Option A — Power series solution of Legendre’s differential equation

Legendre’s differential equation is

[ (1-x^2)y''-2xy'+l(l+1)y=0, ]

where (l) is a constant. The point (x=0) is an ordinary point, so we seek a power series solution of the form

[ y=\sum_{k=0}^{\infty}a_kx^k. ]

Then

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

Substituting in Legendre’s equation,

[ (1-x^2)\sum_{k=2}^{\infty}k(k-1)a_kx^{k-2} -2x\sum_{k=1}^{\infty}ka_kx^{k-1} +l(l+1)\sum_{k=0}^{\infty}a_kx^k=0. ]

This gives

[ \sum_{k=2}^{\infty}k(k-1)a_kx^{k-2} -\sum_{k=2}^{\infty}k(k-1)a_kx^k -2\sum_{k=1}^{\infty}ka_kx^k +l(l+1)\sum_{k=0}^{\infty}a_kx^k=0. ]

Shift the first sum by putting (k=m+2):

[ \sum_{m=0}^{\infty}(m+2)(m+1)a_{m+2}x^m -\sum_{m=2}^{\infty}m(m-1)a_mx^m -2\sum_{m=1}^{\infty}ma_mx^m +l(l+1)\sum_{m=0}^{\infty}a_mx^m=0. ]

Equating the coefficient of (x^m) to zero gives the recurrence relation

[ (m+2)(m+1)a_{m+2}

\bigl[m(m+1)-l(l+1)\bigr]a_m. ]

Hence

[ \boxed{ a_{m+2}

-\frac{(l-m)(l+m+1)}{(m+2)(m+1)},a_m } \qquad (m=0,1,2,\dots). ]

Thus the coefficients are determined in terms of the two arbitrary constants (a_0) and (a_1).

Two independent solutions

For even powers, taking (a_0\neq 0,\ a_1=0),

[ y_1(x)=a_0\left[ 1-\frac{l(l+1)}{2!}x^2 +\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots \right]. ]

For odd powers, taking (a_0=0,\ a_1\neq 0),

[ y_2(x)=a_1\left[ x-\frac{(l-1)(l+2)}{3!}x^3 +\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots \right]. ]

Therefore the general solution is

[ \boxed{ y(x)=a_0y_1(x)+a_1y_2(x) }. ]

These series converge for (|x|<1), since the nearest singular points of Legendre’s equation are at (x=\pm1).

Legendre polynomials

If (l=n), a non-negative integer, one of the above series terminates and gives a polynomial solution. Normalising it by (P_n(1)=1), we get the Legendre polynomial

[ \boxed{ P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n } ]

or explicitly,

[ P_n(x)=\sum_{r=0}^{\lfloor n/2\rfloor} (-1)^r \frac{(2n-2r)!}{2^n r!(n-r)!(n-2r)!} x^{n-2r}. ]

For example,

[ P_0(x)=1,\qquad P_1(x)=x,\qquad P_2(x)=\frac{3x^2-1}{2}. ]

The non-terminating solution is the Legendre function of the second kind, (Q_n(x)). Hence for integer (n),

[ \boxed{ y(x)=A P_n(x)+B Q_n(x) }. ]

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Option B — Curvilinear coordinates and metric elements

Curvilinear coordinates

If the position vector of a point is expressed as

[ \mathbf r=\mathbf r(u_1,u_2,u_3), ]

where (u_1,u_2,u_3) are not necessarily rectangular coordinates, then (u_1,u_2,u_3) are called curvilinear coordinates. The surfaces

[ u_1=\text{constant},\qquad u_2=\text{constant},\qquad u_3=\text{constant} ]

are called coordinate surfaces, and their intersections are coordinate curves.

If the coordinate surfaces intersect at right angles, the system is called an orthogonal curvilinear coordinate system.

Unit tangent vectors and scale factors

Along the coordinate curve of (u_1), the tangent vector is

[ \frac{\partial \mathbf r}{\partial u_1}. ]

Its magnitude is the scale factor

[ h_1=\left|\frac{\partial \mathbf r}{\partial u_1}\right|. ]

Similarly,

[ h_2=\left|\frac{\partial \mathbf r}{\partial u_2}\right|,\qquad h_3=\left|\frac{\partial \mathbf r}{\partial u_3}\right|. ]

The unit tangent vectors are therefore

[ \boxed{ \mathbf e_1=\frac{1}{h_1}\frac{\partial \mathbf r}{\partial u_1},\quad \mathbf e_2=\frac{1}{h_2}\frac{\partial \mathbf r}{\partial u_2},\quad \mathbf e_3=\frac{1}{h_3}\frac{\partial \mathbf r}{\partial u_3}. } ]

For an orthogonal system,

[ \mathbf e_i\cdot \mathbf e_j=\delta_{ij}. ]

Unit normal

The unit normal to the surface (u_1=\text{constant}) is (\mathbf e_1). Similarly, the unit normals to (u_2=\text{constant}) and (u_3=\text{constant}) are (\mathbf e_2) and (\mathbf e_3), respectively.

