Electric and Magnetic Forces in Lagrangian
and Hamiltonian Formalism
Benjamin Hornberger
10/26/01
Phy 505, Classical Electrodynamics, Prof. Goldhaber
Lecture notes from Oct. 26, 2001
(Lecture held by Prof. Weisberger)
1
Introduction
Conservative forces can be derived from a Potential V (q, t). Then, as we
know from classical mechanics, we can write the Lagrangian as
L(q, q̇, t) = T − V,
(1)
where T is the kinetic energy of the system. The Euler-Lagrangian equations of motion are then given by
d
dt
Ã
∂L
∂ q̇i
!
−
∂L
= 0.
∂qi
(2)
In three dimensions with cartesian Coordinates, this can be written as
d ³~ ´ ~
∇~v L − ∇L = 0.
dt
(3)
~ ~v means the gradient with respect to the velocity coordinates.
Here, ∇
Now we generalize V (q, t) to U (q, q̇, t) – this is possible as long as L =
T − U gives the correct equations of motion.
1
2 LORENTZ FORCE LAW
2
2
Lorentz Force Law
The Lorentz force in Gaussian Units is given by:
Ã
F~ = Q
!
~ + ~v × B
~ ,
E
c
(4)
~ x, t) is the electric field and B(~
~ x, t) is
where Q is the electric charge, E(~
~ and
the magnetic field. If the sources (charges or currents) are far away, E
~ solve the homogeneous Maxwell equations. In Gaussian Units, they are
B
given by
~ ·B
~ =0
∇
(5)
~
~ ×E
~ + 1 ∂B = 0
∇
c ∂t
(6)
~ can be derived from a vector potential A:
~
The magnetic field B
~ =∇
~ ×A
~
B
(7)
If we plug this into Eq. (6), we get


~
~ × E
~ + 1 ∂A  = 0
∇
c ∂t
(8)
So the expression in square brackets is a vector field with no curl and can
be written as the gradient of a scalar potential ϕ:
~
~ + 1 ∂ A = −∇ϕ
~
E
c ∂t
(9)
or
~
~ = −∇ϕ
~ − 1 ∂A
E
c ∂t
This we plug into Eq. (4) for the Lorentz force law and we get


(10)

³
´
~
~ ×A
~  .
~ − 1  ∂ A − ~v × ∇
F~ = Q −∇ϕ
c ∂t
(11)
3 LAGRANGIAN FORMALISM
3
If we apply the general general vector relation
³
´
³
´
~a × ~b × ~c = ~b (~a · ~c) − ~a · ~b ~c
(12)
to the triple vector cross product in the square brackets, we get
³
´
³
´
³
´
~ ×A
~ =∇
~ ~v · A
~ − ~v · ∇
~ A.
~
~v × ∇
(13)
So the equation for the Lorentz force law is now



´
³
´
~ ³
~ − 1  ∂ A + ~v · ∇
~ A
~−∇
~ ~v · A
~  .
F~ = Q −∇ϕ
c ∂t
(14)
~ x, t):
Now let’s look at the total time derivative of A(~
X
d ~
∂ ~
∂ ~
A (~x, t) = A
(~x, t) +
vj
A (~x, t)
dt
∂t
∂xj
j
|
{z
(15)
}
~ )A(~
~ x,t)
=(~v ·∇
The right side of the equation corresponds to the first two terms in the
square brackets of Eq. (14), and we can write


³
´
~
~ − 1 dA + 1 ∇
~ ~v · A
~ 
F~ = Q −∇ϕ
c dt
c
3
3.1
(16)
Lagrangian Formalism
The Lorentz Force Law in the Lagrangian Formalism
~ to the LaLet’s try to add a vector potential term UA~ (~x, ~v , t) = − Qc ~v · A
grangian:
1
Q
~
L = mv 2 − Q ϕ(~x, t) + ~v · A
2
c
|
{z
} | {z }
I
(17)
II
If we apply the Euler-Lagrangian equation of motion (Eq. (3)) on part I
of Eq. (17), we get
3 LAGRANGIAN FORMALISM
4
d~v
~ = 0,
+ Q ∇ϕ
dt
and applying it to part II gives
m
(18)
³
´
´
~
d ³~
~ ~v · A
~ =0
~ ~ = Q dA − Q ∇
∇~v UA~ − ∇U
(19)
A
dt
c dt
c
Altogether, the Euler-Lagrangian equation of motion, applied on the Lagrangian of Eq. (17), gives
³
´
~
d~v
~ + Q dA − Q ∇
~ ~v · A
~ =0
+ Q ∇ϕ
(20)
dt
c dt
c
v
If we identify m d~
with the force F~ , given by Newton’s Law, we can solve
dt
Eq. (20) for F~ :
m


