Option 1 — Power series solution of Legendre's differential equation

Legendre's equation is

$(1-x^2)\,y''-2x\,y'+n(n+1)\,y=0 \qquad \cdots(1)$

where $n$ is a real parameter. The points $x=0$ is an ordinary point (the coefficients $1-x^2$ and $-2x$ are analytic at $x=0$), so a power series solution of the form

$y=\sum_{k=0}^{\infty}a_kx^k \qquad \cdots(2)$

exists about $x=0$. Differentiating (2) term by term,

$y'=\sum_{k=1}^{\infty}k\,a_kx^{k-1},\qquad y''=\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2}.$

Substituting these in (1):

$\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2}-x^2\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2}-2x\sum_{k=1}^{\infty}ka_kx^{k-1}+n(n+1)\sum_{k=0}^{\infty}a_kx^k=0$

$\Rightarrow\ \sum_{k=2}^{\infty}k(k-1)a_kx^{k-2}-\sum_{k=2}^{\infty}k(k-1)a_kx^{k}-\sum_{k=1}^{\infty}2ka_kx^{k}+\sum_{k=0}^{\infty}n(n+1)a_kx^{k}=0.$

In the first sum put $k\to k+2$, so that it becomes $\sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^{k}$. The remaining three sums may be written from $k=0$ onward, since the terms with $k=0,1$ vanish automatically. Hence

$\sum_{k=0}^{\infty}\Big[(k+2)(k+1)a_{k+2}-k(k-1)a_k-2ka_k+n(n+1)a_k\Big]x^k=0.$

Equating the coefficient of $x^k$ to zero:

$(k+2)(k+1)a_{k+2}-\big[k(k-1)+2k-n(n+1)\big]a_k=0$

$(k+2)(k+1)a_{k+2}-\big[k(k+1)-n(n+1)\big]a_k=0.$

This gives the recurrence relation

$\boxed{\,a_{k+2}=\frac{k(k+1)-n(n+1)}{(k+1)(k+2)}\,a_k=-\frac{(n-k)(n+k+1)}{(k+1)(k+2)}\,a_k\,}\qquad \cdots(3)$

since $k(k+1)-n(n+1)=(k-n)(k+n+1)$.

Two independent solutions. Relation (3) connects $a_k$ with $a_{k+2}$, so the even coefficients are determined by $a_0$ and the odd coefficients by $a_1$; these two constants are arbitrary. Thus

$y=a_0\,y_1(x)+a_1\,y_2(x),$

where, putting $k=0,2,4,\dots$ and $k=1,3,5,\dots$ in (3),

$y_1(x)=1-\frac{n(n+1)}{2!}x^2+\frac{(n-2)n(n+1)(n+3)}{4!}x^4-\cdots$

$y_2(x)=x-\frac{(n-1)(n+2)}{3!}x^3+\frac{(n-1)(n-3)(n+2)(n+4)}{5!}x^5-\cdots$

In general,

$y_1(x)=\sum_{r=0}^{\infty}(-1)^r\frac{n(n-2)\cdots(n-2r+2)\,(n+1)(n+3)\cdots(n+2r-1)}{(2r)!}\,x^{2r},$

$y_2(x)=\sum_{r=0}^{\infty}(-1)^r\frac{(n-1)(n-3)\cdots(n-2r+1)\,(n+2)(n+4)\cdots(n+2r)}{(2r+1)!}\,x^{2r+1}.$

Convergence. By the ratio test,

$\lim_{k\to\infty}\left|\frac{a_{k+2}x^{k+2}}{a_kx^k}\right|=|x|^2\lim_{k\to\infty}\frac{(n-k)(n+k+1)}{(k+1)(k+2)}=|x|^2,$

so both series converge for $|x|<1$ and diverge for $|x|>1$; the radius of convergence is $1$, the distance from $x=0$ to the nearest singular point $x=\pm1$. Since $y_1$ and $y_2$ are linearly independent (their Wronskian at $x=0$ equals $1$), the general solution for $|x|<1$ is

$\boxed{\,y=A\,y_1(x)+B\,y_2(x)\,}$

Legendre polynomials. If $n$ is a non-negative integer, the factor $(n-k)$ in (3) vanishes when $k=n$, so one of the series terminates and becomes a polynomial of degree $n$; the other series remains infinite. Choosing the arbitrary constant so that the polynomial equals $1$ at $x=1$ gives the Legendre polynomial $P_n(x)$:

$P_0(x)=1,\quad P_1(x)=x,\quad P_2(x)=\tfrac12(3x^2-1),\quad P_3(x)=\tfrac12(5x^3-3x),\ \dots$

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Option 2 — Curvilinear coordinates and the orthogonal coordinate system

Definition. If the Cartesian coordinates $(x,y,z)$ of a point are expressed as single-valued, continuously differentiable functions of three independent variables $q_1,q_2,q_3$,

$x=x(q_1,q_2,q_3),\qquad y=y(q_1,q_2,q_3),\qquad z=z(q_1,q_2,q_3),$

then $(q_1,q_2,q_3)$ are called curvilinear coordinates of the point. The surfaces $q_1=$ const., $q_2=$ const., $q_3=$ const. are the coordinate surfaces, and their intersections are the coordinate curves. The system is orthogonal if the coordinate curves cut one another at right angles.

