Bessel's Differential Equation

Statement. The differential equation

$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-n^2)y=0$

where $n$ is a real (or complex) constant called the order of the equation, is called Bessel's differential equation. In its standard (normal) form it is written as

$\frac{d^2y}{dx^2}+\frac{1}{x}\frac{dy}{dx}+\left(1-\frac{n^2}{x^2}\right)y=0 .$

Explanation.

1. Nature of the equation. It is a linear, homogeneous, second-order ordinary differential equation with variable coefficients. The point $x=0$ is a regular singular point (since $xP(x)=1$ and $x^2Q(x)=x^2-n^2$ are analytic at $x=0$), so the method of Frobenius applies and at least one solution can be obtained as a series about $x=0$.

2. Solutions. Applying the Frobenius method with $y=\sum_{r\ge 0}a_r x^{r+k}$ gives the indicial equation $k^2-n^2=0$, i.e. $k=\pm n$. The solution corresponding to $k=+n$ is the Bessel function of the first kind of order $n$:

$J_n(x)=\sum_{r=0}^{\infty}\frac{(-1)^r}{r!\,\Gamma(n+r+1)}\left(\frac{x}{2}\right)^{n+2r}.$

When $n$ is not an integer, $J_{-n}(x)$ is a second independent solution. When $n$ is an integer, the second independent solution is the Bessel function of the second kind $Y_n(x)$ (Neumann function). Hence the complete solution is

$y=A\,J_n(x)+B\,Y_n(x).$

3. Recurrence relations. Bessel functions of order $n$ satisfy the recurrence relations

$f_{n-1}(x)+f_{n+1}(x)=\frac{2n}{x}f_n(x),\qquad f_{n-1}(x)-f_{n+1}(x)=2f_n'(x),$

where $f_n$ stands for $J_n$ or $Y_n$. These relations are the key to showing that $f_n$ satisfies Bessel's equation, as done below.

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Proof that $f_n(x)$ satisfies Bessel's equation

Let $f_n(x)$ obey the two given recurrence relations:

$f_{n-1}+f_{n+1}=\frac{2n}{x}f_n \quad (1)$

$f_{n-1}-f_{n+1}=2f_n' \quad (2)$

Step 1. Obtain $f_{n-1}$ and $f_{n+1}$ in terms of $f_n$.

Adding (1) and (2):

$2f_{n-1}=\frac{2n}{x}f_n+2f_n' \quad\Longrightarrow\quad f_{n-1}=\frac{n}{x}f_n+f_n' \quad (3)$

Subtracting (2) from (1):

$2f_{n+1}=\frac{2n}{x}f_n-2f_n' \quad\Longrightarrow\quad f_{n+1}=\frac{n}{x}f_n-f_n' \quad (4)$

Step 2. Shift the index in (3). Replacing $n$ by $n+1$ in (3):

$f_n=\frac{n+1}{x}f_{n+1}+f_{n+1}' \quad (5)$

Step 3. Differentiate (4) with respect to $x$ (product rule on $\frac{n}{x}f_n$):

$f_{n+1}'=\frac{n}{x}f_n'-\frac{n}{x^2}f_n-f_n'' \quad (6)$

Step 4. Substitute (4) and (6) into (5):

$f_n=\frac{n+1}{x}\left(\frac{n}{x}f_n-f_n'\right)+\left(\frac{n}{x}f_n'-\frac{n}{x^2}f_n-f_n''\right)$

Expanding,

$f_n=\frac{n(n+1)}{x^2}f_n-\frac{n+1}{x}f_n'+\frac{n}{x}f_n'-\frac{n}{x^2}f_n-f_n'' .$

Step 5. Collect terms.

$f_n=\frac{n(n+1)-n}{x^2}f_n+\left(-\frac{n+1}{x}+\frac{n}{x}\right)f_n'-f_n''$

$f_n=\frac{n^2}{x^2}f_n-\frac{1}{x}f_n'-f_n'' .$

Step 6. Rearrange. Bringing all terms to the left,

$f_n''+\frac{1}{x}f_n'+\left(1-\frac{n^2}{x^2}\right)f_n=0 .$

Multiplying throughout by $x^2$,

$\boxed{x^2f_n''+xf_n'+(x^2-n^2)f_n=0}$

which is exactly Bessel's differential equation of order $n$.

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Conclusion. Since the recurrence relations (1) and (2) are satisfied by the Bessel functions $J_n(x)$ and $Y_n(x)$, the above derivation proves that every such function $f_n(x)$ satisfies Bessel's differential equation $x^2f_n''+xf_n'+(x^2-n^2)f_n=0$, and therefore $J_n$ and $Y_n$ are precisely the solutions of this equation, the general solution being $y=AJ_n(x)+BY_n(x)$.

