EULER-LAGRANGE TO HAMILTON
LANCE D. DRAGER
The goal of these notes is to give one way of getting from the Euler-Lagrange
equations to Hamilton’s equations.
1. Euler-Lagrange to Hamilton
We will often write elements of a Euclidean space R2𝑁 as a pair of 𝑁 -vectors,
(𝑥, 𝑦) = (𝑥1 , . . . , 𝑥𝑁 , 𝑦1 , . . . , 𝑦𝑁 ).
We start with a lemma.
Lemma 1.1. Let 𝐹 : R2𝑁 → R2𝑁 be a diffeomorphism of the form 𝐹 (𝑥, 𝑦) =
(𝑥, 𝐴(𝑥, 𝑦)), for some function 𝐴 : R2𝑁 → R𝑁 . Then the inverse function 𝐺 has
the form 𝐺(𝑢, 𝑣) = (𝑢, Γ(𝑢, 𝑣)) and we have the following identifies:
(1)
𝐴(𝑢, Γ(𝑢, 𝑣)) = 𝑣
(2)
Γ(𝑥, 𝐴(𝑥, 𝑦)) = 𝑦.
Proof. The function 𝐺 can be written as (𝐺1 (𝑢, 𝑣), 𝐺2 (𝑢, 𝑣)), and similarly for 𝐹 .
Since 𝐹 and 𝐺 are inverses, we have
(𝑢, 𝑣) = (𝐹 ∘ 𝐺)𝑢, 𝑣
= 𝐹 (𝐺1 (𝑢, 𝑣), 𝐺2 (𝑢, 𝑣))
= (𝐺1 (𝑢, 𝑣), 𝐴(𝐺1 (𝑢, 𝑣), 𝐺2 (𝑢, 𝑣))).
Comparing the first slots, we find 𝐺1 (𝑢, 𝑣) = 𝑢. Plugging this into the second slots
gives 𝑣 = 𝐴(𝑢, 𝐺2 (𝑢, 𝑣)) and we have 𝐺(𝑢, 𝑣) = (𝑢, 𝐺2 (𝑢, 𝑣)). Set Γ = 𝐺2 , and we
have 𝑣 = 𝐴(𝑢, Γ(𝑢, 𝑣)).
The identity in equation (2) is derived in a similar fashion from 𝐺(𝐹 (𝑥, 𝑦)) =
(𝑥, 𝑦).
Let 𝐿 : R2𝑁 → R be a Lagrangian, which we write as
𝐿(𝑞, 𝑞)
˙ = 𝐿(𝑞1 , . . . , 𝑞𝑁 , 𝑞˙1 , . . . , 𝑞˙𝑁 )
where the 𝑞˙𝑖 are just variables.
If 𝑐(𝑡) is a path in R𝑁 , the position of the system, we follow classical tradition
and write 𝑐(𝑡) = 𝑞(𝑡). If we add the velocity, we get a path 𝑐ˆ(𝑡) = (𝑞(𝑡), 𝑞(𝑡))
˙
in
R2𝑁 . In this notation we do mean that 𝑞(𝑡)
˙ = 𝑞 ′ (𝑡) is the derivative of 𝑞(𝑡).
The path satisfies the Euler-Lagrange equations if
𝜕𝐿
𝑑 𝜕𝐿
(3)
(𝑞(𝑡), 𝑞(𝑡))
˙
=
(𝑞(𝑡), 𝑞(𝑡)),
˙
𝑖 = 1, . . . , 𝑁.
𝑑𝑡 𝜕 𝑞˙𝑖
𝜕𝑞𝑖
For brevity, we sometimes write expressions for partial derivatives like
𝜕𝐿 (︀ 𝜕𝐿
𝜕𝐿 )︀
=
,...,
.
𝜕𝑞
𝜕𝑞1
𝜕𝑞𝑁
Version Time-stamp: ”2014-10-01 15:56:09 drager”.
