Maxwell's Equations & Wave Equation: An Overview

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Homework Help Overview

The discussion revolves around Maxwell's equations and their implications for the wave equation in the context of electromagnetic fields. Participants are exploring how the magnetic and electric fields, represented as vectors, satisfy the wave equation under certain conditions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the wave equation from Maxwell's equations by manipulating the curl and divergence operations. There are discussions about the application of vector calculus identities and the implications of various terms in the equations.

Discussion Status

Some participants have made progress in their derivations, while others are questioning their steps and the correctness of their signs. There is a mix of confusion and clarification as they work through the mathematical relationships, with some guidance offered regarding the treatment of derivatives and signs.

Contextual Notes

Participants express uncertainty about specific steps in their calculations and the implications of their results for satisfying the wave equation. There is acknowledgment of mistakes in sign and notation that may affect their conclusions.

may de vera
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Let respectively b = (b1, b2, b3) and e = (e1, e2, e3) denote the magnetic
and electric field in some medium. They are governed by Maxwell’s equations which look as follows:
(0.1) [tex]\partial[/tex]te = curl b
(0.2) [tex]\partial[/tex]tb = − curl e
(0.3) div e = 0
(0.4) div b = 0.
Show that each bi and each ei satisfies the wave equation
([tex]\partial[/tex]t2 - [tex]\Delta[/tex]) [tex]\varphi[/tex]=0
 
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Try it. It's not that hard. It might help to know curl(curl(A))=grad(div(A))-laplacian(A).
 
taking curl of both side:
curl (dt e) = grad (div e) - laplacian (e)
curl (dt e) = grad (0) - laplacian (e)

i don't know what I'm doing =(
 
dt e=curl b. Taking curl of both sides. curl(dt e)=curl(curl(b))=grad(div B)-laplacian(B)=grad(0)-laplacian(b)=-laplacian(b). Now curl(dt e)=dt(curl e). Continue...
 
curl (dt e) = -laplacian (b)
dt (curl e) = -laplacian (b)
-dt b = -laplacian (b)
dt b = laplacian (b)
0 = laplacian (b)

and then taking curl of 0.2:
dtb = -curl e
curl (dt b) = curl (-curl e)
dt (curl b) = -curl (curl e)
dt (e) = -grad( div e) - laplacian (e)
= -grad (0) - laplacian (e)
dt (e) = - laplacian (e)
0 = laplacian e

how does that satisfy the wave equation? is it because that they both equal to zero?
 
You aren't being very careful with signs and you are dropping a dt. dt(dt(e))=[itex]\partial^2_t E[/itex]. Why are you turning dt(e) and dt(b) into zero? You want to get dt(dt(b))=laplacian(b) and the same for e.
 
i finally got down to: dt(dt(b)) = laplacian (b) and dt(dt(e))= -laplacian e

i still don't understand how that applies to the wave equation.
 
may de vera said:
i finally got down to: dt(dt(b)) = laplacian (b) and dt(dt(e))= -laplacian e

i still don't understand how that applies to the wave equation.

You've STILL got a sign wrong in the e part. What you've quoted in the problem as the wave equation is what we've been writing as dt(dt(phi))-laplacian(phi)=0. Look up those symbols.
 
i found out where i made a mistake with that negative sign. i got it now. thank you so so muchhhhh! happy thanksgiving!
 

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