Lagrangian Dynamics Flashcards
Total energy
E = T + V
Total = Kinetic + Potential
Kinetic energy
T = 1/2 mẋ²
Derive Newton’s 2nd Law using Lagrangian thinking
E = T + V
Energy is conserved, so
dE/dt = 0
dE/dt = ∂E/∂t + ∂E/∂x dx/dt + ∂E/∂ẋ dẋ/dt
= 0 + ∂V/∂x ẋ + mẋẍ
= (mẍ + ∂V/∂x) ẋ
Either ẋ = 0 OR mẍ = -∂V/∂x
Lagrangian
L = T − V
= 1/2 mẋ² − V(x)
Derive Lagrange’s equation from the Lagrangian
L = T − V
= 1/2 mẋ² − V(x)
Derivative w.r.t. x:
∂L/∂x = −∂V/∂x
N2L
∂L/∂x = −∂V/∂x = mẍ = d/dt (∂L/∂ẋ)
Lagrange’s equation:
d/dt (∂L/∂ẋ) = ∂L/∂x
One exists for each degree of freedom
Plane polar coordinates
r = |r| and θ degrees of freedom
x = r cosθ
y = r sinθ
V(r) = V(r,θ)
Lagrangian in plane polar
vᵣ = ṙ
v_θ = rθ̇
v² = vᵣ² + v_θ²
L = T − V
= 1/2 m (ṙ² + r²θ̇²) − V(r,θ)
Varying r with θ fixed:
d/dt (∂L/∂ṙ) = ∂L/∂r
Varying θ with fixed r:
Small increase in θ written δθ.
Infinitesimal displacement along circumference δc = rδθ.
Force along δc is
F_θ = −1/r ∂V/∂θ
Torque force around origin is
G = rF_θ = −∂V/∂θ = ∂L/∂θ
Using Newton,
G = Iθ̈ = mr²θ̈
r is fixed
G = d/dt (mr²θ̇)
Hence,
d/dt (∂L/∂θ̇) = ∂L/∂θ
differentials are angular momentum and torque
{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}
d/dt (∂L/∂ṙ) = ∂L/∂r
d/dt (∂L/∂θ̇) = ∂L/∂θ
General eq for change in energy
dE/dt = ∂E/∂t + ∂E/∂x dx/dt + ∂E/∂ẋ dẋ/dt
= 0 + ∂V/∂x ẋ + mẋẍ
= (mẍ + ∂V/∂x) ẋ
Newton vs Lagrange approach to energy change equation
dE/dt = (mẍ + ∂V/∂x) ẋ
Newton: mẍ = −∂V/∂x, so dE/dt = 0
Lagrange: Energy is conserved, so dE/dt = 0.
Either ẋ = 0, particle is always stationary and nothing
interesting happens.
Or particle moves in such a way that mẍ is always equal to −∂V/∂x.
Newton method:
Mass on a spring
Motion in 1 dimension under potential 1/2 kx²
mẍ = F = −∂V/∂x = −kx
Corresponds to SHM with angular frequency ω = √k/m’
Energy conservation method:
Mass on a spring
Motion in 1 dimension under potential 1/2 kx²
E = 1/2 mẋ² + 1/2 kx² = constant
dE/dt = (mẍ + kx)ẋ = 0
Either ẋ = 0 (BORING!) or mẍ = −kx
Lagrange’s method:
Mass on a spring
Motion in 1 dimension under potential 1/2 kx²
L = 1/2 mẋ² − 1/2 kx²
d/dt (∂L/∂ẋ) = ∂L/∂x
so
d/dt (mẋ) = −kx
mẍ = −kx
3D Lagrange equation
In 3D,
V = V (x, y, z) and
T = 1/2m (ẋ² + ẏ² + ż²)
Three Lagrange’s equations:
d/dt (∂L/∂ẋ) = ∂L/∂x
d/dt (∂L/∂ẏ) = ∂L/∂y
d/dt (∂L/∂ż) = ∂L/∂z
In more compact form,
for xᵢ = x, y, z
d/dt (∂L/∂ẋᵢ) = ∂L/∂xᵢ
Spherical polar coordinates
x = r sinθ cosφ
y = r sinθ sinφ
z = r cosθ
Lagrangian in spherical polar
Lagranian
L = 1/2 m (ṙ² + r²θ̇² + r²sin²θ φ̇²) − V(r,θ,φ)
Equations of motion
d/dt (∂L/∂ṙ) = ∂L/∂r
d/dt (∂L/∂θ̇) = ∂L/∂θ
d/dt (∂L/∂φ̇) = ∂L/∂φ
Each degree of freedom is treated identically.
Lagrangian in generalized coordinates
Generalized coordinates
qᵢ(t)
Generalized velocities
q̇ᵢ(t) = dqᵢ/dt
Lagrangian
L = L (qᵢ(t), q̇ᵢ(t), t)
may be a function of generalized coordinates and velocities, and may have explicit time dependence.
