The Lagrangian
Equation
INTERMEDIATE MECHANICS
WESLEY QUEEN
Developing a new perspective

Hero of Alexandria (70 A.D.) - light reflections takes the shortest path

Pierre de Fermat (1657) - light travels along a path that requires the
least time

Maupertuis (1747) - action is minimized through the “wisdom of
God”


Action is a quantity with dimensions of energy x time
William Hamilton (1834) –
“Of all the possible paths along which a dynamical
system may move from one point to another within a
specified time interval, the actual path followed is that
which minimizes the time integral of the difference
between the kinetic and potential energies.”
The Lagrangian
L=T-U

T = Kinetic Energy

U = Potential Energy

Asserts that for any given system there is a extremum (maximum or
minimum) when the time integral is taken of the difference of Kinetic
and Potential Energies.
𝑡2

𝑡1
𝑇 − 𝑈 𝑑𝑡 = Min or Max (Minimum in almost all dynamic
systems)

Let S =
𝑡2
𝐿
𝑡1
𝑑𝑡
“Action”
Calculus of Variations
Consider a function 𝑥𝑎 𝑡 = 𝑥0 𝑡 + 𝛼𝛽(𝑡) where 𝑥0 𝑡 produces a min value
for S, 𝛼 is a number, and 𝛽(𝑡) is 0 at both end points of our interval.

When this is integrated, the t is integrated out and S becomes a number
dependent on 𝛼, 𝑡1 , & 𝑡2 .
How does S depend on 𝛼?
𝜕
𝑆
𝜕𝛼
𝑥𝑎 𝑡
=
𝜕
𝑆
𝜕𝛼
𝑥𝑎 𝑡
=
𝜕 𝑡2
𝐿
𝜕𝛼 𝑡1
𝑑𝑡 =
𝑡2 𝜕𝐿 𝜕𝑥𝑎
(
𝑡1 𝜕𝑥𝑎 𝜕𝛼
+
𝑡2 𝜕
𝐿
𝑡1 𝜕𝛼
𝑑𝑡
𝜕𝐿 𝜕 𝑥𝑎
)𝑑𝑡
𝜕𝑥𝑎 𝜕𝛼
Use the Chain rule to separate
Calculus of Variations
Use substitution:
From our initial function:
𝑥𝑎 𝑡 = 𝑥0 𝑡 + 𝛼𝛽(𝑡)
𝜕
𝑆
𝜕𝛼
𝜕
𝑆
𝜕𝛼
𝑥𝑎 𝑡
𝑥𝑎 𝑡
𝜕𝑥𝑎
𝜕𝛼
=
𝑡2 𝜕𝐿 𝜕𝑥𝑎
(
𝑡1 𝜕𝑥𝑎 𝜕𝛼
=
𝑡2 𝜕𝐿
(
𝛽
𝑡1 𝜕𝑥𝑎
𝜕𝐿 𝜕 𝑥𝑎
+
)𝑑𝑡
𝜕𝑥𝑎 𝜕𝛼
+
𝜕𝐿
𝛽)𝑑𝑡
𝜕 𝑥𝑎
= 𝛽 &
𝜕 𝑥𝑎
𝜕𝛼
=𝛽
Calculus of Variations
Use integration by parts:
𝜕
𝑆
𝜕𝛼
𝜕
𝑆
𝜕𝛼
𝜕
𝑆
𝜕𝛼
𝑥𝑎 𝑡
=
𝑥𝑎 𝑡 =
𝑥𝑎 𝑡
=
𝑡2 𝜕𝐿
(
𝛽
𝑡1 𝜕𝑥𝑎
𝑡2 𝜕𝐿
𝛽𝑑𝑡
𝑡1 𝜕𝑥𝑎
𝑡2 𝜕𝐿
(
𝑡1 𝜕𝑥𝑎
−
+
𝜕𝐿
𝛽)𝑑𝑡
𝜕 𝑥𝑎
𝜕𝐿
𝑡
+
𝛽|𝑡21
𝜕 𝑥𝑎
−
𝑑 𝜕𝐿
)𝛽𝑑𝑡
𝑑𝑡 𝜕 𝑥𝑎
Euler’s Equation:
𝜕𝐿
𝜕𝑥0
=
𝑡2 𝑑 𝜕𝐿
(
)𝛽𝑑𝑡
𝑡1 𝑑𝑡 𝜕 𝑥𝑎
Since 𝑥0 𝑡 produces a stationary value
𝜕
for S,
𝑆 𝑥𝑎 𝑡 = 0
𝜕𝛼
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝑥0
Calculus of Variations