Differential length

The displacement vector is

[ d\mathbf r

\frac{\partial \mathbf r}{\partial u_1}du_1 +\frac{\partial \mathbf r}{\partial u_2}du_2 +\frac{\partial \mathbf r}{\partial u_3}du_3. ]

Using the unit tangent vectors,

[ d\mathbf r

h_1du_1\mathbf e_1 +h_2du_2\mathbf e_2 +h_3du_3\mathbf e_3. ]

Therefore the square of the differential length is

[ ds^2=d\mathbf r\cdot d\mathbf r

h_1^2du_1^2+h_2^2du_2^2+h_3^2du_3^2. ]

Hence

[ \boxed{ ds=\sqrt{h_1^2du_1^2+h_2^2du_2^2+h_3^2du_3^2}. } ]

Area element

On the surface (u_1=\text{constant}), the area element is formed by displacements along (u_2) and (u_3):

[ d\mathbf S_1

(h_2du_2\mathbf e_2)\times(h_3du_3\mathbf e_3)

h_2h_3du_2du_3,\mathbf e_1. ]

Thus

[ \boxed{ dS_1=h_2h_3,du_2du_3. } ]

Similarly,

[ \boxed{ dS_2=h_3h_1,du_3du_1,\qquad dS_3=h_1h_2,du_1du_2. } ]

Volume element

The volume element is the scalar triple product

[ dV

\left| (h_1du_1\mathbf e_1)\cdot \bigl[(h_2du_2\mathbf e_2)\times(h_3du_3\mathbf e_3)\bigr] \right|. ]

Since (\mathbf e_1,\mathbf e_2,\mathbf e_3) form an orthonormal triad,

[ \mathbf e_1\cdot(\mathbf e_2\times\mathbf e_3)=1. ]

Therefore

[ \boxed{ dV=h_1h_2h_3,du_1du_2du_3. } ]

Examples

Thus in orthogonal curvilinear coordinates, the unit tangents, differential length, area element and volume element 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 (Statement)

Hamilton's principle states that the motion of a dynamical system between two fixed configurations at times $t_1$ and $t_2$ is such that the action integral $S$ has a stationary (extremum) value, i.e., its first variation vanishes:

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

where $L = T - V$ is the Lagrangian of the system. The variation $\delta$ is taken over all possible paths connecting the same end points, with $\delta q_i(t_1) = \delta q_i(t_2) = 0$.

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Derivation of Lagrange's Equation

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

$$L = L(q_1, q_2, \dots, q_n, \dot{q}_1, \dot{q}_2, \dots, \dot{q}_n, t)$$

The action integral is

$$S = \int_{t_1}^{t_2} L(q_i, \dot{q}_i, t) \, dt$$

Taking the variation,

$$\delta S = \int_{t_1}^{t_2} \delta L \, dt = \int_{t_1}^{t_2} \sum_{i=1}^{n} \left( \frac{\partial L}{\partial q_i} \delta q_i + \frac{\partial L}{\partial \dot{q}_i} \delta \dot{q}_i \right) dt = 0$$

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,

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

The boundary term vanishes because $\delta q_i(t_1) = \delta q_i(t_2) = 0$. Hence,

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

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 yield the same dynamics as Newton's laws but are valid in any generalized coordinate system, making them far more convenient for constrained systems.

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OR

Cyclic Coordinates and Conservation Laws in Lagrangian Formulation

Cyclic (Ignorable) Coordinates

A generalized coordinate $q_k$ is called cyclic or ignorable if the Lagrangian does not depend explicitly on it:

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

The coordinate may still appear through its velocity $\dot{q}_k$, but its absence from $L$ itself is the defining property.

Conservation of Generalized Momentum

For a cyclic coordinate $q_k$, Lagrange's equation gives

$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_k} \right) - \frac{\partial L}{\partial q_k} = 0$$

Since $\dfrac{\partial L}{\partial q_k} = 0$,

$$\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_k} \right) = 0$$

Therefore,

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

The quantity $p_k$ is the generalized momentum conjugate to $q_k$, and it is conserved. This is the Lagrangian statement of the conservation of momentum: every cyclic coordinate corresponds to a conserved momentum.

Example: For a free particle, $L = \frac{1}{2}m\dot{x}^2$, so $x$ is cyclic and $p_x = m\dot{x} = \text{constant}$, i.e., linear momentum is conserved. For a particle in a central field, the angular coordinate $\theta$ is cyclic, giving conservation of angular momentum.

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

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

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 ($\dfrac{\partial L}{\partial t} = 0$), then

$$\frac{dh}{dt} = 0 \quad \Rightarrow \quad h = \text{constant}$$

For a conservative system with a time-independent potential, $h = T + V = E$, the total mechanical energy. Thus the conservation of energy follows directly from the time-translation symmetry of the Lagrangian.

One-dimensional wave equation for a stretched string

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

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

where (c) is the wave speed.

Given boundary and initial conditions

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

[ y(x,0)=f(x),\qquad \left(\frac{\partial y}{\partial t}\right)_{t=0}=0 ]

The last condition means the string is initially at rest.

