QuantumPhysics.dvi

For the problems of radiation which we shall be interested inhere, one particularly convenient
choice is thetransverse or radiation gauge,

∇·A= 0 (19.6)

Given fieldsA 0 andA, the explicit gauge transformation Λ that transform∇·Aobeys the following
relation,∇·A−∆Λ = 0, whose general solution is given by,

Λ(t,x) =

∫

d^3 x′

1

4 π|x′−x|

∇·A(t,x′) (19.7)

where we have used the basic Coulomb equation, ∆(1/|x|) =− 4 πδ(3)(x). This gauge transformation
is then to be carried out on bothA 0 andA. In transverse gauge, Gauss’s law, namely∇·E=ρ/ε 0
allow one to solve for the electric potentialA 0 , and we find,

A 0 (t,x) =−

∫

d^3 x′

1

4 πε 0 |x′−x|

ρ(t,x′) (19.8)

Since, in the transverse gauge, the fieldA 0 , and its time-dependence, are completely determined by
the external sourceρ, the fieldA 0 hasno dynamics. The fieldAmay be solved for in an analogous
fashion. ExpressingEandBin terms ofA 0 andA, using (19.4), the general relation

∇×(∇×A) =−∆A−∇(∇·A) (19.9)

as well as the transverse gauge condition (19.6), we find thatAobeys the following wave-like
equation,

1

c^2

∂t^2 A−∆A=μ 0 j−

1

c^2

∂t∇A 0 (19.10)

whereA 0 is now viewed as given in terms ofρby (19.8). Because in∂t∇A 0 , only thet-derivative
ofρenters, we are free to use the charge conservation equation (19.3), and replace∂tρby−∇·j.
The result may be expressed as follows,

1

c^2

∂t^2 A−∆A=μ 0 j⊥ (19.11)

wherej⊥is given by

j⊥(t,x) =j(t,x)−∇

∫

d^3 x′

1

4 π|x′−x|

∇·j(t,x′) (19.12)

Since Maxwell’s equations, for given external sourcesρandjare linear, they may be solved explic-
itly.

The LagrangianLand HamiltonianHare given as integrals over the space of the Lagrangian
densityLand Hamiltonian densityHrespectively as follows,

L=

∫

d^3 xL L=

1

2

ε 0 E^2 −

1

2 μ 0

B^2 +ρA 0 +A·j

H=

∫

d^3 xH H=

1

2

ε 0 E^2 +

1

2 μ 0

B^2 −ρA 0 −A·j (19.13)