Unit 3: Hamiltonian Formalism

1. Canonical Momenta and Hamiltonian

Conceptual Background

In Lagrangian mechanics, the state of a system with n degrees of freedom is specified by generalized coordinates q_i and generalized velocities q̇_i = dq_i/dt for i = 1, 2, ..., n. The Lagrangian L(q_i, q̇_i, t) is a function defined in a 2n-dimensional configuration space plus time.

Hamiltonian mechanics transitions the description from configuration space (q_i, q̇_i) to phase space (q_i, p_i), replacing generalized velocities with generalized (canonical) momenta. This formulation replaces second-order differential equations with first-order differential equations.

Generalized (Canonical) Momentum Definition

Definition: For a system described by Lagrangian L(q_i, q̇_i, t), the generalized momentum p_i conjugate to the generalized coordinate q_i is defined as:
p_i = ∂L / ∂q̇_i

Legendre Transformation and Definition of Hamiltonian

To change the variables from (q_i, q̇_i, t) to (q_i, p_i, t), we apply a mathematical technique called the Legendre Transformation.

The total differential of the Lagrangian L(q_i, q̇_i, t) is given by:

dL = ∑_i=1ⁿ [ (∂L / ∂q_i) dq_i + (∂L / ∂q̇_i) dq̇_i ] + (∂L / ∂t) dt

Using the definition of canonical momentum p_i = ∂L / ∂q̇_i and Lagrange's equation of motion d/dt(∂L / ∂q̇_i) - ∂L / ∂q_i = 0 (which gives ṗ_i = ∂L / ∂q_i), we substitute these into dL:

dL = ∑_i=1ⁿ [ ṗ_i dq_i + p_i dq̇_i ] + (∂L / ∂t) dt

Rearranging the term ∑_i p_i dq̇_i using the product rule d(p_i q̇_i) = p_i dq̇_i + q̇_i dp_i:

dL = ∑_i=1ⁿ ṗ_i dq_i + d(∑_i=1ⁿ p_i q̇_i) - ∑_i=1ⁿ q̇_i dp_i + (∂L / ∂t) dt

d(∑_i=1ⁿ p_i q̇_i - L) = ∑_i=1ⁿ [ q̇_i dp_i - ṗ_i dq_i ] - (∂L / ∂t) dt

Definition of Hamiltonian (H): The Hamiltonian function H(q_i, p_i, t) is defined as:
H = ∑_i=1ⁿ p_i q̇_i - L(q_i, q̇_i, t)

Here, all velocity terms q̇_i in the expression must be expressed completely in terms of momentum p_i and coordinates q_i.

2. Hamilton's Equations of Motion

Derivation of Equations of Motion

Since the Hamiltonian H is explicitly a function of coordinates q_i, canonical momenta p_i, and time t, i.e., H = H(q₁,...,q_n, p₁,...,p_n, t), its total differential is:

dH = ∑_i=1ⁿ [ (∂H / ∂q_i) dq_i + (∂H / ∂p_i) dp_i ] + (∂H / ∂t) dt

From the Legendre transformation derivation, we previously established that:

dH = ∑_i=1ⁿ [ q̇_i dp_i - ṗ_i dq_i ] - (∂L / ∂t) dt

Comparing coefficients of independent differentials dq_i, dp_i, and dt in both total differential expressions yields Hamilton's Canonical Equations of Motion:

Hamilton's Canonical Equations:
1. q̇_i = ∂H / ∂p_i
2. ṗ_i = - ∂H / ∂q_i
3. ∂H / ∂t = - ∂L / ∂t

Properties of Canonical Equations

• They consist of 2n first-order differential equations replacing n second-order Lagrange equations.
• They show high mathematical symmetry, differing only by a minus sign.
• They describe the phase space trajectory of a mechanical system.

Exam Note: Remember that while ṗ_i has a negative sign (-∂H/∂q_i), q̇_i has a positive sign (+∂H/∂p_i). Reversing these signs is a common exam mistake.

3. Physical Significance of Hamiltonian (H)

1. Hamiltonian as Total Energy

Consider a conservative system where kinetic energy T is a homogeneous quadratic function of generalized velocities:

T = ∑_i,j a_i,j q̇_i q̇_j

By Euler's theorem for homogeneous functions of degree 2:

∑_i=1ⁿ q̇_i (∂T / ∂q̇_i) = 2T

If potential energy V is independent of generalized velocities q̇_i (i.e., V = V(q_i, t)), then:

p_i = ∂L / ∂q̇_i = ∂(T - V) / ∂q̇_i = ∂T / ∂q̇_i

Substituting this into the Hamiltonian definition:

H = ∑_i=1ⁿ p_i q̇_i - L = 2T - (T - V) = T + V = E

Physical Significance Rule 1: If the transformation equations between Cartesian and generalized coordinates do not depend explicitly on time (synamical constraints are time-independent) and potential energy is velocity-independent, the Hamiltonian H represents the total mechanical energy (E = T + V) of the system.

