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IGNOU MPH-006 Previous Year Solved Question Paper in Hindi

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Q1. (a) The Hamiltonian of a simple harmonic oscillator is given as : H = p²/2m + ½mω²q². Show that the velocity vector is tangential to the curve defined by the Hamiltonian. (b) Consider two Lagrangians, L(q, q̇, t) and L'(q, q̇, t) such that : L'(q, q̇, t) = L(q, q̇, t) + dF(q,t)/dt. Show that the variation of action is the same for both the Lagrangians that is δS = δS’.

Ans. (a) For a system, the trajectory in phase space lies on a surface of constant energy, defined by H(q, p) = E = constant . The velocity vector in phase space is given by v _phase = (q̇, ṗ). Hamilton’s equations of motion are: q̇ = ∂H/∂p ṗ = -∂H/∂q For the given simple harmonic oscillator Hamiltonian: H = p²/2m + ½mω²q² We calculate the partial derivatives: ∂H/∂p = p/m ∂H/∂q = mω²q Thus, the velocity vector in phase space is: v _phase = (q̇, ṗ) = (p/m, -mω²q) The normal vector to the constant energy surface H = E at any point is given by the gradient of the Hamiltonian: ∇H = (∂H/∂q, ∂H/∂p) = (mω²q, p/m) To show that the velocity vector is tangential to the curve, we must show that the velocity vector is orthogonal to the normal vector. This can be done by calculating their dot product: v _phase ⋅ ∇H = (p/m)(mω²q) + (-mω²q)(p/m) v _phase ⋅ ∇H = pω²q – pω²q = 0 Since the dot product is zero, the velocity vector (q̇, ṗ) is perpendicular to the gradient vector (∇H). Because the gradient vector is always normal to the surface, this implies that the velocity vector is necessarily tangential to the surface . Thus, the trajectory in phase space is tangential to the constant-energy curve defined by the Hamiltonian. (b) The action S is defined as the time integral of the Lagrangian L: S = ∫[t₁ to t₂] L(q, q̇, t) dt The corresponding action S’ for the second Lagrangian L’ is: S’ = ∫[t₁ to t₂] L'(q, q̇, t) dt Substituting the given relation L’ = L + dF/dt: S’ = ∫[t₁ to t₂] (L + dF/dt) dt S’ = ∫[t₁ to t₂] L dt + ∫[t₁ to t₂] (dF/dt) dt The first term is the original action S. The second term can be integrated: ∫[t₁ to t₂] (dF/dt) dt = F(q(t₂), t₂) – F(q(t₁), t₁) Thus, S’ = S + [F] ^t₂ _t₁ Now, we compute the variation of the action, δS’. According to the principle of variation, the variation at the endpoints of the path vanishes, i.e., δq(t₁) = 0 and δq(t₂) = 0 . δS’ = δ(S + F(q(t₂), t₂) – F(q(t₁), t₁)) δS’ = δS + δF(q(t₂), t₂) – δF(q(t₁), t₁) The variation of F(q,t) is δF = (∂F/∂q)δq. At the endpoints: δF(q(t₂), t₂) = (∂F/∂q)| _t₂ δq(t₂) = 0 δF(q(t₁), t₁) = (∂F/∂q)| _t₁ δq(t₁) = 0 Since δF at the endpoints is zero, we get: δS’ = δS + 0 – 0 δS’ = δS This demonstrates that both Lagrangians L and L’ lead to the same equations of motion, because Hamilton’s principle states that δS = 0. If δS = 0, then δS’ = 0 as well.

Q2. (a) Consider a particle of mass m moving under a uniform force F = F₀k̂. If the Lagrangian is given as : L = ½m(ẋ² + ẏ² + ż²) – F₀z Determine the conservation laws under rotation. (b) A particle of mass m has the Lagrangian : L = ½m(ẋ² – kx²) + ½aẋ⁴. Obtain the Hamiltonian of the system and the corresponding Hamilton’s equations of motion. Take k and a as constants.

