PHYS 321A Lecture Notes 26 University of Victoria
Lecture 26: The Principle of Least Action (Hamilton’s
Principle)
Where does F~ = m~a or −∇V~ = m~r¨ come from? Is there a more fundamental reason as to
why these equations hold?
Define the Lagrangian function as
L=T −V
For example, in a conservative system in 1D we would have
1
L(y, ẏ) = mẏ 2 − V (y)
2
Suppose that a particle starts at y1 = y(t1 ) and ends its trajectory at y2 = y(t2 ). The path
taken is a particular (y(t), ẏ(t)) trajectory; the action of this trajectory is given by
Z t2
J= Ldt
t1
Hamilton’s principle states that out of all the infinite family of motions (y(t), ẏ(t)), the ac-
tual motion that takes place is the one for which the action is an extrema:
Z t2
δJ = δ Ldt = 0
t1
i.e., any variation on top of this motion (y(t) + n(t), ẏ + ṅ) with endpoints fixed (n(t1 ) =
n(t2 ) = 0) vanishes in 1st order. The usual situation is that J has a global minimum at the
actual trajectory.
Functional Derivative:
δJ
J = J[y(t)] , =0
δy
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PHYS 321A Lecture Notes 26 University of Victoria
Example: Particle in Free Fall
1
L = mẏ 2 − mgy
2
Z t2 Z t2 hm i
δJ = δ Ldt = δ ẏ 2 − mgy dt
t1 t1 2
Z t2 hm i
= 2ẏδy − mgδy dt
t1 2
Note that δ ẏ = dtd (δy). We integrate the first term by parts udv = uv − vdu. Let u = mẏ
R R
and dtd (δy) = dv. Then we have
Z t2 Z
d t2 d
mẏ (δy)dt = mẏ(δy) t1 − (δy) (mẏ)dt
t1 dt dt
t
Note that mẏ(δy) t21 = 0 since δy(ti ) = 0. Hence
Z t2 Z Z
d d
mẏ (δy)dt = − (δy) (mẏ)dt = − mÿ(δy)dt
t1 dt dt
Z t2
=⇒ δJ = − [mÿ + mg](δy)dt
t1
This = 0 for any arbitrary (δy) if and only if
mÿ = −mg Newton’s 2nd law!
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PHYS 321A Lecture Notes 26 University of Victoria
We proved that δJ = 0 for the actual trajectory. Is J a minimum or a maximum?
1
Actual trajectory: y(t) = − gt2 (y(t = 0) = 0, y(t = t2 ) = − 21 gt2 2 )
2
Assume y(α, t) = y(0, t) + αn(t) where y(0, t) is the actual solution, α is a bookkeeping
constant, and n(t) is some arbitrary function.
Z t2 ẏ(α, t) = ẏ(0, t) + αṅ(t) = (−gt + αṅ)
J(α) = dtL[y(α, t), ẏ(α, t)] T = 12 mẏ 2 = 12 m[−gt + αṅ]2
t1
V = mgy = mg − 21 gt2 + αn
Z t2
1 22 1
m g t − 2gtαṅ + α2 ṅ2 − mg[− gt2 + αn]
J(α) = dt
t1 2 2
Z t2
2 2 1 2 2
= dt mg t − mgα(tṅ + n) + mα ṅ
t1 2
Z t2
2 2 1 2 2
= dt mg t − mgα(−n + n) + mα ṅ
t1 2
t2
mg 2 3
Z
1
= (t2 − t1 3 ) + mα2 dtṅ2 = J0 + J1 α2
3 2 t1
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PHYS 321A Lecture Notes 26 University of Victoria
Hence J(α = 0) is a local minimum in the space of all functions n(t)!
∂J(α)
=0
∂α α=0
Generalized Coordinates
Consider a pendulum in the xy plane. How many degrees of freedom does it have?
(x, y, z) are inter-related. The two constraints are z = 0 and r2 − (x2 + y 2 ) = 0.
=⇒ only one independent degree of freedom. We can choose x, but that’s awkward.
It’s even double valued (we can have the same x with a different configuration).
The natural choice is θ: here we only need a single number to determine the location of the
pendulum.
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PHYS 321A Lecture Notes 26 University of Victoria
Generalized Coordinates are any collection of independent variables (q1 , q2 , ..., qn ) (not
connected by any equation of constraint) that just suffice to specify uniquely the configu-
ration of a system of particles. The number n of open coordinates is equal to the system’s
degree of freedom.
For example, if we use:
(i) Smaller # than n coordinates: System’s motion is indeterminate
(ii) Larger # than n coordinates: Some coordinates are completely given by others
Another Example: Consider two particles connected by a rigid rod.
(
6 coordinates : (x1 , y1 , z1 ) , (x2 , y2 , z2 )
One constraint : d2 − [(x1 − x2 )2 + (y2 − y1 )2 + (z2 − z1 )2 ] = 0
=⇒ 5 degrees of freedom. Natural choice for generalized coordinates?
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PHYS 321A Lecture Notes 26 University of Victoria
In general: N particles require 3N coordinates. Suppose there are m constraints:
fj = (xi , yi , zi , t) = 0 j = 1, 2, ..., m Holonomic Constraint
=⇒ (q1 , q2 , ..., q3N −m ) generalized coordinates
Non-holonomic Constrains: e.g [(x2 + y 2 + z 2 − R2 )] ≥ 0 (we cannot go inside the earth).
This cannot be used to reduce the number of degrees of freedom.
(i) Point in a ball rolling on a table =⇒ still needs 3 coordinates to describe points; con-
straint only binds z ∈ [0, 2R].
(ii) Ball rolling without slipping on a table =⇒ velocity constraint, not coordinate con-
straint! (Angular orientation of ball, position in the plane),
=⇒ There are the coordinates (x, y, z, θ, φ) for
pthe ball. We have the holonomic constraint
z = 0 and the non-holonomic constraint V⊥ = ẋ2 + ẏ 2 = Rθ̇. Hence there are 4 degrees of
freedom.
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