2. Hamiltonian as Constant of Motion

Taking the total time derivative of H(q_i, p_i, t):

dH / dt = ∑_i=1ⁿ [ (∂H / ∂q_i) q̇_i + (∂H / ∂p_i) ṗ_i ] + ∂H / ∂t

Substitute Hamilton's canonical equations (q̇_i = ∂H/∂p_i and ṗ_i = -∂H/∂q_i):

dH / dt = ∑_i=1ⁿ [ (∂H / ∂q_i)(∂H / ∂p_i) + (∂H / ∂p_i)(-∂H / ∂q_i) ] + ∂H / ∂t

dH / dt = ∂H / ∂t

Physical Significance Rule 2: If the Hamiltonian does not depend explicitly on time t (i.e., ∂H / ∂t = 0), then dH / dt = 0, meaning H is a constant of motion (conserved quantity).

4. Application: Free Particle

System Description

Consider a particle of mass m moving freely in three-dimensional space with no external potential acting on it (V = 0).

Step-by-Step Derivation

Step 1: Write Lagrangian
L = T - V = ½ m (ẋ² + ẏ² + ż²)

Step 2: Calculate Canonical Momenta
p_x = ∂L / ∂ẋ = m ẋ → ẋ = p_x / m
p_y = ∂L / ∂ẏ = m ẏ → ẏ = p_y / m
p_z = ∂L / ∂ż = m ż → ż = p_z / m

Step 3: Construct Hamiltonian
H = p_x ẋ + p_y ẏ + p_z ż - L
H = (p_x² + p_y² + p_z²) / m - ½ m [ (p_x/m)² + (p_y/m)² + (p_z/m)² ]
H = (p_x² + p_y² + p_z²) / (2m)

Step 4: Apply Hamilton's Equations
For x-component:
ẋ = ∂H / ∂p_x = p_x / m
ṗ_x = - ∂H / ∂x = 0 → p_x = constant (c₁)
Therefore, ẋ = c₁ / m = constant → x(t) = (c₁/m)t + x₀

Similarly for y and z coordinates, p_y and p_z are constant.

Physical Conclusion

The free particle moves in a straight line with constant velocity, illustrating Newton's First Law of Motion.

5. Application: Simple Harmonic Oscillator

System Description

A mass m attached to a spring with force constant k performing one-dimensional oscillation along the x-axis.

Step-by-Step Derivation

Step 1: Write Lagrangian
T = ½ m ẋ², V = ½ k x²
L = ½ m ẋ² - ½ k x²

Step 2: Calculate Canonical Momentum
p = ∂L / ∂ẋ = m ẋ → ẋ = p / m

Step 3: Construct Hamiltonian
H = p ẋ - L = p (p/m) - [ ½ m (p/m)² - ½ k x² ]
H = p² / (2m) + ½ k x²

Step 4: Apply Hamilton's Equations
1. ẋ = ∂H / ∂p = p / m
2. ṗ = - ∂H / ∂x = - k x

Differentiating equation (1) with respect to time t:
ẍ = ṗ / m
Substitute ṗ = -k x into the above equation:
ẍ = - (k / m) x
ẍ + ω² x = 0 (where ω² = k / m)

Physical Conclusion

The equation obtained is the standard equation of Simple Harmonic Motion with angular frequency ω = √(k/m).

6. Application: Simple Pendulum

System Description

A point mass m suspended by a light, inextensible string of length l swinging in a vertical plane under gravity g. Generalized coordinate is angle θ.

Step-by-Step Derivation

Step 1: Write Lagrangian
Velocity v = l θ̇
Kinetic Energy: T = ½ m l² θ̇²
Potential Energy (relative to equilibrium lowest position): V = m g l (1 - cos θ)
L = ½ m l² θ̇² - m g l (1 - cos θ)

Step 2: Calculate Canonical Momentum
p_θ = ∂L / ∂θ̇ = m l² θ̇ → θ̇ = p_θ / (m l²)

Step 3: Construct Hamiltonian
H = p_θ θ̇ - L = p_θ² / (m l²) - [ ½ p_θ² / (m l²) - m g l (1 - cos θ) ]
H = p_θ² / (2 m l²) + m g l (1 - cos θ)