1
2
LANCE D. DRAGER
Define a map 𝐹 : R2𝑁 → R2𝑁 by
(︀ 𝜕𝐿
𝐹 (𝑞, 𝑞)
˙ = 𝑞,
(𝑞, 𝑞)).
˙
𝜕𝑞
The Lagrangian 𝐿 is called hyperregular if this map is a diffeomorphism.
For the rest of these notes, we deal with hyperregular Lagrangian.
Let 𝐺(𝑞, 𝑝) = (𝑞, Γ(𝑞, 𝑝)) be the inverse of 𝐹 . By our Lemma, we have
(4)
(5)
𝜕𝐿
(𝑞, Γ(𝑞, 𝑝)) = 𝑝
𝜕 𝑞˙
(︀ 𝜕𝐿
)︀
Γ 𝑞,
(𝑞, 𝑞)
˙ = 𝑞˙
𝜕 𝑞˙
The quantify
(6)
𝑝𝑖 =
𝜕𝐿
(𝑞, 𝑞)
˙
𝜕 𝑞˙𝑖
is called the (generalized) momentum conjugate to 𝑞𝑖 .
Given a path 𝑐ˆ(𝑡) = (𝑞(𝑡), 𝑞(𝑡)),
˙
we get a path 𝛾(𝑡) = 𝐹 (𝑞(𝑡), 𝑞(𝑡))
˙
= (𝑞(𝑡), 𝑝(𝑡)),
where
𝑝(𝑡) =
𝜕𝐿
(𝑞(𝑡), 𝑞(𝑡)).
˙
𝜕 𝑞˙
If 𝑐 is a solution of the Euler-Lagrange equations, we want to find the differential
equations satisfied by 𝛾.
From the Euler-Lagrange equations, we have
(7)
𝑑
𝜕𝐿
𝑝(𝑡) =
(𝑞(𝑡), 𝑞(𝑡))
˙
𝑑𝑡
𝜕𝑞
and we have
(8)
𝑑
𝑞(𝑡) = 𝑞(𝑡)
˙ = Γ(𝑞(𝑡), 𝑝(𝑡)),
𝑑𝑡
We need to eliminate the 𝑞(𝑡)
˙ in (7), which we can do using (8),
(9)
𝑑
𝜕𝐿
𝑝(𝑡) =
(𝑞(𝑡), Γ(𝑞(𝑡), 𝑝(𝑡))).
𝑑𝑡
𝜕𝑞
The equations (9) and (8) give us our desired differential equations. A path
(𝑞(𝑡), 𝑝(𝑡)) satisfies these equations if and only if the corresponding path (𝑞(𝑡), 𝑞(𝑡))
˙
satisfies the Euler-Lagrange equations.
However, we can write the equations in a more informative form. Define the
̃︀ 𝑝) by
function 𝐿(𝑞,
̃︀ 𝑝) = 𝐿(𝑞, Γ(𝑞, 𝑝)),
𝐿(𝑞,
the Lagrangian expressed in terms of 𝑝 and 𝑞.