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1. Separation of variables

Assume

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

Substituting in the wave equation,

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

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

For oscillatory solutions, take (\lambda=k^2>0). Then

[ X''+k^2X=0 ]

[ T''+c^2k^2T=0 ]

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2. Solution for (X(x))

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

Using (y(0,t)=0), we get (X(0)=0), so

[ A=0 ]

Thus

[ X(x)=B\sin kx ]

Using (y(l,t)=0), we get (X(l)=0), so

[ B\sin kl=0 ]

For a non-trivial solution, (B\neq 0). Hence

[ \sin kl=0 ]

[ kl=n\pi ]

[ k_n=\frac{n\pi}{l},\qquad n=1,2,3,\dots ]

Therefore,

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

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3. Solution for (T(t))

[ T''+c^2k_n^2T=0 ]

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

Hence the general solution is

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

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4. Use the initial velocity condition

Given

[ \left(\frac{\partial y}{\partial t}\right)_{t=0}=0 ]

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

[ \frac{\partial y}{\partial t}

\sum_{n=1}^{\infty} \left[ -A_n\frac{n\pi c}{l}\sin\frac{n\pi ct}{l} + B_n\frac{n\pi c}{l}\cos\frac{n\pi ct}{l} \right] \sin\frac{n\pi x}{l} ]

At (t=0),

[ 0=\sum_{n=1}^{\infty} B_n\frac{n\pi c}{l}\sin\frac{n\pi x}{l} ]

Therefore,

[ B_n=0 ]

So the solution reduces to

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

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5. Use the initial displacement condition

Given

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

Putting (t=0),

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

This is the Fourier sine series of (f(x)) in (0<x<l). By orthogonality,

[ \int_0^l \sin\frac{n\pi x}{l}\sin\frac{m\pi x}{l},dx

\begin{cases} \dfrac{l}{2}, & n=m\[6pt] 0, & n\neq m \end{cases} ]

Hence,

[ 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\sin\frac{n\pi x}{l} \cos\frac{n\pi ct}{l} } ]

where

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

This is the required solution of the one-dimensional wave equation for a string fixed at both ends, initially displaced into the shape (f(x)) and released from rest.

If the string also has an initial velocity (g(x)), then the general solution is

[ y(x,t)=\sum_{n=1}^{\infty} \left[ A_n\cos\frac{n\pi ct}{l} + B_n\sin\frac{n\pi ct}{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 problem, (g(x)=0), so (B_n=0).

Euler's equations of motion for a rigid body

Consider a rigid body rotating about a fixed point $O$ (or about its centre of mass) with angular velocity $\vec\omega$. Choose body-fixed axes $\hat e_1,\hat e_2,\hat e_3$ along the principal axes, with principal moments of inertia $I_1,I_2,I_3$ and $\vec\omega=\omega_1\hat e_1+\omega_2\hat e_2+\omega_3\hat e_3$.

Step 1: Angular momentum. Along principal axes the inertia tensor is diagonal, so

$\vec L=I_1\omega_1\hat e_1+I_2\omega_2\hat e_2+I_3\omega_3\hat e_3$

Step 2: Rotating-frame relation. Newton's law for rotation holds in the inertial (space) frame: $\vec N=\left(\dfrac{d\vec L}{dt}\right)_{\text{space}}$, where $\vec N$ is the external torque about $O$. For any vector $\vec A$,

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

Hence

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

Step 3: Evaluate the terms. In the body frame the unit vectors are fixed, so

$\left(\frac{d\vec L}{dt}\right)_{\text{body}}=I_1\dot\omega_1\hat e_1+I_2\dot\omega_2\hat e_2+I_3\dot\omega_3\hat e_3$

and

$$\vec\omega\times\vec L=\begin{vmatrix}\hat e_1&\hat e_2&\hat e_3\\ \omega_1&\omega_2&\omega_3\\ I_1\omega_1&I_2\omega_2&I_3\omega_3\end{vmatrix} =(I_3-I_2)\omega_2\omega_3\,\hat e_1+(I_1-I_3)\omega_3\omega_1\,\hat e_2+(I_2-I_1)\omega_1\omega_2\,\hat e_3$$

Step 4: Equate components.

$$\boxed{\begin{aligned} N_1&=I_1\dot\omega_1-(I_2-I_3)\,\omega_2\omega_3\\ N_2&=I_2\dot\omega_2-(I_3-I_1)\,\omega_3\omega_1\\ N_3&=I_3\dot\omega_3-(I_1-I_2)\,\omega_1\omega_2 \end{aligned}}$$

These are Euler's equations of motion of a rigid body.

Special cases.

• Torque-free motion ($\vec N=0$): $I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3$, and cyclically. These give conservation of kinetic energy $T=\tfrac12\sum I_i\omega_i^2$ and of $L^2$.
• Symmetric top ($I_1=I_2\neq I_3$): the third equation gives $\dot\omega_3=0$, so $\omega_3$ is constant, and $\vec\omega$ precesses about the symmetry axis with angular frequency $\Omega=\dfrac{I_3-I_1}{I_1}\,\omega_3$.
• Sphere ($I_1=I_2=I_3$): $\vec N=I\dot{\vec\omega}$, the rotational analogue of $\vec F=m\vec a$.

