Physics 318: Solutions to Homework #3
February 18, 2008
1
E
=
−∇
Φ
−
˙
A
B
=
∇×
A
L
(
x
,
˙x
, t
) =
1
2
m
˙x
2
−
q
Φ(
x
, t
) +
q
˙x
·
A
(
x
, t
)
(1)
a.)
Writing out all the components explicitly, the Lagrangian is given by
L
(
x, y, z,
˙
x,
˙
y,
˙
z, t
) =
m
2
( ˙
x
2
+ ˙
y
2
+ ˙
z
2
)
−
q
Φ(
x, y, z, t
)+
q
[ ˙
xA
x
(
x, y, z, t
)+ ˙
yA
y
(
x, y, z, t
)+ ˙
zA
z
(
x, y, z, t
)]
.
We compute
d
dt
∂
L
∂
˙
x
−
∂
L
∂x
= 0
,
(2)
and similarly for
y
and
z
, remembering that
d
dt
=
∂
∂t
+
˙x
·∇
=
∂
∂t
+ ˙
x
∂
∂x
+ ˙
y
∂
∂y
+ ˙
z
∂
∂z
.
(3)
Using
∂
L
∂
˙
x
=
m
˙
x
+
qA
x
(4)
leads to
m
¨
x
=
q
b
−
˙
A
x
−
( ˙
x
∂
∂x
+ ˙
y
∂
∂y
+ ˙
z
∂
∂z
)
A
x
−
∂
Φ
∂x
+
∂
∂x
( ˙
xA
x
+ ˙
yA
y
+ ˙
zA
z
)
B
(5)
and similarly for
y
and
z
. In vector notation, this becomes
m
¨x
=
q
(
−∇
Φ
−
˙
A
) +
q
[
∇
(
˙x
·
A
)
−
(
˙x
·∇
)
A
]
(6)
=
q
E
+
q
˙x
×
(
∇×
A
)
(7)
=
q
E
+
q
˙x
×
B
(8)
We can simplify the computations leading to this result by writing (
x, y, z
) as (
x
1
, x
2
, x
3
) and intro-
ducing the notation
x
i
= (
x
1
, x
2
, x
3
). We use
x
i
and
x
i
interchangably, there is no signi±cance to the
position of the index, up or down. The Lagrangian can be expressed as
L
(
x
i
,
˙
x
i
, t
) =
1
2
m
3
s
i
=1
˙
x
i
˙
x
i
−
q
Φ(
x
i
, t
) +
q
3
s
i
=1
˙
x
i
A
i
(
x
i
, t
)
.
(9)
1