Generalized momenta
pᵢ(t)
given by
pᵢ = ∂L/∂q̇ᵢ
Lagrange’s equations:
d/dt (∂L/∂q̇ᵢ) = ∂L/∂qᵢ
Lagrange:
Particle mass m, central attractive spherically symmetric potential V(r) = − A/r
Use plane polar:
L = T − V
= 1/2 m (ṙ² + r²θ̇²) + A/r
Angular:
d/dt (∂L/∂θ̇) = d/dt (mr²θ̇) = ∂L/∂θ = 0
mr²θ̇ is a constant of motion = angular momentum, J
Whenever L is independent of a degree of freedom, there is a corresponding conserved quantity.
Radial:
d/dt (∂L/∂ṙ) = ∂L/∂r
mr̈ = mrθ̇² − A/r²
First term is “centrifugal force” Second term is inverse square attractive force (as expected for this potential)
Stable solution can be found with mr̈ = 0.
This corresponds to circular orbit with
r = (mr²θ̇²)²/mA = J²/mA ≡ r₀
Can set r = r₀ + δr and the equation of motion for δr (with δr ≪ r₀)
Lagrange:
Two masses m₁ and m₂ connected by massless, inextensible string that moves without slipping over a pulley of radius a and mass M.
Label distance down from pivot to mass m₁ as x.
Potential energy:
V = gx(m₂ − m₁)
Lagrangian:
L = T − V
= 1/2 (m₁ẋ² + m₂ẋ² + Iθ̇²) − gx(m₂−m₁)
where θ describes rotation of the pulley.
No slipping:
θ̇ = ẋ/a
Moment of inertia of disk of mass M and radius a:
I = 1/2 Ma²
L = 1/2 (m₁ + m₂ + 1/2M)ẋ² − gx(m₂−m₁)
This is a function of only one degree of freedom.
Lagrange equation:
d/dt (∂L/∂ẋ) = ∂L/∂x
⇒ ẍ = (m₁−m₂) / (m₁+m₂+½M) g
Generalized momentum
∂L/∂ẋ mixes terms corresponding to both linear and angular momentum.
Define action
S = ∫L dt
Equation for principle of least action
Seek path x(t) to minimize the integral:
S[x(t)] = ∫t₀→t₁ L(x(t), ẋ(t)) dt
Least action for particle of mass m moving in constant potential
Constant potential implies no force.
N1L, expect solution to be motion at constant velocity.
Lagrangian:
L = T − V
= 1/2 m [ẋ(t)]² − V
where V is constant.
Action:
S = ∫t₀→t₁ L dt
= ∫t₀→t₁ 1/2 m [ẋ(t)]² dt − V(t₁−t₀)
Final term is constant, so irrelevant for minimizing S.
Want to minimize
X ≡ ∫t₀→t₁ [ẋ(t)]² dt
subject to x(t₀) = x₀ and x(t₁) = x₁
Assuming motion at constant velocity
v = x₁−x₀/t₁−t₀
X = ∫t₀→t₁ v² dt = v²∆t
where ∆t = t₁ − t₀
Try simplest non-constant solution, motion at higher speed v + δv for half of the time, ∆t/2.
To arrive at destination at correct time, would have to have motion at v − δv for the remaining ∆t/2:
X(δv) = (v + δv)² ∆t/2 + (v − δv)² ∆t/2
= v²∆t + (δv)²∆t
X, and hence S, is minimized by δv = 0, i.e. it prefers to move at constant speed.
Least action for particle of mass m with potential energy mgx.
S[x(t)] = m ∫t₀→t₁ (½[ẋ(t)]² − gx(t)) dt
To simplify, set t₀ = 0 and x₀ = 0
Expect motion with constant acceleration a = −g
Calculate motion with arbitrary a
x(t) = ut + ½at²
Particle arrives at x = x₁ at time t = t₁
u = x₁/t₁ − ½at₁
ẋ(t) = (x₁/t₁ − ½at₁) + at
Action:
S(a) = m (1/24 (a²+2ag)t₁³ − ½gx₁t₁ + x₁²/2t₁)
This is minimized for a = −g, so principle of least action is consistent with Newton’s laws.
Demonstrate Calculus of Variations
Consider function f(t).
At some general point t, if t is varied slightly, the function f will vary linearly in response. At minima/maxima/turning points, varying t slightly varies f quadratically (df/dt is zero so first non-zero term in Taylor expansion comes from d²f/dt²). Apply same principle to functions.
Path that minimizes action is x(t).
It can be deformed slightly to non-minimal path x(t) + δx(t).
δx(t) is completely arbitrary, but assumed to be very small w.r.t. x(t) and obeys same boundary conditions as original.