Relativistic momentum–energy relation

Let a particle of rest mass (m_0) move with velocity (v) in an inertial frame. Define

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

The relativistic momentum is

[ \vec p=\gamma m_0\vec v=\frac{m_0\vec v}{\sqrt{1-v^2/c^2}}, ]

and the total relativistic energy is

[ E=\gamma m_0 c^2=\frac{m_0 c^2}{\sqrt{1-v^2/c^2}}. ]

The rest energy is

[ E_0=m_0 c^2, ]

and the kinetic energy is

[ K=E-E_0=(\gamma-1)m_0 c^2. ]

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Derivation of the relation

Consider the quantity

[ E^2-p^2c^2. ]

Substitute the expressions for (E) and (p):

[ E^2-p^2c^2

\gamma^2 m_0^2 c^4-\gamma^2 m_0^2 v^2 c^2. ]

Taking (\gamma^2 m_0^2 c^4) common,

[ E^2-p^2c^2

\gamma^2 m_0^2 c^4\left(1-\frac{v^2}{c^2}\right). ]

But

[ 1-\frac{v^2}{c^2}=\frac{1}{\gamma^2}. ]

Therefore,

[ E^2-p^2c^2

\gamma^2 m_0^2 c^4\cdot \frac{1}{\gamma^2}

m_0^2 c^4. ]

Hence,

[ \boxed{E^2=p^2c^2+m_0^2c^4} ]

or equivalently,

[ \boxed{E=\sqrt{p^2c^2+m_0^2c^4}}. ]

This is the relativistic momentum–energy relation.

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Interpretation

1. Relativistic dispersion relation
   The relation connects the total energy (E), momentum (p), and rest mass (m_0) of a free relativistic particle. It replaces the non-relativistic relation (E=p^2/2m_0).

2. Lorentz invariant
   The quantity

   [ E^2-p^2c^2=m_0^2c^4 ]

   is invariant under Lorentz transformation. It has the same value in all inertial frames. This invariant defines the rest mass of the particle.

3. Geometrical meaning
   The relation can be written as

   [ E^2=(pc)^2+(m_0c^2)^2. ]

   Thus (E), (pc), and (m_0c^2) form a right triangle in energy–momentum space. Here (E) is the hypotenuse, while (pc) and (m_0c^2) are the two perpendicular sides.

4. Non-relativistic limit
   For (v\ll c),

   [ E=m_0c^2+\frac{p^2}{2m_0}+\cdots ]

   Hence the kinetic energy becomes

   [ K=E-m_0c^2\approx \frac{p^2}{2m_0}, ]

   which is the classical result.

5. Ultra-relativistic limit
   For (v\to c), the rest mass term becomes negligible compared with (pc), so

   [ E\approx pc. ]

   This is the behaviour of highly relativistic particles.

6. Massless particles
   For a photon or any massless particle, (m_0=0), so

   [ E=pc. ]

   Hence for a photon,

   [ p=\frac{E}{c}=\frac{h\nu}{c}=\frac{h}{\lambda}. ]

7. Relation with kinetic energy
   From (E=K+m_0c^2), we get

   [ (K+m_0c^2)^2=p^2c^2+m_0^2c^4. ]

   Simplifying,

   [ \boxed{p^2c^2=K^2+2Km_0c^2} ]

   or

   [ \boxed{K=\sqrt{p^2c^2+m_0^2c^4}-m_0c^2}. ]

   This relation is widely used in nuclear and particle physics to calculate momentum from kinetic energy.

8. Rest energy
   When the particle is at rest, (p=0), and the relation gives

   [ E=m_0c^2. ]

   This shows that mass itself is a form of energy.

D'Alembert's Principle

D'Alembert's principle states that for a system of particles in motion, the sum of the applied forces and the inertial forces (equal to the rate of change of momentum, taken with a negative sign) acting on each particle is zero. In other words, if $\vec{F}_i$ is the total applied force on the $i$-th particle of mass $m_i$ and $\vec{a}_i$ its acceleration, then

$\vec{F}_i - m_i\vec{a}_i = 0 \quad \text{for each particle } i.$

Equivalently, the total virtual work done by the applied forces together with the inertial forces, for any virtual displacement $\delta\vec{r}_i$ consistent with the constraints, is zero:

$\sum_i (\vec{F}_i - m_i\vec{a}_i)\cdot\delta\vec{r}_i = 0.$

This principle converts a dynamical problem into a statical one by treating $-m_i\vec{a}_i$ as an additional (inertial) force.