Begun by Newton

Developed by Johann Bernoulli, Jakob Bernoulli, and Leonhard Euler

Important contributions made by Joseph Lagrange, Hamilton, and
Jacobi
Leonhard Euler
(1707 - 1783)
A falling object
Consider a falling object (no motion in the x or z dimensions)
1
2
𝑇 = 𝑚𝑦 2
Apply Euler’s Equation
𝑑
−𝑚𝑔 = (𝑚𝑦)
𝑑𝑡
1
2
𝐿 = 𝑚𝑦 2 - 𝑚𝑔𝑦
𝑈 = 𝑚𝑔𝑦
𝜕𝐿
𝜕𝑦
=
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝑦
−𝑚𝑔 = 𝑚𝑦
𝑦 = −g
The Lagrangian gives us Newton’s second law with respect to gravity.
A projectile
Consider a projectile launched in the positive x direction at some angle 𝜃,
where 0 < 𝜃 < 90.
1
1
𝑇 = 𝑚𝑥 2 + 𝑚𝑦 2
𝑈 = 𝑚𝑔𝑦
2
2
1
1
2
𝐿 = 𝑚𝑥 + 𝑚𝑦 2 − 𝑚𝑔𝑦
2
2
X dimension
𝜕𝐿
𝜕𝑥
=
0=
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝑥
𝑑
(𝑚𝑥)
𝑑𝑡
Y dimension
𝜕𝐿
𝜕𝑦
=
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝑦
−𝑚𝑔 =
𝑑
(𝑚𝑦)
𝑑𝑡
0 = 𝑚𝑥
−𝑚𝑔 = 𝑚𝑦
𝑥=0
𝑦 = −g
An Orbiting body
y
Consider a body orbiting about a another body with a central force
acting on it.
Position
𝑥 = 𝑟𝑐𝑜𝑠 𝜃
y = 𝑟𝑠𝑖𝑛 𝜃
Velocity
M
𝑥 = 𝑟𝑐𝑜𝑠 𝜃 − 𝑟𝜃sin(𝜃) 𝑦 = 𝑟𝑠𝑖𝑛 𝜃 − 𝑟𝜃𝑐𝑜𝑠(𝜃)
The Lagrangian
1
2
𝑇 = 𝑚(𝑥 2 + 𝑦 2 )
1
2
L = 𝑚(𝑥 2 + 𝑦 2 ) +
1
2
r
𝑈=−
𝐺𝑀𝑚
𝑟
L = 𝑚(𝑟 2 + 𝑟 2 𝜃 2 ) +
𝐺𝑀𝑚
𝑟
𝐺𝑀𝑚
𝑟
m
𝜃
x
An Orbiting body
y
Consider a body orbiting about a another body with a central force
acting on it.
The Lagrangian
1
2
L = 𝑚(𝑟 2 + 𝑟 2 𝜃 2 ) +
𝐺𝑀𝑚
𝑟
r
M
Apply Euler’s Equations
𝜕𝐿
𝜕𝜃
𝑑 𝜕𝐿
= 𝑑𝑡 𝜕𝜃
𝑑
𝜕𝐿
𝜕𝑟
𝑑 𝜕𝐿
= 𝑑𝑡 𝜕𝑟
0 = 𝑑𝑡(m𝑟 2 𝜃)
𝑚𝑟𝜃 2 −
𝐿 = m𝑟 2 𝜃
𝑚𝑟𝜃 2 −
𝑟=−
Shows conservation of
Angular momentum
𝐺𝑀𝑚
𝑟2
𝐺𝑀𝑚
𝑟2
𝐺𝑀
𝑟2
𝑑
= 𝑑𝑡 (m𝑟)
= 𝑚𝑟