Transformation law of Christoffel symbols

Let $x^i$ be the original coordinates and $\bar x^i$ the new ones. The metric transforms as

$\bar g_{ab}=\frac{\partial x^p}{\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}\,g_{pq}$

The Christoffel symbols are

$[ab,c]=\tfrac12\left(\partial_b g_{ac}+\partial_a g_{bc}-\partial_c g_{ab}\right)\ \ (\text{first kind}),\qquad \Gamma^c_{ab}=g^{cd}[ab,d]\ \ (\text{second kind})$

First kind

Differentiate $\bar g_{ab}$ with respect to $\bar x^c$ (write $x^p_{,a}\equiv\partial x^p/\partial\bar x^a$):

$\frac{\partial\bar g_{ab}}{\partial\bar x^c}=\frac{\partial^2x^p}{\partial\bar x^c\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}g_{pq}+\frac{\partial x^p}{\partial\bar x^a}\frac{\partial^2x^q}{\partial\bar x^c\partial\bar x^b}g_{pq}+\frac{\partial x^p}{\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}\frac{\partial x^r}{\partial\bar x^c}\frac{\partial g_{pq}}{\partial x^r}$

Form the combination $\dfrac{\partial\bar g_{ac}}{\partial\bar x^b}+\dfrac{\partial\bar g_{bc}}{\partial\bar x^a}-\dfrac{\partial\bar g_{ab}}{\partial\bar x^c}$ by cyclic interchange of $a,b,c$ in the above.

• The terms containing $\partial g_{pq}/\partial x^r$ combine to $\dfrac{\partial x^p}{\partial\bar x^a}\dfrac{\partial x^q}{\partial\bar x^b}\dfrac{\partial x^r}{\partial\bar x^c}\,2[pq,r]$.
• Among the second-derivative terms, two of them are equal and add to $2\,\dfrac{\partial^2x^p}{\partial\bar x^a\partial\bar x^b}\dfrac{\partial x^q}{\partial\bar x^c}g_{pq}$; the remaining four cancel in pairs because $g_{pq}=g_{qp}$ and mixed partial derivatives commute.

Multiplying by $\tfrac12$:

$\boxed{\overline{[ab,c]}=\frac{\partial x^p}{\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}\frac{\partial x^r}{\partial\bar x^c}\,[pq,r]+g_{pq}\,\frac{\partial^2x^p}{\partial\bar x^a\partial\bar x^b}\frac{\partial x^q}{\partial\bar x^c}}$

Second kind

Start from $\bar\Gamma^c_{ab}=\bar g^{cd}\,\overline{[ab,d]}$ with $\bar g^{cd}=\dfrac{\partial\bar x^c}{\partial x^s}\dfrac{\partial\bar x^d}{\partial x^t}g^{st}$.

First term. Using $\dfrac{\partial\bar x^d}{\partial x^t}\dfrac{\partial x^r}{\partial\bar x^d}=\delta^r_t$ and $g^{sr}[pq,r]=\Gamma^s_{pq}$:

$\frac{\partial\bar x^c}{\partial x^s}\frac{\partial x^p}{\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}\,g^{sr}[pq,r]=\frac{\partial\bar x^c}{\partial x^s}\frac{\partial x^p}{\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}\,\Gamma^s_{pq}$

Second term. Using $\dfrac{\partial\bar x^d}{\partial x^t}\dfrac{\partial x^q}{\partial\bar x^d}=\delta^q_t$ and $g^{st}g_{pt}=\delta^s_p$:

$\frac{\partial\bar x^c}{\partial x^s}g^{st}g_{pq}\,\delta^q_t\,\frac{\partial^2x^p}{\partial\bar x^a\partial\bar x^b}=\frac{\partial\bar x^c}{\partial x^p}\frac{\partial^2x^p}{\partial\bar x^a\partial\bar x^b}$

Hence

$\boxed{\bar\Gamma^c_{ab}=\frac{\partial\bar x^c}{\partial x^s}\frac{\partial x^p}{\partial\bar x^a}\frac{\partial x^q}{\partial\bar x^b}\,\Gamma^s_{pq}+\frac{\partial\bar x^c}{\partial x^p}\frac{\partial^2x^p}{\partial\bar x^a\partial\bar x^b}}$

Remarks.

• The first term is the ordinary tensor law (one contravariant, two covariant indices); the second, inhomogeneous term shows that Christoffel symbols are not tensors.
• The inhomogeneous term vanishes only under linear transformations. This is why $\Gamma$ can be made zero at a point (as in Cartesian coordinates) but not in general.
• An equivalent form of the second term is $-\dfrac{\partial x^p}{\partial\bar x^a}\dfrac{\partial x^q}{\partial\bar x^b}\dfrac{\partial^2\bar x^c}{\partial x^p\partial x^q}$.

Brachistochrone Problem

The brachistochrone problem (from the Greek brachistos = shortest, chronos = time) is a classical problem in the calculus of variations. It asks:

Among all plane curves joining two fixed points A and B (B lower than A, not directly below it), along which a particle slides frictionlessly under gravity alone, which curve gives the shortest time of descent?

Statement of the problem

Let $A(0,0)$ and $B(x_1, y_1)$ be two points in a vertical plane, with $B$ below $A$. A bead slides from rest at $A$ along a smooth curve to $B$ under gravity. Find the shape of the curve $y = y(x)$ for which the total time of descent is minimum.

Formulation

Take $x$ horizontal and $y$ vertically downward. By conservation of energy, if the particle starts from rest at $A$,

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

The time to travel an element of arc $ds = \sqrt{1 + y'^2}\,dx$ is $dt = \dfrac{ds}{v}$. Hence the total time is the functional

$T[y] = \int_{0}^{x_1} \frac{\sqrt{1 + y'^2}}{\sqrt{2gy}}\,dx, \qquad y' = \frac{dy}{dx}$

We must minimise $T[y]$ subject to $y(0) = 0$ and $y(x_1) = y_1$.