Let $\vec r=x\hat i+y\hat j+z\hat k$ be the position vector. Then

$\vec r=\vec r(q_1,q_2,q_3),\qquad d\vec r=\frac{\partial\vec r}{\partial q_1}dq_1+\frac{\partial\vec r}{\partial q_2}dq_2+\frac{\partial\vec r}{\partial q_3}dq_3.$

Scale factors and unit tangent vectors. Define

$h_i=\left|\frac{\partial\vec r}{\partial q_i}\right|,\qquad i=1,2,3 .$

The vectors $\partial\vec r/\partial q_i$ are tangent to the $q_i$-curves. Hence the unit tangent vectors are

$\boxed{\ \hat e_i=\frac{1}{h_i}\frac{\partial\vec r}{\partial q_i}\ },\qquad\text{so that}\qquad \frac{\partial\vec r}{\partial q_i}=h_i\hat e_i .$

For an orthogonal system $\hat e_1,\hat e_2,\hat e_3$ form a right-handed triad, i.e. $\hat e_i\cdot\hat e_j=\delta_{ij}$, which is equivalent to

$\frac{\partial\vec r}{\partial q_i}\cdot\frac{\partial\vec r}{\partial q_j}=0\quad (i\ne j).$

Differential length (line element). Since $d\vec r=h_1dq_1\hat e_1+h_2dq_2\hat e_2+h_3dq_3\hat e_3$,

$ds^2=d\vec r\cdot d\vec r=h_1^2dq_1^2+h_2^2dq_2^2+h_3^2dq_3^2$

$\boxed{\ ds=\sqrt{h_1^2dq_1^2+h_2^2dq_2^2+h_3^2dq_3^2}\ }$

Volume element. The volume of the parallelepiped with edges $h_1dq_1\hat e_1,\ h_2dq_2\hat e_2,\ h_3dq_3\hat e_3$ is the absolute value of their scalar triple product. As the triad is orthonormal,

$dV=\left|\frac{\partial(x,y,z)}{\partial(q_1,q_2,q_3)}\right|dq_1dq_2dq_3=(\hat e_1h_1dq_1)\cdot\big[(\hat e_2h_2dq_2)\times(\hat e_3h_3dq_3)\big]$

$\boxed{\ dV=h_1h_2h_3\,dq_1dq_2dq_3\ }$

Area element. On the coordinate surface $q_3=$ const., the element is the parallelogram with sides $h_1dq_1\hat e_1$ and $h_2dq_2\hat e_2$:

$dA_3=|h_1dq_1\hat e_1\times h_2dq_2\hat e_2|=h_1h_2\,dq_1dq_2 .$

Similarly,

$\boxed{\ dA_1=h_2h_3\,dq_2dq_3,\qquad dA_2=h_3h_1\,dq_3dq_1,\qquad dA_3=h_1h_2\,dq_1dq_2\ }$

Illustrations.

Thus in an orthogonal curvilinear system the unit tangent vectors, the differential length, the volume element and the area elements are completely determined by the three scale factors $h_1,h_2,h_3$.

Hamilton's Principle and Derivation of Lagrange's Equation of Motion

Hamilton's Principle (Principle of Least Action)

Hamilton's principle states that the actual path followed by a dynamical system between two specified configurations at times $t_1$ and $t_2$ is the one for which the time integral of the Lagrangian (called the action) is an extremum (usually a minimum).

If $L = T - V$ is the Lagrangian of the system, the action is defined as

$S = \int_{t_1}^{t_2} L\, dt$

Hamilton's principle requires

$\delta S = \delta \int_{t_1}^{t_2} L\, dt = 0$

where the variation $\delta$ is taken with the end points fixed, i.e. $\delta q_i(t_1) = \delta q_i(t_2) = 0$ for all generalized coordinates $q_i$.

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Derivation of Lagrange's Equation from Hamilton's Principle

Let the system be described by $n$ generalized coordinates $q_1, q_2, \dots, q_n$. The Lagrangian is

$L = L(q_1, q_2, \dots, q_n,\; \dot q_1, \dot q_2, \dots, \dot q_n,\; t)$

According to Hamilton's principle,

$\delta S = \delta \int_{t_1}^{t_2} L\, dt = 0$

Interchanging $\delta$ and integration,

$\int_{t_1}^{t_2} \delta L\, dt = 0$

Now

$\delta L = \sum_{i=1}^{n} \frac{\partial L}{\partial q_i}\,\delta q_i + \sum_{i=1}^{n} \frac{\partial L}{\partial \dot q_i}\,\delta \dot q_i$