δx(t₀) = δx(t₁) = 0
On true path
L = L(x(t), ẋ(t))
On modified path
L(x(t) + δx(t), ẋ(t) + d/dt (δx(t))) ≡ L + δL
δx(t) assumed small, can Taylor expand L:
δL = ∂L/∂x δx(t) + ∂L/∂ẋ d/dt (δx(t))
Amount action changes:
δS = ∫t₀→t₁ δL dt
= ∫t₀→t₁ [∂L/∂x δx(t) + ∂L/∂ẋ d/dt (δx(t))] dt
Leave first term as is, second term integrate by parts: define u = ∂L/∂ẋ and dv = d(δx(t))
∫t₀→t₁ ∂L/∂ẋ d/dt (δx(t)) dt
= [∂L/∂ẋ δx(t)]t₀→t₁ − ∫t₀→t₁ d/dt (∂L/∂ẋ) δx(t) dt
First term is zero, boundary conditions ensure all paths pass through same end-points, so δx(t₀) = δx(t₁) = 0.
Sub second term into δS equation, and extract common factor δx(t)
δS = ∫t₀→t₁ δL dt
= ∫t₀→t₁ [∂L/∂x − d/dt (∂L/∂ẋ)] δx(t) dt
Want path x(t) to be such that small variations do not affect action, corresponding to δS = 0. This means either integrand is +ve and -ve in such a way to integrate exactly to zero, or integrand is zero everywhere. Want δx(t) to be arbitrary, so cannot accept first possibility.
It must be the case that
d/dt (∂L/∂ẋ) = ∂L/∂x
Principle of Least Action directly implies that x(t) obey’s Lagrange’s equation.
Calculus of Variations:
Shortest path between two points
Infinitesimal step
dD = √dx² + dy² ‘
= √1 + (dy/dx)² ‘ dx
= √1 + y’² ‘ dx
Total distance
D = ∫x₀→x₁ √1 + y’² ‘ dx
Want to minimize D.
Euler-Lagrange equations tell us y(x) is such that
d/dx (∂f/∂y’) = ∂f/∂y
where f = √1 + y’² ‘
d/dx (y’/√(1+y’²)’) = 0
⇒ y’/√(1+y’²)’ = constant
⇒ y’ = constant
⇒ y = mx + c
Shows shortest path between two points is a straight line.
Also note that
y’/√(1+y’²)’ = (dy/dx) / √1+(dy/dx)²’
= dy/√dx²+dy²’ = sinθ
Calculus of Variations:
Lifeguard problem
Lifeguard watching a beach, can run at v₁ on sand and swim at v₂ < v₁ in sea.
Initially at position (0, 0), swimmer drowning at (x₂, y₂) (shoreline at x = x₁).
Travel time
T = ∫0→x₂ dD/v
dD = √(1+y’²)’ dx
where dD is infinitesimal distance and v is a function of x.
From this, can obtain
y’/v√(1+y’²)’ = constant
Wherever v is constant, y’ is also, so in sea and on sand the shortest time is given by a straight line.
Gradients for sand and sea are not the same, because v is not the same.
Must have
y₁’/v₁√(1+y₁’²)’ = y₂’/v₂√(1+y₂’²)’
Can be written as:
sinθ₁/v₁ = sinθ₂/v₂
ibn-Sahl law!
Fermat’s Principle of Least Time!
Lagrangian:
Bead on a rotating wire
Point-like bead of mass m, free to slide frictionlessly along straight wire inclined at angle θ, rotating at constant angular speed ω about the vertical.
Parametrize position, r, in spherical polar co-ords: θ is constant and φ̇ = ω is constant
T = ½m (ṙ²+ r²sin²θ ω²)
V = mgr cosθ
L = T − V
= ½m (ṙ²+ r²sin²θ ω²) − mgr cosθ
r is the only dynamic degree of freedom, so only have one Lagrange equation:
d/dt (∂L/∂ṙ) = mr̈ = ∂L/∂r = mrsin²θ ω² − mgcosθ
Acceleration is therefore
r̈ = r sin²θ ω² − g cosθ
Equilibrium point (r̈ = 0, ṙ = 0) at
r ≡ r₀ = gcosθ/sin²(θ)ω²
What kind of equilibrium?
Consider r = r₀ + δr
r̈ = δr̈ = δr sin²θ ω²
Unstable equilibrium: if m starts ‘at rest’ close to equilibrium position, will accelerate away.
Note that energy is not conserved (motor has to do work on the wire to keep it turning at constant speed)
[Exercise: write E = T + V , calculate dE/dt and interpret your result in terms of power exerted by the motor, P = ωd/dt (Iω)]
Lagrangian:
Particle in a conical vase
Particle of mass m, sliding frictionlessly inside conical vase of opening angle θ. Similar to bead on wire but φ̇ is dynamical.
L = T − V
= ½mṙ² + ½mr²sin²θ φ̇² − mgrcosθ
Two degrees of freedom, r and φ.
d/dt (∂L/∂φ̇) = d/dt (mr²sin²θ φ̇) = ∂L/∂φ = 0
⇒ mr² sin²θ φ̇ = constant = J
(Angular momentum)
d/dt (∂L/∂ṙ) = mr̈ = ∂L/∂r = mrsin²θ φ̇² − mgcosθ
Substitute in constant J,
φ̇ = J/mr²sin²θ
⇒ r̈ = J²/m²r³sin²θ − gcosθ
Equilibrium position (r̈ = 0, ṙ = 0)
is at r = r₀,
r₀³ = J²/m²gsin²θcosθ
What kind of equilibrium?