Solution by the calculus of variations

The integrand $f(y, y') = \dfrac{\sqrt{1 + y'^2}}{\sqrt{2gy}}$ does not contain $x$ explicitly, so the Beltrami identity applies:

$f - y'\frac{\partial f}{\partial y'} = C \quad (\text{constant})$

Substituting and simplifying gives

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

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

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

Result

The curve of quickest descent is a cycloid — the path traced by a point on the rim of a circle of radius $a$ rolling along the horizontal line $y = 0$. The constant $a$ is fixed by the condition that the cycloid passes through $B(x_1, y_1)$.

Physical significance

• The cycloid is not the straight line (shortest distance) nor the arc of steepest initial slope; it is the curve that trades steepness at the start for speed gained early, minimising total time.
• The problem was posed by Johann Bernoulli in 1696 and solved by Newton, Leibniz, L'Hôpital and the Bernoullis. Its solution marked the birth of the calculus of variations and led to the Euler–Lagrange equation.
• The same cycloid is also the tautochrone: the time of descent to the lowest point is the same for all starting positions on the curve.

(a) Coriolis force

Definition. When a body moves with velocity $\vec v$ in a frame of reference rotating with constant angular velocity $\vec\omega$, it experiences an additional inertial (pseudo) force, called the Coriolis force, given by

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

where $m$ is the mass of the body. It is a fictitious force, because it does not arise from any physical interaction but only from the acceleration of the rotating frame; it must be added to the real forces when Newton's laws are applied in that frame.

Properties.

• It is proportional to the mass $m$, to the speed $v$ and to $\omega$.
• It is perpendicular to both $\vec\omega$ and $\vec v$, so it can do no work and cannot change the speed of the body — it only deflects its path.
• It vanishes when $\vec v=0$ or when $\vec v$ is parallel to $\vec\omega$ (e.g. motion along the axis of rotation).
• Its direction is given by the vector product $\vec v\times\vec\omega$ (right-hand rule).

Examples. Deflection of winds and ocean currents (to the right in the northern hemisphere, to the left in the southern hemisphere), formation of cyclones, the eastward deflection of a freely falling body, and the slow rotation of the plane of a Foucault pendulum.

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OR — Scattering cross-section

When a beam of particles is incident on a target, the cross-section measures the probability of scattering.

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

$\frac{d\sigma}{d\Omega}=\frac{\text{number scattered into }d\Omega \text{ per unit time}}{N\,d\Omega}$

Total cross-section is the integral over all directions:

$\sigma_{\text{tot}}=\int \frac{d\sigma}{d\Omega}\,d\Omega=\int_0^{2\pi}\!\!\int_0^{\pi}\frac{d\sigma}{d\Omega}\sin\theta\,d\theta\,d\phi$

Relation with impact parameter. For a central potential, particles with impact parameter between $b$ and $b+db$ are scattered between $\theta$ and $\theta+d\theta$, so $d\sigma=2\pi b\,db$, giving

$\frac{d\sigma}{d\Omega}=\frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right|$

Unit. $1\ \text{barn}=10^{-28}\ \text{m}^2$. For Rutherford scattering of $\alpha$-particles,

$\frac{d\sigma}{d\Omega}=\left(\frac{Z_1Z_2e^2}{4\pi\varepsilon_0}\right)^{2}\frac{1}{16E^{2}\sin^{4}(\theta/2)}=\left(\frac{Ze^2}{8\pi\varepsilon_0E}\right)^{2}\frac{1}{\sin^{4}(\theta/2)}$

where $Z_1=2$ (charge of the $\alpha$-particle), $Z_2=Z$ (charge number of the nucleus) and $E$ is the kinetic energy of the $\alpha$-particle.

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(b) D'Alembert's principle and vertical work

D'Alembert's principle. Newton's second law for the $i$-th particle of a system is $\vec F_i=\dot{\vec p_i}$, i.e. $\vec F_i-\dot{\vec p_i}=0$. Multiplying by an arbitrary virtual displacement $\delta\vec r_i$ (consistent with the constraints) and summing over all particles,

$\sum_i\left(\vec F_i-\dot{\vec p_i}\right)\cdot\delta\vec r_i=0$

This is D'Alembert's principle. The term $-\dot{\vec p_i}=-m_i\vec a_i$ is the reversed effective force (inertial force). The principle thus converts a problem of dynamics into one of statics: the applied forces together with the reversed effective forces form a system in equilibrium for any virtual displacement. For ideal constraints the constraint forces do no virtual work, so they drop out, and the principle leads directly to Lagrange's equations.

Concept of vertical work done. Work is said to be done when a force produces a displacement in its own direction: $dW=\vec F\cdot d\vec s$. For a vertical displacement, the work done against gravity in raising a body of mass $m$ through a height $h$ is

$W=\int_0^{h} mg\,dh=mgh$

Key points:

• The work is positive when the body is lifted and negative when it falls freely under gravity ($W=-mgh$).
• Gravity is a conservative force, so this work depends only on the initial and final heights, not on the path followed; it equals the change in potential energy, $W=-\Delta U$.
• If the body is carried horizontally at constant height, no work is done against gravity ($\vec F\perp d\vec s$).
• The virtual work of the applied forces for a virtual displacement $\delta\vec r$ is $\delta W=\sum\vec F_i\cdot\delta\vec r_i$; for equilibrium of an ideal system $\delta W=0$.

