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
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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.