Therefore,

$\int_{t_1}^{t_2} \left[ \sum_{i=1}^{n} \frac{\partial L}{\partial q_i}\,\delta q_i + \sum_{i=1}^{n} \frac{\partial L}{\partial \dot q_i}\,\delta \dot q_i \right] dt = 0$

Since $\delta \dot q_i = \dfrac{d}{dt}(\delta q_i)$, the second term becomes

$\int_{t_1}^{t_2} \frac{\partial L}{\partial \dot q_i}\,\frac{d}{dt}(\delta q_i)\,dt$

Integrating by parts,

$\int_{t_1}^{t_2} \frac{\partial L}{\partial \dot q_i}\,\frac{d}{dt}(\delta q_i)\,dt = \left[ \frac{\partial L}{\partial \dot q_i}\,\delta q_i \right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_i}\right)\delta q_i\,dt$

Since the end points are fixed, $\delta q_i(t_1) = \delta q_i(t_2) = 0$, so the integrated term vanishes. Hence,

$\int_{t_1}^{t_2} \sum_{i=1}^{n} \left[ \frac{\partial L}{\partial q_i} - \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_i}\right) \right] \delta q_i\,dt = 0$

Since the variations $\delta q_i$ are arbitrary and independent, each coefficient must vanish separately:

$\boxed{\;\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_i}\right) - \frac{\partial L}{\partial q_i} = 0\;}, \qquad i = 1, 2, \dots, n$

These are the Lagrange equations of motion for the system. They form a set of $n$ second-order differential equations whose solution gives $q_i(t)$, i.e. the motion of the system.

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OR

Cyclic Co-ordinates, Conservation of Momentum and Energy

Cyclic Co-ordinates

A generalized coordinate $q_k$ is said to be cyclic (or ignorable) if it does not appear explicitly in the Lagrangian, i.e. if

$\frac{\partial L}{\partial q_k} = 0$

although $\dot q_k$ may appear in $L$. For such a coordinate, Lagrange's equation gives

$\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_k}\right) - \frac{\partial L}{\partial q_k} = 0 \;\Rightarrow\; \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_k}\right) = 0$

Hence,

$p_k = \frac{\partial L}{\partial \dot q_k} = \text{constant}$

The quantity $p_k$ is called the generalized momentum conjugate to $q_k$. Thus, the generalized momentum conjugate to a cyclic coordinate is conserved.

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Conservation of Linear Momentum

For a free particle, the Lagrangian in Cartesian coordinates is

$L = \frac{1}{2}m(\dot x^2 + \dot y^2 + \dot z^2) - V$

If the potential $V$ does not depend on $x$ (i.e. the system is invariant under translation along $x$), then $x$ is cyclic:

$\frac{\partial L}{\partial x} = 0 \;\Rightarrow\; p_x = \frac{\partial L}{\partial \dot x} = m\dot x = \text{constant}$

This is the law of conservation of linear momentum: if the Lagrangian is invariant under translation in a given direction (space is homogeneous), the corresponding component of linear momentum is conserved.

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Conservation of Angular Momentum

For a particle moving in a plane, using polar coordinates $(r, \theta)$,

$L = \frac{1}{2}m(\dot r^2 + r^2\dot\theta^2) - V(r)$

Since $V$ depends only on $r$, the coordinate $\theta$ is cyclic:

$\frac{\partial L}{\partial \theta} = 0 \;\Rightarrow\; p_\theta = \frac{\partial L}{\partial \dot\theta} = m r^2 \dot\theta = \text{constant}$

This is the law of conservation of angular momentum: if the Lagrangian is invariant under rotation (space is isotropic), angular momentum is conserved.

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Conservation of Energy

The total time derivative of the Lagrangian is

$\frac{dL}{dt} = \sum_i \frac{\partial L}{\partial q_i}\dot q_i + \sum_i \frac{\partial L}{\partial \dot q_i}\ddot q_i + \frac{\partial L}{\partial t}$

Using Lagrange's equation, $\dfrac{\partial L}{\partial q_i} = \dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial \dot q_i}\right)$, we get

$\frac{dL}{dt} = \sum_i \left[ \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot q_i}\right)\dot q_i + \frac{\partial L}{\partial \dot q_i}\ddot q_i \right] + \frac{\partial L}{\partial t}$

$= \frac{d}{dt}\!\left( \sum_i \frac{\partial L}{\partial \dot q_i}\dot q_i \right) + \frac{\partial L}{\partial t}$

Rearranging,

$\frac{d}{dt}\!\left( \sum_i \frac{\partial L}{\partial \dot q_i}\dot q_i - L \right) = -\frac{\partial L}{\partial t}$

The quantity in brackets is the Jacobian integral (generalized energy):

$h = \sum_i p_i \dot q_i - L$

If the Lagrangian does not depend explicitly on time, $\dfrac{\partial L}{\partial t} = 0$, so

$\frac{dh}{dt} = 0 \;\Rightarrow\; h = \sum_i p_i \dot q_i - L = \text{constant}$

For a conservative system with a potential $V$ independent of velocities, $L = T - V$, and one can show $h = T + V = E$. Hence,

$T + V = E = \text{constant}$

This is the law of conservation of energy: if the Lagrangian is invariant under translation in time (time is homogeneous), the total energy of the system is conserved.

