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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II December 18, 2012 Prof. Alan Guth

FORMULA SHEET FOR FINAL EXAM Exam Date: December 19, 2012 ∗∗∗

Some sections below are marked with asterisks, as this section is. The asterisks indicate that you won’t need this material for the quiz, and need not understand it. It is included, however, for completeness, and because some people might want to make use of it to solve problems by methods other than the intended ones. Index Notation:  i = ǫijk Aj Bk ,  ×B A

 ·B  = Ai B i , A Rotation of a Vector:

ǫijk ǫpqk = δip δjq − δiq δjp det A = ǫi1 i2 ···in A1,i1 A2,i2 · · · An,in

Ai′ = Rij Aj ,

Orthogonality: Rij Rik = δjk j =1

Rotation about z-axis by φ: Rz (φ)ij Rotation about axis n ˆ by φ:∗∗∗

Gradient:

 ϕ)i = ∂i ϕ , (∇

Divergence:

 ·A  ≡ ∂i Ai ∇

Laplacian:

∂i ≡

∂ ∂xi

 ×A )i = ǫijk ∂j Ak (∇

2  ϕ) = ∂ ϕ ∇2 ϕ =  ∇ · (∇ ∂xi ∂xi

Fundamental Theorems of Vector Calculus: Gradient:



b

 a

Divergence:



V

Curl:

j =3

0 0 1

      

R(ˆ n, φ)ij = δij cos φ + n ˆin ˆj (1 − cos φ) − ǫijk ˆnk sin φ .

Vector Calculus:

Curl:

j =2

 i=1  cos φ − sin φ   = i=2  sin φ cos φ   i=3 0 0

(RT T = I)



S

 ϕ · dℓ = ϕ(b) − ϕ(a) ∇  ·A  d3 x = ∇



S

 · da A

where S is the boundary of V     · dℓ (∇ × A) · da = A P

where P is the boundary of S

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 2

Delta Functions:   ′ ′ ϕ(x)δ(x − x ) dx = ϕ(x ) , ϕ(r )δ 3 (r − r ′ ) d3 x = ϕ(r ′ )   d dϕ  ′ ϕ(x) δ(x − x ) dx = − dx dx x=x′  δ( x − xi ) δ(g(x)) = , g(xi ) = 0 |g ′( xi )| i   1 r − r ′  · ∇ = −∇2 = 4πδ 3 (r − r ′ ) ′ 3 ′  r  r  r  r | − | | − |       rˆj xj 1 δij − 3ˆ ri rˆj 4π ≡ ∂i 3 = − ∂i ∂j + ∂i = δij δ 3(r) 3 2 r r 3 r r  r − d 8π  · 3(d · rˆ)ˆ  )δ 3 (r ) = − (d · ∇ ∇ r3 3  r − d 4π  × 3(d · rˆ)ˆ  δ 3 (r ) ∇ = − d × ∇ 3 3 r

Electrostatics:  = qE  , where F   (r − r ′ ) qi 1 (r − r ′ ) 1  r) = = r ′ ) d3 x′ E( 3 3 ρ( ′ ′ 4πǫ0 4πǫ |r − r | 0 |r − r | i

ǫ0 =permittivity of free space = 8.854 × 10−12 C2 /(N·m2 ) 1 = 8.988 × 109 N· m2 /C2 4πǫ0  r  1 ρ(r ′ ) 3 ′ ′ ′   E(r ) · dℓ = V (r ) = V (r 0 ) − d x 4πǫ0 |r − r ′ |  r0 ρ  ·E =  ×E  = 0,  = −∇ V ∇ E ∇ , ǫ0 ρ ∇2 V = − (Poisson’s Eq.) , ρ = 0 =⇒ ∇2 V = 0 (Laplace’s Eq.) ǫ0 Laplacian Mean Value Theorem (no generally accepted name): If ∇2 V = 0, then the average value of V on a spherical surface equals its value at the center. Energy: 1 1  qi qj 1 1 W = = 2 4πǫ0 ij rij 2 4πǫ0 1 W = 2



 i=j

d3 xρ(r )V (r ) =

1 ǫ0 2





d3 x d3x′  2 3 E  d x

ρ(r )ρ(r ′ ) |r − r ′ |

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 3

Conductors: σ Just outside, E = n ˆ ǫ0 Pressure on surface:

1  2 σ| E|outside

Two-conductor system with charges Q and −Q: Q = CV , W = 12 CV 2 N isolated conductors: Vi =



Pij Qj ,

Pij = elastance matrix, or reciprocal capacitance matrix

Cij Vj ,

Cij = capacitance matrix

j

Qi =

 j

Image charge in sphere of radius a: Image of Q at R is q = −

a a2 Q, r = R R

Separation of Variables for Laplace’s Equation in Cartesian Coordinates: V =



cos αx sin αx



cos βy sin βy



cosh γz sinh γz



where γ 2 = α2 + β 2

Separation of Variables for Laplace’s Equation in Spherical Coordinates: Traceless Symmetric Tensor expansion:   1 ∂ 1 2 ∂ϕ ∇ ϕ(r, θ, φ) = 2 + 2 ∇2θ ϕ = 0 , r r ∂r r ∂r where the angular part is given by   1 ∂ ∂ϕ 1 ∂ 2ϕ 2 ∇θ ϕ ≡ sin θ + sin θ ∂θ ∂θ sin2 θ ∂φ2 2

(ℓ) ˆ iℓ = −ℓ(ℓ + 1)C(ℓ) ˆ i1 n ˆ i2 . . . n ∇θ2 Ci 1 i2 ...i ℓ n ˆ i1 ˆni2 . . . n ˆ iℓ , i1 i2 ...iℓ n (ℓ)

where Ci1 i2 ...iℓ is a symmetric traceless tensor and n ˆ = sin θ cos φ eˆ1 + sin θ sin φ eˆ2 + cos θ eˆ3 . General solution to Laplace’s equation:   ′( ℓ) ∞  C (ℓ) i2 ...iℓ V (r ) = Ci 1 i2...iℓ rℓ + i1ℓ+1 ˆri1 rˆi2 . . . rˆiℓ , r ℓ=0

where r = rrˆ

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 4

Azimuthal Symmetry:  ∞   Bℓ ℓ V (r ) = Aℓ r + ℓ+1 { zˆi1 . . . zˆiℓ } rˆi1 . . . rˆi ℓ r ℓ=0 where { . . . } denotes the traceless symmetric part of . . . . Special cases:

{1} = 1 { zˆi } = zˆi

{ zˆi zˆj } = zˆi zˆj − 13 δij

  zˆi δjk + zˆj δik + zˆk δij  { zˆi zˆj zˆk zˆm } = zˆi zˆj zˆk zˆm − 17 zˆi zˆj δkm + zˆi zˆk δmj + zˆi zˆm δjk + zˆj zˆk δim  1  + zˆj zˆm δik + zˆk zˆm δij + 35 δij δkm + δik δjm + δim δjk { zˆi zˆj zˆk } = zˆi zˆj zˆk −

1 5

Legendre Polynomial / Spherical Harmonic expansion: General solution to Laplace’s equation:   ∞  ℓ  Bℓm ℓ V (r ) = Aℓm r + ℓ+1 Yℓm (θ, φ) r ℓ=0 m=−ℓ Orthonormality:



2π

dφ 0



π 0

sin θ dθ Yℓ∗′ m′ (θ, φ ) Y ℓm(θ, φ ) = δ ℓ′ ℓδ m′ m

Azimuthal Symmetry:  ∞   Bℓ ℓ V (r ) = Aℓ r + ℓ+1 Pℓ (cos θ) r ℓ=0

Electric Multipole Expansion: First several terms:  1  Q p · ˆr 1 rˆi rˆj V (r ) = + 2 + · · · , where Q + ij 2 r3 4πǫ0 r r    3 3 Q = d x ρ(r ) , pi = d x ρ(r ) xi Qij = d3 x ρ(r )(3xi xj − δij |r |2 ) , 1  Edip (r ) = − ∇ 4πǫ0

 × Edip (r ) = 0 , ∇



p · rˆ r2



1 3(p · rˆ)ˆ r − p 1 − pi δ 3 (r ) 3ǫ0 4πǫ0 r3 ∇ · E  dip (r ) = 1 ρdip( r ) = − 1 p · ∇δ  3 (r ) ǫ0 ǫ0 =

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 5

Traceless Symmetric Tensor version: V (r) =

∞ 1  1 (ℓ) rˆi . . . rˆiℓ , C 4πǫ0 rℓ+1 i1 ...iℓ 1 ℓ=0

where (ℓ) Ci1 ...iℓ

(2ℓ − 1)!! = ℓ!