Set r = r₀ + δr and consider limit δr≪r₀. Taylor expand 1/r³ about r=r₀, keep only terms linear in δr.
δr̈ = −3gcosθ/r₀ δr
Stable equilibrium: slight displacement from equilibrium, mass will oscillate about it, with angular frequency
ω = √3gcosθ/r₀’
Legendre transformation
- Define x∗ to be the Legendre conjugate of x
x∗ ≡ ∂f/∂x - Define Legendre transformation of f(x)
g(x∗) = x x∗ − f(x)
Derive Hamilton’s equations
System with one degree of freedom, q, and velocity, q̇, Lagrangian is L(q, q̇, t)
dL = ∂L/∂q dq + ∂L/∂q̇ dq̇ + ∂L/∂t dt
Want to find function H(q, p, t), where p = ∂L/∂q̇, which depends only on q and p but not on q̇.
dH = ∂H/∂q dq + ∂H/∂p dp + ∂H/∂t dt
p is Legendre conjugate to q̇ and H is Legendre transformation of L:
H(q, p, t) = pq̇ − L(q, q̇, t)
H is the Hamiltonian
General change in H is
dH = p dq̇ + q̇ dp − dL
dH = p dq̇ + q̇ dp − ∂L/∂q dq − ∂L/∂q̇ dq̇ − ∂L/∂t dt
Fourth term, replace ∂L/∂q̇ by p.
Third term, use Lagrange’s equation,
∂L/∂q = d/dt (∂L/∂q̇) = d/dt (p) = ṗ
dH = q̇ dp − ṗ dq − ∂L/∂t dt
Compare coefficients in dH equations.
∂H/∂p = q̇
∂H/∂q = −ṗ
∂H/∂t = −∂L/∂t
Hamiltonian:
Particle in a conical vase
Particle of mass m sliding frictionlessly inside a conical vase of opening angle θ.
Lagrangian is
L = ½mṙ² +½ mr²sin²θ φ̇² − mgrcosθ
Two degrees of freedom, r and φ, with momenta
pᵣ = ∂L/∂ṙ = mṙ
and
p_φ = ∂L/∂φ̇ = mr²sin²θ φ̇
These can be trivially inverted,
ṙ = 1/m pᵣ
φ̇ = 1/mr²sin²θ p_φ
so Lagrangian can be rewritten
L = pᵣ²/2m + p_φ²/2mr²sin²θ − mgrcosθ
Hamiltonian is given by
H = Σi=r,φ pᵢq̇ᵢ − L = pᵣṙ + p_φφ̇ − L
= pᵣ²/2m + p_φ²/2mr²sin²θ + mgrcosθ
Apply Hamilton’s equations:
∂H/∂φ = 0 = −ṗ_φ
so p_φ is a constant
∂H/∂r = −p_φ²/mr³sin²θ + mgcosθ
= −ṗᵣ = −mr̈
so
r̈ = p_φ²/m²r³sin²θ − gcosθ
Applying Hamilton’s other two equations, obtain:
ṙ = ∂H/∂pᵣ = pᵣ/m
φ̇ = ∂H/∂pφ = p_φ/mr²sin²θ
Useful cross check:
ensure expression for the Hamiltonian depends only on qᵢ and pᵢ, no terms that still depend on q̇ᵢ
When is the Hamiltonian conserved?
In general, H changes with time according to
dH/dt = Σᵢ (∂H/∂pᵢ ṗᵢ + ∂H/∂qᵢ q̇ᵢ) + ∂H/∂t
Use Hamilton’s equations to replace ṗᵢ by −∂H/∂qᵢ and q̇ᵢ by ∂H/∂pᵢ
dH/dt = Σᵢ (−∂H/∂pᵢ ∂H/∂qᵢ + ∂H/∂qᵢ ∂H/∂pᵢ) + ∂H/∂t
i.e.
dH/dt = ∂H/∂t
We know that ∂H/∂t = −∂L/∂t.
For any system that is explicitly time independent, the Hamiltonian is conserved.
Hamiltonian:
Bead on a rotating wire
Bead of mass m on straight wire at angle θ to the vertical, being rotated at constant angular speed ω.
Lagrangian:
L = ½mṙ² + ½mr²sin²θ ω² − mgrcosθ
Only one degree of freedom, r, with momentum
pᵣ = ∂L/∂ṙ = mṙ
Lagrangian rewritten
L = pᵣ²/2m + ½mr²sin²θ ω² − mgrcosθ
Hamiltonian given by
H = pᵣṙ − L
= pᵣ²/m − (pᵣ²/2m + ½mr²sin²θ ω² − mgrcosθ)
= pᵣ²/2m − ½mr²sin²θ ω² + mgrcosθ
Note: NOT equal to total energy of the system.