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OR — Lorentz transformation

The Lorentz transformation relates the space and time coordinates $(x,y,z,t)$ of an event in an inertial frame $S$ to the coordinates $(x',y',z',t')$ in another inertial frame $S'$ moving with uniform velocity $v$ along the common $x$-axis. It is the transformation that leaves the speed of light $c$ invariant and preserves the form of the laws of physics in all inertial frames.

Transformation equations:

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

Inverse transformation (replace $v$ by $-v$):

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

where the Lorentz factor is

$\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}=\frac{1}{\sqrt{1-\beta^{2}}},\qquad \beta=\frac{v}{c}$

Basis. The transformation is linear (uniform motion must map to uniform motion), and it must leave the space–time interval invariant:

$s^{2}=x^{2}+y^{2}+z^{2}-c^{2}t^{2}=x'^{2}+y'^{2}+z'^{2}-c^{2}t'^{2}$

Consequences.

• For $v\ll c$, $\gamma\to 1$ and the equations reduce to the Galilean transformation $x'=x-vt,\ t'=t$.
• Length contraction: $L=L_0/\gamma$ (lengths parallel to the motion contract).
• Time dilation: $\Delta t=\gamma\,\Delta t_0$ (moving clocks run slow).
• Relativity of simultaneity: events simultaneous in $S$ are not simultaneous in $S'$.
• Velocity addition: $u_x'=\dfrac{u_x-v}{1-u_xv/c^{2}}$, which never exceeds $c$.

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

The half-range cosine series is

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

Constant term:

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

Coefficients:

$a_n=\frac{2}{\pi}\int_0^{\pi}x^2\cos nx\,dx$

Using $\displaystyle\int x^2\cos nx\,dx=\frac{x^2\sin nx}{n}+\frac{2x\cos nx}{n^2}-\frac{2\sin nx}{n^3}$ and applying the limits $0$ to $\pi$ (the $\sin n\pi$ terms vanish):

$\int_0^{\pi}x^2\cos nx\,dx=\frac{2\pi\cos n\pi}{n^2}=\frac{2\pi(-1)^n}{n^2}$

$\therefore\ a_n=\frac{2}{\pi}\cdot\frac{2\pi(-1)^n}{n^2}=\frac{4(-1)^n}{n^2}$

Hence 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 $(1+\cos 2t)$

By linearity,

$\mathcal{L}\{1+\cos 2t\}=\mathcal{L}\{1\}+\mathcal{L}\{\cos 2t\}$

Using $\mathcal{L}\{1\}=\dfrac{1}{s}$ and $\mathcal{L}\{\cos at\}=\dfrac{s}{s^2+a^2}$ with $a=2$:

$\boxed{\mathcal{L}\{1+\cos 2t\}=\frac{1}{s}+\frac{s}{s^2+4},\qquad s>0}$

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(c) Eigenvalues of a Hermitian matrix are real

Let $A$ be Hermitian, so $A^{\theta}=A$ (conjugate transpose equals itself). Let $\lambda$ be an eigenvalue with corresponding non-zero eigenvector $X$:

$AX=\lambda X \quad (1)$

Premultiplying (1) by $X^{\theta}$ (conjugate transpose of $X$):

$X^{\theta}AX=\lambda\,X^{\theta}X \quad (2)$

Now take the conjugate transpose of the scalar $X^{\theta}AX$. Since the conjugate transpose of a scalar is its complex conjugate,

$\overline{X^{\theta}AX}=(X^{\theta}AX)^{\theta}=X^{\theta}A^{\theta}X=X^{\theta}AX$

(the last step uses $A^{\theta}=A$). Hence $X^{\theta}AX$ equals its own conjugate, so $X^{\theta}AX$ is real.

Also $X^{\theta}X=|x_1|^2+|x_2|^2+\cdots+|x_n|^2$ is a positive real number, since $X\neq 0$.

From (2),

$\lambda=\frac{X^{\theta}AX}{X^{\theta}X}=\frac{\text{real}}{\text{positive real}}$

$\boxed{\therefore\ \lambda \text{ is real}}$

Thus every eigenvalue of a Hermitian matrix is real.

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(d) Central force

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

If $\hat{r}$ is the unit vector along the position vector $\vec{r}$ of the particle from the centre, then a central force is written as

$\vec{F}=f(r)\,\hat{r}$

where $f(r)$ is a function of $r$ only. The force is attractive if $f(r)<0$ and repulsive if $f(r)>0$.

Examples: gravitational force, electrostatic (Coulomb) force, and the force due to a stretched spring fixed at one end. A central force is conservative, and under it the angular momentum of the particle about the centre remains conserved.

Given: Distance of Mars from the Sun, (r_M = 1.5,r_E) Time of revolution of Earth, (T_E = 1\ \text{year})

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

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

Result: [ \boxed{T_M \approx 1.84\ \text{years} \approx 671\ \text{days}} ]

So, the time of revolution of Mars about the Sun is approximately 1.84 years.

