambda(\vec x,t)$ is zero. Hence, option (C) is correct.
2. Consider the matrix $M=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}$
1. The eigenvalues of $M$ are
1. 0, 1, 2
2. 0, 0, 3
3. 1, 1, 1
4. -1, 1, 3
2. The exponential of M simplifies to ($I$ is the $3\times3$ identity matrix)
1. $e^M=I+\left(\frac{e^3-1}{3}\right)M$
2. $e^M=I+M+\frac{M^2}{2!}$
3. $e^M=I+3^3M$
4. $e^M=(e-1)M$
(A) $\begin{vmatrix} 1-\lambda&1&1\\1&1-\lambda&1\\1&1&1-\lambda \end{vmatrix}=0$ gives \begin{align*} &(1-\lambda)\left\{(1-\lambda)^2-1\right\}\\ &-\left\{(1-\lambda)-1\right\}+\left\{1-(1-\lambda)\right\}=0 \end{align*} $$\text{i.e.}\quad \lambda^2(3-\lambda)=0$$ Hence, $\lambda=0, 0,3$
(B)$$e^M=I+M+\frac{M^2}{2!}+\frac{M^3}{3!}+\dots$$ But $M^2=3M$, $M^3=3^2M$, $M^4=3^2M$, $M^n=3^{n-1}M$ $$e^M=I+M+M\frac{3^1}{2!}+M\frac{3^2}{3!}+\dots$$ $$e^M=I+M\left(1+\frac{3^1}{2!}+\frac{3^2}{3!}+\dots\right)$$ $$e^M=I+\frac{M}{3}\left(3+\frac{3^2}{2!}+\frac{3^3}{3!}+\dots\right)$$ \begin{align*} e^M&=I\\ &+\frac{M}{3}\left(\left[1+3+\frac{3^2}{2!}+\frac{3^3}{3!}+\dots\right]-1\right)\\ &=I+\frac{M}{3}\left(e^3-1\right) \end{align*} Hence, answer is (i)
3. The magnetic field due to the $TE_{11}$ mode in a rectangular wave guide aligned along Z-axis is given by $H_z=H_1\cos{(0.3\:\pi x)}\cos{(0.4\:\pi y)}$, where $x$ and $y$ are in cm.
1. The dimensions of the rectangular wave guide are
1. $a=3.33$ cm, $b=2.50$ cm
2. $a=0.40$ cm, $b=0.30$
3. $a=0.80$ cm, $b=0.60$ cm
4. $a=1.66$ cm and $1.25$ cm
2. The entire range of frequencies $f$ for which the $TE_{11}$ mode will propagate is
1. 6.0GHz < $f$ < 7.5GHz
2. 7.5GHz< $f$ < 9.0GHz
3. 7.5GHz< $f$ < 12.0GHz
4. 7.5GHz< $f$
(A) In a rectangular wave guide the longitudinal magnetic field is given by $$H_z(x,y)=H_0\cos{(k_xx)}\cos{(k_yy)}$$ where $k_x=\frac{n\pi}{a}$ and $k_y=\frac{m\pi}{b}$. For $TE_{11}$ mode $n=m=1$.
Hence, $k_x=\frac{\pi}{a}$ and $k_y=\frac{\pi}{b}$
Hence, $a=\frac{\pi}{k_x}=\frac{1}{0.3}=3.33$ cm and $b=\frac{\pi}{k_y}=\frac{1}{0.4}=2.5$ cm
(B) The cutoff frequency for $TE_{mn}$ mode is given by $$f_{c_{mn}}=\frac{c}{2n}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}$$ The cutoff frequency for $TE_{mn}$ mode is the operating frequency below which attenuaiion occurs and above which propagation takes place. <\br> For air filled waveguide $n=1$ \begin{align*} f_{c_{11}}&={\scriptstyle\frac{3\times10^{10}}{2}\sqrt{\left(\frac{1}{3.33}\right)^2+\left(\frac{1}{2.5}\right)^2}}\\ &=7.5\:GHz \end{align*} Hence $7.5\:GHz < f$, hence, answer is (D).
4. Consider two independently diffusing non-interacting particles in 3-dimensional space, both placed at the origin at time $t = 0$. These particles have different diffusion constants $D_1$ and $D_2$. The quantity $\left<\left[\vec R_1(t)- \vec R_2(t)\right]^2\right>$ where $\vec R_1(t)$ and $\vec R_2(t)$ are the positions of the particles at time $t$, behaves as:
1. $6t(D_1+D_2)$
2. $6t|D_1+D_2|$
3. $6t\sqrt{D_1^2+D_2^2}$
4. $6t\sqrt{D_1D_2}$
For 3-dimensional diffusion, according to Einstein-Smoluchowski relation or Einstein relation, diffusion coefficient is given by $$ D=\frac{L^2}{6t}\quad \text{Einstein-Smoluchowski}$$ where, $L$ is mean square displacement or diffusion length,and $t$ is diffusion time. $$D=\frac{< r^2 >}{g*t}\quad \text{Einstein}$$ where, $< r^2 >$ is the average of the square of the mean displacement, $g=1,2\text{ or }6$ depending if diffusion dimension is 1, 2, or 3.
Now, $\left[\vec R_1-\vec R_2\right]^2=R_1^2+R_2^2-2\vec R_1\cdot\vec R_2$ \begin{align*} &\left < \left[\vec R_1(t)- \vec R_2(t)\right]^2\right > \\ &=\left< \vec R_1(t)^2\right > + \left<\vec R_2(t)^2\right>\\ &-2\left < \vec R_1(t)\cdot\vec R_2(t)\right >\\ &=\left< \vec R_1(t)^2\right > + \left<\vec R_2(t)^2\right>\\ &=6t(D_1+D_2) \end{align*}