Ans. (a) The given Lagrangian L = ½m(ẋ² + ẏ² + ż²) – F₀z describes a particle moving under a uniform force (like gravity) in the negative z-direction. Here, the kinetic energy is T = ½m(ẋ² + ẏ² + ż²) and the potential energy is V = F₀z. We check for conservation laws under rotation using Noether’s theorem , which states that if a Lagrangian is invariant under a continuous symmetry, a corresponding quantity is conserved. Rotation about the z-axis: Consider a rotation by an angle φ about the z-axis. The coordinates transform as: x’ = x cosφ – y sinφ y’ = x sinφ + y cosφ z’ = z The sum of the squares of the velocities remains invariant: ẋ’² + ẏ’² = (ẋ cosφ – ẏ sinφ)² + (ẋ sinφ + ẏ cosφ)² = ẋ² + ẏ² Since z’ = z and ż’ = ż, the kinetic energy T’ = ½m(ẋ’² + ẏ’² + ż’²) = ½m(ẋ² + ẏ² + ż²) = T, is invariant. The potential energy V = F₀z depends only on z. Since z’ = z under this rotation, the potential energy is also invariant: V’ = F₀z’ = F₀z = V. Because both T and V are invariant under rotations about the z-axis, the Lagrangian L = T – V is also invariant. According to Noether’s theorem, there must be a conserved quantity corresponding to this symmetry. For rotation about the z-axis, the conserved quantity is the component of angular momentum along the z-axis (L _z ) . L _z = x p _y – y p _x = m(xẏ – yẋ) Rotation about x- or y-axis: If we rotate about the x-axis or y-axis, the z coordinate will change (e.g., z’ = y sinθ + z cosθ for rotation about x-axis). Since the potential energy V = F₀z depends on z, V and therefore L are not invariant under these rotations. Consequently, the x and y components of angular momentum, L _x and L _y , are not conserved. Thus, the only conservation law under rotation is the conservation of the z-component of angular momentum . (b) The given Lagrangian is: L = ½mẋ² – ½mkx² + ½aẋ⁴. This is a non-standard Lagrangian due to the term with the fourth power of velocity (ẋ⁴). 1. Obtain Canonical Momentum: The generalized coordinate is x. The canonical momentum p conjugate to x is: p = ∂L/∂ẋ = ∂/∂ẋ (½mẋ² – ½mkx² + ½aẋ⁴) p = mẋ + 2aẋ³ 2. Obtain the Hamiltonian: The Hamiltonian H is defined by the Legendre transformation: H = pẋ – L. H = (mẋ + 2aẋ³)ẋ – (½mẋ² – ½mkx² + ½aẋ⁴) H = mẋ² + 2aẋ⁴ – ½mẋ² + ½mkx² – ½aẋ⁴ H = ½mẋ² + (3/2)aẋ⁴ + ½mkx² This is the energy function. The Hamiltonian should typically be expressed as a function of coordinates and momenta (x, p). However, the equation p = mẋ + 2aẋ³ is not easily solvable for ẋ. We therefore leave H in terms of x and ẋ, with the understanding that ẋ is treated as an implicit function of p, ẋ(p). 3. Hamilton’s Equations of Motion: Hamilton’s equations are: ẋ = ∂H/∂p ṗ = -∂H/∂x To calculate ṗ, we use H(x, p) = pẋ(p) – L(x, ẋ(p)). ṗ = -∂H/∂x = -[∂(pẋ)/∂x – ∂L/∂x] = ∂L/∂x (because in H(x, p), x and p are independent variables, so ∂ẋ/∂x = 0 and ∂p/∂x = 0) ∂L/∂x = ∂/∂x (½mẋ² – ½mkx² + ½aẋ⁴) = -mkx Thus, ṗ = -mkx . For the ẋ equation: ẋ = ∂H/∂p = ∂/∂p (pẋ – L) Using the chain rule: ∂H/∂p = ẋ + p(∂ẋ/∂p) – (∂L/∂ẋ)(∂ẋ/∂p) Since p = ∂L/∂ẋ, we get: ∂H/∂p = ẋ + p(∂ẋ/∂p) – p(∂ẋ/∂p) = ẋ Thus, ẋ = ẋ(p) , where p = mẋ + 2aẋ³ . In summary, the equations of motion are:

• ẋ = ẋ(p) , which is implicitly defined by p = mẋ + 2aẋ³ .
• ṗ = -mkx

Q3. (a) Write the bilinear invariant condition for canonical transformation. Using the condition, show whether the transformation: Q = -p/√2, P = q²/√2 is a canonical transformation. (b) Consider a dynamical system described by a Hamiltonian : H = p²/2m + ½mω²q². Obtain the new Hamiltonian K(Q, P) using the generating function, F₁ = (mω/2)q² cot(Q).

