Wolfram Alpha:

```Laplace's and Poisson's Equations
---------------------------------

Maxwells Equations:

∇ x E = -∂B/∂t

∇.E = -ρ/ε0

∇.B = 0

∇ x B = μ0J + μ0ε0∂E/∂t

Static Case (electrostatics)
----------------------------

Electric Field:

∇ x E = 0

Using the identity that ∇ x (-∇φ) = 0 we can imply that,

E = -∇φ where φ is the ELECTRIC or SCALAR POTENTIAL

Therefore, ∇.E = ρ/ε0 becomes

∇.(∇φ) = -ρ/ε0

∇2φ = -ρ/ε0

This is Poisson's equation.  It has the solution,

φ = (1/4πε0)∫(ρ/r)dv where v is the volume

In a charge free space

∇2φ = 0

This is Laplace's equation.

Magnetic Field:

∇.B = 0

Using the identity that ∇.(∇ x A) = 0 we can imply that B = ∇ x A
where we define A as the MAGNETIC or VECTOR POTENTIAL.

∇ x B = μ0J

∇ x (∇ x A) =  μ0J

∇(∇.A) - ∇2A = μ0J

We can simplify this equation by 'fixing' the gauge. If we set ∇.A = 0
(Coulomb gauge) we get,

∇2A = -μ0J

Comparing this to the form of the Poisson equation, leads to the following
solution,

A = (μ0/4π)∫(J/r)dv

Dynamic Case (electrodynamics)
------------------------------

Electric Field:

∇ x E = -∂B/∂t

= -∂(∇ x A)/∂t

∇ x E + ∂(∇ x A)/∂t = 0

Which can be written as,

∇ x (E + ∂A/∂t) = 0

Using the identity that ∇ x (∇φ) = 0 we can imply that

∇φ ≡ (E + ∂A/∂t)

Therefore,

E = -∇φ - ∂A/∂t

or

∇φ = -E - ∂A/∂t

∇.(∇φ) = -∇.E - ∇.(∂A/∂t)

∇2φ = -ρ/ε0 - ∂(∇.A)/∂t

Magnetic Field:

∇ x B = μ0J + μ0ε0∂E/∂t

∇ x (∇ x A) = μ0J + μ0ε0∂(-∇φ - ∂A/∂t)/∂t

∇2A = μ0J - μ0ε0∇(∂φ/∂t) - μ0ε0∂2A/∂t2

∇2A - μ0ε0∂2A/∂t2 - ∇(∇.A + μ0ε0∂φ/∂t) = -μ0J

We can simplify this equation by 'fixing' the gauge. If we set
∇.A + μ0ε0∂φ/∂t = 0 (Lorentz gauge) we get,

∇2A - μ0ε0∂2A/∂t2 = -μ0J

We can also apply the same Lorentz gauge condition to,

∇2φ = -ρ/ε0 - ∂(∇.A)/∂t that we derived above.  We get,

∇2φ = -ρ/ε0 - ∂(-μ0ε0∂φ/∂t)/∂t

∇2φ = -ρ/ε0 + μ0ε0∂2φ/∂t2

In summary,

∇2A - μ0ε0∂2A/∂t2 = -μ0J

and

∇2φ = -ρ/ε0 + μ0ε0∂2φ/∂t2

are simpler and more symmetric forms of Maxwell's equations and embody everything in
electrodynamics.  Notice that they reduce to the familiar static equations when A and φ
do not depend on time.

Gauge Transformations
---------------------
The process changing φ and A in such a way that E and B do not change is
called a gauge transformation. Gauge transformations have been used to make
calculations 'easier' in electromagnetic theory. Consider,

φ' = φ + constant

∇φ' = ∇φ

and

A' = A + ∇λ

∇ x A' = ∇ x A + ∇ x (∇λ)

∇ x A' = ∇ x A

∇.A' = ∇.A + ∇.(∇λ)

= ∇.A + ∇2λ

Similarly,

E = -∇φ - ∂A/∂t

= -∇φ - ∂(A + ∇λ)/∂t

= -∇φ - ∂A/∂t - ∇∂λ/∂t

= -∇(φ + ∂λ/∂t) - ∂A/∂t

which changes E!  The only way does not change is if φ -> φ - ∂λ/∂t

In summary,

Gauge transformation for the scalar potential: φ' = φ - ∂λ/∂t

Gauge transformation for the vector potential: A' = A + ∇λ

Coulomb Gauge
-------------
From above,

∇.A' = ∇.A + ∇2λ

If we choose λ such that ∇2λ = -∇.A then,

making ∇.A' = 0

This is referred to as the COULOMB GAUGE.  The Coulomb gauge is particularly useful
in electrostatics because of the simple relations it leads to for A and φ

Lorentz Gauge
-------------

∇.A + μ0ε0∂φ/∂t = 0

∇.A' + μ0ε0∂φ'/∂t = ∇.A + ∇2λ + μ0ε0∂φ/∂t - μ0ε0∂2λ/∂t2 = 0

∴ ∇2λ - μ0ε0∂2λ/∂t2 = ∇.A + μ0ε0∂φ/∂t

So as long as we choose λ that meets the requirement that,

∇2λ - μ0ε0∂2λ/∂t2 = 0

the Lorentz requirement will be satisfied.

The Lorentz gauge is particularly usefull in electrodynamics because of the simple
relations it leads to for A and φ
```