Problem set 55

1. The mass $m$ of a moving particle is $\frac{2m_0}{\sqrt{3}}$, where $m_0$ is its rest mass. The linear momentum of the particle is
1. $2m_0c$
2. $\frac{2m_0c}{\sqrt{3}}$
3. $m_0c$
4. $\frac{m_0c}{\sqrt{3}}$

Given $m=\frac{2m_0}{\sqrt{3}}$. Relativistic mass is given by $$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$ Hence, $$\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{2m_0}{\sqrt{3}}$$ Hence, $$\sqrt{1-\frac{v^2}{c^2}}=\frac{\sqrt{3}}{2}$$ $$1-\frac{v^2}{c^2}=\frac{3}{4}$$ $$\frac{v^2}{c^2}=1-\frac{3}{4}=\frac{1}{4}$$ $$v=\frac{c}{2}$$ Momentum is given by $$\begin{align*} p&=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}\\ &=\frac{2m_0}{\sqrt{3}}\frac{c}{2}\\ &=\frac{m_0c}{\sqrt{3}} \end{align*}$$

Hence, answer is (D)

2. For the given transformations (i) $Q = p$, $P = -q$ and (ii) $Q = p$, $P = q$, where $p$ and $q$ are canonically conjugate variables, which one of the following statements is true?
1. Both (i) and (ii) are canonical
2. Only (i) is canonical
3. Only (ii) is canonical
4. Neither (i) nor (ii) is canonical
A transformation is said be canonical if it satisfies Poisson Bracket $\left\{Q,P\right\}_{q,p}=1$ i.e. $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$ For (i) $Q = p$, $P = -q$ $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=0-1\times-1=1$$ Hence, transformation is canonical

For (ii) $Q = p$, $P = q$ $$\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=0-1\times1=-1$$ Hence, transformation is not canonical

Hence, answer is (B)

3. Consider three inertial frames of reference A, B and C. The frame B moves with a velocity $c/2$ with respect to $A$, and $C$ moves with velocity $c/10$ with respect to B in the same direction. The velocity of C as measured in A is
1. $\frac{3c}{7}$
2. $\frac{4c}{7}$
3. $\frac{c}{7}$
4. $\frac{\sqrt{3}c}{7}$

Let $V$ be the velocity of frame $B$ with respect to frame A, $u_x$ be the velocity of frame C with respect to frame A, and $u_x'$ be the velocity of frame C with respect to frame B, then we have $${\textstyle u_x'=\frac{u_x-V}{1-u_xV/c^2}\quad\text{and}\quad u_x=\frac{u_x'+V}{1+u_x'V/c^2}}$$ $$\begin{align*} u_x&=\frac{u_x'+V}{1+u_x'V/c^2}\\ &=\frac{\frac{c}{10}+\frac{c}{2}}{1+\frac{c}{10}\frac{c}{2}\frac{1}{c^2}}\\ &=\frac{4}{7}c \end{align*}$$

Hence, answer is (B)

4. A system of four particles is in $x$-$y$ plane. Of these, two particles each of mass $m$ are located at $( -1, 1)$ and $(1, -1)$. The remaining two particles each of mass $2m$ are located at $(1, 1)$ and $( -1, -1)$. The $xy$-component of the moment of inertia tensor of this system of particles is
1. $10m$
2. $-10m$
3. $2m$
4. $-2m$

Inertia tensor is given by $${\scriptscriptstyle{I}=\begin{pmatrix} \sum\limits_\alpha\! m_\alpha\left(x_{\alpha,2}^2+x_{\alpha,3}^2\right)&\!-\!\sum\limits_\alpha\! m_\alpha x_{\alpha,1}x_{\alpha,2}&\!-\!\sum\limits_\alpha\! m_\alpha x_{\alpha,1}x_{\alpha,3}\\ -\sum\limits_\alpha\! m_\alpha x_{\alpha,2}x_{\alpha,1}&\!\sum\limits_\alpha\! m_\alpha\left(x_{\alpha,1}^2+x_{\alpha,3}^2\right)&\!-\!\sum\limits_\alpha\! m_\alpha x_{\alpha,2}x_{\alpha,3}\\ -\sum\limits_\alpha\! m_\alpha x_{\alpha,3}x_{\alpha,1}&\!-\!\sum\limits_\alpha\! m_\alpha x_{\alpha,3}x_{\alpha,2}&\!\sum\limits_\alpha\! m_\alpha\left(x_{\alpha,1}^2+x_{\alpha,2}^2\right) \end{pmatrix} }$$ where, $\alpha$ runs over number of particles. Here, $x_{\alpha,1}$ means x-coordinate of particle number $\alpha$, $x_{\alpha,2}$ means y-coordinate of particle number $\alpha$ and $x_{\alpha,3}$ means z-coordinate of particle number $\alpha$

For a two-dimensional case, we have $${\scriptstyle{I}=\begin{pmatrix} \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)&-\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}&\sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right) \end{pmatrix} }$$ $$\begin{align*} I_{11}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,2}^2\right)\\ &=m+m+2m+2m\\ &=6m \end{align*}$$ $$\begin{align*} &I_{12}= -\sum\limits_\alpha m_\alpha x_{\alpha,1}x_{\alpha,2}\\ &={\scriptscriptstyle -(m(-1)(1)+m(1)(-1)+2m(1)(1)+2m(-1)(-1))}\\ &=-(-m-m+2m+2m)\\ &=-2m \end{align*}$$ $$\begin{align*} &I_{21}= -\sum\limits_\alpha m_\alpha x_{\alpha,2}x_{\alpha,1}\\ &={\scriptscriptstyle -(m(1)(-1)+m(-1)(1)+2m(1)(1)+2m(-1)(-1))}\\ &=-2m \end{align*}$$ $$\begin{align*} I_{22}&= \sum\limits_\alpha m_\alpha\left(x_{\alpha,1}^2\right)\\ &=m+m+2m+2m\\ &=6m \end{align*}$$ $${I}=\begin{pmatrix} 6m&-2m\\ -2m&6m \end{pmatrix}$$ Hence, $xy$-component of moment of inertia is $-2m$

Hence, answer is (D)

5. A particle of mass $m$ is represented by the wavefunction $\psi(x) = Ae^{ikx}$, where $k$ is the wavevector and $A$ is a constant. The magnitude of the probability current density of the particle is
1. $|A|^2\frac{\hbar k}{m}$
2. $|A|^2\frac{\hbar k}{2m}$
3. $|A|^2\frac{\left(\hbar k\right)^2}{m}$
4. $|A|^2\frac{\left(\hbar k\right)^2}{2m}$

$$j=\frac{i\hbar}{2m}\left(\psi\frac{\partial\psi^*}{\partial x}-\psi^*\frac{\partial\psi}{\partial x}\right)$$ $$\psi(x) = Ae^{ikx}$$ $$\psi^*(x) = A^*e^{-ikx}$$ $$\frac{\partial\psi}{\partial x} = ikAe^{ikx}$$ $$\frac{\partial\psi^*}{\partial x} = -ikA^*e^{-ikx}$$ $$\psi^*\frac{\partial\psi}{\partial x} = A^*e^{-ikx} ikAe^{ikx}=ik|A|^2$$ $$\begin{align*} \psi\frac{\partial\psi^*}{\partial x} &= Ae^{ikx}\left(-ikA^*e^{-ikx}\right)\\ &=-ik|A|^2 \end{align*}$$ $$\begin{align*} j&=\frac{i\hbar}{2m}\left(-ik|A|^2-ik|A|^2 \right)\\ &=|A|^2\frac{\hbar k}{m} \end{align*}$$

Hence, answer is (A)