Step 4: Apply Hamilton's Equations
1. θ̇ = ∂H / ∂p_θ = p_θ / (m l²)
2. ṗ_θ = - ∂H / ∂θ = - m g l sin θ

Differentiating equation (1) with respect to t:
θ̈ = ṗ_θ / (m l²)
Substitute ṗ_θ = - m g l sin θ:
θ̈ = - (m g l sin θ) / (m l²)
θ̈ + (g / l) sin θ = 0

For small angular displacements (sin θ ≈ θ):
θ̈ + (g / l) θ = 0

7. Application: Compound Pendulum

System Description

A rigid body of mass M pivoted about a horizontal axis through a point O, oscillating under gravity. Let I be the moment of inertia about the axis of rotation and h be the distance from the pivot O to the center of mass C.

Step-by-Step Derivation

Step 1: Write Lagrangian
Kinetic Energy: T = ½ I θ̇²
Potential Energy: V = M g h (1 - cos θ)
L = ½ I θ̇² - M g h (1 - cos θ)

Step 2: Calculate Canonical Momentum
p_θ = ∂L / ∂θ̇ = I θ̇ → θ̇ = p_θ / I

Step 3: Construct Hamiltonian
H = p_θ θ̇ - L = p_θ (p_θ / I) - [ ½ I (p_θ / I)² - M g h (1 - cos θ) ]
H = p_θ² / (2 I) + M g h (1 - cos θ)

Step 4: Apply Hamilton's Equations
1. θ̇ = ∂H / ∂p_θ = p_θ / I
2. ṗ_θ = - ∂H / ∂θ = - M g h sin θ

Differentiating equation (1) with respect to t:
θ̈ = ṗ_θ / I = - (M g h / I) sin θ
I θ̈ + M g h sin θ = 0

For small amplitudes (sin θ ≈ θ):
θ̈ + (M g h / I) θ = 0

The frequency of oscillation is ω = √(Mgh / I), and equivalent length L_eq = I / (Mh).

8. Application: Particle in a Central Force Field

System Description

A particle of mass m moving under a central potential V(r) depending only on the distance from a central origin. Motion is confined to a plane, described by plane polar coordinates (r, θ).

Step-by-Step Derivation

Step 1: Write Lagrangian
T = ½ m (ṙ² + r² θ̇²)
L = ½ m (ṙ² + r² θ̇²) - V(r)

Step 2: Calculate Canonical Momenta
p_r = ∂L / ∂ṙ = m ṙ → ṙ = p_r / m
p_θ = ∂L / ∂θ̇ = m r² θ̇ → θ̇ = p_θ / (m r²)

Step 3: Construct Hamiltonian
H = p_r ṙ + p_θ θ̇ - L
H = p_r (p_r/m) + p_θ (p_θ/(m r²)) - [ ½ m ((p_r/m)² + r² (p_θ/(m r²))²) - V(r) ]
H = p_r² / (2m) + p_θ² / (2 m r²) + V(r)

Step 4: Apply Hamilton's Equations
1. ṙ = ∂H / ∂p_r = p_r / m
2. θ̇ = ∂H / ∂p_θ = p_θ / (m r²)
3. ṗ_r = - ∂H / ∂r = p_θ² / (m r³) - ∂V / ∂r
4. ṗ_θ = - ∂H / ∂θ = 0

Key Observations

• Since θ does not appear explicitly in H, θ is a cyclic (ignorable) coordinate.
• ṗ_θ = 0 implies p_θ = constant = L_z (angular momentum is conserved).
• Equation for radial motion: m r̈ = p_θ² / (m r³) - ∂V / ∂r.

9. Application: Charged Particle in an Electromagnetic Field

System Description

A particle with charge q and mass m moving in an electromagnetic field described by scalar potential ϕ(r, t) and vector potential A(r, t).

Step-by-Step Derivation

Step 1: Generalized Velocity-Dependent Potential and Lagrangian
The velocity-dependent potential is U = q ϕ - q v · A.
L = T - U = ½ m v² - q ϕ + q (v · A)
In Cartesian components:
L = ½ m (ẋ² + ẏ² + ż²) - q ϕ + q (ẋ A_x + ẏ A_y + ż A_z)

Step 2: Canonical Momentum
p_x = ∂L / ∂ẋ = m ẋ + q A_x → m ẋ = p_x - q A_x
In vector form: p = m v + q A

Important Note: The canonical momentum p is not equal to mechanical momentum p_mech = mv; it includes the field term qA.

