Charge in Electromagnetic Fields
Uploaded by
lad.kocbCharge in Electromagnetic Fields
Uploaded by
lad.kocbCharge in Electric and Magnetic
Fields
Based on
Bransden and Joachain: Physics of Atoms and
Molecules
and
Goldstein: Classical Mechanics
09.11.2013
Lagrange equations
d
dt
L
ri
L
=0
ri
i = 1, 2, ....
For conservative systems, i.e. with usual forces from potential energy
L(ri , ri , t) = T (ri ) V (ri )
but the Lorentz force depends on velocity. Lagrange function L(ri , ri , t) must
be modified.
Lorentz force
1
F = q E + [v B]
c
With and A the scalar and vector potentials
E =
1 A
c t
B=A
the Lorentz force becomes
1 A 1
F = q
+ v [ A]
c t
c
with (r, t) and A(r, t) describing the fields
2
With only the scalar potential (r, t) the Lagrange function would be
L(r, r , t) =
m
r r q(r, t) =
2
1
m v2 q(r, t)
2
and the Lagrange equation would lead to the electrostatic
mr = q
It can be shown that the Newton equation with the electromagnetic Lorentz
force
1 A 1
mr = q +
+ v [ A]
c t
c
can be derived from a surprisingly simple Lagrange function (see below)
L(r, r , t) =
1
q
m v2 q(r, t) + v A(r, t)
2
c
when inserted into the three Lagrange equations (r1 x, r2 y , r3 z)
d L
L
=0
i = 1, 2, 3
dt ri
ri
This is because the term v[ A] can be expressed without vector products.
3
Transforming the term v [ A]
This contains time derivatives as well as the x, y, z derivatives
v [ A] r [ A]
consider first the total time derivative A,
= dA = A + A dx + A dy + A dz = A + vx A + vy A + vz A
A
dt
t
x dt
y dt
z dt
t
x
y
z
Take now its x-component and re-arrange
Ax
Ax
Ax
dAx Ax
= vx
+ vy
+ vz
dt
t
x
y
z
Take now x-component of v [ A]
Ay Ax
Ax Az
vz
{v [ A]}x = vy
x
y
z
x
and now re-arrange - adding and subtracting (and compare with above)
vx
Ax
Ay
Az
+ vy
+ vz
x
x
x
vx
Ax
Ax
Ax
vy
vz
x
y
z
Comparing those expressions we can replace the second term
{v [ A]}x =
dAx Ax
(vx Ax + vy Ay + vz Az )
+
x
dt
t
and now it can be written for all components as
v [ A] = (v A)
dA A
+
dt
t
Now we insert this expression into the Lorentz force
1 A 1
+ v [ A]
F = q
c t
c
So that the Lorentz force becomes
1 dA 1
F = q
+ (v A)
c dt
c
This can be also written as
1d
1
F = q
v (v A) + (v A)
c dt
c
Lagrange equations in vector form
d L
L
= 0 i = 1, 2, ....
dt ri
ri
1
q
m r 2 q(r, t) + r A(r, t)
2
c
q
q
m r + A
r L = qr + r (r A)
c
c
L(r, r , t) =
r L =
d
(r L) (r L) = 0
dt
Hamiltonian formalism results from a Legendre transform from xi and its
derivative x i to a pair of conjugated variables xi and pi where
pi =
L
ri
p = r L
when L(xi , x i , t) is replaced by H(xi , pi , t) with
X
H(xi , pi , t) =
(x i pi ) L(xi , x i , t)
H = r p L
We have from above
L=
q
1
m r 2 q + r A
2
c
p = r L
p=
m r +
q
A
c