Let the transverse displacement of the string be (y(x,t)). The one-dimensional wave equation is

[ \frac{\partial^2 y}{\partial t^2}=c^2\frac{\partial^2 y}{\partial x^2}, \qquad 0<x<l,\quad t>0, ]

where (c=\sqrt{T/\rho}) is the wave speed.

Boundary conditions:

[ y(0,t)=0,\qquad y(l,t)=0. ]

Initial conditions: taking the initial displacement as (f(x)) and initial velocity zero,

[ y(x,0)=f(x),\qquad y_t(x,0)=0. ]

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1. Separation of variables

Assume

[ y(x,t)=X(x)T(t). ]

Substituting into the wave equation,

[ X T''=c^2 X'' T ]

[ \frac{T''}{c^2 T}=\frac{X''}{X}=-\lambda^2. ]

Thus,

[ X''+\lambda^2 X=0, ]

[ T''+c^2\lambda^2 T=0. ]

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2. Solve the (X)-equation

[ X''+\lambda^2 X=0. ]

Its general solution is

[ X(x)=A\cos \lambda x+B\sin \lambda x. ]

Using (y(0,t)=0),

[ X(0)T(t)=0 \implies X(0)=0. ]

So,

[ A=0. ]

Hence,

[ X(x)=B\sin \lambda x. ]

Using (y(l,t)=0),

[ X(l)T(t)=0 \implies X(l)=0. ]

Therefore,

[ B\sin \lambda l=0. ]

For a nontrivial solution, (B\neq 0), so

[ \sin \lambda l=0. ]

Thus,

[ \lambda l=n\pi,\qquad n=1,2,3,\dots ]

[ \lambda_n=\frac{n\pi}{l}. ]

The eigenfunctions are

[ X_n(x)=\sin \frac{n\pi x}{l}. ]

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3. Solve the (T)-equation

For (\lambda=\lambda_n),

[ T''+c^2\lambda_n^2 T=0. ]

So,

[ T_n(t)=C_n\cos \frac{n\pi c t}{l}+D_n\sin \frac{n\pi c t}{l}. ]

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4. General solution

By the principle of superposition,

[ y(x,t)=\sum_{n=1}^{\infty} \left[ A_n\cos \frac{n\pi c t}{l} + B_n\sin \frac{n\pi c t}{l} \right] \sin \frac{n\pi x}{l}. ]

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5. Apply initial conditions

Given

[ y_t(x,0)=0. ]

Differentiate (y(x,t)) with respect to (t):

[ y_t(x,t)=\sum_{n=1}^{\infty} \left[ -A_n\frac{n\pi c}{l}\sin \frac{n\pi c t}{l} + B_n\frac{n\pi c}{l}\cos \frac{n\pi c t}{l} \right] \sin \frac{n\pi x}{l}. ]

At (t=0),

[ y_t(x,0)=\sum_{n=1}^{\infty} B_n\frac{n\pi c}{l}\sin \frac{n\pi x}{l}=0. ]

Hence,

[ B_n=0. ]

So,

[ y(x,t)=\sum_{n=1}^{\infty} A_n\cos \frac{n\pi c t}{l} \sin \frac{n\pi x}{l}. ]

Now apply

[ y(x,0)=f(x). ]

Then,

[ f(x)=\sum_{n=1}^{\infty} A_n\sin \frac{n\pi x}{l}. ]

This is the Fourier sine series of (f(x)) on ((0,l)). Therefore,

[ A_n=\frac{2}{l}\int_0^l f(x)\sin \frac{n\pi x}{l},dx. ]

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Final solution

[ \boxed{ y(x,t)=\sum_{n=1}^{\infty} A_n \cos \frac{n\pi c t}{l} \sin \frac{n\pi x}{l} } ]

where

[ \boxed{ A_n=\frac{2}{l}\int_0^l f(x)\sin \frac{n\pi x}{l},dx } ]

and

[ c=\sqrt{\frac{T}{\rho}}. ]

If the initial velocity is also nonzero, say (y_t(x,0)=g(x)), then the general solution is

[ y(x,t)=\sum_{n=1}^{\infty} \left[ A_n\cos \frac{n\pi c t}{l} + B_n\sin \frac{n\pi c t}{l} \right] \sin \frac{n\pi x}{l}, ]

with

[ A_n=\frac{2}{l}\int_0^l f(x)\sin \frac{n\pi x}{l},dx, ]

[ B_n=\frac{2}{n\pi c}\int_0^l g(x)\sin \frac{n\pi x}{l},dx. ]

For the given case (g(x)=0), so (B_n=0), giving the solution above.

