Subfield: Mechanics

A The Extremum Principle in Mechanics

Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts).

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Figure 1
Context question:
A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.
Context answer:
\boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$}


Context question:
A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$.
Context answer:
\boxed{$v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$}


Extra Supplementary Reading Materials:

We define a quantity called action $A=m \int v(s) d s$, where $d s$ is the infinitesimal length along the trajectory of a particle of mass $m$ moving with speed $v(s)$. The integral is taken over the path. As an example, for a particle moving with constant speed $v$ on a circular path of radius $R$, the action $A$ for one revolution will be $2 \pi m R v$. For a particle with constant energy $E$, it can be shown that of all the possible trajectories between two fixed points, the actual trajectory is the one on which $A$ defined above is an extremum (minimum or maximum). Historically this is known as the Principle of Least Action (PLA).
Context question:
A.3 PLA implies that the trajectory of a particle moving between two fixed points in a region of constant <br> potential will be a straight line. Let the two fixed points $\mathrm{O}$ and $\mathrm{P}$ in Fig. 1 have coordinates $(0,0)$ and <br> $\left(x_{0}, y_{0}\right)$ respectively and the boundary point where the particle transits from region I to region II have <br> coordinates $\left(x_{1}, \alpha\right)$. Note that $x_{1}$ is fixed and the action depends on the coordinate $\alpha$ only. State the <br> expression for the action $A(\alpha)$. Use PLA to obtain the relationship between $v_{1} / v_{2}$ and these coordinates.
Context answer:
\boxed{$\frac{v_{1}}{v_{2}}=\frac{(y_{0}-\alpha)(x_{1}^{2}+\alpha^{2})^{1 / 2}}{\alpha[(x_{0}-x_{1})^{2}+(y_{0}-\alpha)^{2}]^{1 / 2}}$}


Extra Supplementary Reading Materials:

B The Extremum Principle in Optics

A light ray travels from medium I to medium II with refractive indices $n_{1}$ and $n_{2}$ respectively. The two media are separated by a line parallel to the $x$-axis. The light ray makes an angle $i_{1}$ with the $y$-axis in medium I and $i_{2}$ in medium II (see Fig. 2). To obtain the trajectory of the ray, we make use of another extremum (minimum or maximum) principle known as Fermat's principle of least time.

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Figure 2
Context question:
B.1 The principle states that between two fixed points, a light ray moves along a path such that time taken between the two points is an extremum. Derive the relation between $\sin i_{1}$ and $\sin i_{2}$ on the basis of Fermat's principle.
Context answer:
\boxed{$n_{1} \sin i_{1}=n_{2} \sin i_{2}$}


Extra Supplementary Reading Materials:

Shown in Fig. 3 is a schematic sketch of the path of a laser beam incident horizontally on a solution of sugar in which the concentration of sugar decreases with height. As a consequence, the refractive index of the solution also decreases with height.

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Figure 3: Tank of Sugar Solution
Context question:
B.2 Assume that the refractive index $n(y)$ depends only on $y$. Use the equation obtained in B1 to obtain the expression for the slope $d y / d x$ of the beam's path in terms of refractive index $n_{0}$ at $y=0$ and $n(y)$.
Context answer:
\boxed{$\frac{d y}{d x}=-\sqrt{(\frac{n(y)}{n_{0} \sin i_{0}})^{2}-1}$}


Question: B.3 The laser beam is directed horizontally from the origin $(0,0)$ into the sugar solution at a height $y_{0}$ from the Obtain an expression for $x$ in terms of $y$ and related quantities for the actual trajectory of the laser beam.[^0]

You may use: $\int \sec \theta d \theta=\ln (\sec \theta+\tan \theta)+$ constant, where $\sec \theta=1 / \cos \theta$ or <br> $\int \frac{d x}{\sqrt{x^{2}-1}}=\ln \left(x+\sqrt{x^{2}-1}\right)+$ constant

Solution:
$$
\int \frac{d y}{\sqrt{\left(\frac{n_{0}-k y}{n_{0} \sin i_{0}}\right)^{2}-1}}=-\int d x
$$

Note $i_{0}=90^{\circ}$ so $\sin i_{0}=1$.



Method I We employ the substitution

$$
\begin{gathered}
\xi=\frac{n_{0}-k y}{n_{0}} \\
\int \frac{d \xi\left(-\frac{n_{0}}{k}\right)}{\sqrt{\xi^{2}-1}}=-\int d x
\end{gathered}
$$

Let $\xi=\sec \theta$. Then

$$
\frac{n_{0}}{k} \ln (\sec \theta+\tan \theta)=x+c
$$

Or METHOD II 

We employ the substition

$$
\begin{gathered}
\xi=\frac{n_{0}-k y}{n_{0}} \\
\int \frac{d \xi\left(-\frac{n_{0}}{k}\right)}{\sqrt{\xi^{2}-1}}=-\int d x \\
-\frac{n_{0}}{k} \ln \left(\frac{n_{0}-k y}{n_{0}}+\sqrt{\left(\frac{n_{0}-k y}{n_{0}}\right)^{2}-1}\right)=-x+c
\end{gathered}
$$

Now continuing

Considering the substitutions and boundary condition, $x=0$ for $y=0$ we obtain that the constant $c=0$.

Hence we obtain the following trajectory:

$$
x=\frac{n_{0}}{k} \ln \left(\frac{n_{0}-k y}{n_{0}}+\sqrt{\left(\frac{n_{0}-k y}{n_{0}}\right)^{2}-1}\right)
$$

Final answer: $x=\frac{n_{0}}{k} \ln (\frac{n_{0}-k y}{n_{0}}+\sqrt{(\frac{n_{0}-k y}{n_{0}})^{2}-1})$