³
´
~
~ ~v · A
~ 
~ − 1 dA + 1 ∇
F~ = Q −∇ϕ
c dt
c
(21)
which is just the correct expression for the Lorentz Force Law, given by
Eq. (16).
3.2
How does a gauge transformation affect this Lagrangian?
~ and B
~ fields are invariant under gauge transformations
As we know, E
~ (~x, t) → A
~0 = A
~ + ∇Λ
~ (~x, t)
A
(22)
1
(23)
ϕ (~x, t) → ϕ0 = ϕ − Λ̇ (~x, t) ,
c
where Λ(~x, t) is an arbitrary scalar function. If we plug these new scalar
and vector potentials into the Lagrangian (Eq. (17)), it changes to
´
Q³
~ (~x, t)
Λ̇ (~x, t)) + ~v · ∇Λ
(24)
c
The expression in brackets is just the total time derivative of Λ(~x, t), so
we get
L → L0 = L +
4 HAMILTONIAN FORMALISM
L0 = L +
5
Q d
Λ (~x, t)
c dt
(25)
.
But as we know, adding to the Lagrangian a total time derivative of a
function of ~x and t does not change the equations of motion.
So, the Lagrangian for a particle in an electromagnetic field is given by
1
Q
~
L = mv 2 − Q ϕ + ~v · A
2
c
4
(26)
Hamiltonian Formalism
4.1
The Hamiltonian for the EM-Field
We know the canonical momentum from classical mechanics:
pi =
∂L
∂ ẋi
(27)
Using the Lagrangian from Eq. (26), we get
pi = mvi +
Q
Ai
c
(28)
The Hamiltonian is then given by
H=
X
i
1
pi ẋi − L = mv 2 + Q ϕ,
2
(29)
where v resp. ẋ must be replaced by p: Solving Eq. (28) for vi and
plugging into Eq. (29) gives
¯
¯
1 ¯¯
Q ~ ¯¯2
H=
A +Q ϕ
p
~
−
2m ¯
c ¯
So the kinetic momentum in is in this case given by
Q ~
A
P~ = m~v = p~ −
c
(30)
(31)
4 HAMILTONIAN FORMALISM
6
Example: Uniform constant magnetic field
~ in z-direction:
We assume B


0

~
B = B · ẑ =  0 

B
(32)
The vector potential can then be written as
~=1 B
~ × ~r
A
(33)
2
This is an arbitrary choice, but it is easy to prove that it gives the correct
~ Now suppose the particle is bound in a strong central potential
result for B.
~
and B is relatively weak. If we plug the vector potential (Eq. (33)) into the
Hamiltonian (Eq. (30)), we get
H=
2
³
´ ³
´
|~p|2
Q
~ × ~r + Q
~ × ~r · B
~ × ~r
+Q ϕ−
p~ · B
B
2m
2mc
8m2 c2 |
{z
}
~ 2~
~ r)2
B
r2 −(B·~
(34)
The last term in this equation can be neglected for a bound particle in a
weak field. For the mixed scalar / cross product in the second term, we can
write
~ × ~r = ~r × p~ · B
~ =L
~ · B,
~
p~ · B
(35)
~ is the angular momentum. So the Hamiltonian is
where L
|~p|2
Q ~ ~
+Q ϕ−
L·B
(36)
2m
2mc
The last term is this Hamiltonian causes the ordinary Zeeman Effect.
H'
4.2
Hamiltonian Equations of Motion
The Hamiltonian equations of motion are given by
ẋi =
∂H
∂pi
and
ṗi = −
∂H
.
∂xi
(37)
4 HAMILTONIAN FORMALISM
7
If we apply these equations on the Hamiltonian (Eq. (30)), we get
·
Q
1
ẋi =
pi −
Ai
m
c
¸
(38)


µ
¶
X
1 
Q
∂ϕ
Q ∂Aj 
ṗi =
pj −
−Q
Aj
m
c
j
c ∂xi
∂xi
(39)
Example: Uniform constant magnetic field
Again we look at a constant magnetic field in z-direction (no other potential):


0

~ = B · ẑ = 
B
 0 
B
(40)
For the vector potential, we choose


0


~
A = x · B · ŷ =  x · B 
0
(41)
~ We
This is again an arbitrary choice which gives the correct result for B.
put this vector potential into the Hamiltonian and get
"
µ
1
QB
H=
p2z + p2x + py −
x
2m
c
¶2 #
= Hz + H⊥ .
(42)
The second part H⊥ of the Hamiltonian can be written as
·
¸
"
#2
p2
m QB 2
c
H⊥ = x +
x−
py
2m
2 | mc
QB
{z } |
{z
}
2
:=ωL
(43)
:=q12
where we define the Larmor Frequency ωL := QB
and introduce a new
mc
c
py ). Furthermore, we set px = p1 , py = p2 and
coordinate q1 := (x − QB
pz = p3 .
H⊥ =
1
p21
+ m ωL2 q12
2m 2
(44)
4 HAMILTONIAN FORMALISM
8
This is just the Hamiltonian for a harmonic oscillator. In Quantum Mechanics, we can use the commutator
[q1 , p1 ] = ih̄,
(45)
and for the harmonic oscillator, the energy eigenvalues are
µ
¶
1
En = n +
h̄ ωL
2
(46)