Setting up. Let the body axes $1,2,3$ be the principal axes at the fixed point (or centre of mass), with principal moments $I_1, I_2, I_3$ and angular velocity $\vec\omega=(\omega_1,\omega_2,\omega_3)$ measured along these axes. The angular momentum is $\vec L=(I_1\omega_1, I_2\omega_2, I_3\omega_3)$.

Derivation. In the inertial frame, torque equals the rate of change of angular momentum. The principal axes rotate with the body, so the time derivative in the inertial frame is the derivative relative to the body axes plus $\vec\omega\times\vec L$:

$\vec\tau=\left(\frac{d\vec L}{dt}\right)_{\text{body}}+\vec\omega\times\vec L$

The first component is

$\tau_1=I_1\dot\omega_1+\omega_2L_3-\omega_3L_2=I_1\dot\omega_1+\omega_2(I_3\omega_3)-\omega_3(I_2\omega_2)$

Taking the other two components in cyclic order gives Euler's equations of motion:

$I_1\dot\omega_1-(I_2-I_3)\,\omega_2\omega_3=\tau_1$ $I_2\dot\omega_2-(I_3-I_1)\,\omega_3\omega_1=\tau_2$ $I_3\dot\omega_3-(I_1-I_2)\,\omega_1\omega_2=\tau_3$

Torque-free motion ($\tau_i=0$):

$I_1\dot\omega_1=(I_2-I_3)\,\omega_2\omega_3,\quad I_2\dot\omega_2=(I_3-I_1)\,\omega_3\omega_1,\quad I_3\dot\omega_3=(I_1-I_2)\,\omega_1\omega_2$

These are nonlinear, coupled first-order equations. They conserve the kinetic energy $T=\tfrac12\sum I_i\omega_i^2$ and the magnitude $L^2=\sum I_i^2\omega_i^2$.

Special case. For a symmetric body with $I_1=I_2$, the third equation gives $\dot\omega_3=0$, so $\omega_3$ is constant. The other two equations then describe $\omega_1,\omega_2$ rotating about the symmetry axis with angular frequency $\Omega=\dfrac{(I_1-I_3)}{I_1}\,\omega_3$.

Setup. Let $x^i$ and $\bar x^i$ be two coordinate systems, with $A^i_{\ \bar m}=\frac{\partial x^i}{\partial\bar x^m},\qquad A^{\bar m}_{\ i}=\frac{\partial\bar x^m}{\partial x^i}.$

1. Christoffel symbols of the first kind

Definition: $[ij,k]=\tfrac12\left(\frac{\partial g_{jk}}{\partial x^i}+\frac{\partial g_{ik}}{\partial x^j}-\frac{\partial g_{ij}}{\partial x^k}\right)$

The metric is a covariant tensor of rank 2: $\bar g_{mn}=g_{ij}\,A^i_{\ \bar m}A^j_{\ \bar n}$

Differentiating with respect to $\bar x^p$: $\frac{\partial\bar g_{mn}}{\partial\bar x^p}=\frac{\partial g_{ij}}{\partial x^k}A^k_{\ \bar p}A^i_{\ \bar m}A^j_{\ \bar n}+g_{ij}\left(\frac{\partial^2x^i}{\partial\bar x^m\partial\bar x^p}A^j_{\ \bar n}+A^i_{\ \bar m}\frac{\partial^2x^j}{\partial\bar x^n\partial\bar x^p}\right)$

Form $[\bar m\bar n,\bar p]=\tfrac12\left(\partial_{\bar n}\bar g_{mp}+\partial_{\bar m}\bar g_{np}-\partial_{\bar p}\bar g_{mn}\right)$ and substitute:

• The terms containing $\partial g/\partial x$ combine to $[ij,k]\,A^i_{\ \bar m}A^j_{\ \bar n}A^k_{\ \bar p}$.
• The second-derivative terms use the symmetry of $g_{ij}$ and of mixed second derivatives. The terms with $\partial^2x/\partial\bar x^m\partial\bar x^n$ cancel in pairs, and the terms with $\partial^2x/\partial\bar x^m\partial\bar x^p$ cancel in pairs, leaving only $g_{ij}A^i_{\ \bar m}\,\partial^2x^j/\partial\bar x^n\partial\bar x^p$.

$\boxed{[\bar m\bar n,\bar p]=[ij,k]\frac{\partial x^i}{\partial\bar x^m}\frac{\partial x^j}{\partial\bar x^n}\frac{\partial x^k}{\partial\bar x^p}+g_{ij}\frac{\partial x^i}{\partial\bar x^m}\frac{\partial^2x^j}{\partial\bar x^n\partial\bar x^p}}$

2. Christoffel symbols of the second kind

Definition: $\Gamma^i_{\ jk}=g^{i\ell}[jk,\ell]$. The inverse metric transforms as $\bar g^{\,i\ell}=g^{ab}A^{\bar i}_{\ a}A^{\bar\ell}_{\ b}$.

