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Laplace's and Poisson's Equations
---------------------------------
Maxwells Equations:
∇ x E = -∂B/∂t
∇.E = -ρ/ε_{0}
∇.B = 0
∇ x B = μ_{0}J + μ_{0}ε_{0}∂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/ε_{0} becomes
∇.(∇φ) = -ρ/ε_{0}
∇^{2}φ = -ρ/ε_{0}
This is Poisson's equation. It has the solution,
φ = (1/4πε_{0})∫(ρ/r)dv where v is the volume
In a charge free space
∇^{2}φ = 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 = μ_{0}J
∇ x (∇ x A) = μ_{0}J
∇(∇.A) - ∇^{2}A = μ_{0}J
We can simplify this equation by 'fixing' the gauge. If we set ∇.A = 0
(Coulomb gauge) we get,
∇^{2}A = -μ_{0}J
Comparing this to the form of the Poisson equation, leads to the following
solution,
A = (μ_{0}/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)
∇^{2}φ = -ρ/ε_{0} - ∂(∇.A)/∂t
Magnetic Field:
∇ x B = μ_{0}J + μ_{0}ε_{0}∂E/∂t
∇ x (∇ x A) = μ_{0}J + μ_{0}ε_{0}∂(-∇φ - ∂A/∂t)/∂t
∇^{2}A = μ_{0}J - μ_{0}ε_{0}∇(∂φ/∂t) - μ_{0}ε_{0}∂^{2}A/∂t^{2}
∇^{2}A - μ_{0}ε_{0}∂^{2}A/∂t^{2} - ∇(∇.A + μ_{0}ε_{0}∂φ/∂t) = -μ_{0}J
We can simplify this equation by 'fixing' the gauge. If we set
∇.A + μ_{0}ε_{0}∂φ/∂t = 0 (Lorentz gauge) we get,
∇^{2}A - μ_{0}ε_{0}∂^{2}A/∂t^{2} = -μ_{0}J
We can also apply the same Lorentz gauge condition to,
∇^{2}φ = -ρ/ε_{0} - ∂(∇.A)/∂t that we derived above. We get,
∇^{2}φ = -ρ/ε_{0} - ∂(-μ_{0}ε_{0}∂φ/∂t)/∂t
∇^{2}φ = -ρ/ε_{0} + μ_{0}ε_{0}∂^{2}φ/∂t^{2}
In summary,
∇^{2}A - μ_{0}ε_{0}∂^{2}A/∂t^{2} = -μ_{0}J
and
∇^{2}φ = -ρ/ε_{0} + μ_{0}ε_{0}∂^{2}φ/∂t^{2}
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 + ∇^{2}λ
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 + ∇^{2}λ
If we choose λ such that ∇^{2}λ = -∇.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 + μ_{0}ε_{0}∂φ/∂t = 0
∇.A' + μ_{0}ε_{0}∂φ'/∂t = ∇.A + ∇^{2}λ + μ_{0}ε_{0}∂φ/∂t - μ_{0}ε_{0}∂^{2}λ/∂t^{2} = 0
∴ ∇^{2}λ - μ_{0}ε_{0}∂^{2}λ/∂t^{2} = ∇.A + μ_{0}ε_{0}∂φ/∂t
So as long as we choose λ that meets the requirement that,
∇^{2}λ - μ_{0}ε_{0}∂^{2}λ/∂t^{2} = 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