Electric and Magnetic Forces in Lagrangian and Hamiltonian Formalism

stewsystemElectronics - Devices

Oct 18, 2013 (4 years and 22 days ago)

84 views

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;˙q;t) = T ¡V;(1)
where T is the kinetic energy of the system.The Euler-Lagrangian equa-
tions of motion are then given by
d
dt
Ã
@L
@ ˙q
i
!
¡
@L
@q
i
= 0:(2)
In three dimensions with cartesian Coordinates,this can be written as
d
dt
³
~
r
~v
L
´
¡
~
rL = 0:(3)
Here,
~
r
~v
means the gradient with respect to the velocity coordinates.
Now we generalize V (q;t) to U(q;˙q;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:
~
F = Q
Ã
~
E +
~v
c
£
~
B
!
;(4)
where Q is the electric charge,
~
E(~x;t) is the electric field and
~
B(~x;t) is
the magnetic field.If the sources (charges or currents) are far away,
~
E and
~
B solve the homogeneous Maxwell equations.In Gaussian Units,they are
given by
~

~
B = 0 (5)
~

~
E +
1
c
@
~
B
@t
= 0 (6)
The magnetic field
~
B can be derived from a vector potential
~
A:
~
B =
~

~
A (7)
If we plug this into Eq.(6),we get
~

2
4
~
E +
1
c
@
~
A
@t
3
5
= 0 (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':
~
E +
1
c
@
~
A
@t
= ¡
~
r'(9)
or
~
E = ¡
~
r'¡
1
c
@
~
A
@t
(10)
This we plug into Eq.(4) for the Lorentz force law and we get
~
F = Q
0
@
¡
~
r'¡
1
c
2
4
@
~
A
@t
¡~v £
³
~

~
A
´
3
5
1
A
:(11)
3 LAGRANGIAN FORMALISM 3
If we apply the general general vector relation
~a £
³
~
b £~c
´
=
~
b (~a ¢ ~c) ¡
³
~a ¢
~
b
´
~c (12)
to the triple vector cross product in the square brackets,we get
~v £
³
~

~
A
´
=
~
r
³
~v ¢
~
A
´
¡
³
~v ¢
~
r
´
~
A:(13)
So the equation for the Lorentz force law is now
~
F = Q
0
@
¡
~
r'¡
1
c
2
4
@
~
A
@t
+
³
~v ¢
~
r
´
~