Hamilton’s equation gives:
∂H/∂r = −mrsin²θ ω² + mgcosθ = −ṗᵣ = −mr̈
so
r̈ = rsin²θ ω² − gcosθ
dE/dt ≠ 0, but dH/dt = 0.
Energy not conserved, but Hamiltonian is.
Why do we use the Legendre Transformation
Can’t just replace velocity with momentum, but momentum is more fundamental as a conserved quantity
4 axioms of Galileo’s relativity
There is no absolute or preferred:
1. origin of time
2. position in space
3. orientation in space
4. inertial frame of reference
Hamiltonian consequence:
No absolute origin of time
EoMs unchanged by displacement in time
t → t + δt
Requires ∂H/∂t = 0, i.e. no explicit time dependence.
This implies H is a conserved quantity, and for an isolated system total energy E is conserved.
Hamiltonian consequence:
No absolute position in space
EoMs unchanged by displacement of entire system in space
rⱼ → rⱼ + δr, ∀ j
Can only be the case if the Hamiltonian depends only on relative displacements,
rⱼₖ ≡ rⱼ − rₖ
since rⱼₖs are unchanged by translations of the entire system.
Hamiltonian consequence:
No absolute orientation in space
Since rⱼₖ has direction, if Hamiltonian depended directly on it, the EoMs would have a preferred direction. Thus Hamiltonian can depend only on scalar quantities, such as
rⱼₖ² = |rⱼₖ|² = rⱼₖ · rⱼₖ
or rⱼₖ · rₗₘ
Poisson bracket
Any two observables, F and G, their Poisson
bracket is
[F, G] = Σᵢ (∂F/∂qᵢ ∂G/∂pᵢ − ∂F/∂pᵢ ∂G/∂qᵢ)
Poisson bracket w.r.t. H
(Time dependence)
General expression for rate of change of any observable F:
dF/dt = Σᵢ (∂F/∂qᵢ q̇ᵢ+ ∂F/∂pᵢ ṗᵢ) + ∂F/∂t
= Σᵢ (∂F/∂qᵢ ∂H/∂pᵢ − ∂F/∂pᵢ ∂H/∂qᵢ) + ∂F/∂t
With Poisson bracket
dF/dt = [F, H] + ∂F/∂t
Poisson bracket of H with any observable F generates implicit time dependence of F.
“H generates a displacement in time”
Note if ∂F/∂t = 0 and [F, H] = 0, F is a conserved quantity
Since [H, H] = 0, if ∂H/∂t = 0 then H is a conserved quantity, principle of relativity guarantees ∂H/∂t = 0.
Poisson bracket w.r.t. momentum
(Spacial dependence)
Displacement of N particles by distance δx in x direction:
qₓⱼ → qₓⱼ + δx for all j = 1, … , N
Observable F may change,
F → F + δF
Assuming δx is small, Taylor expand to calculate change in F:
δF = Σⱼ ∂F/∂qₓⱼ δx
Total x-momentum
Pₓ = Σₖ pₓₖ
Poisson bracket:
[F, Pₓ] = Σₛ Σⱼ [ ∂F/∂qₛⱼ ∂/∂pₛⱼ (Σₖ pₓₖ) − ∂F/∂pₛⱼ ∂/∂qₛⱼ (Σₖ pₓₖ) ]
Recall that the 3N qᵢs and 3N pᵢs are independent variables. All of the partial derivatives ∂pₓₖ/∂qₛⱼ are therefore zero. Likewise, most partial derivatives ∂pₓₖ/∂pₛⱼ are zero, with the exception of case s = x, k = j, where ∂pₓₖ/∂pₓₖ = 1
∂/∂pₛⱼ (Σₖ pₓₖ) = δₛₓ
[F, Pₓ] = Σⱼ ∂F/∂qₓⱼ
and
δF = [F, Pₓ] δx
“Pₓ generates a displacement in the x direction of the entire system”
Applying Pₓ to the Hamiltonian, to calculate how much it is changed by displacement in x:
δH = [H, Pₓ] δx
Principle of Relativity, spatial displacements cannot change the Hamiltonian
δH = 0, so [H, Pₓ] = 0
Time dependence of any function, e.g. Pₓ, is
driven by the Poisson bracket with the Hamiltonian, dPₓ/dt = [Pₓ, H]
From definition of Poisson bracket,
[A, B] = −[B, A]
and hence
dPₓ/dt = [Pₓ, H] = −[H, Pₓ] = 0
so total linear momentum is conserved .
