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Field Theory
Answer for 5th Set of Problems
SPECIAL THEORY OF RELATIVITY
1. (a) The field strength tensors are given by

0
Ex
Ey
 −Ex
0
−B
z
Fαβ = 
 −Ey Bz
0
−Ez −By Bx


Ez
By 

−Bx 
0
while their raised index counterparts are

0 −Ex −Ey −Ez
 Ex
0
−Bz By
αβ
F =
 Ey Bz
0
−Bx
Ez −By Bx
0
and Fαβ
0
 −Bx
=
 −By
−Bz


and F αβ



0
 Bx
=
 By
Bz
Bx
0
−Ez
Ey
−Bx
0
−Ez
Ey
By
Ez
0
−Ex
−By
Ez
0
−Ex

Bz
−Ey 
,
Ex 
0

−Bz
−Ey 
.
Ex 
0
Thus
F αβ Fαβ
~2 − B
~2 ,
= −2 E
~2 − B
~2 ,
= 2 E
F αβ Fαβ
~ · B.
~
= −4E
F αβ Fαβ
~2 − B
~ 2 must be the same in all frames. On the
(b) The value of the scalar invariant F αβ Fαβ = −2 E
other hand, a purely electric field would lead to F αβ Fαβ < 0, while a purely magnetic field would yield
F αβ Fαβ > 0. Obviously, only one of these conditions can be satisfied at the same time. Thus it is impossible
to have an electromagnetic field that is purely electric in on frame and purely magnetic in another.
(c) If there is an inertial frame in which there is no electric field, we would have
~2 − B
~ 2 = 2B
~ 2 ≥ 0,
F αβ Fαβ = −2 E
F αβ Fαβ
~ ·B
~ = 0,
= −4E
~ and B
~ must satisfy the conditions
Hence in any inertial frame, E
~2 ≤ B
~2
E
~ ·B
~ = 0.
and E
(d) The macroscopic field strength tensor Gαβ is given by