Multiply the first-kind law (with $p$ replaced by $q$) by $\bar g^{\,iq}$ and sum over $q$: $\bar\Gamma^{\,i}_{\ mn}=g^{ab}A^{\bar i}_{\ a}A^{\bar q}_{\ b}\left([jk,c]A^j_{\ \bar m}A^k_{\ \bar n}A^c_{\ \bar q}+g_{jk}A^j_{\ \bar m}\frac{\partial^2x^k}{\partial\bar x^n\partial\bar x^q}\right)$

• First term. $A^{\bar q}_{\ b}A^c_{\ \bar q}=\delta^c_b$ and $g^{ab}[jk,b]=\Gamma^a_{\ jk}$, so it becomes $\Gamma^a_{\ jk}A^{\bar i}_{\ a}A^j_{\ \bar m}A^k_{\ \bar n}$.
• Second term. $g^{ab}g_{jk}A^j_{\ \bar m}$ with $A^{\bar q}_{\ b}$ gives $A^{\bar q}_{\ b}A^j_{\ \bar m}\,g^{ab}g_{jk}$. Using $g^{ab}g_{bk}=\delta^a_k$ and $A^{\bar q}_{\ j}A^j_{\ \bar m}=\delta^{\bar q}_{\ \bar m}$, it becomes $A^{\bar i}_{\ k}\,\dfrac{\partial^2x^k}{\partial\bar x^m\partial\bar x^n}$.

$\boxed{\bar\Gamma^{\,i}_{\ mn}=\Gamma^a_{\ bc}\frac{\partial\bar x^i}{\partial x^a}\frac{\partial x^b}{\partial\bar x^m}\frac{\partial x^c}{\partial\bar x^n}+\frac{\partial\bar x^i}{\partial x^k}\frac{\partial^2x^k}{\partial\bar x^m\partial\bar x^n}}$

3. Remarks

1. In each law the first term is the ordinary tensor rule. The second term is inhomogeneous, so Christoffel symbols are not tensors.
2. They vanish in Cartesian coordinates but not in polar coordinates, which confirms this.
3. The inhomogeneous terms cancel in the Riemann–Christoffel curvature tensor, so that is a genuine tensor.
4. The difference of two sets of Christoffel symbols (from two connections on the same manifold) is a tensor of type $(1,2)$, because the inhomogeneous term is the same for both and cancels.

Brachistochrone Problem

The brachistochrone problem (from the Greek brachistos = shortest, chronos = time) is one of the most celebrated problems in the calculus of variations. It asks:

Among all possible curves joining two fixed points A and B, where B lies below A but not directly beneath it, find the curve along which a particle, starting from rest at A and sliding under gravity alone (friction and air resistance neglected), reaches B in the shortest possible time.

Statement of the Problem

Let A and B be two fixed points in a vertical plane, with A above B. A bead slides from rest at A to B along a smooth curve under gravity. The problem is to determine the shape of the curve $y = y(x)$ for which the time of descent is minimum.

Formulation

Take A as the origin, with the x-axis horizontal and the y-axis vertically downward. Let the coordinates of B be $(x_1, y_1)$.

If $v$ is the speed of the particle at any point $(x, y)$, then by the conservation of energy (starting from rest at A):

$\frac{1}{2}mv^2 = mgy \quad \Rightarrow \quad v = \sqrt{2gy}$

The time taken to traverse an element of arc length $ds$ is:

$dt = \frac{ds}{v} = \frac{\sqrt{dx^2 + dy^2}}{\sqrt{2gy}} = \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}\,dx$

where $y' = \dfrac{dy}{dx}$. Hence the total time of descent is the functional:

$T = \int_0^{x_1} \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}\,dx$

The brachistochrone problem is therefore to find the function $y(x)$ that minimizes this integral, subject to the boundary conditions $y(0) = 0$ and $y(x_1) = y_1$.

Solution

The integrand $F = \dfrac{\sqrt{1 + y'^2}}{\sqrt{2gy}}$ does not contain $x$ explicitly, so we use Beltrami's identity:

$F - y'\frac{\partial F}{\partial y'} = c \quad \text{(constant)}$

Substituting and simplifying gives:

$y(1 + y'^2) = 2a \quad \text{(where } 2a \text{ is a constant)}$

Solving this differential equation by the substitution $y = a(1 - \cos\theta)$ yields the parametric equations:

$x = a(\theta - \sin\theta), \qquad y = a(1 - \cos\theta)$

These are the parametric equations of a cycloid — the curve traced by a point on the circumference of a circle of radius $a$ rolling along the horizontal line $y = 0$.

Conclusion

The curve of quickest descent is not a straight line (as one might naively expect) but an inverted cycloid (also called a brachistochrone curve). The particle initially falls steeply to gain speed quickly, then travels along the flatter portion of the curve to cover the remaining horizontal distance rapidly. The constant $a$ is determined from the condition that the cycloid passes through B.

