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Relativistic trajectory in a constant null crossed field
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Electromagnetism
Relativistic Lorentz force
Null crossed electromagnetic field
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B
=
B
e
z
,
E
=
c
B
e
x
,
Ω
=
m
qB
.
(9)
A particle released from rest at the origin has the proper-time trajectory
t
(
τ
)
=
τ
+
6
Ω
2
τ
3
,
x
(
τ
)
=
2
c
Ω
τ
2
,
y
(
τ
)
=
−
6
c
Ω
2
τ
3
,
z
(
τ
)
=
0.
(10)
Ancestors
(6)
Null crossed electromagnetic field
Relativistic Lorentz force
Electromagnetism
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Physics
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(1)
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