Given:
Rest mass of proton, (m_0 = 1.67 \times 10^{-27}\ \text{kg})
Speed of proton, (v = 0.707c)
Speed of light, (c = 3 \times 10^8\ \text{m/s})

So, [ \beta = \frac{v}{c} = 0.707 ]

Relativistic factor, [ \gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{1}{\sqrt{1-(0.707)^2}} \approx 1.414 ]

1. Momentum

Relativistic momentum is [ p = \gamma m_0 v ] [ p = 1.414 \times 1.67 \times 10^{-27} \times 0.707 \times 3 \times 10^8 ] [ p \approx 5.01 \times 10^{-19}\ \text{kg m/s} ]

2. Total energy

Total relativistic energy is [ E = \gamma m_0 c^2 ] First, [ m_0 c^2 = 1.67 \times 10^{-27} \times (3 \times 10^8)^2 = 1.503 \times 10^{-10}\ \text{J} ] Then, [ E = 1.414 \times 1.503 \times 10^{-10} ] [ E \approx 2.125 \times 10^{-10}\ \text{J} ]

In MeV, [ E \approx \frac{2.125 \times 10^{-10}}{1.602 \times 10^{-13}} \approx 1326\ \text{MeV} ]

Answer: [ \boxed{p \approx 5.01 \times 10^{-19}\ \text{kg m/s}} ] [ \boxed{E \approx 2.125 \times 10^{-10}\ \text{J} \approx 1326\ \text{MeV}} ]

Let
[ \mathcal{L}{y(t)}=Y(s) ]

Given: [ y''-y'-6y=0,\qquad y(0)=1,\quad y'(0)=-1 ]

Using Laplace transform formulas: [ \mathcal{L}{y'}=sY(s)-y(0)=sY(s)-1 ] [ \mathcal{L}{y''}=s^2Y(s)-sy(0)-y'(0)=s^2Y(s)-s+1 ]

Substitute into the differential equation: [ (s^2Y(s)-s+1)-(sY(s)-1)-6Y(s)=0 ]

Simplify: [ (s^2-s-6)Y(s)-s+2=0 ] [ (s^2-s-6)Y(s)=s-2 ] [ Y(s)=\frac{s-2}{s^2-s-6} ]

Factor the denominator: [ s^2-s-6=(s-3)(s+2) ] So, [ Y(s)=\frac{s-2}{(s-3)(s+2)} ]

Now use partial fractions: [ \frac{s-2}{(s-3)(s+2)}=\frac{A}{s-3}+\frac{B}{s+2} ]

Multiplying by ((s-3)(s+2)): [ s-2=A(s+2)+B(s-3) ]

Put (s=3): [ 1=5A \implies A=\frac15 ]

Put (s=-2): [ -4=-5B \implies B=\frac45 ]

Hence, [ Y(s)=\frac15\cdot \frac1{s-3}+\frac45\cdot \frac1{s+2} ]

Taking inverse Laplace transform: [ y(t)=\frac15 e^{3t}+\frac45 e^{-2t} ]

[ \boxed{y(t)=\frac15 e^{3t}+\frac45 e^{-2t}} ]

Verification: [ y(0)=\frac15+\frac45=1 ] [ y'(t)=\frac35 e^{3t}-\frac85 e^{-2t} ] [ y'(0)=\frac35-\frac85=-1 ]

So the required solution is
[ \boxed{y(t)=\frac15 e^{3t}+\frac45 e^{-2t}} ]

The Bessel function of the first kind of order (\nu) is

[ J_\nu(x)=\sum_{r=0}^{\infty}\frac{(-1)^r}{r!,\Gamma(\nu+r+1)} \left(\frac{x}{2}\right)^{\nu+2r}. ]

We use the duplication formula

[ \Gamma\left(r+\frac12\right)=\frac{(2r)!}{4^r r!}\sqrt{\pi}, ]

and hence

[ \Gamma\left(r+\frac32\right) =\frac{(2r+1)!}{4^r r!}\frac{\sqrt{\pi}}{2}. ]

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(a) Prove that (J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x)

Put (\nu=\frac12) in the series:

[ J_{1/2}(x)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!,\Gamma\left(r+\frac32\right)} \left(\frac{x}{2}\right)^{2r+\frac12}. ]

Using

[ \Gamma\left(r+\frac32\right) =\frac{(2r+1)!}{4^r r!}\frac{\sqrt{\pi}}{2}, ]

we get

[ J_{1/2}(x) =\frac{2}{\sqrt{\pi}}\sum_{r=0}^{\infty} \frac{(-1)^r4^r}{(2r+1)!} \left(\frac{x}{2}\right)^{2r+\frac12}. ]

Now,

[ 4^r\left(\frac{x}{2}\right)^{2r+\frac12} =\sqrt{\frac{x}{2}},x^{2r}. ]

Hence

[ J_{1/2}(x) =\frac{2}{\sqrt{\pi}}\sqrt{\frac{x}{2}} \sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r+1)!}. ]

But

[ \frac{2}{\sqrt{\pi}}\sqrt{\frac{x}{2}} =\sqrt{\frac{2x}{\pi}}. ]

Therefore,

[ J_{1/2}(x) =\sqrt{\frac{2x}{\pi}} \sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r+1)!}. ]

Multiplying numerator and denominator inside the square root by (x),

[ J_{1/2}(x) =\sqrt{\frac{2}{\pi x}} \sum_{r=0}^{\infty}\frac{(-1)^r x^{2r+1}}{(2r+1)!}. ]