EULER-LAGRANGE TO HAMILTON
3
Using the chain rule, we get
̃︀
]︀
𝜕𝐿
𝜕 [︀
(𝑞, 𝑝) =
𝐿(𝑞, Γ(𝑞, 𝑝))
𝜕𝑞𝑖
𝜕𝑞𝑖
𝑁
∑︁
𝜕𝐿
𝜕Γ𝑗
𝜕𝐿
=
(𝑞, Γ(𝑞, 𝑝)) +
(𝑞, Γ(𝑞, 𝑝)))
(𝑞, 𝑝)
𝜕𝑞𝑖
𝜕 𝑞˙𝑗
𝜕𝑞𝑖
𝑗=1
=
𝑁
∑︁
𝜕𝐿
𝜕Γ𝑗
(𝑞, Γ(𝑞, 𝑝)) +
(𝑞, 𝑝),
𝑝𝑗
𝜕𝑞𝑖
𝜕𝑞𝑖
𝑗=1
using (4). We can solve this equation for 𝜕𝐿/𝜕𝑞𝑖 to get
𝑁
∑︁
̃︀
𝜕𝐿
𝜕Γ𝑗
𝜕𝐿
(𝑞, Γ(𝑞, 𝑝)) =
(𝑞, 𝑝) −
𝑝𝑗
(𝑞, 𝑝)
𝜕𝑞𝑖
𝜕𝑞𝑖
𝜕𝑞𝑖
𝑗=1
=−
𝜕
𝜕𝑞𝑖
{︂ ∑︁
𝑁
}︂
̃︀ 𝑞)
𝑝𝑗 Γ𝑗 (𝑞, 𝑝) − 𝐿(𝑝,
𝑗=1
𝜕𝐻
=−
(𝑞, 𝑝),
𝜕𝑞𝑖
where
(10)
𝐻(𝑝, 𝑞) =
𝑁
∑︁
̃︀ 𝑞)
𝑝𝑗 Γ𝑗 (𝑞, 𝑝) − 𝐿(𝑝,
𝑗=1
is the Hamiltonian of the problem. The minus sign is conventional.
Thus, by (7), we have
(11)
𝑑𝑝𝑖
𝜕𝐻
(𝑡) = −
(𝑞(𝑡), 𝑝(𝑡))
𝑑𝑡
𝜕𝑞𝑖
which is one of Hamilton’s equations. This equation holds only when the EulerLagrange equations are satisfied.
For the other equation, we can calculate as follows:
]︂
[︂ 𝑁
𝜕𝐻
𝜕 ∑︁
̃︀
(𝑞, 𝑝) =
𝑝𝑗 Γ𝑗 (𝑞, 𝑝) − 𝐿(𝑝, 𝑞)
𝜕𝑝𝑖
𝜕𝑝𝑖 𝑗=1
(12)
𝑁
∑︁
̃︀
𝜕Γ𝑗
𝜕𝐿
= Γ𝑖 (𝑝, 𝑞) +
𝑝𝑗
(𝑞, 𝑝) −
(𝑞, 𝑝).
𝜕𝑝𝑖
𝜕𝑝𝑖
𝑗=1
̃︀ we have
To calculate the partial derivatives of 𝐿,
̃︀
]︀
𝜕𝐿
𝜕 [︀
(𝑞, 𝑝) =
𝐿(𝑞, Γ(𝑞, 𝑝))
𝜕𝑝𝑖
𝜕𝑝𝑖
𝑁
∑︁
𝜕𝐿
𝜕Γ𝑗
=
(𝑞, Γ(𝑞, 𝑝))
(𝑝, 𝑞)
𝜕
𝑞
˙
𝜕𝑝𝑖
𝑗
𝑗=1
=
𝑁
∑︁
𝑗=1
𝑝𝑗
𝜕Γ𝑗
(𝑝, 𝑞),
𝜕𝑝𝑖
4
LANCE D. DRAGER
using equation (4). Plugging this into (12), we have
𝑁
∑︁
̃︀
𝜕𝐻
𝜕𝐿
𝜕Γ𝑗
(𝑞, 𝑝) = Γ𝑖 (𝑝, 𝑞) +
(𝑞, 𝑝) −
(𝑞, 𝑝)
𝑝𝑗
𝜕𝑝𝑖
𝜕𝑝𝑖
𝜕𝑝𝑖
𝑗=1
𝑁
∑︁
𝜕Γ𝑗
𝜕Γ𝑗
(𝑞, 𝑝) −
(𝑝, 𝑞)
= Γ𝑖 (𝑝, 𝑞) +
𝑝𝑗
𝑝𝑗
𝜕𝑝
𝜕𝑝𝑖
𝑖
𝑗=1
𝑗=1
𝑁
∑︁
= Γ𝑖 (𝑞, 𝑝),
in other words
𝜕𝐻
= Γ(𝑞, 𝑝),
𝜕𝑝
(13)
so (8) becomes
𝜕𝐻
𝑑𝑞
(𝑡) =
(𝑞(𝑡), 𝑝(𝑡)).
𝑑𝑡
𝜕𝑝
(14)
This equation holds for any path.
Putting this together, we have Hamilton’s Equations
𝜕𝐻
𝑑𝑞
(𝑡) =
(𝑞(𝑡), 𝑝(𝑡))
𝑑𝑡
𝜕𝑝
𝜕𝐻
𝑑𝑝
(𝑡) = −
(𝑞(𝑡), 𝑝(𝑡)),
𝑑𝑡
𝜕𝑞
(15)
or, to put in the indices,
(16)
𝑑𝑞𝑖
𝜕𝐻
(𝑡) =
(𝑞(𝑡), 𝑝(𝑡))
𝑑𝑡
𝜕𝑝𝑖
𝑑𝑝𝑖
𝜕𝐻
(𝑡) = −
(𝑞(𝑡), 𝑝(𝑡)),
𝑑𝑡
𝜕𝑞𝑖
𝑖 = 1, . . . , 𝑁.
In the classical literature, the formula for 𝐻,
𝐻(𝑝, 𝑞) =
𝑁
∑︁
̃︀ 𝑞)
𝑝𝑗 Γ𝑗 (𝑞, 𝑝) − 𝐿(𝑝,
𝑗=1
is often written as
𝐻(𝑞, 𝑝) =
𝑁
∑︁
̃︀ 𝑝),
𝑝𝑗 𝑞˙𝑗 − 𝐿(𝑞,
𝑗=1
in view of our formula 𝑞˙ = Γ(𝑞, 𝑝), but this is a bit obscure, since you have to
remember that 𝑞˙ here is a function of 𝑞 and 𝑝, not an independent variable.
Example 1.2. Consider a particle moving in the plane, subject to a force that
comes from a potential. We’ll use the coordinates (𝑞1 , 𝑞2 ) on the plane and follow
the notation above. The Lagrangian is
𝐿(𝑞, 𝑞)
˙ =
1
𝑚(𝑞˙12 + 𝑞˙22 ) − 𝑉 (𝑞1 , 𝑞2 ).
2
EULER-LAGRANGE TO HAMILTON
5
The conjugate momenta are
𝜕𝐿
(𝑞, 𝑞)
˙ = 𝑚𝑞˙1
𝜕 𝑞˙1
𝜕𝐿
(𝑞, 𝑞)
˙ = 𝑚𝑞˙2
𝑝2 =
𝜕 𝑞˙2
𝑝1 =
We can easily solve this for the 𝑞’s
˙ in terms of the 𝑝’s:
𝑞˙1 = 𝑝1 /𝑚
𝑞˙2 = 𝑝2 /𝑚,
in other words
Γ(𝑞, 𝑝) =
1
𝑝.
𝑚
Substituting in the Lagrangian, we get
̃︀ 𝑝) = 𝐿(𝑞, Γ(𝑞, 𝑝))
𝐿(𝑞,
[︂(︂ )︂2 (︂ )︂2 ]︂
𝑝1
𝑝2
1
+
− 𝑉 (𝑞)
= 𝑚
2
𝑚
𝑚
1 2
=
(𝑝 + 𝑝22 ) − 𝑉 (𝑞).
2𝑚 1
Then we get
𝐻(𝑞, 𝑝) =
𝑁
∑︁
̃︀ 𝑝)
𝑝𝑗 𝑞˙𝑗 − 𝐿(𝑞,
𝑗=1
̃︀ 𝑝)
= 𝑝1 𝑞˙1 + 𝑝2 𝑞˙2 − 𝐿(𝑞,
{︂
}︂
𝑝1
1 2
𝑝2
2
= 𝑝1 + 𝑝2 −
(𝑝 + 𝑝2 ) − 𝑉 (𝑞)
𝑚
𝑚
2𝑚 1
1 2
=
(𝑝 + 𝑝22 ) + 𝑉 (𝑞)
2𝑚 1
If we write this on the Lagrangian side as
𝐸(𝑞, 𝑞)
˙ = (𝐻 ∘ 𝐹 )(𝑞, 𝑞)
˙
1
=
((𝑚𝑞˙1 )2 + (𝑚𝑞˙2 )2 ) + 𝑉 (𝑞)
2𝑚
1
= 𝑚(𝑞˙12 + 𝑞˙22 ) + 𝑉 (𝑞),
2
we get the kinetic energy plus the potential energy, so 𝐸(𝑞, 𝑞)
˙ is the total energy of
the system and 𝐻 is the total energy of the system expressed in terms of position
and momentum.
Going back to the Hamiltonian side, Hamilton’s equations would be
𝜕𝐻
𝜕𝑉
𝑑𝑝𝑖
(𝑡) = −
(𝑞, 𝑝) = −
𝑑𝑡
𝜕𝑞𝑖
𝜕𝑞𝑖
𝑑𝑞𝑖
𝜕𝐻
(𝑡) =
(𝑞, 𝑝) = 𝑝𝑖 /𝑚
𝑑𝑡
𝜕𝑝𝑖
We’ll leave it as an exercise to substitute 𝑝𝑖 = 𝑚𝑞˙𝑖 and put the equations into one
second order equation to get Newton’s equations.
6
LANCE D. DRAGER
Since 𝐻 or 𝐸 represents the total energy, we expect that it should be conserved
(when the Lagrangian and Hamiltonian are not time dependent, as in our discussion,
i.e., we don’t have 𝐿(𝑡, 𝑞, 𝑞)).
˙
Theorem 1.3. If (𝑞(𝑡), 𝑝(𝑡)) is a solution of Hamilton’s equations (16), then 𝐻(𝑞(𝑡), 𝑝(𝑡))
is a constant function of 𝑡, i.e., “Energy is conserved.”
Proof. We just differentiate 𝐻(𝑞(𝑡), 𝑝(𝑡)) and plug in Hamilton’s equations:
𝑁
𝑁
∑︁
∑︁
𝑑
𝑑𝑞𝑗
𝑑𝑝𝑗
𝜕𝐻
𝜕𝐻
𝐻(𝑞(𝑡), 𝑝(𝑡)) =
(𝑞(𝑡), 𝑝(𝑡))
(𝑡) +
(𝑞(𝑡), 𝑝(𝑡))
(𝑡)
𝑑𝑡
𝜕𝑞
𝑑𝑡
𝜕𝑝
𝑑𝑡
𝑗
𝑗
𝑗=1
𝑗=1
=
𝑁
∑︁
𝜕𝐻
𝑗=1
𝜕𝑞𝑗
(𝑞(𝑡), 𝑝(𝑡))
𝜕𝐻
(𝑞(𝑡), 𝑝(𝑡))+
𝜕𝑝𝑗
𝑁
∑︁
𝜕𝐻
𝑗=1
[︂
]︂
𝜕𝐻
(𝑞(𝑡), 𝑝(𝑡)) −
(𝑞(𝑡), 𝑝(𝑡))
𝜕𝑝𝑗
𝜕𝑞𝑗
= 0.
The derivative of 𝐻(𝑞(𝑡), 𝑝(𝑡)) is zero, so it’s a constant function.
Exercise 1.4. Show directly from the Euler-Lagrange equations that the energy
𝐸(𝑞(𝑡), 𝑞(𝑡))
˙
is constant on solutions (𝑞(𝑡), 𝑞(𝑡))
˙
of the Euler-Lagrange equations.
Exercise 1.5. What happens for time dependent Lagrangians 𝐿(𝑡, 𝑞, 𝑞)?
˙
Department of Mathematics and Statistics, Texas Tech University, Lubbock, TX
79409-1042
E-mail address: lance.drager@ttu.edu