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  • 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