Download Chapter 1 Lagrangian field theory II

Chapter 1
Lagrangian field theory II
In the previuous chapter we showed that the Lagrangian for a complex scalar field is invariant
under global phase transformations i.e. transformations of the form
Φ → eıα Φ; ,
Φ∗ → e−ıα Φ∗ ,
(1.1)
(1.2)
where α is a constant phase, i. e. independent of spacetime coordinates. This symmetry
implies the conservation of particle number or conservation of charge.
Let us now generalize the transformations to local phase transformations, i.e. transformations where α = α(x). The Lagrangian then transforms as
L → L − iφ∗ (∂ µ φ) (∂µ α) + iφ (∂µ φ∗ ) (∂µ α) + Φ∗ Φ (∂µ α) (∂ µ α) .
(1.3)
The Lagrangian is therefore not invariant under local phase transformations. However, this
can be remidied by introducing another field Aµ , called a gauge field. The Lagrangian must
be modified and is given by
L = (∂µ − iqAµ Φ∗ ) (∂ µ + iqAµ Φ) − m2 Φ∗ Φ ,
(1.4)
where e is a constant (the charge) and Dµ = ∂µ − ieAµ is called the covariant derivative.
The field Aµ must transform as
1
Aµ → Aµ − ∂µ α .
q
(1.5)
The transformation (1.5) is called a gauge transformation. The Lagrangian is invariant under
the simultaneous transformation (1.5) and
Φ → eıα(x) Φ; ,
Φ∗ → e−ıα(x) Φ∗ .
1
(1.6)
(1.7)
This can be seen by inserting the transformed field into the Lagrangian:
L = (∂µ − iqAµ Φ∗ ) (∂ µ + iqAµ Φ) − m2 Φ∗ Φ
0
→ L
"
#"
#
1
1
−iα(x) ∗
iα(x) ∗
= ∂µ − iq(Aµ − ∂µ α)e
Φ
∂µ + iq(Aµ − ∂µ α)e
Φ
q
q
= L
(1.8)
Electromagnetism
We next consider the Maxwell field. The electric and magnetic fields can be expressed in
~ as
terms of the gauge field Aµ = (φ, A)
~
~ = −∇Φ − ∂ A ,
E
∂t
~
~
B = ∇×A.
(1.9)
(1.10)
~ and B
~ are gauge invariant, i.e. invariant under the
We recall that the physical fields E
µ
transformation (1.5) of 4-vector field A . The Lagrangian we are interested in must give
rise to Maxwell equations as field equations. The Lagrangian can not simply be a found by
replacing φ by Aµ in the Lagrangian for the scalar field. Doing so, yields 1
L =
1
1
(∂µ Aν ) (∂ µ Aν ) − m2 Aµ Aµ .
2
2
(1.11)
The problem is that none of these terms are gauge invariant. In particular the mass term is
not gauge invariant and this explains why the photon is massless: Maxwell theory is gauge
theory and gauge invariance forbids a mass term in the Lagrangian. However, the field
strength F µν defined by
F µν = ∂ µ Aν − ∂ ν Aµ ,
(1.12)
is gauge invariant (show it!). The field strength is an antisymmetric tensor of rank two. Now
recall that a Lagrangian must be a Lorentz scalar and therefore we must contract the two
indices upstairs. We can contract it with Fµν and try
1
L = − Fµν F µν .
4
(1.13)
This Lagrangian is Lorentz invariant and gauge invariant and is gives rise to Maxwell’s
equation in the vacuum. If there are charges and currents present, we must generalize the
1
~
Note that Aµ = (Φ, −A).
result. The charge density ρ and current density ~j can be written as a four vector j µ . We
therefore add the term jµ Aµ to the Lagrangian
1
L = − Fµν F µν − jµ Aµ
4
1
~.
= − Fµν F µν − ρΦ + ~j · A
4
(1.14)
The field strength tensor reads




F µν = 
0 −Ex −Ey −Ez
Ex
0
−Bz By 


Ey Bz
0
−Bx 
Ez −By Bx
0

(1.15)
For example
∂A1 ∂A0
−
∂x0
∂x1
1
∂A
+ (∇Φ)x
=
∂t
= −Ex .
F 01 =
Similarly, we find
F
12
∂A1 ∂A2
=
−
∂x2
∂x1
∂Ay ∂Ax
~ z
=
−
(∇ × A)
∂x
∂y
= Bz .
(1.16)
Maxwell’s equations in the vacuum can then be written as
∂µ F µν = 0 .
(1.17)
~ it is straightforward to express Fµν in terms of E
~ and
Using the fact that Aµ = (φ, −A)
~ It can then be shown that
B.
~2 − B
~ 2) .
F µν Fµν = −2(E
(1.18)
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Exercises
The dual tensor F̃ µν is defined by
F̃ µν =
1 µναβ
ε
Fαβ ,
2
(1.19)
where εµναβ is the Levi-Civita symbol. εµναβ is totally antisymmetric under permutation of
indiced and ε0123 = +1. Find the matrix F̃ µν and show that
~ ·B
~ .
F̃ µν Fµν = −4E
(1.20)
Finally, find an expression for F̃ µν F̃µν .
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