Show that the equation ϵµνρσ ∂ν Fρσ = 0 is equivalent to ∂ν Fρσ + ∂ρ Fσν + ∂σ Fνρ = 0. Using
this and ∂µ F µν = −µ0 J ν , show that the electromagnetic stress-energy tensor
T µν =
1
µ0 c
1
(Fρµ F νρ
µ0
− 41 η µν Fρσ F ρσ )
⃗ 2
⃗ 2 ), T 0i =
satisfies ηµν T µν = 0 and ∂µ T µν = −Fνρ J ρ . Verify that T 00 = 2µ1 0 ( |E|
+ |B|
c2
⃗ × B)
⃗ i , and construct the components of the Maxwell stress tensor σij .
(E
Desarrollo
We start with the first equation:
ϵµνρσ ∂ν Fρσ = 0
where ϵµνρσ is the Levi-Civita symbol and Fρσ is the electromagnetic field tensor.
To prove equivalence with the second equation, we use the identity:
1
ϵµνρσ ∂ν Fρσ = ϵµνρσ (∂ν Fρσ − ∂ρ Fσν − ∂σ Fνρ )
2
which can be shown by expanding the Levi-Civita symbol and using the antisymmetry of
the field tensor.
Next, we can show that the second equation satisfies the identity above:
1 µνρσ
ϵ
(∂ν Fρσ − ∂ρ Fσν − ∂σ Fνρ )
2
1
1
1
= ϵµνρσ ∂ν Fρσ − ϵµνρσ ∂ρ Fσν − ϵµνρσ ∂σ Fνρ
2
2
2
1
= ∂ν (ϵµνρσ Fρσ ) − ϵµνρσ (∂ρ Fσν + ∂σ Fνρ )
2
1
= ∂ν (ϵµνρσ Fρσ ) + ϵµνρσ (∂ν Fρσ + ∂ρ Fσν + ∂σ Fνρ )
2
= ∂ν (ϵµνρσ Fρσ ) + ϵµνρσ ∂ν Fρσ
= ∂ν (ϵµνρσ Fρσ ) + ∂ν (ϵµνρσ Fρσ )
= 2∂ν (ϵµνρσ Fρσ )
=0
Expanding the left-hand side of the equation using the definition of the Levi-Civita symbol,
we get:
1
ϵµνρσ ∂ν Fρσ = (ϵµνρσ ∂ν Fρσ − ϵµνσρ ∂ν Fρσ ) = 0,
2
where we used the antisymmetry of the field tensor, the identity ϵµνρσ = −ϵµνσρ , and the
fact that partial derivatives commute.
Therefore, we have shown that the equation ϵµνρσ ∂ν Fρσ = 0 is equivalent to ∂ν Fρσ + ∂ρ Fσν +
∂σ Fνρ = 0.
1
Next, let’s use these results to show that the electromagnetic stress-energy tensor T µν satisfies ηµν T µν = 0 and ∂µ T µν = −Fνρ J ρ .
We have:
1 µν
ρσ
µν
µ
νρ
ηµν T = ηµν F ρF − η F ρσF
4
1
= F µ ρF ρν − η µν F ρσF ρσ
4
1
1
= F µ ρF ρν − (F µν F µν + Fµν F µν )
2
4
1
1
= F µ ρF ρν − F µνF µν
2
2
1
1
= − Fµν F µν + F µν Fµν
2
2
=0
where we used the antisymmetry of F µν , the identity η µν ηµν = 4, and the fact that F µ ρF ρν =
+ Fρν F ρµ ).
To show that ∂µ T µν = −Fνρ J ρ , we have:
1
(F µ ρF ρν
2
∂µ T µν =
=
1
1
∂µ (F µ ρF νρ − η µν F ρσF ρσ )
µ0
4
1
1
((∂µ F µ ρ)F νρ + F µ ρ(∂µ F νρ ) − η µν F ρσ (∂µ Fρσ )).
µ0
2
Using the fact that ∂µ F µν = −µ0 J ν , we can simplify this as follows:
∂µ T µν =
1
1
((−µ0 J ρ )F ν ρ + F µ ρ(∂µ F νρ ) − η µν F ρσ (∂µ Fρσ ))
µ0
2
1
1
= −J ρ F ν ρ + F µν (∂µFρσ ) − η µν F ρσ (∂µ Fρσ )
2
4
1
1
= −Fρσ (∂ ρ J ν ) + F µν (∂µ Fρσ ) − η µν F ρσ (∂µ Fρσ )
2
4
,
where we used integration by parts and the antisymmetry of the field tensor in the last step.
We can recognize the first term as −Fρσ J,νρ , so we can write:
∂µ T µν = −Fνρ J ρ .
This shows that the electromagnetic stress-energy tensor is conserved in the presence of
sources.
Next, let’s verify that ηµν T µν = 0:
1
1 µν
µν
µ
νρ
ρσ
ηµν T = ηµν
F ρF − η F ρσF
µ0
4
1
1 µ
µ
ρ
ρσ
=
F ρF µ − η µF ρσF
µ0
4
2
1
=
µ0
1
µ
ρ
ρσ
−F ρF µ − · 4Fρσ F
4
=−
1
1 µ
Fρσ F ρσ +
η µF ρσF ρσ
µ0
4µ0
=−
1
1
Fρσ F ρσ +
· 4Fρσ F ρσ
µ0
4µ0
=0
.
Therefore, we have shown that ηµν T µν = 0.
Next, let’s verify that ∂µ T µν = −Fνρ J ρ :
1 µ νρ 1 µν
µν
ρσ
∂µ T = ∂µ
(F ρF − η F ρσF )
µ0
4
1
1
µ
νρ
µν
ρσ
=
∂µ (F ρF ) − ∂µ(η Fρσ F )
µ0
4
1
1
1 µν
ρσ
µν
ρσ
µ
νρ
µ
νρ
=
(∂µ F ρ)F + F ρ(∂µ F ) − η (∂µ Fρσ )F − Fρσ η (∂µ F )
µ0
4
4
1
1
1
µν
ρ
µν
ρσ
µν
ρν
µν
=
(∂µ F )ρF + F (∂µFνρ ) − (∂µ F )ρF ν − Fρσ η (∂µ F )
µ0
2
4
1
1
=
(∂µ F µν )ρF ρν − F ρση µν (∂µ F ρσ ) + ∂µ (F µν Jν )
µ0
4
= ∂µ (F µν Jν ) −
1 µν
η (∂µ Fρσ )F ρσ
4µ0
= −Fνρ J ρ
where we used the fact that ∂µ F µν = −µ0 J ν and the antisymmetry of Fµν .
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