Step 3: Construct Hamiltonian
H = p · v - L
H = p · [ (p - q A) / m ] - [ ½ m ((p - q A) / m)² - q ϕ + q ((p - q A)/m) · A ]
Simplifying terms:
H = (p - q A)² / (2m) + q ϕ

Step 4: Hamilton's Canonical Equations
1. ṙ = ∇_p H = (p - q A) / m
2. ṗ = - ∇ H = - ∇ [ (p - q A)² / (2m) + q ϕ ]

Evaluating Hamilton's equations yields the standard Lorentz Force Law:
F = m a = q (E + v × B)
where E = -∇ϕ - ∂A/∂t and B = ∇ × A.

10. Application: Particle Moving Near the Surface of Earth

System Description

A particle of mass m moving near Earth's surface where acceleration due to gravity g is assumed constant along the vertical z-axis (z pointing vertically upwards).

Step-by-Step Derivation

Step 1: Write Lagrangian
Position: (x, y, z)
Kinetic Energy: T = ½ m (ẋ² + ẏ² + ż²)
Potential Energy: V = m g z
L = ½ m (ẋ² + ẏ² + ż²) - m g z

Step 2: Calculate Canonical Momenta
p_x = ∂L / ∂ẋ = m ẋ → ẋ = p_x / m
p_y = ∂L / ∂ẏ = m ẏ → ẏ = p_y / m
p_z = ∂L / ∂ż = m ż → ż = p_z / m

Step 3: Construct Hamiltonian
H = p_x ẋ + p_y ẏ + p_z ż - L
H = (p_x² + p_y² + p_z²) / (2m) + m g z

Step 4: Apply Hamilton's Equations
For x and y components:
ẋ = p_x / m, ṗ_x = - ∂H / ∂x = 0 → p_x = constant
ẏ = p_y / m, ṗ_y = - ∂H / ∂y = 0 → p_y = constant

For vertical z component:
ż = p_z / m
ṗ_z = - ∂H / ∂z = - m g

Differentiating ż with respect to t:
z̈ = ṗ_z / m = - (m g) / m = - g

Physical Results

• Motion in x and y directions is uniform unaccelerated motion.
• Motion along the z direction is uniformly accelerated downward motion with acceleration -g.
• Coordinates x and y are cyclic; hence p_x and p_y are conserved constants of motion.

11. Conservation Theorems in Hamiltonian Formalism

1. Cyclic Coordinates and Conservation of Canonical Momentum

Definition: A generalized coordinate q_k is said to be cyclic or ignorable if it does not explicitly appear in the Hamiltonian H, i.e., ∂H / ∂q_k = 0.

Using Hamilton's second canonical equation for a cyclic coordinate q_k:

ṗ_k = - ∂H / ∂q_k = 0 → p_k = constant

Theorem: The canonical momentum conjugate to a cyclic coordinate is a constant of motion (conserved quantity).

2. Conservation of Linear Momentum

Consider a closed system unaffected by external forces. If the entire system is translated along a direction (e.g., x-axis) by an infinitesimal displacement, the potential energy and Hamiltonian remain invariant.

Since H is invariant under translation along x, the coordinate x is cyclic (∂H / ∂x = 0).

From Hamilton's equation:
ṗ_x = - ∂H / ∂x = 0 → p_x = constant

Conclusion: Translational invariance of a system implies conservation of total linear momentum along that direction.

3. Conservation of Angular Momentum

Consider a system invariant under rotation about a specific axis (e.g., z-axis). If the orientation angle θ about this axis does not appear in H, then θ is a cyclic coordinate.

∂H / ∂θ = 0

From Hamilton's equation:
ṗ_θ = - ∂H / ∂θ = 0 → p_θ = constant

Since p_θ is the canonical angular momentum L_z about the rotation axis:
L_z = constant

Conclusion: Rotational invariance of a system implies conservation of angular momentum about the symmetry axis.

4. Conservation of Energy

The total time derivative of Hamiltonian H(q_i, p_i, t) is:

dH / dt = ∑_i=1ⁿ [ (∂H / ∂q_i) q̇_i + (∂H / ∂p_i) ṗ_i ] + ∂H / ∂t

Substituting q̇_i = ∂H / ∂p_i and ṗ_i = - ∂H / ∂q_i gives:

dH / dt = ∂H / ∂t

If the system's constraints and potential are time-independent, H does not depend explicitly on time t (∂H / ∂t = 0).

Therefore:
dH / dt = 0 → H = constant

If, additionally, coordinates are independent of explicit time, H equals total mechanical energy E, giving total energy conservation E = constant.

12. Summary Table of Applications