Poisson bracket w.r.t. angular momentum
(Spacial orientation)
Rotation of N particles by angle δφ:
qᵩⱼ → qᵩⱼ + δφ for all j = 1, … , N
φ component of total angular momentum:
Pᵩ = Σⱼ pᵩⱼ
Poisson bracket:
[F, Pᵩ] = Σⱼ Σₛ [ ∂F/∂qₛⱼ ∂/∂pₛⱼ (Σₖ pᵩₖ) − ∂F/∂pₛⱼ ∂/∂qₛⱼ (Σₖ pᵩₖ) ]
= Σⱼ Σₛ (∂F/∂qₛⱼ Σₖδⱼₖ δₛᵩ)
= Σⱼ Σₛ (∂F/∂qₛⱼ δₛᵩ)
= Σⱼ ∂F/∂qᵩⱼ
Change in F is
δF = (Σⱼ ∂F/∂qᵩⱼ) δφ = [F, Pᵩ] δφ
Applying Pᵩ to the Hamiltonian:
δH = [H, Pᵩ] δx = 0 (from relativity)
Pᵩ has no explicit time dependence, and so is conserved.
Lagrangian addition of a total derivative
Can add a term to the Lagrangian
L → L + df({qᵢ}, t)/dt
without changing the physics.
f can be a function of co-ordinates and time, but not of velocities.
To see this doesn’t affect physics, use principle of least action.
S → ∫t₀→t₁ (L + df/dt) dt
= S + [ f({qᵢ}, t) ] t₀→t₁
= S + f( {qᵢ(t₁)}, t₁ ) − f( {qᵢ(t₀)}, t₀ )
Extra terms depend only on end-points, so are fixed during minimization. Path minimizing new action is the same one that minimized original action. Therefore does not change EoMs.
Lagrangian rescaling
Lagrangian appears both sides of EoM.
Can multiply it by a constant factor
L → k L
without changing the physics
Lagrangian addition of constant
Lagrangian only appears in derivatives, can add a constant (not dependent on qᵢ, q̇ᵢ, or t)
L → L + c
without changing the physics.
Lagrangian parity
Parity transformation reverses direction of coordinate axes,
x → −x , y → −y , z → −z
Reversing direction of odd number of axes changes a right-handed coordinate system into a left-handed system, and vice versa.
Classical mechanics, cannot tell whether we are looking at a valid physical system or its parity reflection or mirror image.
Principle of relativity w.r.t. translations and rotations implies the Hamiltonian can depend only on scalar combinations of relative positions, such as
rⱼₖ² = |rⱼₖ|² = rⱼₖ · rⱼₖ , or rⱼₖ · rₗₘ
Considering combination
(rⱼₖ × rₗₘ) · rₙₒ
Invariant under translations and rotations, but changes sign under reflection of coordinate axes. Thus, in classical physics, Hamiltonian can only depend on its square or absolute value.
Prove kinetic energy using boosts
Centre of mass position
X = 1/M Σⱼ mⱼqₓⱼ
where M = Σⱼ mⱼ
Not conserved unless in centre-of-mass frame.
In general, moves at constant velocity
Ẋ = 1/M Σⱼ mⱼq̇ₓⱼ = 1/M Σⱼ pₓⱼ = Pₓ/M
where Pₓ is total linear momentum in x direction.
Construct a conserved quantity:
B = M (X − Ẋt) = Σⱼ (mⱼqₓⱼ − pₓⱼt)
To calculate the transformation this corresponds to, take Poisson bracket
of Bε with generalized coordinates qₛⱼ and momenta pₛⱼ:
qₛⱼ → qₛⱼ + [qₛⱼ, B]ε
= qₛⱼ + Σᵤ Σₖ (∂qₛⱼ/∂qᵤₖ ∂B/∂pᵤₖ − ∂qₛⱼ/∂pᵤₖ ∂B/∂qᵤₖ) ε
pₛⱼ → pₛⱼ + [pₛⱼ, B]ε
= pₛⱼ + Σᵤ Σₖ (∂pₛⱼ/∂qᵤₖ ∂B/∂pᵤₖ − ∂pₛⱼ/∂pᵤₖ ∂B/∂qᵤₖ) ε
Now have
∂qₛⱼ/∂qᵤₖ = δₛᵤ δⱼₖ
∂qₛⱼ/∂pᵤₖ = 0
∂pₛⱼ/∂qᵤₖ = 0
∂pₛⱼ/∂pᵤₖ = δₛᵤ δⱼₖ
and
∂B/∂pᵤₖ = ∂/∂pᵤₖ Σⱼ (mⱼqₓⱼ − tpₓⱼ) = −δₓₛ t
∂B/∂qᵤₖ = ∂/∂qᵤₖ Σⱼ (mⱼqₓⱼ − tpₓⱼ) = mₖδₓₛ
Therefore,
qₛⱼ → qₛⱼ − δₓₛ ε t
which gives
qₓⱼ → qₓⱼ − εt
qᵧⱼ → qᵧⱼ
q₂ⱼ → q₂ⱼ
Similarly,
pₓⱼ → pₓⱼ − mⱼ ε
With ε = v, these are the Galilean transformations between two frames moving with relative velocity v in the x direction.