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Derivation of Lagrange's Equation of Motion

Let the system be described by $n$ generalized coordinates $q_1, q_2, \dots, q_n$, so that the position of the $i$-th particle is

$\vec{r}_i = \vec{r}_i(q_1, q_2, \dots, q_n, t).$

Step 1: Virtual displacement in generalized coordinates

The virtual displacement of the $i$-th particle is

$\delta\vec{r}_i = \sum_{j=1}^{n}\frac{\partial \vec{r}_i}{\partial q_j}\,\delta q_j.$

Step 2: Apply D'Alembert's principle

$\sum_i (\vec{F}_i - m_i\vec{a}_i)\cdot\delta\vec{r}_i = 0.$

Substituting $\delta\vec{r}_i$:

$\sum_i \vec{F}_i\cdot\delta\vec{r}_i - \sum_i m_i\vec{a}_i\cdot\delta\vec{r}_i = 0. \quad (1)$

Step 3: The first term — generalized force

Define the generalized force

$Q_j = \sum_i \vec{F}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}.$

Then

$\sum_i \vec{F}_i\cdot\delta\vec{r}_i = \sum_j Q_j\,\delta q_j. \quad (2)$

Step 4: The second term — transform the inertial part

$\sum_i m_i\vec{a}_i\cdot\delta\vec{r}_i = \sum_i m_i\ddot{\vec{r}}_i\cdot\sum_j\frac{\partial \vec{r}_i}{\partial q_j}\,\delta q_j = \sum_j\left[\sum_i m_i\,\ddot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}\right]\delta q_j. \quad (3)$

Now use the identity

$\ddot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j} = \frac{d}{dt}\!\left(\dot{\vec{r}}_i\cdot\frac{\partial \vec{r}_i}{\partial q_j}\right) - \dot{\vec{r}}_i\cdot\frac{d}{dt}\!\left(\frac{\partial \vec{r}_i}{\partial q_j}\right).$

Since $\dot{\vec{r}}_i = \sum_k \dfrac{\partial \vec{r}_i}{\partial q_k}\dot{q}_k + \dfrac{\partial \vec{r}_i}{\partial t}$, we have the two standard relations

$\frac{\partial \dot{\vec{r}}_i}{\partial \dot{q}_j} = \frac{\partial \vec{r}_i}{\partial q_j}, \qquad \frac{d}{dt}\!\left(\frac{\partial \vec{r}_i}{\partial q_j}\right) = \frac{\partial \dot{\vec{r}}_i}{\partial q_j}.$

Using these, equation (3) becomes

$\sum_i m_i\vec{a}_i\cdot\delta\vec{r}_i = \sum_j\left[\frac{d}{dt}\!\left(\sum_i m_i\dot{\vec{r}}_i\cdot\frac{\partial \dot{\vec{r}}_i}{\partial \dot{q}_j}\right) - \sum_i m_i\dot{\vec{r}}_i\cdot\frac{\partial \dot{\vec{r}}_i}{\partial q_j}\right]\delta q_j.$

Since the kinetic energy is $T = \dfrac{1}{2}\sum_i m_i\,\dot{\vec{r}}_i^{\,2}$,

$\sum_i m_i\dot{\vec{r}}_i\cdot\frac{\partial \dot{\vec{r}}_i}{\partial \dot{q}_j} = \frac{\partial T}{\partial \dot{q}_j}, \qquad \sum_i m_i\dot{\vec{r}}_i\cdot\frac{\partial \dot{\vec{r}}_i}{\partial q_j} = \frac{\partial T}{\partial q_j}.$

Hence

$\sum_i m_i\vec{a}_i\cdot\delta\vec{r}_i = \sum_j\left[\frac{d}{dt}\!\left(\frac{\partial T}{\partial \dot{q}_j}\right) - \frac{\partial T}{\partial q_j}\right]\delta q_j. \quad (4)$

Step 5: Combine and use independence of $\delta q_j$

Substituting (2) and (4) into (1):

$\sum_j\left[Q_j - \frac{d}{dt}\!\left(\frac{\partial T}{\partial \dot{q}_j}\right) + \frac{\partial T}{\partial q_j}\right]\delta q_j = 0.$

Since the generalized coordinates are independent, the virtual displacements $\delta q_j$ are arbitrary and independent. Therefore each coefficient must vanish separately:

$\frac{d}{dt}\!\left(\frac{\partial T}{\partial \dot{q}_j}\right) - \frac{\partial T}{\partial q_j} = Q_j, \qquad j = 1, 2, \dots, n. \quad (5)$

Step 6: Conservative forces and the Lagrangian

If the forces are conservative, they derive from a potential $V$:

$Q_j = -\frac{\partial V}{\partial q_j}.$

Defining the Lagrangian $L = T - V$ (with $V$ independent of $\dot{q}_j$), equation (5) takes the standard form

$\boxed{\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_j}\right) - \frac{\partial L}{\partial q_j} = 0, \qquad j = 1, 2, \dots, n.}$

These are Lagrange's equations of motion for a conservative system with $n$ degrees of freedom.

Symmetric and skew-symmetric tensors

A tensor is symmetric in a pair of indices if interchanging them leaves the components unchanged, and skew-symmetric (antisymmetric) if interchanging them changes the sign. For two covariant indices,

$A_{ij}=A_{ji}\ \ (\text{symmetric}),\qquad A_{ij}=-A_{ji}\ \ (\text{skew-symmetric})$

and similarly for $A^{ij}$. For a skew-symmetric tensor the diagonal components vanish: $A_{ii}=-A_{ii}\Rightarrow A_{ii}=0$ (no sum).