0
Dx
 −Dx
0
Gαβ = 
 −Dy Hz
−Dz −Hy
Dy
−Hz
0
Hx

Dz
Hy 
.
−Hx 
0
This leads to invariants quadratic in Gαβ
G αβ Gαβ
~ 2 − H~ 2 ,
= −2 D
~2 −H
~2 ,
= 2 D
G αβ Gαβ
~ · H,
~
= −4D
Gαβ Gαβ
1
Combining F and G yields the invariants
F αβ Gαβ
F αβ Gαβ
F αβ Gαβ
F αβ Gαβ
~ ·D
~ −B
~ ·H
~ ,
−2 E
~ ·D
~ −B
~ ·H
~ ,
= 2 E
~ ·H
~ +B
~ ·D
~ ,
= −2 E
~ ·H
~ +B
~ ·D
~ .
= −2 E
=
2. (a) The rest frame is defined as the frame where the 4-velocity has the form U 0µ = (c, 0). This allows us to
express
0µν 0
~0 .
the electric field as a contraction of the Maxwell field strength tensor with the 4-velocity: F Uν = 0, cE
Thus we obtain the relation
~ 0 = σ F 0µν Uν0 .
0, σ E
c
(1)
Next we express the 3-current vector J~ in terms of the 4-current J µ . Considering that J 0ν Uν0 = c2 ρ0 , we
can write
1
0, J~0 = J 0µ − 2 (J 0ν Uν0 ) U 0µ .
(2)
c
~ 0 , we find that the covariant form
Thus by substituting the equations (1) and (2) into Ohm’s law J~0 = σ E
of Ohm’s law is given by
σ
1
(3)
J µ − 2 (J ν Uν ) U µ = F µν Uν ,
c
c
where we have dropped all primes, since this equation is in covariant form, i.e. frame independent.
(b) Working in the lab frame, we write
U µ = (γc, γ~v ) ,
J µ = cρ, J~ .
Substituting this into equation (3), the time component of equation (3) becomes
~
−v 2 γ 2 ρ + γ 2~v · J~ = σγ~v · E,
(4)
~ 2 . The space components of equation (3) become
where we use the relation J ν Uν /c2 = γ ρ − ~v · J/c
γ 2 ~
1
2
~
~
~
J − γ ρ~v + 2 ~v ~v · J = σγ E + ~v × B .
c
c
(5)
Thus solving equation (4) for ~v · J~ and substituting this into equation (5) gives
h
i
~·E
~ + β~ × B
~ − β~ β
~ + ρ~v .
J~ = σγ E
(c) If the medium is uncharged in its rest frame, the 4-current must satisfy the relation J ν Uν = 0. In this case,
the covariant version (3) of Ohm’s law reduces to
Jµ =
σ µν
F Uν .
c
Replacing J µ = cρ, J~ and U µ = (cγ, ~v γ) in the equation above directly results in
ρ=
σγ ~ ~ β·E
c
2
~ + β~ × B
~ .
J~ = σγ E
3. (a) Since the particle is moving along the ~ez direction, we can write β~ = (v/c)~ez . In this case, the fields in the
rest frame are given by
0
~ 0 = 0.
~ 0 = q~r ,
B
E
r03
On the other hand, the fields in the lab frame can be described by
2
~ × E.
~ =β
~
~ = γE
~ 0 − γ β~ β~ · E
~0 ,
B
E
γ+1
If we now consider the explicit boost transformation given by
x0 = x,
y 0 = y,
z 0 = γ(z − vt),
we obtain the relation ~r0 = ~r⊥ + γ(z − vt)~ez . As a result, the electric field in the lab frame is given by
~ = γq (~r⊥ + (z − vt)~ez ) ,
E
2 + γ 2 (z − vt)2 )3/2
(r⊥
which is an exact expression for any value of β. If we use the fact that
lim
γ→∞
(A2
γ
1
ξ
2
= 2 lim
= 2 δ(B),
2
2
3/2
2
2
3/2
A ξ→∞ (1 + ξ B )
A
+B γ )
then taking A = r⊥ and B = z − vt results in
~ = 2q (~r⊥ + (z − vt)~ez ) δ(z − vt) = 2q~r⊥ δ(z − vt).
E
2
2
r⊥
r⊥
By taking the limit v → c, we finally arrive at
~ = 2q ~r⊥ δ(z − ct),
E
2
r⊥
ˆ
~ = β~ × E
~ = 2q ~v × ~r⊥ δ(z − ct).
B
2
r⊥
(b) The time component of the Maxwell equations, ∂µ F µν = (4π/c)J ν , gives
4π 0 ~ ~
~ ⊥ · ~r⊥ δ(z − ct) = 4πqδ (2) (~r⊥ )δ(z − ct),
J = ∇ · E = 2q ∇
2
c
r⊥
(6)
where we use the following representation of the 2-dimensional delta-function
~r⊥
~
∇⊥ ·
= 2πδ (2) (~r⊥ ).
2
r⊥
On the other hand, the space components of Maxwell equations give
~
1 ∂E
~ ×B
~
+∇
c ∂t
~r⊥
~ × ~ez × ~r⊥ δ(z − ct)
= 2q 2 δ 0 (z − ct) + 2q ∇
r⊥
r2
⊥ ~r ~r⊥
~r⊥
~r⊥
⊥
0
~
~
δ
(z
−
ct)
+
2q
~
e
∇
·
−
~
e
·
∇
δ(z − ct)
= 2q 2 δ 0 (z − ct) + 2q~ez × ~ez ×
z
z
2
2
2
r⊥
r⊥
r⊥
r⊥
4π ~
J =
c
=
=
−
4πq~ez δ (2) (~r⊥ )δ(z − ct),
4πq~vˆδ (2) (~r⊥ )δ(z − ct),
(7)
~ = ∂z and ~vˆ = ~ez . Thus equations (6) and (7) give
where we use the fact that ~ez · ∇
J µ = cqv µ δ (2) (~r⊥ )δ(z − ct),
where v µ = (1, ~vˆ).
3
~ ⊥ = 0, we have
(c) For the first case where A
~
E
=
~ 0−
−∇A
~
1 ∂A
c ∂t
2q ∂
~ez [δ(z − ct) ln (λr⊥ )]
c ∂t
~ ⊥ ln (λr⊥ ) + 2q~ez δ 0 (z − ct) ln (λr⊥ ) − 2q~ez δ 0 (z − ct) ln (λr⊥ )
= 2qδ(z − ct)∇
~r⊥
= 2qδ(z − ct) 2
r⊥
=
~ [δ(z − ct) ln (λr⊥ )] +
2q ∇
and
~
B
~ ×A
~
= ∇
~
= −2q ∇ × [~ez δ(z − ct) ln (λr⊥ )]
~ ⊥ × [δ(z − ct) ln (λr⊥ ) ~ez ]
= −2q ∇
~r⊥
= −2q 2 × ~ez δ(z − ct)
r⊥
ˆ
~v × ~r⊥
= 2q
δ(z − ct),
2
r⊥
which agrees with the fields specified in part (a).
We now consider the second case for which A0 = 0 = Az . Since
~ ⊥ = −2qΘ(ct − z)∇
~ ⊥ ln (λr⊥ ) = −2qΘ(ct − z) ~r⊥ ,
A
2
r⊥
we have
~
~ = − 1 ∂ A = 2qδ(ct − z) ~r⊥
E
2
c ∂t
r⊥
and
~ =∇
~ ×A
~
B
~r⊥
~
= −2q ∇ × Θ(ct − z) 2
r⊥
~r⊥
~r⊥
~
= 2qδ(ct − z)~ez × 2 − 2qΘ(ct − z)∇ ×
2
r⊥
r⊥
~vˆ × ~r⊥
δ(ct − z),
= 2q
2
r⊥
2
~ ln (λr⊥ ) is curl-free. Thus this result also agrees with
where we use the fact that the function ~r⊥ /r⊥
=∇
the fields specified in part (a).
Since the above two vector potentials are necessarily related by a gauge transformation χ through
A0µ = Aµ − ∂ µ χ,
we find that
∂ µ χ = Aµ − A0µ
~ ⊥ ln (λr⊥ ) − δ(ct − z) ln (λr⊥ ) ~ez
= 2q −δ(ct − z) ln (λr⊥ ) , Θ(ct − z)∇
1 ∂
∂
~
= 2q −
Θ(ct − z) ln (λr⊥ ) , Θ(ct − z)∇⊥ ln (λr⊥ ) + ~ez Θ(ct − z) ln (λr⊥ )
c ∂t
∂z
∂
~ Θ(ct − z) ln (λr⊥ ) .
= −2q
, −∇
∂x0
Thus the gauge function χ is
χ = −2qΘ(ct − z) ln (λr⊥ ) .
4