Ans. (a) Bilinear Invariant Condition: A transformation (q, p) → (Q, P) is canonical if the Jacobian matrix M of the transformation satisfies M ^T JM = J, where J is the standard symplectic matrix. An alternative condition, the bilinear invariant condition, states that the expression ∑ᵢ (pᵢδqᵢ – PᵢδQᵢ) must be an exact differential of a function F. For one degree of freedom, this condition becomes: pδq – PδQ = dF A more direct way to test this transformation is by using the fundamental Poisson bracket . A transformation is canonical if and only if {Q, P} _q,p = 1. {Q, P} _q,p = (∂Q/∂q)(∂P/∂p) – (∂Q/∂p)(∂P/∂q) For the given transformation: Q = -p/√2 P = q²/√2 We calculate the required partial derivatives: ∂Q/∂q = 0 ∂Q/∂p = -1/√2 ∂P/∂q = 2q/√2 = √2q ∂P/∂p = 0 Substituting these values into the Poisson bracket formula: {Q, P} = (0)(0) – (-1/√2)(√2q) {Q, P} = q Since {Q, P} = q ≠ 1 (it is a function of q and not equal to 1), the given transformation is not canonical . (b) [Note: The generating function printed in the question paper is ambiguous. A standard generating function that simplifies the simple harmonic oscillator problem is the F₁-type function F₁(q,Q) = (mω/2)q² cot(Q). We will proceed with this function as it is the most likely intent.] The given Hamiltonian is: H = p²/2m + ½mω²q² The generating function is: F₁(q, Q) For a type-F₁ generating function, the transformation equations are: p = ∂F₁/∂q P = -∂F₁/∂Q For the given F₁(q,Q) = (mω/2)q² cot(Q): p = ∂/∂q [(mω/2)q² cot(Q)] = mωq cot(Q) P = -∂/∂Q [(mω/2)q² cot(Q)] = -(mω/2)q² [-csc²(Q)] = (mω/2)q² csc²(Q) Now we express the old variables (q, p) in terms of the new ones (Q, P): From the equation for P: q² = 2P / (mω csc²(Q)) = (2P/mω) sin²(Q). From the equation for p, p² = m²ω²q² cot²(Q). Substituting the expression for q²: p² = m²ω² [(2P/mω) sin²(Q)] cot²(Q) p² = 2mωP sin²(Q) (cos²(Q)/sin²(Q)) = 2mωP cos²(Q) Now we substitute these expressions for q² and p² into the original Hamiltonian H: H = p²/2m + ½mω²q² K = (2mωP cos²(Q)) / (2m) + ½mω² [(2P/mω) sin²(Q)] K = ωP cos²(Q) + ωP sin²(Q) K = ωP (cos²(Q) + sin²(Q)) K(Q, P) = ωP Since the generating function F₁ has no explicit time dependence, the new Hamiltonian K is equal to the old Hamiltonian H, expressed in the new coordinates. The new Hamiltonian is K = ωP. In this new form, Q is a cyclic coordinate as it does not appear in K, and the equations of motion become trivial: Ṗ = -∂K/∂Q = 0 (P is a constant) and Q̇ = ∂K/∂P = ω (Q increases linearly with time).

Q4. Write down the fundamental Poisson brackets for two functions f(q, p, t) and g(q, p, t). If the functions f(q, p, t) and g(q, p, t) are constants of motion, show that the Poisson bracket of the two constants of motion is also a constant of motion.

Ans. Fundamental Poisson Brackets: For canonical coordinates qᵢ and momenta pᵢ, the fundamental Poisson brackets are defined as:

• {qᵢ, qⱼ} = 0
• {pᵢ, pⱼ} = 0
• {qᵢ, pⱼ} = δᵢⱼ

where δᵢⱼ is the Kronecker delta (1 if i = j, 0 otherwise).