Historical Note

The problem was posed by Johann Bernoulli in 1696 as a challenge to the mathematicians of Europe. It was solved independently by Newton (who solved it overnight), Leibniz, L'Hôpital, and Johann Bernoulli himself. Its solution led to the development of the calculus of variations by Euler and Lagrange.

(a) Define Coriolis force. [3]

Coriolis force is the fictitious (inertial) force that appears to act on a body moving with velocity $\vec v$ in a frame of reference rotating with constant angular velocity $\vec\omega$. It is defined as

$\vec F_{\text{cor}} = -2m\,(\vec\omega \times \vec v) \qquad\text{or}\qquad \vec F_{\text{cor}} = 2m\,(\vec v \times \vec\omega)$

Its magnitude is $F = 2mv\omega\sin\theta$, where $\theta$ is the angle between $\vec v$ and $\vec\omega$. The force acts perpendicular to both $\vec v$ and $\vec\omega$, so it does no work on the body.

Key points:

• It is a pseudo/inertial force — it arises only because the observer's frame is rotating (non-inertial), not from any physical interaction.
• It vanishes when $\vec v = 0$, when $\vec v \parallel \vec\omega$, and when $\omega = 0$.
• On the Earth it deflects moving bodies to the right in the Northern Hemisphere and to the left in the Southern Hemisphere; it is responsible for the circulation of cyclones, the trade winds and the erosion of river banks.

OR

Scattering cross-section [3]

When a beam of particles strikes a target, the fraction scattered into a given direction measures the "effective area" presented by the target. This is the scattering cross-section.

Differential cross-section $\dfrac{d\sigma}{d\Omega}$ is defined as the number of particles scattered per unit time into a solid angle $d\Omega$ in the direction $(\theta,\phi)$, divided by the incident flux (particles per unit area per unit time):

$\frac{d\sigma}{d\Omega}=\frac{\text{particles scattered into } d\Omega \text{ per unit time}}{\text{incident flux}\times d\Omega}$

Total cross-section is the integral over all directions:

$\sigma=\int_{0}^{4\pi}\frac{d\sigma}{d\Omega}\,d\Omega=\int_{0}^{\pi}\int_{0}^{2\pi}\frac{d\sigma}{d\Omega}\sin\theta\,d\theta\,d\phi$

It has the dimensions of area; the practical unit is the barn ($1\ \text{barn}=10^{-28}\ \text{m}^2$). For a rigid sphere of radius $R$, $\sigma=\pi R^2$. The Rutherford formula gives $\dfrac{d\sigma}{d\Omega}=\left(\dfrac{Ze^2}{4\pi\varepsilon_0}\right)^2\dfrac{1}{(4E)^2\sin^4(\theta/2)}$ for Coulomb scattering.

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(b) Explain D'Alembert's principle and concepts of vertical work. [3]

D'Alembert's principle states that for a system of particles, the sum of the virtual work done by the applied forces and the reversed effective (inertial) forces is zero for every virtual displacement consistent with the constraints:

$\sum_i\left(\vec F_i - m_i\vec a_i\right)\cdot\delta\vec r_i = 0$

Here $\vec F_i$ is the applied force and $-m_i\vec a_i$ is the reversed effective force (an inertial force equal to $ma$ acting opposite to the acceleration).

Significance:

• It converts a dynamical problem into a statically equivalent problem — the particle is treated as if in equilibrium under $\vec F_i$ and $-m_i\vec a_i$.
• Forces of constraint do no virtual work (for rigid constraints), so they drop out automatically, leaving only the applied forces.
• It is the starting point for Lagrange's equations of motion, $\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot q}\right)-\dfrac{\partial L}{\partial q}=0$.

Concept of vertical work: Work done when a body is displaced in the vertical direction. If a body of mass $m$ is raised through a height $h$,

$W=\int_0^h F\,dz = mgh \quad(\text{slowly, } a=0)$

If it is lifted with upward acceleration $a$, the applied force is $m(g+a)$ and $W=m(g+a)h$. The work done by gravity is $-mgh$, which appears as the change in gravitational potential energy, $W_{\text{grav}}=-\Delta U$. In D'Alembert's scheme the reversed effective force $-ma$ is added to the weight, so vertical work terms balance exactly as in statics.

OR

Lorentz transformation [3]

The Lorentz transformation gives the relation between the space and time coordinates $(x,y,z,t)$ of an event in an inertial frame $S$ and the coordinates $(x',y',z',t')$ in another inertial frame $S'$ moving with uniform velocity $v$ along the common $x$-axis. It follows from the two postulates of special relativity — the principle of relativity and the constancy of the speed of light.