+ 𝑟𝜃 2
Indicates the acceleration in the direction r is the
central force acceleration + the tangential
acceleration
m
𝜃
x
Charge interaction
𝑞𝑏(𝑥𝑏 ,𝑦𝑏 )
Consider 3 charges interacting in a 2 dimensional plane
1
2
1
2
1
2
𝐿 = 𝑚𝑥𝑎2 + 𝑚𝑦𝑎2 −
𝑈=
𝑘𝑞𝑎 𝑞𝑏
(𝑥𝑏 −𝑥𝑎 )2 +(𝑦𝑏 −𝑦𝑎 )
𝑘𝑞𝑎 𝑞𝑏
−
2
(𝑥𝑏 −𝑥𝑎 )2 +(𝑦𝑏 −𝑦𝑎 )
=
(𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2
𝑘𝑞𝑎 𝑞𝑐
𝜕𝐿
𝜕𝑦𝑎
𝑑 𝜕𝐿
𝑑𝑡 𝜕 𝑥𝑎
𝑘𝑞𝑎 𝑞𝑏 (𝑥𝑏 −𝑥𝑎 )
2
2 3/2
𝑏 −𝑥𝑎 ) +(𝑦𝑏 −𝑦𝑎 ) )
-
𝑘𝑞𝑎 𝑞𝑐 (𝑥𝑐 −𝑥𝑎 )
((𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2 )3/2
= 𝑑𝑡 𝑚𝑥𝑎
-(
𝑘𝑞𝑎 𝑞𝑏 (𝑥𝑏 −𝑥𝑎 )
2
2 3/2
𝑏 −𝑥𝑎 ) +(𝑦𝑏 −𝑦𝑎 ) )
-
𝑘𝑞𝑎 𝑞𝑐 (𝑥𝑐 −𝑥𝑎 )
((𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2 )3/2
= 𝑚𝑥𝑎
-
- ((𝑥
- ((𝑥
𝑞𝑐(𝑥𝑐 ,𝑦𝑐 )
(𝑥𝑐 −𝑥𝑎 )2 +(𝑦𝑐 −𝑦𝑎 )2
𝑞𝑎(𝑥𝑎,𝑦𝑎)
Y dimension
X dimension
𝜕𝐿
𝜕𝑥𝑎
𝑘𝑞𝑎 𝑞𝑐
+
2
𝑑
=
y
x
1
2
𝑇 = 𝑚𝑥𝑎2 + 𝑚𝑦𝑎2
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝑦𝑎
𝑘𝑞𝑎 𝑞𝑏 (𝑦𝑏 −𝑦𝑎 )
𝑥𝑏 −𝑥𝑎 2 + 𝑦𝑏 −𝑦𝑎 2 )3/2
𝑘𝑞𝑎 𝑞𝑏 (𝑦𝑏 −𝑦𝑎 )
( 𝑥𝑏 −𝑥𝑎 2 + 𝑦𝑏 −𝑦𝑎 2 )3/2
−(
−
𝑘𝑞𝑎 𝑞𝑐 (𝑦𝑐 −𝑦𝑎 )
𝑥𝑐 −𝑥𝑎 2 + 𝑦𝑐 −𝑦𝑎 2 )3/2
𝑘𝑞𝑎 𝑞𝑐 (𝑦𝑐 −𝑦𝑎 )
( 𝑥𝑐 −𝑥𝑎 2 + 𝑦𝑐 −𝑦𝑎 2 )3/2
𝑑
= 𝑑𝑡 (𝑚𝑦𝑎 )
= 𝑚𝑦𝑎
A rotating pendulum
y
Consider a pendulum of length 𝑙 moving on a massless ring
of radius r and constant angular velocity ω.
r
Position
𝑥 = 𝑟𝑐𝑜𝑠 ω𝑡 + 𝑙𝑠𝑖𝑛(θ) y = 𝑟𝑠𝑖𝑛 ω𝑡 − 𝑙𝑐𝑜𝑠(θ)
ωt
x
θ
Velocity
𝑙
𝑥 = −𝑟ω𝑠𝑖𝑛 ω𝑡 + 𝑙 𝜃𝑐𝑜𝑠(𝜃) 𝑦 = 𝑟𝜔𝑐𝑜𝑠 ω𝑡 + 𝑙 𝜃𝑠𝑖𝑛(𝜃)
The Lagrangian
1
2
𝑇 = 𝑚(𝑥 2 + 𝑦 2 )
1
2
𝑈 = 𝑚𝑔𝑦
1
2
𝐿 = 𝑚(𝑥 2 + 𝑦 2 ) - 𝑚𝑔𝑦
𝐿 = 𝑚[𝑟 2 𝜔2 + 𝑙 2 𝜃 2 + 2𝑟𝜔𝑙 𝜃𝑠𝑖𝑛(𝜃 − 𝜔𝑡)] - 𝑚𝑔(𝑟𝑠𝑖𝑛 ω𝑡 − 𝑙𝑐𝑜𝑠 θ )
m
A rotating pendulum
y
Consider a pendulum of length 𝑙 moving on a massless ring
