Physics 321
Hour 15
The Calculus of Variations
Bottom Line
• A simple derivative is useful for finding the value of a
variable that maximizes or minimizes a function.
• The calculus of variations is useful for finding the
function that maximizes or minimizes a quantity that
depends on the function.
• You don’t need to reproduce this section, but it is an
important tool in physics that you should put forth
some real effort to understand.
The Life Guard Problem
You are a life guard on a beach and see someone
needing your help. You want to get there as fast as
possible, where do you jump in the water?
Example
SnellsLaw.nb
Finding the Shortest Path between Points
The path length is
(x1,y1)
𝑃2
𝐿=
P1
y
𝑑𝑠
P2
𝑃1
𝑑𝑠 =
𝑑𝑥 2
+ 𝑑𝑦 2
y=y(x)
(x2,y2)
𝑑𝑦
= 𝑑𝑥 1 +
𝑑𝑥
𝑃2
1 + 𝑦′2 𝑑𝑥
𝐿=
𝑃1
2
= 𝑑𝑥 1 + 𝑦′2
x
Finding the Shortest Path between Points
The path length is
𝑃2
y
P1
(x1,y1)
y=y(x)
1 + 𝑦′2 𝑑𝑥
𝐿=
𝑃1
Take another function 𝑌 𝑥 = 𝑦 𝑥 + 𝛼η(𝑥)
subject to η 𝑥1 = η 𝑥2 = 0.
What does that tell us about Y(x)?
P2
(x2,y2)
x
Finding the Shortest Path between Points
Is y the shortest path?
L(α)
α=0
α
Finding the Shortest Path between Points
Is y the shortest path?
L(α)
α=0
α
It is – if the same thing works for every function η 𝑥
subject to η 𝑥1 = η 𝑥2 = 0.
Finding the Shortest Path between Points
If we choose an arbitrary path 𝑌 𝑥 = 𝑦 𝑥 + 𝛼η(𝑥) ,
where η 𝑥1 = η 𝑥2 = 0, then 𝑦 𝑥 is the shortest
path if
𝜕𝐿
=0
𝜕𝛼 𝛼→0
for all functions η(𝑥).
A Useful Theorem
Integration by parts give us:
𝐵
𝑢𝑑𝑣 = 𝑢𝑣
𝐴
Let 𝑑𝑢 =
𝑑𝑢
𝑑𝑥
𝑑𝑥
𝐵
𝐴
−
𝑣𝑑𝑢
𝐴
= 𝑢′ 𝑑𝑥 𝑑𝑣 = 𝑣 ′ 𝑑𝑥
𝐵
𝑢𝑣 ′ 𝑑𝑥 = 𝑢𝑣
𝐴
𝐵
𝐵
𝐴
𝐵
𝑣𝑢′ 𝑑𝑥
−
𝐴
Distance Between Two Points
𝑃2
𝐿=
𝑑𝑠 =
𝑑𝑠
𝑃1
𝑑𝑥 2 + 𝑑𝑦 2 =
𝑑𝑥 1 + 𝑦′2
Let 𝑌 𝑥 = 𝑦 𝑥 + 𝛼𝜂 𝑥
where 𝜂 𝑥1 = 𝜂 𝑥2 = 0
𝑥2
𝐿 𝛼 =
𝑥1
1 + 𝑌′(𝑥)2 𝑑𝑥
Distance Between Two Points
𝑥2
𝐿 𝛼 =
1 + 𝑌′(𝑥)2 𝑑𝑥
𝑥1
𝑥2
1 + (𝑦 ′ + 𝛼𝜂′)2 𝑑𝑥
=
𝑥1
𝑥2
≡
𝑓(𝛼) 𝑑𝑥
𝑥1
𝜕𝑓
𝜕𝑓
𝑁𝑜𝑡𝑒
= 𝜂′
𝜕𝛼
𝜕𝑦′
Distance Between Two Points
𝜕𝐿
=
𝜕𝛼
𝑥2
𝜕𝑓
= 𝜂
−
𝜕𝑦′ 𝑥
1
𝑥2
𝜕𝑓
𝜂′
𝑑𝑥
𝜕𝑦′
𝑥1
𝑥2
𝑑 𝜕𝑓
𝜂
𝑑𝑥 = 0
′
𝑑𝑥 𝜕𝑦
𝑥1
0
𝑑 𝜕𝑓
→
=0
′
𝑑𝑥 𝜕𝑦
Distance Between Two Points
𝜕𝑓
𝜕𝑦′
=
𝛼→0
=
2(𝑦 ′ + 𝛼𝜂′)
2 1 + (𝑦 ′ + 𝛼𝜂′)2
𝑦′
= constant
1 + (𝑦′)2
→ 𝑦′ 𝑥 = constant
→ 𝑦 𝑥 = 𝑚𝑥 + 𝑏
𝛼→0
The Action Integral
𝑡2
𝑆=
𝑡2
𝑆(𝛼) =
ℒ(𝑥, 𝑥, 𝑡) 𝑑𝑡
𝑡1
ℒ(𝑥 + 𝛼𝜂, 𝑥 + 𝛼𝜂, 𝑡) 𝑑𝑡
𝑡1
r
s
𝑡2
𝜕𝑆
𝜕ℒ 𝜕𝑟 𝜕ℒ 𝜕𝑠
=
−
𝑑𝑡 = 0
𝜕𝛼
𝜕𝑠 𝜕𝛼
𝑡1 𝜕𝑟 𝜕𝛼
𝑡2
𝜕𝑆
𝜕ℒ
𝜕ℒ
=
𝜂−
𝜂 𝑑𝑡 = 0
𝜕𝛼
𝜕𝑥
𝑡1 𝜕𝑥
The Action Integral
𝑡2
𝜕𝑆
𝜕ℒ
𝜕ℒ
=
𝜂−
𝜂 𝑑𝑡 = 0
𝜕𝛼
𝜕𝑥
𝑡1 𝜕𝑥
𝑡2
𝜕𝑆
𝜕ℒ
𝑑 𝜕ℒ
=
𝜂−
𝜂 𝑑𝑡 = 0
𝜕𝛼
𝑑𝑡 𝜕𝑥
𝑡1 𝜕𝑥
𝑡2
𝜕𝑆
𝜕ℒ 𝑑 𝜕ℒ
=
𝜂
−
𝑑𝑡 = 0
𝜕𝛼
𝜕𝑥 𝑑𝑡 𝜕𝑥
𝑡1
𝜕ℒ
𝑑 𝜕ℒ
→
=
𝜕𝑥 𝑑𝑡 𝜕𝑥