# Maxwell’s equations • Wave equations • Plane · PDF file• Maxwell’s...

date post

02-May-2018Category

## Documents

view

220download

2

Embed Size (px)

### Transcript of Maxwell’s equations • Wave equations • Plane · PDF file• Maxwell’s...

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 1Massachusetts Institute of Technology RF Cavity and Components for Accelerators USPAS 2010

Maxwells equations Wave equations Plane Waves Boundary conditions

A. Nassiri - ANL

Lecture 1- Review

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 2

Maxwells Equations

The general form of the time-varying Maxwells equations can be written in differential form as:

0==

=

+=

BD

BE

DJH

t

t

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 3

A few other fundamental relationships

H/m104 F/m,108548with

ity)(permeabil and ity)(permittiv here

ips"relationsh veconstituti"

equation" continuity"

law" sOhm'"

70

120

00

==

==

==

=

=

.

HBED

J

EJ

rr

t

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 4

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 5

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 6

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 7

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 8

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 9

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 10

A few other fundamental relationships

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 11

Integral form of the equations

= D

=S

QsdD

tBE

=

=

SCsd

tBdE

0= B

=S

sdB 0

tDJH

+=

+=SC

sdtDJdH

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 12

Wave Equations

In any problem with unknown E, D, B, H we have 12 unknowns. To solve for these we need 12 scalar equations. Maxwells equations provide 3 each for the two curl equations. and 3 each for both constitutive relations (difficult task).

Instead we anticipate that electromagnetic fields propagate as waves. Thus if we can find a wave equation, we could solve it to find out the fields directly.

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 13

Wave equations

Take the curl of the first Maxwell:

( )

( )

2

2

t

tt

t

t

=

+=

+=

+=

HJ

HJ

EJ

EJH

( ) LHS on the use Now 2 HHH 0

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 14

Wave Equations

The result is:

JHH

=

2

22

t

Similarly, the same process for the second Maxwell produces

+

=

ttJEE

2

22

Note how in both case we have a wave equation (2nd order PDE) for both E and H with fields to the left of the = sign and sourcesto the right. These two wave equations are completely equivalentto the Maxwell equations.

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 15

Solutions to the wave equations

Consider a region of free space ( = 0) where there are no sources(J = 0). The wave equations become homogeneous:

022

2 =

tEE

022

2 =

tHH

Actually there are 6 equations; we will only consider one component:e.g. Ex(z,t)

2

00

22

2

22

2 1 e wher01 cvtE

vzE xx ===

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 16

Solutions to the wave equation

Try a solution of the form f(z-vt) e.g. sin[(z-vt)]. By differentiatingtwice and substituting back into the scalar wave equation, we find that it satisfies!

f(z) t=0z

f(z-vt1) t=t1 z

f(z-vt2) t=t2 z

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 17

Plane Waves

First treat plane waves in free space. Then interaction of plane waves with media. We assume time harmonic case, and source free situation.

0202 =+ EkE

We require solutions for E and H (which are solutions to thefollowing PDE) in free space

No potentials here!(no sources)

Note that this is actually three equations:

zyxiEkzE

yE

xE

iiii ,,==+

+

+ 0202

2

2

2

2

2

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 18

How do we find a solution?

Usual procedure is to use Separation of Variables (SOV). Take one component for example Ex.

( ) ( ) ( )

0

0

20

20

=+

+

+

=+++

=

khh

gg

ff

fghkhfggfhfgh

zhygxfEx

Functions of a single variable sum = constant = -k02

ckkkkk

khhk

ggk

ff

zyx

zyx

===++

=

=

=

2 with so and

020

222

222 ;;

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 19

Mathematical Solution

We note we have 3 ODEs now.

zjkz

yjky

xjkx

z

y

x

ehhkdz

hd

egkdy

gd

effkdx

fd

==+

==+

==+

issolution 0

g issolution 0

issolution 0

22

2

22

2

22

2

( )zkykxkjx

zyxeAE ++=

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 20

But, what does it mean physically?

( )zkykxkjx

zyxeAE ++=This represents the x-component of the travelling wave E-field(like on a transmission line) which is travelling in the direction of the propagation vector, with Amplitude A. The direction ofpropagation is given by

zkykxkk zyx ++=

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 21

Physical interpretation

The solution represents a wave travelling in the +z direction withvelocity c. Similarly, f(z+vt) is a solution as well. This latter solution represents a wave travelling in the -z direction.So generally,

( ) ( )( )( )[ ]vtzvtyvtxftzEx =,In practice, we solve for either E or H and then obtain theother field using the appropriate curl equation.

What about when sources are present? Looks difficult!

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 22

Generalize for all components

If we define the normal 3D position vector as:

rkjz

rkjy

rkjx

zyx

CeE

BeE

AeE

zkykxkrk

zzyyxxr

=

=

=

++=

++=

similarly

so

then

zCyBxAEeEE rkj ++== 00 where

General expressionfor a plane wave

+sign droppedhere

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 23

Properties of plane waves

For source free propagation we must have E = 0. If we satisfythis requirement we must have kE0= 0. This means that E0is perpendicular to k.

The corresponding expression for H can be found by substitution of the solution for E into the E equation. The result is:

EnkH

= 0

0

Where n is a unit vector in the k direction.

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 24

Transverse Electromagnetic (TEM) wave

Note that H is also perpendicular to k and also perpendicular toE. This can be established from the expression for H.

E

H

Direction of propagationk,n

Note that:

knHE or =E and H lie on theplane of constantphase (kr = const)

Massachusetts Institute of Technology RF Cavity and Components for Accelerators 25

Plane waves at interfaces (normal incidence)

Consi

*View more*