Poisson Bracket of Two Functions:

For any two functions f(q, p, t) and g(q, p, t) in phase space, the Poisson bracket is defined as:

{f, g} = ∑ᵢ (∂f/∂qᵢ ∂g/∂pᵢ – ∂f/∂pᵢ ∂g/∂qᵢ)

Proof of the Theorem (Poisson’s Theorem):

We need to show that if f and g are constants of motion, then their Poisson bracket {f, g} is also a constant of motion.

A function F is a constant of motion if its total time derivative is zero:

dF/dt = 0

The total time derivative of any function F(q, p, t) is given by:

dF/dt = ∂F/∂t + {F, H}

where H is the Hamiltonian of the system.

Since f and g are constants of motion, we have:

1. df/dt = ∂f/∂t + {f, H} = 0

2. dg/dt = ∂g/∂t + {g, H} = 0

We want to show that d{f,g}/dt = 0. The total time derivative of the Poisson bracket {f, g} is:

d{f,g}/dt = ∂{f,g}/∂t + {{f, g}, H}

Now, we evaluate each term separately.

The first term, the partial derivative with respect to time, can be shown, using the product rule and rearranging terms, to be:

∂{f,g}/∂t = {∂f/∂t, g} + {f, ∂g/∂t}

The second term, {{f, g}, H}, can be expanded using the

Jacobi identity

:

{{A,B},C} + {{B,C},A} + {{C,A},B} = 0

Letting A=f, B=g, C=H:

{{f,g},H} + {{g,H},f} + {{H,f},g} = 0

Rearranging this gives:

{{f,g},H} = -{{g,H},f} – {{H,f},g} = {f, {g,H}} + {g, {f,H}}

(since {A,B} = -{B,A})

Now combine these two results into the expression for d{f,g}/dt:

d{f,g}/dt = (∂{f,g}/∂t) + ({{f,g},H})

d{f,g}/dt = [{∂f/∂t, g} + {f, ∂g/∂t}] + [{f, {g,H}} + {g, {f,H}}]

Regrouping the terms:

d{f,g}/dt = [{∂f/∂t + {f,H}}, g] + [{f, ∂g/∂t + {g,H}}]

This is valid due to the linearity of the Poisson bracket. We now recognize the terms in the square brackets as df/dt and dg/dt:

d{f,g}/dt = {df/dt, g} + {f, dg/dt}

Since f and g are constants of motion, df/dt = 0 and dg/dt = 0. Therefore:

d{f,g}/dt = {0, g} + {f, 0} = 0 + 0 = 0

This proves that {f, g} is also a constant of motion. This result is known as

Poisson’s Theorem

.

Q5. The Hamiltonian for a system of two degrees of freedom is given as : H(q₁, q₂, p₁, p₂) = (p₁²/2m + p₂²/2m) + cos(q₁ + 4q₂). Obtain the constants of motion for the system in Q, P variables. Hence obtain the new Hamiltonian for Q₁ = q₁ + 4q₂ and Q₂ = q₂.

Ans. The given Hamiltonian is: H = (p₁²/2m + p₂²/2m) + cos(q₁ + 4q₂). Constants of Motion: First, we note that the Hamiltonian H does not depend explicitly on time (∂H/∂t = 0). Therefore, the total energy E = H is itself a constant of motion. Second, we use a canonical transformation to look for another constant. The coordinates q₁ or q₂ are not individually cyclic because they both appear in the potential term cos(q₁ + 4q₂). We define new coordinates with the given transformation: Q₁ = q₁ + 4q₂ Q₂ = q₂ We must find the corresponding new momenta P₁ and P₂ such that the transformation (q₁,q₂,p₁,p₂) → (Q₁,Q₂,P₁,P₂) is canonical. We use the Poisson bracket conditions {Qᵢ, Pⱼ} = δᵢⱼ. Let’s assume the new momenta are linear combinations of the old momenta: P₁ = ap₁ + bp₂, P₂ = cp₁ + dp₂.

• {Q₁, P₁} = {q₁+4q₂, ap₁+bp₂} = a{q₁,p₁} + 4b{q₂,p₂} = a+4b = 1
• {Q₂, P₁} = {q₂, ap₁+bp₂} = b{q₂,p₂} = b = 0
• {Q₁, P₂} = {q₁+4q₂, cp₁+dp₂} = c{q₁,p₁} + 4d{q₂,p₂} = c+4d = 0
• {Q₂, P₂} = {q₂, cp₁+dp₂} = d{q₂,p₂} = d = 1

Solving these equations gives: b=0, d=1, a=1, c=-4.

