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Monday, October 31: Relativistic Charged
Particles
As I was saying, before the midterm exam intervened, in an inertial frame of
reference
K
there exists an electric field
~
E
and a magnetic field
~
B
at a space
time location (
~
r, t
). Another inertial frame of reference
K
0
is moving relative
to the first at a constant velocity
~v
=
~
βc
. In the
K
0
frame of reference, the
spacetime location (
~
r, t
) transforms to (
~
r
0
, t
0
) by the Lorentz transformation.
In the
K
0
frame of reference,
~
E
0
k
=
~
E
k
(1)
and
~
B
0
k
=
~
B
k
,
(2)
where
~
E
k
and
~
B
k
are the components of the
E
and
B
fields parallel to the
relative velocity vector
~
β
. The transformations for the components perpen
dicular to
β
are more complicated, and thus more interesting.
~
E
0
⊥
=
γ
(
~
E
⊥
+
~
β
×
~
B
)
(3)
and
~
B
0
⊥
=
γ
(
~
B
⊥

~
β
×
~
E
)
,
(4)
where
γ
is the usually relativistic “gamma factor”,
γ
≡
(1

β
2
)

1
/
2
. We can,
of course, also the transformations from the primed frame
K
0
to the unprimed
frame
K
, by switching primed and unprimed quantities and changing the sign
of
~
β
:
~
E
⊥
=
γ
(
~
E
0
⊥

~
β
×
~
B
0
)
(5)
and
~
B
⊥
=
γ
(
~
B
0
⊥
+
~
β
×
~
E
0
)
.
(6)
If there exists a single charged particle that moves with a constant velocity
(as seen from an inertial frame of reference), transforming its electromagnetic
field between one inertial frame and another is a fairly straightforward task,
outlined in section 4.6 of the textbook.
One interesting application is a
charged particle which is moving past you (an inertial observer) on a straight
line at a constant speed. The relative speed of you and the charged particle
is
v
=
βc
.
The distance of closest approach between you and the particle
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(the
impact parameter
) is
b
. Choose a coordinate system so that you, in the
K
frame of reference, are at
~
r
=
b
ˆ
e
y
, and the particle is moving along the
x
axis, with
~
r
=
vt
ˆ
e
x
.
(Thus, I am choosing the moment
t
= 0 to be the
moment of closest approach between you and the charged particle.) In the
K
0
frame of reference, in which the particle is at rest, the
~
B
0
field is zero,
and the
~
E
0
field is a simple inverse square relation:
~
E
0
=
q~
r
0
(
r
0
)
3
.
(7)
In the
K
frame of reference, in which you are at rest, the
~
B
field is no longer
zero and the
vecE
field is no longer isotropic around the particle.
In the
limit of highly relativistic motion (
β
≈
1,
γ
1), the nonzero components
of the electromagnetic field are
E
x
≈

qγct
(
γ
2
c
2
t
2
+
b
2
)
3
/
2
(8)
E
y
≈
qγb
(
γ
2
c
2
t
2
+
b
2
)
3
/
2
(9)
B
z
≈
E
y
.
(10)
The
E
field is maximized when
t
= 0, and the point charge is at its closest
approach. At this instant,
E
y
≈
B
z
≈
γ
q
b
2
,
(11)
larger by a factor of
γ
than the value of
E
y
you would measure in the non
relativistic case. On the other hand,
E
y
only has a high value when
γc

t
 ¿
b
,
or

t
 ¿
1
γ
b
c
.
(12)
In the more leisurely nonrelativistic case,
E
y
has a high value when
v

t
 ¿
b
,
or

t
 ¿
b
v
∼
1
β
b
c
.
(13)
In the highly relativistic case,
γ
1, and the encounter time is brief.
In
the nonrelativistic case,
β
¿
1, and the encounter time is long. A highly
relativistic charged particle moving past you will cause a much higher electric
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 Fall '05
 RYDEN
 Space, Electron, Special Relativity, Electric charge, inertial frame, instantaneous rest frame

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