$x'=\gamma(x-vt),\qquad y'=y,\qquad z'=z,\qquad t'=\gamma\left(t-\frac{vx}{c^2}\right)$

where $\gamma=\dfrac{1}{\sqrt{1-v^2/c^2}}$ is the Lorentz factor. The inverse transformation is obtained by replacing $v$ with $-v$:

$x=\gamma(x'+vt'),\qquad t=\gamma\left(t'+\frac{vx'}{c^2}\right)$

Important consequences:

• Length contraction: $L=L_0\sqrt{1-v^2/c^2}$.
• Time dilation: $\Delta t=\dfrac{\Delta t_0}{\sqrt{1-v^2/c^2}}$.
• Relativity of simultaneity and the relativistic velocity-addition formula $u_x'=\dfrac{u_x-v}{1-u_xv/c^2}$.
• The space–time interval $s^2=x^2+y^2+z^2-c^2t^2$ is invariant, and the transformation leaves the wave equation $\Box^2\psi=0$ unchanged.
• For $v\ll c$, $\gamma\to1$ and it reduces to the Galilean transformation $x'=x-vt,\ t'=t$.

(a) Cosine series for (f(x)=x^2,\ 0<x<\pi)

The half-range cosine series is

[ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx ]

where

[ a_0=\frac{2}{\pi}\int_0^\pi x^2,dx =\frac{2}{\pi}\cdot \frac{\pi^3}{3} =\frac{2\pi^2}{3} ]

and

[ a_n=\frac{2}{\pi}\int_0^\pi x^2\cos nx,dx. ]

Now,

[ \int x^2\cos nx,dx

\frac{x^2\sin nx}{n} + \frac{2x\cos nx}{n^2}

\frac{2\sin nx}{n^3} ]

Evaluating from (0) to (\pi), since (\sin n\pi=0) and (\cos n\pi=(-1)^n),

[ \int_0^\pi x^2\cos nx,dx

\frac{2\pi(-1)^n}{n^2} ]

Hence,

[ a_n=\frac{2}{\pi}\cdot \frac{2\pi(-1)^n}{n^2}

\frac{4(-1)^n}{n^2} ]

Therefore the required cosine series is

[ \boxed{ x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx, \qquad 0<x<\pi } ]

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(b) Laplace transform of (1+\cos 2t)

[ \mathcal{L}{1+\cos 2t}

\mathcal{L}{1}+\mathcal{L}{\cos 2t} ]

Using standard results,

[ \mathcal{L}{1}=\frac{1}{s}, \qquad \mathcal{L}{\cos at}=\frac{s}{s^2+a^2} ]

For (a=2),

[ \mathcal{L}{\cos 2t}=\frac{s}{s^2+4} ]

Thus,

[ \boxed{ \mathcal{L}{1+\cos 2t}

\frac{1}{s}+\frac{s}{s^2+4}

\frac{2s^2+4}{s(s^2+4)} } ]

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(c) Eigenvalues of a Hermitian matrix are real

Let (A) be a Hermitian matrix, so

[ A^\dagger=A. ]

Let (\lambda) be an eigenvalue of (A) and (x\neq 0) be the corresponding eigenvector. Then

[ Ax=\lambda x. ]

Taking the conjugate transpose,

[ x^\dagger A^\dagger=\overline{\lambda}x^\dagger. ]

Since (A^\dagger=A),

[ x^\dagger A=\overline{\lambda}x^\dagger. ]

Multiplying on the right by (x),

[ x^\dagger A x=\overline{\lambda}x^\dagger x. ]

But from (Ax=\lambda x), multiplying on the left by (x^\dagger),

[ x^\dagger A x=\lambda x^\dagger x. ]

Therefore,

[ \lambda x^\dagger x=\overline{\lambda}x^\dagger x. ]

Since (x\neq 0), we have (x^\dagger x>0). Hence,

[ \lambda=\overline{\lambda}. ]

A number equal to its conjugate is real. Therefore,

[ \boxed{\text{Every eigenvalue of a Hermitian matrix is real.}} ]

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(d) Define central force

A force acting on a particle is called a central force if its line of action always passes through a fixed point, called the centre of force, and its magnitude depends only on the distance of the particle from that fixed point.

In vector form, if the fixed point is taken as the origin and (\vec r) is the position vector of the particle, then

[ \boxed{\vec F=f(r)\hat r} ]

or

[ \boxed{\vec F=F(r)\frac{\vec r}{r}} ]

where (r=|\vec r|). Examples are gravitational force and electrostatic force. A central force is conservative and produces zero torque about the centre of force.

Given: Distance of Mars from the Sun,
$r_M = 1.5\,r_E$ where (r_E) is the Earth’s distance from the Sun.

Time period of Earth,
$T_E = 1\ \text{year}$

Formula (Kepler’s third law): $T^2 \propto r^3$ So, $\frac{T_M^2}{T_E^2} = \frac{r_M^3}{r_E^3}$

Substitution: $\frac{T_M^2}{(1)^2} = \left(\frac{1.5\,r_E}{r_E}\right)^3$ $T_M^2 = (1.5)^3 = 3.375$ $T_M = \sqrt{3.375} = 1.837\ \text{years}$

Result: $\boxed{T_M \approx 1.84\ \text{years}}$

So, the time of revolution of Mars about the Sun is approximately 1.84 years (about 671 days).