So, the new momenta are:

P₁ = p₁

P₂ = p₂ – 4p₁

From this transformation, we see that

P₂ = p₂ – 4p₁ is a conserved quantity.

We can verify this using Hamilton’s equations: Ṗ₂ = {P₂, H}.

{P₂, H} = {p₂ – 4p₁, (p₁²/2m + p₂²/2m) + cos(q₁ + 4q₂)}

= {p₂, cos(q₁+4q₂)} – 4{p₁, cos(q₁+4q₂)}

= -∂/∂q₂(cos(q₁+4q₂)) + 4∂/∂q₁(cos(q₁+4q₂))

= -(-sin(q₁+4q₂)

4) + 4(-sin(q₁+4q₂)

1)

= 4sin(q₁+4q₂) – 4sin(q₁+4q₂) = 0

Since {P₂, H} = 0 and P₂ has no explicit time dependence, P₂ is a constant of motion.

Obtain the New Hamiltonian K(Q, P):

We need to invert the transformation to express the old variables (q, p) in terms of the new ones (Q, P):

q₂ = Q₂

q₁ = Q₁ – 4q₂ = Q₁ – 4Q₂

p₁ = P₁

p₂ = P₂ + 4p₁ = P₂ + 4P₁

Now substitute these expressions into the original Hamiltonian H. Since the transformation is time-independent, the new Hamiltonian K is just H expressed in the new coordinates.

K = (p₁²/2m + p₂²/2m) + cos(q₁ + 4q₂)

K = (P₁²/2m + (P₂ + 4P₁)²/2m) + cos(Q₁)

K = (1/2m) [P₁² + (P₂² + 8P₁P₂ + 16P₁²)] + cos(Q₁)

K = (1/2m) [17P₁² + P₂² + 8P₁P₂] + cos(Q₁)

Thus, the new Hamiltonian is:

K(Q₁, Q₂, P₁, P₂) = (17P₁² + P₂² + 8P₁P₂)/2m + cos(Q₁)

As expected, the new Hamiltonian does not depend on Q₂, confirming that Q₂ is a cyclic coordinate and its conjugate momentum P₂ is a conserved quantity (constant of motion).

Q6. The Lagrangian of a damped harmonic oscillator is given as : L = e ^γt (½mẋ² – ½mω²x²). (a) Obtain the Hamiltonian of the system and write the Hamilton-Jacobi (H-J) equation. (b) Using the generating function : F₂(q, P, t) = e ^γt/2 qP, obtain the new Hamiltonian K(Q, P). Hence write the H-J equation associated with the new Hamiltonian.