~
r
³
~v ¢
~
A
´
3
5
1
A
:(14)
Now let’s look at the total time derivative of
~
A(~x;t):
d
dt
~
A(~x;t) =
@
@t
~
A(~x;t) +
X
j
v
j
@
@x
j
~
A(~x;t)
|
{z
}
=
(
~v¢
~
r
)
~
A(~x;t)
(15)
The right side of the equation corresponds to the first two terms in the
square brackets of Eq.(14),and we can write
~
F = Q
0
@
¡
~
r'¡
1
c
d
~
A
dt
+
1
c
~
r
³
~v ¢
~
A
´
1
A
(16)
3 Lagrangian Formalism
3.1 The Lorentz Force Law in the Lagrangian Formal-
ism
Let’s try to add a vector potential term U
~
A
(~x;~v;t) = ¡
Q
c
~v ¢
~
A to the La-
grangian:
L =
1
2
mv
2
¡Q'(~x;t)
|
{z
}
I
+
Q
c
~v ¢
~
A
|
{z
}
II
(17)
If we apply the Euler-Lagrangian equation of motion (Eq.(3)) on part I
of Eq.(17),we get
3 LAGRANGIAN FORMALISM 4
m
d~v
dt
+Q
~
r'= 0;(18)
and applying it to part II gives
d
dt
³
~
r
~v
U
~
A
´
¡
~
rU
~
A
=
Q
c
d
~
A
dt
¡
Q
c
~
r
³
~v ¢
~
A
´
= 0 (19)
Altogether,the Euler-Lagrangian equation of motion,applied on the La-
grangian of Eq.(17),gives
m
d~v
dt
+Q
~
r'+
Q
c
d
~
A
dt
¡
Q
c
~
r
³
~v ¢
~
A
´
= 0 (20)
If we identify m
d~v
dt
with the force
~
F,given by Newton’s Law,we can solve
Eq.(20) for
~
F:
~
F = Q
0
@
¡
~
r'¡
1
c
d
~
A
dt
+
1
c
~
r
³
~v ¢
~
A
´
1
A
(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 La-
grangian?
As we know,
~
E and
~
B fields are invariant under gauge transformations
~
A(~x;t)!
~
A
0
=
~
A+
~
rΛ(~x;t) (22)
'(~x;t)!'
0
='¡
1
c
˙
Λ(~x;t);(23)
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
L!L
0
= L +
Q
c
³
˙
Λ(~x;t)) +~v ¢
~
rΛ(~x;t)
´
(24)
The expression in brackets is just the total time derivative of Λ(~x;t),so
we get
4 HAMILTONIAN FORMALISM 5
L
0
= L+
Q
c
d
dt
Λ(~x;t) (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
L =
1
2
mv
2
¡Q'+
Q
c
~v ¢
~
A (26)
4 Hamiltonian Formalism
4.1 The Hamiltonian for the EM-Field
We know the canonical momentum from classical mechanics:
p
i
=
@L
@ ˙x
i
(27)
Using the Lagrangian from Eq.(26),we get
p
i
= mv
i
+
Q
c
A
i
(28)
The Hamiltonian is then given by
H =
X
i
p
i
˙x
i
¡L =
1
2
mv
2
+Q';(29)
where v resp.˙x must be replaced by p:Solving Eq.(28) for v
i
and
plugging into Eq.(29) gives
H =
1
2m
¯
¯
¯
¯
~p ¡
Q
c
~
A
¯
¯
¯
¯
2
+Q'(30)
So the kinetic momentum in is in this case given by
~
P = m~v = ~p ¡
Q
c
~
A (31)
4 HAMILTONIAN FORMALISM 6
Example:Uniform constant magnetic field
We assume
~
B in z-direction:
~
B = B ¢ ˆz =
0
B
@
0
0
B
1
C
A
(32)
The vector potential can then be written as
~
A =
1
2
~
B £~r (33)
This is an arbitrary choice,but it is easy to prove that it gives the correct
result for
~
B.Now suppose the particle is bound in a strong central potential
and
~
B is relatively weak.If we plug the vector potential (Eq.(33)) into the
Hamiltonian (Eq.(30)),we get
H =
j~pj
2
2m
+Q'¡
Q
2mc
~p ¢
~
B £~r +
Q
2
8m
2
c
2
³
~
B £~r
´
¢
³
~
B £~r
´
|
{z
}
~
B
2
~r
2
¡
(
~
B¢~r
)
2
(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
~p ¢
~
B £~r =~r £~p ¢
~
B =
~
L ¢
~
B;(35)
where
~
L is the angular momentum.So the Hamiltonian is
H'
j~pj
2
2m
+Q'¡
Q
2mc
~
L ¢
~
B (36)
The last term is this Hamiltonian causes the ordinary Zeeman Effect.
4.2 Hamiltonian Equations of Motion
The Hamiltonian equations of motion are given by
˙x
i
=
@H
@p
i
and ˙p
i
= ¡
@H
@x
i
:(37)
4 HAMILTONIAN FORMALISM 7
If we apply these equations on the Hamiltonian (Eq.(30)),we get
˙x
i
=
1
m
·
p
i
¡
Q
c
A
i
¸
(38)
˙p
i
=
1
m
0
@
X
j
µ
p
j
¡
Q
c
A
j

Q
c
@A
j
@x
i
1
A
¡Q
@'
@x
i
(39)
Example:Uniform constant magnetic field
Again we look at a constant magnetic field in z-direction (no other potential):
~
B = B ¢ ˆz =
0
B
@
0
0
B
1
C
A
(40)
For the vector potential,we choose
~
A = x ¢ B ¢ ˆy =
0
B
@
0
x ¢ B
0
1
C
A
(41)
This is again an arbitrary choice which gives the correct result for
~
B.We
put this vector potential into the Hamiltonian and get
H =
1
2m
"
p
2
z
+p
2
x
+
µ
p
y
¡
QB
c
x

2
#
= H
z
+H
?
:(42)
The second part H
?
of the Hamiltonian can be written as
H
?
=
p
2
x
2m
+
m
2
·
QB
mc
¸
2
|
{z
}
:=!
2
L
"
x ¡
c
QB
p
y
#
2
|
{z
}
:=q
2
1
(43)
where we define the Larmor Frequency!
L
:=
QB
mc
and introduce a new
coordinate q
1
:= (x ¡
c
QB
p
y
).Furthermore,we set p
x
= p
1
,p
y
= p
2
and
p
z
= p
3
.
H
?
=
p
2
1
2m
+
1
2
m!
2
L
q
2
1
(44)
4 HAMILTONIAN FORMALISM 8
This is just the Hamiltonian for a harmonic oscillator.In Quantum Me-
chanics,we can use the commutator
[q
1
;p
1
] = i¯h;(45)
and for the harmonic oscillator,the energy eigenvalues are
E
n
=
µ
n +
1
2

¯h!
L
(46)