B, the position of the centre of mass at time 0, is the generator of boosts in the x direction.
dB/dt = [B, H] + ∂B/∂t = 0
Since
∂B/∂t = −Σⱼ pₓⱼ
can conclude
[B, H] = Pₓ
Change in Hamiltonian when boosted by infinitesimal velocity −dẊ is
dH = −[H, B] dẊ = [B, H] dẊ = Pₓ dẊ = M Ẋ dẊ
where Ẋ is velocity of the centre of mass.
Starting from the centre-of-mass frame, where Ẋ = 0 and Hamiltonian is H꜀.ₒ.ₘ, and boosting in steps of dẊ to a frame where centre of mass has velocity V,
H = H꜀.ₒ.ₘ + 1/2 MV²
Kinetic energy babeyy!
What are normal modes
A mode in which all parts of the system are oscillating at the same frequency. They are “normal” to each other, orthogonal in vector space. This means there is no coupling or exchange of energy between them: if the system is oscillating in one of its normal modes, it stays in that mode.
Normal modes of two pendulums connected by a spring
Two identical pendula consisting of point masses m suspended by massless strings of length l a distance L apart. Masses are connected together by a massless spring of natural length L and spring constant k.
Equilibrium position is x₁ = 0, x₂ = 0, spring neither compressed nor stretched. Consider small oscillations about this equilibrium, x₁ ∼ x₂ ≪ l.
If a system obeys a symmetry, the normal modes are eigenfunctions of that symmetry.
This case has mirror symmetry, so the normal modes are either symmetric or antisymmetric.
Antisymmetric normal mode:
Both pendula oscillate with same amplitude in the same direction.
The spring never extends or compresses as if it weren’t there.
Frequency is that of the individual pendulum,
ω² = g/l
Symmetric normal mode:
Both oscillate in opposite directions.
Centre of spring doesn’t move, each pendulum is restored by the sum of gravity and a spring of spring constant 2k
ω² = g/l + 2k/m
Lagrangian and EoMs of a two-pendulum system
L = T − V
Kinetic energy
T = ½ m(ẋ₁² + ẋ₂²)
V:
Gravitational potential energy
Small x displacement corresponds to angle
θ ≈ x/l
Height of mass below the pivot is
l cosθ ≈ l(1−½θ²) = l − ½x²/l
Neglecting overall constant term, have potential energy
mg/2l (x₁²+x₂²)
Spring potential
Masses displaced by x₁ and x₂, spring stretched to length
x₁ − x₂ + L
with potential energy
½ k(x₂−x₁)²
Overall Lagrangian
L = ½ m(ẋ₁²+ẋ₂²) − mg/2l (x₁²+x₂²) − ½k(x₂−x₁)²
EoMs
corresponding to x₁ and x₂ respectively
d/dt (∂L/∂ẋ₁) = mẍ₁ = ∂L/∂x₁ = −mg/l x₁ + k(x₂−x₁)
d/dt (∂L/∂ẋ₂) = mẍ₂ = ∂L/∂x₂ = −mg/l x₂ + k(x₁−x₂)
Informal method for Normal Modes of two pendula
Lagrangian:
* L = ½ m(ẋ₁²+ẋ₂²) − mg/2l (x₁²+x₂²) − ½k(x₂−x₁)²
EoMs:
* d/dt (∂L/∂ẋ₁) = mẍ₁ = ∂L/∂x₁ = −mg/l x₁ + k(x₂−x₁)
* d/dt (∂L/∂ẋ₂) = mẍ₂ = ∂L/∂x₂ = −mg/l x₂ + k(x₁−x₂)
{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}
Adding EoMs
m(ẍ₁+ẍ₂) = −mg/l (x₁+x₂)
or
{x₁+x₂}¨ = −g/l (x₁+x₂)
- combination x₁ + x₂ oscillates with ω² = g/l
Subtracting EoMs
m(ẍ₁−ẍ₂) = −mg/l (x₁−x₂) − 2k(x₁−x₂)
or
{x₁−x₂}¨ = −(g/l + 2k/m) (x₁−x₂)
- combination x₁ − x₂ oscillates with ω² = g/l + 2k/m
Formal method for Normal Modes of two pendula
Lagrangian:
* L = ½ m(ẋ₁²+ẋ₂²) − mg/2l (x₁²+x₂²) − ½k(x₂−x₁)²
EoMs:
* d/dt (∂L/∂ẋ₁) = mẍ₁ = ∂L/∂x₁ = −mg/l x₁ + k(x₂−x₁)
* d/dt (∂L/∂ẋ₂) = mẍ₂ = ∂L/∂x₂ = −mg/l x₂ + k(x₁−x₂)
{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}
Normal mode implies all components oscillate at same frequency, but perhaps with different amplitude or phase.
Assume in nth normal mode, frequency is ωₙ²
Displacement of each component can be written
xⱼ = Cⱼexp(iωₙt), j = 1, 2
where the Cⱼ’s are constants.