Mixed tensors. The mixed tensor $A^i{}_j$ cannot have its upper and lower index interchanged directly, since the two are of different kind. Symmetry in a covariant-contravariant pair is therefore defined by lowering the contravariant index with the metric:

$A_{ij}=g_{ik}A^k{}_j$

The tensor $A^i{}_j$ is called symmetric (skew-symmetric) when $A_{ij}=A_{ji}$ ($A_{ij}=-A_{ji}$). Equivalently, $g_{ik}A^k{}_j=g_{jk}A^k{}_i$ for symmetry.

Decomposition

Let $A_{ij}$ be any tensor with two covariant indices (for a mixed tensor, take $A_{ij}=g_{ik}A^k{}_j$). Define

$S_{ij}=\tfrac12\left(A_{ij}+A_{ji}\right),\qquad T_{ij}=\tfrac12\left(A_{ij}-A_{ji}\right)$

1. $S_{ij}$ is symmetric: $S_{ji}=\tfrac12(A_{ji}+A_{ij})=S_{ij}$.

2. $T_{ij}$ is skew-symmetric: $T_{ji}=\tfrac12(A_{ji}-A_{ij})=-T_{ij}$.

3. They add up to $A_{ij}$:

$S_{ij}+T_{ij}=\tfrac12(A_{ij}+A_{ji})+\tfrac12(A_{ij}-A_{ji})=A_{ij}$

Both $S_{ij}$ and $T_{ij}$ are tensors, since they are linear combinations of the tensors $A_{ij}$ and $A_{ji}$. Hence

$\boxed{A_{ij}=\tfrac12(A_{ij}+A_{ji})+\tfrac12(A_{ij}-A_{ji})}$

4. Uniqueness. If also $A_{ij}=P_{ij}+Q_{ij}$ with $P$ symmetric and $Q$ skew-symmetric, then $A_{ji}=P_{ij}-Q_{ij}$. Adding and subtracting gives $P_{ij}=S_{ij}$ and $Q_{ij}=T_{ij}$, so the decomposition is unique.

Mixed form. Raising the index with $g^{ki}$, a mixed tensor splits as

$A^k{}_j=g^{ki}S_{ij}+g^{ki}T_{ij}=S^k{}_j+T^k{}_j$

where $S^k{}_j$ is symmetric and $T^k{}_j$ is skew-symmetric in the covariant-contravariant pair, in the metric sense above.

Therefore every tensor can be written as the sum of a symmetric and a skew-symmetric tensor.

Motion of a Symmetrical Top

Theory

A symmetrical top is a rigid body of revolution that spins about its axis of symmetry (the figure axis) while that axis itself rotates about the vertical. Let:

• $I_3$ = moment of inertia about the symmetry axis (spin axis)
• $I_1 = I_2 = I$ = moment of inertia about any axis perpendicular to it
• $\omega_3$ = spin angular velocity about the symmetry axis
• $\omega$ = angular velocity of precession of the symmetry axis about the vertical

Angular momentum. The angular momentum $\vec{L}$ has two components: one along the symmetry axis, $I_3\omega_3$, and one along the precession axis, $I\omega$. Since $I_3\omega_3 \gg I\omega$ for a fast top, the resultant $\vec{L}$ lies very nearly along the symmetry axis.

Torque and precession. When the symmetry axis is tilted at angle $\theta$ to the vertical, gravity acting at the centre of mass (distance $l$ from the point of support) exerts a torque

$\vec{\tau} = \vec{r}\times M\vec{g}, \qquad \tau = Mgl\sin\theta$

This torque is perpendicular to $\vec{L}$. Since $\vec{\tau} = d\vec{L}/dt$, the change $d\vec{L}$ is perpendicular to $\vec{L}$, so the magnitude of $\vec{L}$ stays constant while its direction rotates. The axis therefore sweeps out a cone about the vertical — this steady rotation is precession.

Rate of precession. In time $dt$ the angular momentum turns through $d\phi$, and

$dL = L\sin\theta\, d\phi = \tau\, dt = Mgl\sin\theta\, dt$

Hence the precession angular velocity is

$\omega_p = \frac{d\phi}{dt} = \frac{Mgl}{L} = \frac{Mgl}{I_3\omega_3}$

This is slow precession (fast spin $\omega_3$, small $\omega_p$). The faster the spin, the slower the precession.

Motion of the Top

1. Spin (rotation): The top spins rapidly about its symmetry axis with angular velocity $\omega_3$; the angular momentum is essentially along this axis.

2. Precession: The symmetry axis sweeps out a cone of semi-vertical angle $\theta$ about the vertical with angular velocity $\omega_p = Mgl/I_3\omega_3$. The direction of precession is the same as that of the spin.

