Browse & Filter

\(\newcommand{\Prime}{^\prime}\)For a four vector \(x=(\vec{x},x_4)\equiv(x_1,x_2,x_3,x_4)\), define a \(2\times2\) matrix \(M\) by \[M = x_\mu\sigma_\mu= x_4 + \vec{x}\cdot\vec{\sigma}\] where \(\sigma_4\) is \(2\times2\) identity matrix and \(\vec{\sigma}\) are Pauli matrices given by \begin{equation*} \sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad \sigma_1=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad \sigma_1=\begin{pmatrix}1&0\\0&-1\end{pmatrix} \end{equation*} Also define a matrix \(U\) by \begin{equation*} U = \cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2}(\hat{n}\cdot\vec{\sigma}) \end{equation*} where \(\hat{n}=(n_1,n_2,n_3)\) is a unit vector.

In Newtonian mechanics,\(x_4\) can be identified with time.

Comment The last equation shows that to every rotation in three  dimensions there are two \(SU(2)\) matrices given by \(\pm U\).

Using definition of Levi-Civita symbol, prove the following identities. \begin{eqnarray} \epsilon_{i\,j\,k}\,\epsilon_{i\,j\,k} \ &=&\ 6 \\ \epsilon_{i\,j\,k}\,\epsilon_{l\,j\,k} \ &=&\ 2\,\delta_{il} \\ \epsilon_{i\,j\,k}\,\epsilon_{l\,m\,k}\ &=&\ (\,\delta_{il}\,\delta_{jm}\,-\,\delta_{im}\,\delta_{jl}) \\ \epsilon_{i\,j\,k}\,\epsilon_{l\,m\,n}\ &=&\ \left(\begin{array}{ccc} \delta_{il}&\delta_{im}&\delta_{in}\\ \delta_{jl}&\delta_{jm}&\delta_{jn}\\ \delta_{kl}&\delta_{km}&\delta_{kn} \end{array} \right) \end{eqnarray}

Use result on \(\vec{A}\times(\vec{B}\times{\vec{C})}\) to give an alternate proof of identity (3) .

\(\newcommand{\Prime}{^\prime}\)Drive \(3\times 3\) rotation matrix corresponding to a rotation about axis \(\hat{n}\) by an angle \(\theta\) using the formula
\begin{equation}
 \vec{x}\Prime = (\vec{x}\cdot\hat{n})\hat{n} -
(\hat{n}\times\vec{x})\sin\theta -
\hat{n}\times(\hat{n}\times\vec{x})\cos\theta.
\end{equation}

Under what conditions can the problem of the motion of two charged particles in a uniform magnetic field be separated into the problem of the centre of mass motion and the relative motion problem of one body with reduced mass?

For a \(N\)- particle system, express the kinetic energy, the linear momentum, and the angular momentum of a system of \(N\) particles in terms of the Jacobi coordinates \begin{eqnarray} \vec{R}_j&=&\frac{m_1\vec{r}_1 + \ldots + m_j \vec{r}_j}{m_1+\ldots+m_j}-\vec{r}_{j+1}, \qquad j=1,\ldots,(N-1).\\ \vec{R}_N &=& \frac{m_1\vec{r}_1 + \ldots + m_N \vec{r}_N}{m_1+\ldots+m_N} \end{eqnarray}

Consider  a particle of mass \(\mu\) moving in a potential \[ V(r) = \frac{1}{2}\mu\omega^2 r^2  +\frac{\lambda^2}{2\mu r^2}. \] Does there exist a circular orbit for \(L=0\)? Assume  \(L=0, E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0   \text{ and } \dot{\theta(t)}\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here.

AKK

Comet Halley approaches the Sun to within 0.570 AU and its period is 75.6 years. How far from the Sun will this comet travel before it starts to return journey. Assume mass of Halley's comet is much smaller than the mass of the Sun.
[Given : 1 AU $= 1.5 \times 10^{11}$ m, $G= 6.63\times10^{-11}$ N.m\(^2\) /kg\(^2\), Mass of Sun=\( 1.99\times10^{30}\) kg ].

Consider a particle of mass \(\mu\) moving in a potential \[ V(r) = \frac{1}{2}\mu\omega^2 r^2 +\frac{\lambda^2}{2\mu r^2}. \] (a) Find condition on energy \(E\) and angular momentum \(L\) for circular orbits to exist.
(b) Does there exist a circular orbit for \(L=0\)?
(c) Assume \(L=0, E= \frac{25}{2}\mu\omega^2a^2\), \(\lambda=12\mu\omega^2\) Use initial conditions \[ r(t)\big|_{t=0} = 4a;\quad \dot{r}(t)\big|_{t=0} = 0 \text{ and } \dot{\theta(t)}\big|_{t=0}=0 \] solve the equations of motion and obtain \(r\), \(\theta\) as function of time. Describe the motion that takes place under conditions specified here.

KApoor

Find rotation matrix for a rotation by an angle \(\alpha\) about the axis \(1,2,1\) where \(\cos\alpha=\frac{3}{5}, \sin\alpha =\frac{4}{5}\).