Take derivative
ẍⱼ = −ωₙ²Cⱼexp(iωₙt)
Sub into EoMs
−mωₙ²C₁exp(iωₙt) = −mg/l C₁exp(iωₙt) + k(C₂exp(iωₙt) − C₁exp(iωₙt))
−mωₙ²C₂exp(iωₙt) = −mg/l C₂exp(iωₙt) + k(C₁exp(iωₙt) − C₂exp(iωₙt))
Exponentials cancel, get ωₙ² from first eq
ωₙ² = g/l − k/m (C₂/C₁ − 1)
Sub into second eq, rearrange and get
C₂² = C₁²
Two solutions:
C₂ = C₁
or
C₂ = −C₁
which correspond to
ω₁² = g/l
or
ω₂² = 2k/m + g/l
Normal mode coordinates
q₁ = 1/√2’ (x₁+x₂)
q₂ = 1/√2’ (x₁−x₂)
Lagrangian
L = m/2 (ẋ₁²+ẋ₂²) − k/2 (x₁−x₂)² − mg/2l (x₁²+x₂²)
in normal mode coordinates
q₁ = 1/√2’ (x₁+x₂)
q₂ = 1/√2’ (x₁−x₂)
L = [m/2 q̇₁² − mg/2l q₁²] + [m/2 q̇₂² − ½(2k + mg/l)q₂²]
Lagrangian decouples:
L = L₁(q₁², q̇₁²) + L₂(q₂², q̇₂²)
EoM for two pendula in normal mode coordinates
q̈₁ = −g/l q₁
SHO with ω₁² = g/l
q̈₂ = −(2k/m + g/l) q₂
SHO with ω₂² = 2k/m + g/l
Solving a matrix equation
For an n × n matrix, A, there are up to n solutions of the equation
Ax = λx
Only has non-trivial solutions if
|A − λI| = 0
Each value of λ = λ⁽ᵏ⁾ corresponds to a different solution for x = x⁽ᵏ⁾.
Found by explicitly solving
(A − λ⁽ᵏ⁾I)x⁽ᵏ⁾ = 0 (Eq.3)
Only interested in the case where n solutions exist. Some may be degenerate:
λ⁽ᵏ⁾ = λ⁽ʲ⁾
for k ≠ j
In cases where solution λ = λ⁽ᵏ⁾ occurs m times, can choose any m vectors that satisfy Eq.3 to be eigenvectors. Is convenient to choose them to be perpendicular,
x⁽ᵏ⁾ · x⁽ʲ⁾ = 0, if λ⁽ᵏ⁾ = λ⁽ʲ⁾ and k ≠ j
Equations of Motion in Matrix Form
Rearrange EoMs of two coupled pendula
mẍ₁ = −(k + mg/l)x₁ − (−k)x₂ = −mω²x₁
mẍ₂ = −(−k)x₁ − (k + mg/l)x₂ = −mω²x₂
Rewrite as a matrix equation:
MẌ = −KX = −ω²MX
where
* X =
(x₁
x₂)
- M =
(m 0
0 m) - K =
(k + mg/l | −k
−k | k + mg/l)
ω² is just a number, can write RHS as
ω²MX = M(ω²X)
Multiply both sides by M⁻¹
EoM in matrix form is
M⁻¹KX = ω²X
Find normal modes of two pendula using matrices
EoM in matrix form (general EoM for any st=ystem close to equilibrium)
M⁻¹KX = ω²X
Rearrange
[M⁻¹K − ωₙ²I]X = 0
where I is the identity matrix.
Only solutions are X = 0, unless
det[M⁻¹K − ωₙ²I] = 0
In two pendula case,
M⁻¹ = 1/m (¹₀⁰₁)
⇒ M⁻¹K =
(k/m + g/l | −k/m
−k/m | k/m + g/l)
0 = det[M⁻¹K − ωₙ²I] =
|k/m + g/l − ωₙ² | −k/m|
|−k/m | k/m + g/l − ωₙ²|
= (k/m + g/l − ωₙ²)² − (k/m)²
∴ k/m + g/l − ωₙ² = ±k/m
∴ ω₁² = g/l
and ω₂² = g/l + 2k/m
To find eigenvectors, sub ω² back into EoMs:
−(k + mg/l)x₁ + kx₂ = −mω₁²x₁ = −mg/l x₁
⇒ x₂ = x₁
or, in matrix form,
Q₁ = 1/√2’ (¹₁)
Q₂ = 1/√2’ (¹₋₁)
To specify actual motion, must include absolute amplitude and phase of each normal mode:
Displacement of mass j due to normal mode n:
xⱼ⁽ⁿ⁾ = Qₙ(j) Aₙexp(i(ωₙt+φₙ))
Total displacement of mass j, sum over all normal modes
xⱼ = Σₙ (Qₙ(j) Aₙexp(i(ωₙt+φₙ))
Initial conditions determine values of Aₙ and φₙ. (2N conditions needed for N degrees of freedom)
Lagrangian in matrix form
L = ½ ẊᵀMẊ − ½ XᵀKX
