Deriving Helmholtz eqn
Assume a material with a lossy, linear, isotropic, homogeneous charge free dielectric
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Lossy - 𝜎 ≠ 0
Linear - 𝑃𝑜𝑙𝑎𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝛼 𝐸 𝑓𝑖𝑒𝑙𝑑 so if I doubled the E field the polarization density also doubles thus
the linear relationship
Isotropic - Behaves the same in every direction
Homogeneous - Material properties are constant throughout the material so 𝜀 𝑎𝑛𝑑 𝜇 remain constant
throughout the material
Charge free - 𝜌𝑣 = 0
Now imagine a sinusoidal excitation to our material such that:
𝐸(𝑡) = Re[𝐸0 𝑒 𝑗𝜔𝑡 ]
𝐵(𝑡) = Re[𝐵0 𝑒 𝑗𝜔𝑡 ]
Going to maxwell’s 3rd equation we would get:
𝛻×𝐸 =−
𝜕𝐵
𝜕𝑡
But since we know what B is we can simplify this expression:
𝜕𝐵
𝜕
= 𝐵0 (𝑒 𝑗𝜔𝑡 ) = 𝑗𝜔(𝐵0 𝑒 𝑗𝜔𝑡 ) = 𝑗𝜔𝐵
𝜕𝑡
𝜕𝑡
Hence:
𝛻 × 𝐸 = −𝑗𝜔𝐵
Taking the del cross of each side:
𝛻 × (𝛻 × 𝐸) = 𝛻 × (−𝑗𝜔𝐵)
𝛻 × (𝛻 × 𝐸) = −𝑗𝜔(𝛻 × 𝐵)
Use this identity 𝛻 × (𝛻 × 𝑉) = 𝛻(𝛻 ⋅ 𝑉) − 𝛻 2 𝑉 where V is a vector
𝛻(𝛻 ⋅ 𝐸) − 𝛻 2 𝐸 = −𝑗𝜔(𝛻 × 𝐵)
But 𝛻(𝛻 ⋅ 𝐸) = 0 since we are in a charge free zone (𝜌𝑣 = 0) and maxwell’s first equation tells us how this
equates to zero (𝛻 ⋅ 𝐸 = 𝜌𝑣 = 0 → 𝛻(0) = 0)
𝛻 2 𝐸 = 𝑗𝜔(𝛻 × 𝐵)
Now coming to maxwell’s fourth equation in phasor form:
𝛻 × 𝐵 = 𝐽 + 𝑗𝜔𝜀𝐸 = 𝜎𝐸 + 𝑗𝜔𝜀𝐸 = 𝐸(𝜎 + 𝑗𝜔𝜀)
Replacing it gives:
𝛻 2 𝐸 = 𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀)𝐸 where 𝐽 = 𝜎𝐸 ≠ 0
So that our constant 𝛾 2 = 𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀) because if we opened the brackets, we would get order 2 terms
therefore using 𝛾 2 would make it easier to analyze
𝛻2𝐸 = 𝛾2 𝐸 → 𝛻2𝐻 = 𝛾2 𝐻
If the dielectric was lossless then 𝛾 2 = 𝜔2 𝜇𝜀𝐸 since 𝐽 = 0 (𝑛𝑜 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛)