(a) A force F⃗ is given by F⃗ = x²y x̂ + zy² ŷ + xz² ẑ. Determine whether or not the force is conservative. 10 marks (b)…

(a) A force is conservative if its curl vanishes everywhere in a simply connected region, i.e. if ∇×F = 0. Here

F = x²y x̂ + zy² ŷ + xz² ẑ.

So the components are:

Fₓ = x²y, F_y = zy², F_z = xz².

Using the curl formula,

(∇×F)ₓ = ∂F_z/∂y - ∂F_y/∂z = ∂(xz²)/∂y - ∂(zy²)/∂z = 0 - y² = -y².

The y-component is

(∇×F)_y = ∂Fₓ/∂z - ∂F_z/∂x = ∂(x²y)/∂z - ∂(xz²)/∂x = 0 - z² = -z².

The z-component is

(∇×F)_z = ∂F_y/∂x - ∂Fₓ/∂y = ∂(zy²)/∂x - ∂(x²y)/∂y = 0 - x² = -x².

Hence

∇×F = -y² x̂ - z² ŷ - x² ẑ.

This is not zero for general x, y, z. Therefore the force is not conservative. The test is valid in the simply connected domain of all space; since the curl does not vanish, no scalar potential V can exist such that F = -∇V.

(b) The gravitational self-energy of a uniform sphere of mass M and radius R is obtained by assembling the sphere shell by shell.

Let the density be uniform:

ρ = M/(4πR³/3).

When mass has been assembled up to radius r, the mass inside is

m(r) = (4πr³/3)ρ = M r³/R³.

Now bring a thin shell of thickness dr at radius r. Its mass is

dm = 4πr²ρ dr = 3M r²/R³ dr.

The work done in bringing this shell from infinity to radius r against the gravitational field of the already assembled mass m(r) is

dU = -G m(r) dm / r.

Thus

dU = -G (M r³/R³)(3M r²/R³ dr)/r = -(3GM²/R⁶) r⁴ dr.

Integrating from r = 0 to r = R,

U = -(3GM²/R⁶) ∫₀ᴿ r⁴ dr = -(3GM²/R⁶)(R⁵/5) = -3GM²/(5R).

Using G = 6.67×10⁻¹¹ N m² kg⁻², M = 6×10²⁴ kg, R = 6.4×10⁶ m,

U = -3(6.67×10⁻¹¹)(6×10²⁴)²/[5(6.4×10⁶)] = -2.25×10³² J.

Final answer: gravitational self-energy of the Earth = -2.25×10³² J. The magnitude or binding energy is 2.25×10³² J. This assumes a spherically symmetric Earth of uniform density and Newtonian gravitation.

(c) Let S′ be a frame moving with velocity v along the x-axis relative to S. The Lorentz transformation is

x′ = γ(x - vt), t′ = γ(t - vx/c²), y′ = y, z′ = z,

where γ = 1/√(1 - v²/c²). Its consequences for length and time are:

• Time dilation: A clock at rest in S measures proper time Δτ. An observer in S′ measures

Δt′ = γ Δτ = Δτ/√(1 - v²/c²).

Since γ > 1, moving clocks run slower. This is observed in the increased lifetime of fast-moving muons.

• Length contraction: A rod at rest in S has proper length L₀ along the direction of motion. An observer in S′ measures

L = L₀/γ = L₀√(1 - v²/c²).

Thus moving lengths contract along the direction of relative motion. Transverse dimensions are unchanged.

• Relativity of simultaneity: If two events are simultaneous in S, with Δt = 0 but separated by Δx, then in S′

Δt′ = -γvΔx/c².

So simultaneity is not absolute. Events simultaneous in one inertial frame may not be simultaneous in another.

• Limiting behaviour: For v << c, γ ≈ 1, and the classical Galilean results are recovered. For v → c, γ → ∞, so time dilation and length contraction become extreme.

These effects are symmetric: each observer sees the other’s clock running slow and the other’s rod contracted. The results are valid for inertial frames moving at constant relativistic velocity, with v < c.

(d) Huygens’ principle states that every point on a wavefront acts as a source of secondary spherical wavelets, and the new wavefront is the common tangent envelope of these wavelets.

Let XY be the plane interface between medium 1 and medium 2. Let AB be the incident plane wavefront, with A on the interface. Let the disturbance from B reach C on the interface in time t. Then

BC = v₁t,

where v₁ is the wave velocity in medium 1.

During the same time t, the secondary wavelet from A travels into medium 2 a distance

AD = v₂t,

where v₂ is the wave velocity in medium 2. Draw CD tangent to this secondary wavelet. Then CD is the refracted wavefront.

Let i be the angle of incidence and r the angle of refraction. From the geometry of the incident wavefront, in triangle ABC,

BC = AC sin i.

Therefore

v₁t = AC sin i.

From the geometry of the refracted wavefront, in triangle ACD,

AD = AC sin r.

Therefore

v₂t = AC sin r.

Dividing the two equations,

(v₁t)/(v₂t) = (AC sin i)/(AC sin r),

so

sin i/sin r = v₁/v₂.

Now the refractive index of a medium is μ = c/v, where c is the speed of light in vacuum. Hence

v₁/v₂ = (c/μ₁)/(c/μ₂) = μ₂/μ₁.

Therefore,

sin i/sin r = v₁/v₂ = μ₂/μ₁.

For medium 2 denser than medium 1, μ₂ > μ₁ and v₂ < v₁, so sin r < sin i, giving r < i. Thus the refracted ray bends towards the normal. This derivation assumes a plane interface, isotropic media, and a monochromatic plane wave.

(e) In a laser, lasing requires a population inversion: the number of atoms in the upper laser level must exceed that in the lower laser level. Pumping raises atoms to higher levels; fast non-radiative decays then populate a metastable upper laser level.

Three-level pumping scheme: The levels are the ground level E₀, a metastable upper laser level E₁, and a pump band E₂. Atoms are pumped from E₀ to E₂. They decay rapidly and non-radiatively to E₁. Lasing occurs from E₁ to E₀.

Schematic:

E₂ (pump band) ───────────── ↑ pump ↓ fast non-radiative E₁ (metastable upper) ───────────── → laser E₀ (ground/lower) ─────────────

Here the lower laser level is the ground state. Therefore more than half the atoms must be raised to E₁ to achieve inversion. This requires a high pump threshold. Ruby laser is a common example.

Four-level pumping scheme: The levels are ground level E₀, lower laser level E₁, metastable upper laser level E₂, and pump band E₃. Atoms are pumped from E₀ to E₃, decay rapidly to E₂, lase from E₂ to E₁, and then E₁ decays rapidly to E₀.

Schematic:

E₃ (pump band) ───────────── ↑ pump ↓ fast non-radiative E₂ (metastable upper) ───────────── → laser E₁ (lower laser level) ───────────── ↓ fast non-radiative E₀ (ground) ─────────────

Because E₁ empties rapidly, its population remains small. Hence population inversion between E₂ and E₁ is easily maintained, and the pump threshold is much lower than in a three-level system. Nd:YAG and He-Ne lasers are examples.

In both schemes, after inversion is established, a spontaneously emitted photon stimulates further emission from the upper level. The emitted photons have the same frequency, phase, direction, and polarization. Optical feedback by mirrors causes amplification, and when gain exceeds losses, coherent laser output is obtained. The upper laser level must be metastable, and the non-radiative decays must be fast for efficient pumping.