The series is the Maclaurin series for (\sin x):

[ \sin x=\sum_{r=0}^{\infty}\frac{(-1)^r x^{2r+1}}{(2r+1)!}. ]

Thus,

[ \boxed{J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x}. ]

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(b) Prove that (J_{-1/2}(x)=\sqrt{\frac{2}{\pi x}}\cos x)

Put (\nu=-\frac12) in the Bessel series:

[ J_{-1/2}(x)=\sum_{r=0}^{\infty} \frac{(-1)^r}{r!,\Gamma\left(r+\frac12\right)} \left(\frac{x}{2}\right)^{2r-\frac12}. ]

Using

[ \Gamma\left(r+\frac12\right) =\frac{(2r)!}{4^r r!}\sqrt{\pi}, ]

we obtain

[ J_{-1/2}(x) =\frac{1}{\sqrt{\pi}}\sum_{r=0}^{\infty} \frac{(-1)^r4^r}{(2r)!} \left(\frac{x}{2}\right)^{2r-\frac12}. ]

Now,

[ 4^r\left(\frac{x}{2}\right)^{2r-\frac12} =\sqrt{\frac{2}{x}},x^{2r}. ]

Hence

[ J_{-1/2}(x) =\frac{1}{\sqrt{\pi}}\sqrt{\frac{2}{x}} \sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r)!}. ]

But

[ \frac{1}{\sqrt{\pi}}\sqrt{\frac{2}{x}} =\sqrt{\frac{2}{\pi x}}. ]

Therefore,

[ J_{-1/2}(x) =\sqrt{\frac{2}{\pi x}} \sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r)!}. ]

The series is the Maclaurin series for (\cos x):

[ \cos x=\sum_{r=0}^{\infty}\frac{(-1)^r x^{2r}}{(2r)!}. ]

Thus,

[ \boxed{J_{-1/2}(x)=\sqrt{\frac{2}{\pi x}}\cos x}. ]

Hence both results are proved.

Significance of Gradient, Divergence and Curl

1. Gradient ($\nabla \phi$)

The gradient of a scalar field $\phi(x,y,z)$ is a vector field given by

$\nabla \phi = \frac{\partial \phi}{\partial x}\,\hat{i} + \frac{\partial \phi}{\partial y}\,\hat{j} + \frac{\partial \phi}{\partial z}\,\hat{k}$

Significance:

• It gives the direction and magnitude of the maximum rate of increase of the scalar field $\phi$ at a point. Its magnitude $|\nabla \phi|$ equals the greatest rate of change of $\phi$.
• It is always perpendicular to the level surface $\phi = \text{constant}$ at every point.
• In physics, the gradient of a scalar potential gives the corresponding vector field: $\vec{E} = -\nabla V$ (electric field from potential), $\vec{F} = -\nabla U$ (force from potential energy), and $\vec{g} = -\nabla \Omega$ (gravitational field from gravitational potential).
• In heat conduction, $\vec{q} = -k\,\nabla T$ shows that heat flows from high to low temperature, i.e. opposite to the temperature gradient.

2. Divergence ($\nabla \cdot \vec{A}$)

The divergence of a vector field $\vec{A}$ is a scalar given by

$\nabla \cdot \vec{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$

Significance:

• It measures the net outward flux of $\vec{A}$ per unit volume at a point, i.e. the strength of the source or sink there.
• If $\nabla \cdot \vec{A} > 0$, the point behaves as a source (field lines diverge); if $\nabla \cdot \vec{A} < 0$, it behaves as a sink (field lines converge); if $\nabla \cdot \vec{A} = 0$, the field is solenoidal (no source or sink, field lines are continuous).
• In physics, Gauss's law $\nabla \cdot \vec{E} = \rho/\varepsilon_0$ relates the divergence of the electric field to charge density, and $\nabla \cdot \vec{B} = 0$ expresses that magnetic monopoles do not exist.
• In fluid flow, $\nabla \cdot \vec{v} = 0$ is the equation of continuity for an incompressible fluid.

3. Curl ($\nabla \times \vec{A}$)

The curl of a vector field $\vec{A}$ is a vector given by

$\nabla \times \vec{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}$

Significance:

• It measures the maximum circulation of $\vec{A}$ per unit area at a point, i.e. the rotational tendency (vorticity) of the field.
• If $\nabla \times \vec{A} = 0$, the field is irrotational and can be expressed as the gradient of a scalar potential, $\vec{A} = \nabla \phi$ (conservative field).
• In physics, Faraday's law $\nabla \times \vec{E} = -\partial \vec{B}/\partial t$ and Ampère–Maxwell law $\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0\varepsilon_0\,\partial \vec{E}/\partial t$ express how changing fields produce circulating fields.
• In fluid dynamics, $\nabla \times \vec{v}$ is the vorticity, equal to twice the local angular velocity of a fluid element ($\vec{\omega}=\tfrac12\nabla\times\vec{v}$), so it tells whether the flow is rotational or irrotational.

Conclusion

In short, the gradient converts a scalar field into a vector field showing the direction of greatest increase, the divergence converts a vector field into a scalar measuring its source/sink strength, and the curl converts a vector field into another vector measuring its rotation. Together they form the mathematical backbone of field theory, electromagnetism, fluid mechanics and gravitational theory.