3. Nutation (nodding): In general the angle $\theta$ is not constant; the axis also nods up and down between two limiting values while precessing. This slow periodic wobble superimposed on precession is called nutation. It arises because the total energy and the vertical component of angular momentum are both conserved, so the axis traces a wavy path on a sphere.

4. Steady precession: If the top is given exactly the right spin and initial push, the nutation vanishes and the axis precesses at a constant angle $\theta$ — the ideal steady motion.

Special cases:

• If $\omega_3$ is very large, $\omega_p$ is very small and the top appears almost upright.
• If the spin is too small, the top cannot precess steadily and it topples over.
• When the top is vertical ($\theta = 0$), the torque is zero and the axis stays fixed; a slight tilt then produces precession.

Thus a symmetrical top executes a combination of fast spin about its symmetry axis, slow precession of that axis about the vertical, and (in general) a superimposed nodding motion called nutation.

Sketch. The graph of $f(x)=x^2$ on $[-\pi,\pi]$ is an upward parabola with vertex at the origin, passing through $(-\pi,\pi^2)$ and $(\pi,\pi^2)$. It is symmetric about the $y$-axis, so $f$ is an even function, and its Fourier series contains only cosine terms ($b_n=0$).

Fourier series. For the interval $(-\pi,\pi)$,

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

Coefficients.

$a_0=\frac1\pi\int_{-\pi}^{\pi}x^2\,dx=\frac1\pi\cdot\frac{2\pi^3}{3}=\frac{2\pi^2}{3}$

$a_n=\frac1\pi\int_{-\pi}^{\pi}x^2\cos nx\,dx$

Integrating by parts twice,

$a_n=\frac1\pi\left[\frac{x^2\sin nx}{n}+\frac{2x\cos nx}{n^2}-\frac{2\sin nx}{n^3}\right]_{-\pi}^{\pi}=\frac1\pi\cdot\frac{4\pi\cos n\pi}{n^2}=\frac{4(-1)^n}{n^2}$

$b_n=\frac1\pi\int_{-\pi}^{\pi}x^2\sin nx\,dx=0\quad(\text{odd integrand})$

Result.

$\boxed{x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx=\frac{\pi^2}{3}-4\left(\cos x-\frac{\cos2x}{2^2}+\frac{\cos3x}{3^2}-\cdots\right)}$

Check at $x=\pi$. Here $\cos n\pi=(-1)^n$, so every term becomes $4/n^2$:

$\pi^2=\frac{\pi^2}{3}+4\left(1+\frac1{2^2}+\frac1{3^2}+\cdots\right)=\frac{\pi^2}{3}+4\cdot\frac{\pi^2}{6}=\pi^2$

This also yields $\displaystyle\sum_{n=1}^{\infty}\frac1{n^2}=\frac{\pi^2}{6}$.

(a) Kepler's laws of motion

1. Law of orbits (Law of ellipses). Every planet revolves around the Sun in an elliptical orbit, with the Sun situated at one of the two foci of the ellipse. (Hence the planet–Sun distance varies; nearest point = perihelion, farthest = aphelion.)

2. Law of areas (Law of equal areas). The radius vector drawn from the Sun to a planet sweeps out equal areas in equal intervals of time. That is, the areal velocity $\dfrac{dA}{dt}$ is constant. This is equivalent to the conservation of the planet's angular momentum, $mr^2\dfrac{d\theta}{dt} =$ constant, so a planet moves faster near perihelion and slower near aphelion.

3. Law of periods (Harmonic law). The square of the period of revolution $T$ of a planet is directly proportional to the cube of the semi-major axis $a$ of its orbit: $T^2 \propto a^3 \quad\Rightarrow\quad T^2 = \frac{4\pi^2}{GM}\,a^3$ where $M$ is the mass of the Sun. This law is used to determine the mass of the Sun or of any central body.

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OR — Linear operators

An operator $\hat A$ is a mathematical rule (instruction) which, when applied to a function, transforms it into another function: $\hat A\, f(x) = g(x)$

An operator is said to be linear if it satisfies the two conditions of superposition:

1. Additivity: $\hat A\,[f_1(x) + f_2(x)] = \hat A f_1(x) + \hat A f_2(x)$
2. Homogeneity (scaling): $\hat A\,[c\,f(x)] = c\,\hat A f(x)$, where $c$ is any constant.

Combining the two, the defining condition is $\hat A\,[c_1 f_1 + c_2 f_2] = c_1 \hat A f_1 + c_2 \hat A f_2$

Examples of linear operators: the differential operator $\dfrac{d}{dx}$, the integral operator $\displaystyle\int dx$, the multiplication operator $\hat x$ (multiply by $x$), the Laplacian $\nabla^2$, and the momentum operator $\hat p = -i\hbar\dfrac{\partial}{\partial x}$.

Examples of non-linear operators: the squaring operator $\hat A f = f^2$, the square-root operator $\hat A f = \sqrt{f}$, and the logarithmic operator $\hat A f = \ln f$.

Significance: In quantum mechanics every physical observable (position, momentum, energy, angular momentum) is represented by a linear Hermitian operator, because linearity guarantees that the superposition principle holds for wave functions and that the operators possess real eigenvalues.

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(b) Concepts of scalar and vector fields

A field is a physical quantity that has a well-defined value at every point of a region of space. Depending on the nature of the quantity, fields are of two types.

1. Scalar field

If a scalar (a quantity having magnitude only) is assigned to every point of a region, the region is called a scalar field. It is written as a scalar function of position: $\phi = \phi(x, y, z)$

Examples:

• Temperature distribution in the atmosphere, $T = T(x,y,z)$
• Atmospheric pressure at different heights
• Electric potential $V$ and gravitational potential $\Omega$ around a charge or mass
• Density of a non-uniform body

A scalar field is represented graphically by contour lines (curves of equal value) or equipotential surfaces. The gradient of a scalar field, $\nabla\phi$, gives a vector field whose direction is that of the steepest increase of $\phi$.

2. Vector field

If a vector (a quantity having both magnitude and direction) is assigned to every point of a region, the region is called a vector field. It is written as $\vec V = V_x(x,y,z)\,\hat i + V_y(x,y,z)\,\hat j + V_z(x,y,z)\,\hat k$

Examples:

• Velocity field $\vec v(x,y,z)$ in a flowing liquid or gas
• Gravitational field $\vec E_g = -\dfrac{GM}{r^2}\hat r$
• Electric field $\vec E$ and magnetic field $\vec B$ of charges and currents
• Wind velocity field in meteorology

A vector field is represented by field lines (streamlines), whose tangent at any point gives the direction of the field. Vector fields are analysed using divergence ($\nabla\cdot\vec V$, giving the flux source per unit volume) and curl ($\nabla\times\vec V$, giving the circulation per unit area). A field with $\nabla\times\vec V = 0$ is irrotational or conservative, and can be derived from a scalar potential, $\vec V = -\nabla\phi$ — this is the link between the two types of field.

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OR — Significance of Foucault's pendulum

A Foucault pendulum is a heavy bob suspended by a long wire from a fixed support, free to swing in any vertical plane. Foucault demonstrated it in Paris (1851) to prove the rotation of the Earth.

Significance:

1. Direct proof of Earth's rotation. The plane of oscillation of the pendulum is fixed in space (in an inertial frame) because no horizontal torque acts on it. But an observer on the rotating Earth sees the plane of oscillation slowly rotate relative to the ground. This relative rotation is direct experimental evidence that the Earth spins about its polar axis.

2. Quantitative measure of latitude. The angular speed of rotation of the plane of oscillation is $\omega_p = \omega_E \sin\lambda = 15^\circ \sin\lambda \ \text{per hour}$ where $\lambda$ is the latitude. At the poles the plane turns through $15^\circ$ per hour (a full turn in 24 h); at the equator it does not rotate at all. Thus the pendulum also measures the latitude of the place.

3. Demonstration of the Coriolis effect. The apparent rotation arises from the same rotating-frame physics that produces the Coriolis force, which explains the deflection of winds, ocean currents, and the rotation of cyclones.

4. Distinguishes inertial and non-inertial frames. It shows that the Earth is a non-inertial (rotating) frame, so Newton's laws must be applied with fictitious forces (centrifugal and Coriolis) when working in the Earth's frame.

5. Direction of rotation. In the Northern Hemisphere the plane rotates clockwise (viewed from above); in the Southern Hemisphere, anticlockwise — confirming the sense of the Earth's rotation.

(a) Meaning of Christoffel's symbols

Meaning. In curvilinear coordinates the base vectors $\mathbf e_i=\partial\mathbf r/\partial x^i$ change direction and magnitude from point to point. Christoffel symbols are the coefficients that express this change:

$\frac{\partial \mathbf e_j}{\partial x^i}=\Gamma^k_{\;ij}\,\mathbf e_k$

So $\Gamma^k_{\;ij}$ measures how much the $j$-th base vector changes as we move along the $i$-th coordinate, resolved along $\mathbf e_k$. They are also the "connection coefficients" appearing in the covariant derivative, $\nabla_i V^j=\partial_iV^j+\Gamma^j_{\;ik}V^k$, and in the geodesic equation

$\frac{d^2x^k}{ds^2}+\Gamma^k_{\;ij}\frac{dx^i}{ds}\frac{dx^j}{ds}=0 .$

They are computed from the metric as

$\Gamma^k_{\;ij}=\tfrac12 g^{kl}\left(\partial_i g_{jl}+\partial_j g_{il}-\partial_l g_{ij}\right).$

Key point. Christoffel symbols are not tensors — they are symmetric in $i,j$ and can be made to vanish at any chosen point by going to local Cartesian coordinates. Physically they represent the "inertial" (centrifugal, Coriolis) effects of a non-inertial or curved coordinate system.

Examples.

• Cartesian coordinates: $g_{ij}=\delta_{ij}$, so all $\Gamma^k_{\;ij}=0$.
• Plane polar coordinates $(r,\theta)$: $ds^2=dr^2+r^2d\theta^2$, giving $\Gamma^r_{\;\theta\theta}=-r,\qquad \Gamma^\theta_{\;r\theta}=\Gamma^\theta_{\;\theta r}=\frac1r,\qquad \text{others zero.}$
• Spherical surface $(\theta,\phi)$: $\Gamma^\theta_{\;\phi\phi}=-\sin\theta\cos\theta,\quad \Gamma^\phi_{\;\theta\phi}=\Gamma^\phi_{\;\phi\theta}=\cot\theta$.

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(b) Hermitian matrix

A square matrix $A$ is Hermitian (or self-adjoint) if it equals its own conjugate transpose:

$A=A^{\dagger},\qquad\text{i.e.}\qquad a_{ij}=\overline{a_{ji}} .$

Thus the element in row $i$, column $j$ is the complex conjugate of the element in row $j$, column $i$. In particular the diagonal elements must be real ($a_{ii}=\overline{a_{ii}}$), and a real Hermitian matrix is simply a symmetric matrix.

Important properties: all eigenvalues are real; eigenvectors belonging to distinct eigenvalues are orthogonal; the matrix is diagonalisable by a unitary transformation. Hermitian matrices represent physical observables in quantum mechanics.

Example. The Pauli matrix

$$\sigma_y=\begin{pmatrix}0&-i\\ i&0\end{pmatrix},\qquad \sigma_y^{\dagger}=\begin{pmatrix}0&\overline{i}\\ \overline{-i}&0\end{pmatrix}=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}=\sigma_y .$$

Also $\begin{pmatrix}2&3+i\\ 3-i&5\end{pmatrix}$ is Hermitian, while $\begin{pmatrix}0&i\\ i&0\end{pmatrix}$ is not.

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(c) Dirac-delta function

The Dirac delta $\delta(x)$ is a generalized function (distribution) used to represent a sharp spike at a point. It is defined by the two properties

$\delta(x)=0\ \ (x\neq0),\qquad \int_{-\infty}^{\infty}\delta(x)\,dx=1,$

together with the sifting (sampling) property

$\int_{-\infty}^{\infty} f(x)\,\delta(x-a)\,dx=f(a),$

for any well-behaved function $f(x)$. It is not a function in the ordinary sense (its value at $x=0$ is not finite); it is obtained as the limit of a sequence of peaked functions, e.g.

$$\delta(x)=\lim_{\varepsilon\to0}\frac{1}{2\varepsilon}\ \ \text{for }|x|<\varepsilon,\qquad \delta(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}\,dk .$$

Useful properties:

$$\delta(-x)=\delta(x),\qquad \delta(ax)=\frac{1}{|a|}\delta(x),\qquad \int f(x)\,\delta'(x-a)\,dx=-f'(a),$$

and in three dimensions $\delta^3(\mathbf r)=\delta(x)\delta(y)\delta(z)$, with $\int\delta^3(\mathbf r)\,d^3r=1$. It is widely used in physics to represent point charges, point masses, impulse forces and orthonormality of continuous eigenfunctions.

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(d) Rotational vs irrotational vectors

A general vector field can be decomposed as $\mathbf F=-\nabla\phi+\nabla\times\mathbf A$ (Helmholtz theorem); a field may be irrotational in one region and rotational in another (e.g. $\mathbf B$ of a wire is irrotational everywhere except on the wire itself).

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(e) Rutherford scattering

Rutherford scattering is the scattering of alpha particles by the Coulomb field of atomic nuclei, studied by Geiger and Marsden (1909) under Rutherford's guidance to test the structure of the atom.

Experiment. A narrow beam of $\alpha$-particles (energy $\sim5$ MeV) from a radioactive source was directed at a very thin gold foil, and the scattered particles were counted with a rotatable ZnS scintillation detector.

Observations.

• Most $\alpha$-particles passed almost undeviated through the foil.
• A small fraction were deflected through large angles, and about 1 in 8000 was scattered back through more than $90^\circ$.

Conclusion. Large-angle scattering cannot be explained by Thomson's uniformly distributed positive charge. It requires the atom to consist of a very small, massive, positively charged nucleus ($\sim10^{-15}$ m) surrounded by mostly empty space, the scattering being due to the Coulomb repulsive force between the $\alpha$-particle and the nucleus.

Result. For a nucleus of charge $Ze$ and $\alpha$-particle of charge $2e$ and energy $E$, the impact parameter is

$b=\frac{Ze^2}{4\pi\varepsilon_0 E}\cot\frac{\theta}{2},$

so the differential cross-section obeys the Rutherford formula

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

i.e. the number scattered at angle $\theta$ varies as $N(\theta)\propto \dfrac{1}{\sin^4(\theta/2)}$, independent of the target material's chemical state — verified experimentally.

Limitations. The formula holds only for $\alpha$-energies below about $25$ MeV; at higher energies the $\alpha$-particle penetrates the nucleus and the nuclear force causes deviations. It also assumes a point nucleus, a stationary heavy nucleus (no recoil) and non-relativistic motion.

For a rigid body, the general number of degrees of freedom in space is

[ N = 6 ]

consisting of 3 translational and 3 rotational degrees of freedom.

(a) Rigid body moving freely in space

No constraints are imposed. Therefore,

[ \boxed{N = 6} ]

These are:

• 3 coordinates to fix the centre of mass,
• 3 angles to fix the orientation of the body.

(b) Rigid body with two fixed points

Let the two fixed points be (A) and (B). Since the body is rigid, the distance (AB) is fixed. The only possible motion left is rotation about the line (AB).

Thus, the body has only one independent rotational degree of freedom.

[ \boxed{N = 1} ]

Final Answer:

[ \boxed{\text{(a) }6 \qquad \text{(b) }1} ]

Given: Total relativistic energy, (E = 10\ \text{GeV}) Relativistic momentum, (p = 8\ \text{GeV}/c)

Formula: The relativistic energy–momentum relation is

$E^2 = p^2c^2 + m_0^2c^4$

where (m_0) is the rest mass.

Calculation: Here,

$pc = \left(8\ \frac{\text{GeV}}{c}\right)c = 8\ \text{GeV}$

So,

$m_0^2c^4 = E^2 - p^2c^2$

$m_0^2c^4 = (10)^2 - (8)^2 = 100 - 64 = 36\ \text{GeV}^2$

$m_0c^2 = \sqrt{36} = 6\ \text{GeV}$

Therefore,

$m_0 = \frac{6\ \text{GeV}}{c^2} = 6\ \text{GeV}/c^2$

Result: The rest mass of the particle is

$\boxed{m_0 = 6\ \text{GeV}/c^2}$

In SI units,

$m_0 = 6 \times 1.7827 \times 10^{-27}\ \text{kg} \approx 1.07 \times 10^{-26}\ \text{kg}$

Given
[ \vec v=(x+3y)\hat i+(y-2z)\hat j+(x+az)\hat k ]

The divergence of (\vec v) is
[ \nabla\cdot \vec v=\frac{\partial}{\partial x}(x+3y)+\frac{\partial}{\partial y}(y-2z)+\frac{\partial}{\partial z}(x+az) ]

[ =1+1+a=2+a ]

Given (\nabla\cdot \vec v=0),
[ 2+a=0 ]

[ \boxed{a=-2} ]

So the required constant value is (\boxed{-2}).

Given
$A=\begin{pmatrix}1&3\\2&1\end{pmatrix}$

For a $2\times2$ matrix
$A=\begin{pmatrix}a&b\\c&d\end{pmatrix},$ the inverse is
$A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix},\qquad ad-bc\ne 0.$

Here,
$a=1,\quad b=3,\quad c=2,\quad d=1.$

So,
$|A|=ad-bc=(1)(1)-(3)(2)=1-6=-5\ne 0.$

Hence $A^{-1}$ exists.

$A^{-1}=\frac{1}{-5}\begin{pmatrix}1&-3\\-2&1\end{pmatrix}$

$A^{-1}=\begin{pmatrix}-\frac15&\frac35\\[4pt]\frac25&-\frac15\end{pmatrix}$

Verification:

$AA^{-1}=\begin{pmatrix}1&3\\2&1\end{pmatrix}\begin{pmatrix}-\frac15&\frac35\\[4pt]\frac25&-\frac15\end{pmatrix}$

$=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I$

Hence,
$\boxed{A^{-1}=\begin{pmatrix}-\frac15&\frac35\\[4pt]\frac25&-\frac15\end{pmatrix}}$

Laplace Transform of (f(x)=e^x)

The Laplace transform of a function (f(x)) is defined as

[ L{f(x)}=\int_0^\infty e^{-sx}f(x),dx, ]

where (s) is a parameter.

Given

[ f(x)=e^x, ]

we have

[ L{e^x}=\int_0^\infty e^{-sx}e^x,dx ]

[ =\int_0^\infty e^{-(s-1)x},dx ]

Now,

[ \int_0^\infty e^{-(s-1)x},dx

\left[\frac{e^{-(s-1)x}}{-(s-1)}\right]_0^\infty ]

[

\frac{1}{s-1}, \qquad \text{provided } s>1. ]

Hence,

[ \boxed{L{e^x}=\frac{1}{s-1}}, \qquad s>1. ]

More generally,

[ L{e^{ax}}=\frac{1}{s-a}, ]

so putting (a=1), we get the required result.