# Derivation of the wave equation satisfied by E and B

## Homework Statement

given a medium in which p=0, j=0 but where the polarization vector P=P(r,t). Derive the wave equation satisfied by E and B.

## Homework Equations

i started with the 4 basic Maxwells equations
∇ · D = ρ (1)
∇ · B = 0 (2)
∇ × E = −∂B/∂t (3)
∇ × H = J + ∂D/∂t (4)

and with the relation
D = ɛE + P (5)
H = 1/µB + M (6)

## The Attempt at a Solution

i took the curl of both sides of 3 and simplified both sides and got the eqaution
laplacian of E = µ∂H/∂t (7)
i assumed that M=0 --> B=µH
taking the curl of both sides of (7) and using (4)
curl of (laplacian of E) = (µ∂/∂t)∂(∇ × D)/∂t (8)

substituting (5) to (8)
laplacian of E = (µɛ∂/∂t)∂E/∂t + µ∂/∂t)∂P/∂t (9)

my final answer is for the wave equation of E is (9), i want to know if my answer is correct before trying to solve B

thank you in advance

## Answers and Replies

i took the curl of both sides of 3 and simplified both sides and got the eqaution
laplacian of E = µ∂H/∂t (7)

That doesn't look right, how did you "simplify" the curls?

That doesn't look right, how did you "simplify" the curls?
hm.. i might have written the wrong thing in my post but here is my handwritten solution

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Yeah there's a problem because the divergence of E is not zero here, only the divergence of D...

Yeah there's a problem because the divergence of E is not zero here, only the divergence of D...

yes thank you

i reworked that part of the solution

divergene of E = (1/epsilon) divergence of (D - P)
= (1/epsilon) {divergence of D - divergence of P}
= 1/epsilon (p_free - (-p_bound))
since p_free + p_bound = p , and it is given that p = 0
thus divergence of E = 0

is that the only correction?

Well, your statement says that you are given a medium in which ρ = 0 but in general the charge density you are given is the free one (i.e. ρfree) because this is the one you can control in an experiment.
This would make ∇⋅E = –∇⋅P/ε = ρbound≠ 0

Well, your statement says that you are given a medium in which ρ = 0 but in general the charge density you are given is the free one (i.e. ρfree) because this is the one you can control in an experiment.
This would make ∇⋅E = –∇⋅P/ε = ρbound≠ 0

oh ok i get it now. ill rework my whole solution

Well, your statement says that you are given a medium in which ρ = 0 but in general the charge density you are given is the free one (i.e. ρfree) because this is the one you can control in an experiment.
This would make ∇⋅E = –∇⋅P/ε = ρbound≠ 0

thank you very much by the way