of radius r and constant angular velocity ω.
The Lagrangian
r
1
2
𝐿 = 𝑚[𝑟 2 𝜔2 + 𝑙 2 𝜃 2 + 2𝑟𝜔𝑙 𝜃𝑠𝑖𝑛(𝜃 − 𝜔𝑡)] - 𝑚𝑔(𝑟𝑠𝑖𝑛 ω𝑡 − 𝑙𝑐𝑜𝑠 θ )
𝜕𝐿
𝜕𝜃
=
𝑙
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝜃
= 𝑚𝑟𝜔𝑙 𝜃𝑐𝑜𝑠(𝜃 − 𝜔𝑡) - 𝑚𝑔𝑙𝑠𝑖𝑛 θ
𝑑
[𝑚𝑙 2 𝜃
𝑑𝑡
x
θ
Apply Euler’s Equation
𝜕𝐿
𝜕𝜃
ωt
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝜃
=
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝜃
= 𝑚𝑙 2 𝜃 + 𝑚𝑟𝜔𝑙 𝜃𝑐𝑜𝑠 (𝜃 − 𝜔𝑡)- 𝑚𝑟𝜔2 𝑙𝑐𝑜𝑠 (𝜃 − 𝜔𝑡)
+ 𝑚𝑟𝜔𝑙𝑠𝑖𝑛(𝜃 − 𝜔𝑡)]
m
A rotating pendulum
y
Consider a pendulum of length 𝑙 moving on a massless ring
of radius r and constant angular velocity ω.
Apply Euler’s Equation
r
𝜕𝐿
𝜕𝜃
=
𝑑 𝜕𝐿
𝑑𝑡 𝜕𝜃
ωt
𝑚𝑟𝜔𝑙𝜃𝑐𝑜𝑠(𝜃 − 𝜔𝑡) − 𝑚𝑔𝑙𝑠𝑖𝑛 θ = 𝑚𝑙2 𝜃 + 𝑚𝑟𝜔𝑙𝜃𝑐𝑜𝑠 (𝜃 − 𝜔𝑡)- 𝑚𝑟𝜔2 𝑙𝑐𝑜𝑠 (𝜃 − 𝜔𝑡)
𝑚𝑙 2 𝜃 = 𝑚𝑟𝜔2 𝑙𝑐𝑜𝑠 (𝜃 − 𝜔𝑡) − 𝑚𝑔𝑙𝑠𝑖𝑛 θ
x
θ
𝑙
Solve for 𝜃
m
𝜃=
𝑟𝜔2 𝑐𝑜𝑠 (𝜃−𝜔𝑡) 𝑔𝑠𝑖𝑛 θ
−
𝑙
𝑙
Reduces to simple pendulum when ω = 0.
Lagrangian Benefits

Deals with energy which is invariant to coordinate transformations.

Can greatly simplify complicated systems.

Allows us to understand mechanical systems where all of the forces
cannot be stated explicitly.

Provides an alternative view of a mechanical system: rather than
seeing only cause and effect, we now see the purpose of the
system which is to minimize the action.
Sources

http://www.people.fas.harvard.edu/~djmorin/chap6.pdf

https://en.wikipedia.org/wiki/Leonhard_Euler (photo)

Marion, Thornton. Classical Dynamics of Particles and Systems, 4th
edition,1995. Harcourt Brace & Co.

http://physicsinsights.org/rotating_polar_1.html