Ans. (a) Hamiltonian and Hamilton-Jacobi Equation The given Lagrangian is: L = e ^γt (½mẋ² – ½mω²x²) 1. Canonical Momentum (p): p = ∂L/∂ẋ = e ^γt (mẋ) From this we can express ẋ in terms of p: ẋ = (p/m)e ^-γt 2. Hamiltonian (H): H = pẋ – L H = p[(p/m)e ^-γt ] – e ^γt [½m((p/m)e ^-γt )² – ½mω²x²] H = (p²/m)e ^-γt – e ^γt [½m(p²/m²)e ^-2γt – ½mω²x²] H = (p²/m)e ^-γt – (p²/2m)e ^-γt + ½mω²x²e ^γt H(x, p, t) = (1/2m)p²e ^-γt + ½mω²x²e ^γt 3. Hamilton-Jacobi (H-J) Equation: The H-J equation is H(q, ∂S/∂q, t) + ∂S/∂t = 0. Here q=x and p = ∂S/∂x. Substituting p with ∂S/∂x in the Hamiltonian, we find: (1/2m)e ^-γt (∂S/∂x)² + ½mω²x²e ^γt + ∂S/∂t = 0 This is the Hamilton-Jacobi equation for the given system. (b) New Hamiltonian and Associated H-J Equation The given generating function is of type F₂: F₂(q, P, t) = e ^γt/2 qP 1. Transformation Equations: For F₂, we have p = ∂F₂/∂q and Q = ∂F₂/∂P. (Here q=x) p = ∂/∂q(e ^γt/2 qP) = e ^γt/2 P Q = ∂/∂P(e ^γt/2 qP) = e ^γt/2 q 2. New Hamiltonian (K): When the generating function depends explicitly on time, the new Hamiltonian is K = H + ∂F₂/∂t. First, we express H in terms of the new variables (Q, P): p = e ^γt/2 P ⇒ p² = e ^γt P² q = e ^-γt/2 Q ⇒ q² = e ^-γt Q² H = (1/2m)p²e ^-γt + ½mω²x²e ^γt H = (1/2m)(e ^γt P²)e ^-γt + ½mω²(e ^-γt Q²)e ^γt H = P²/2m + ½mω²Q² Now, we calculate ∂F₂/∂t: ∂F₂/∂t = ∂/∂t(e ^γt/2 qP) = (γ/2)e ^γt/2 qP To express this in new variables, we note that Q = e ^γt/2 q, so e ^γt/2 q = Q. ∂F₂/∂t = (γ/2)QP Thus, the new Hamiltonian K is: K = (P²/2m + ½mω²Q²) + (γ/2)QP K(Q, P) = P²/2m + ½mω²Q² + (γ/2)QP 3. H-J Equation Associated with the New Hamiltonian: This refers to the H-J equation for a system that would have K as its Hamiltonian. Since K(Q,P) does not explicitly depend on time, the energy E’ = K is conserved. The time-independent H-J equation is K(Q, ∂S’/∂Q) = E’. Replacing P with ∂S’/∂Q: (1/2m)(∂S’/∂Q)² + ½mω²Q² + (γ/2)Q(∂S’/∂Q) = E’ Where S’ is the time-independent part of the action in the new coordinates and E’ is the energy in the new system.

Q7. The Hamiltonian of a particle of mass m in a two-dimensional simple harmonic potential is: H = (pₓ²/2m + pᵧ²/2m) + ½m(ωₓ²x² + ωᵧ²y²). Determine the action-angle variables Jₓ and Jᵧ (in integral form) if the total energy (E) of the system is conserved.

Ans. The given Hamiltonian describes a two-dimensional anisotropic harmonic oscillator: H = (pₓ²/2m + pᵧ²/2m) + ½m(ωₓ²x² + ωᵧ²y²) This Hamiltonian is separable in the x and y coordinates. We can write it as a sum of two independent one-dimensional Hamiltonians: H = Hₓ(x, pₓ) + Hᵧ(y, pᵧ) where: Hₓ = pₓ²/2m + ½mωₓ²x² Hᵧ = pᵧ²/2m + ½mωᵧ²y² Since Hₓ depends only on x and pₓ, and Hᵧ depends only on y and pᵧ, they are both constants of motion separately. We denote them as energies Eₓ and Eᵧ respectively. Hₓ = Eₓ Hᵧ = Eᵧ The total energy of the system is E = Eₓ + Eᵧ, which is conserved. The Action Variable Jᵢ is defined as the integral of the canonical momentum pᵢ over one full period of motion in phase space: Jᵢ = ∮ pᵢ dqᵢ Determination of Jₓ: For the x-motion, the action variable is Jₓ: Jₓ = ∮ pₓ dx We solve for pₓ from Eₓ = pₓ²/2m + ½mωₓ²x²: pₓ² = 2m(Eₓ – ½mωₓ²x²) pₓ = ±√(2m(Eₓ – ½mωₓ²x²)) The integration is performed over a complete cycle of oscillation. The particle oscillates along the x-axis. The turning points are where pₓ = 0, which occurs at x _max = ±√(2Eₓ / mωₓ²). The particle moves from -x _max to +x _max (where pₓ is positive) and then back to -x _max (where pₓ is negative). Therefore, the action variable Jₓ in integral form is: Jₓ = ∮ √(2m(Eₓ – ½mωₓ²x²)) dx This integral represents the area of the ellipse in the x-pₓ plane of phase space. Determination of Jᵧ: The procedure for the y-motion is completely analogous. The action variable is Jᵧ: Jᵧ = ∮ pᵧ dy From Eᵧ = pᵧ²/2m + ½mωᵧ²y², we solve for pᵧ: pᵧ = ±√(2m(Eᵧ – ½mωᵧ²y²)) Thus, the action variable Jᵧ in integral form is: Jᵧ = ∮ √(2m(Eᵧ – ½mωᵧ²y²)) dy The question only requires the integral form, so these are the final answers. If these integrals were evaluated, the results would be Jₓ = 2πEₓ/ωₓ and Jᵧ = 2πEᵧ/ωᵧ, meaning the Hamiltonian could be expressed in terms of the action variables as H = (Jₓωₓ + Jᵧωᵧ)/2π.

Q8. Write the Euler’s equation for a rigid body. Give the conditions for a torque free motion of a symmetric top. Hence obtain the equations of motion of the rigid body.

Ans. Euler’s Equations: Euler’s equations describe the rotational motion of a rigid body in a rotating reference frame attached to the body (usually the principal axes frame). Let ω = (ω₁, ω₂, ω₃) be the components of the angular velocity vector along the principal axes, and I₁, I₂, I₃ be the corresponding principal moments of inertia. If N = (N₁, N₂, N₃) are the components of the external torque, then Euler’s equations are:

• I₁ dω₁/dt + (I₃ – I₂) ω₂ω₃ = N₁
• I₂ dω₂/dt + (I₁ – I₃) ω₃ω₁ = N₂
• I₃ dω₃/dt + (I₂ – I₁) ω₁ω₂ = N₃

Torque-Free Motion of a Symmetric Top:

Conditions:

There are two main conditions for this motion:

1.

Torque-free:

There is no external torque, so

N

= 0. This means N₁ = N₂ = N₃ = 0.

2.

Symmetric Top:

The body has an axis of rotational symmetry, which means two of the principal moments of inertia are equal. By convention, we take

I₁ = I₂ ≠ I₃

. The axis 3 is called the symmetry axis of the body.

Equations of Motion:

Substituting these conditions into Euler’s equations:

1. I₁ dω₁/dt + (I₃ – I₁) ω₂ω₃ = 0

2. I₁ dω₂/dt + (I₁ – I₃) ω₃ω₁ = 0 (since I₂=I₁)

3. I₃ dω₃/dt + (I₁ – I₁) ω₁ω₂ = 0 ⇒ I₃ dω₃/dt = 0

Solution of the Equations:

From equation (3), we immediately conclude that:

dω₃/dt = 0 ⇒

ω₃ = constant

The component of the angular velocity along the symmetry axis remains constant.

Now we solve the first two equations:

dω₁/dt = – [(I₃ – I₁)/I₁] ω₂ω₃ = [(I₁ – I₃)/I₁] ω₃ ω₂

dω₂/dt = – [(I₁ – I₃)/I₁] ω₃ω₁

Let us define a constant frequency

Ω

:

Ω = [(I₁ – I₃)/I₁] ω₃

Since I₁, I₃, and ω₃ are all constants, Ω is also a constant. This is called the

precession frequency

.

The equations now become:

dω₁/dt = Ω ω₂

dω₂/dt = -Ω ω₁

This is a standard system of coupled differential equations. To solve it, differentiate the first equation with respect to t:

d²ω₁/dt² = Ω (dω₂/dt)

Substitute the second equation into this:

d²ω₁/dt² = Ω (-Ω ω₁) = -Ω²ω₁

d²ω₁/dt² + Ω²ω₁ = 0

This is the equation for simple harmonic motion. Its solution is:

ω₁(t) = A cos(Ωt + φ)

where A and φ are constants of integration.

To find ω₂(t), we substitute this solution into dω₁/dt = Ω ω₂:

-AΩ sin(Ωt + φ) = Ω ω₂

ω₂(t) = -A sin(Ωt + φ)

Description of the Motion:

These solutions show that:

• The component of angular velocity ω₃ is constant.
• The components of angular velocity in the 1-2 plane (the plane perpendicular to the symmetry axis), ω₁ and ω₂, form a vector ω _⊥ of constant magnitude, since ω₁² + ω₂² = A²cos²(…) + A²sin²(…) = A² = constant.
• This vector ω _⊥ rotates in the body’s 1-2 plane with angular frequency Ω.

As a result, the total angular velocity vector

ω

= (ω₁, ω₂, ω₃)

precesses

around the body’s axis of symmetry (the 3-axis), sweeping out a cone. This motion is called body-cone precession.