ρ(r ) { xi1 . . . xiℓ } d3 x

(r ≡ rrˆ ≡ xi eˆi )

∞  1 (2ℓ − 1)!! r′ℓ = { rˆi1 . . . rˆiℓ } rˆ′i1 . . . rˆ′iℓ , ′ |r − r | ℓ! rℓ+1

for r′ < r

ℓ=0

(2ℓ − 1)!! ≡ (2ℓ − 1)(2ℓ − 3)(2ℓ − 5) . . . 1 =

(2ℓ)! , with (−1)!! ≡ 1 . 2ℓ ℓ!

Reminder: { . . . } denotes the traceless symmetric part of . . . . Griffiths version: V (r ) =

 ∞ 1  1 ℓ r′ ρ(r ′ )Pℓ (cos θ ′ ) d 3 x 4πǫ0 ℓ=0 rℓ+1

where θ ′ = angle between r and r ′ . ∞

 rℓ 1 < = P (cos θ ′ ) , ℓ+1 ℓ |r − r ′ | r ℓ=0 > 1 Pℓ (x) = ℓ 2 ℓ! Pℓ (1) = 1



d dx

ℓ

(x2 − 1)ℓ ,

∞

√

 1 = λℓ Pℓ (x) 2 1 − 2λx + λ ℓ=0

(Rodrigues’ formula)

ℓ

Pℓ (−x) = (−1) Pℓ (x)



1

dx P ℓ′ (x)Pℓ (x) = −1

2 δℓ′ ℓ 2ℓ + 1

Spherical Harmonic version:∗∗∗ ∞ ℓ 1   4π qℓm Y ℓm (θ, φ) V (r ) = 2ℓ + 1 rℓ+1 4πǫ0 =−ℓ ℓ=0 m

where q ℓm =



∗ ′ℓ r ρ(r ′ ) d3 x′ Y ℓm

∞  ℓ  1 4π r′ℓ ∗ ′ ′ = Y (θ , φ )Yℓm (θ, φ) , |r − r ′ | 2ℓ + 1 rℓ+1 ℓm ℓ=0 m=−ℓ

for r′ < r

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 6

Electric Fields in Matter: Electric Dipoles:  p = d3 x ρ(r ) r

ρdip (r ) = −p · ∇r δ 3 (r − r d ) , where r d = position of dipole  = (p · ∇)E =∇  (p · E)  F (force on a dipole)  (torque on a dipole)  = p × E U = −p · E

Electrically Polarizable Materials:  (r ) = polarization = electric dipole moment per unit volume P ˆ ρbound = −∇ · P , σ bound = P · n  +P  ,  ≡ ǫ0 E D

 ·D  = ρfree , ∇

Boundary conditions:

⊥ ⊥ Eab ove − Ebe low =

σ ǫ0

  Eab ove − E below = 0

 ×E  = 0 (for statics) ∇

⊥ D⊥ above − Dbe low = σfree

    D above − Dbelow = Pabove − Pbelow

Linear Dielectrics:  = ǫ0 χeE,  P χe = electric susceptibility  = ǫE  ǫ ≡ ǫ0 (1 + χe) = permittivity, D ǫ ǫr = = 1 + χe = relative permittivity, or dielectric constant ǫ0

N α/ǫ 0 , where N = number density of atoms 1 − Nα 3ǫ0  = αE)  or (nonpolar) molecules, α = atomic/molecular polarizability ( P  1  d3 x  ·E D (linear materials only) Energy: W = 2  = −∇  W (Even if one or more potential differences are Force on a dielectric: F held fixed, the force can be found by computing the gradient with the total charge on each conductor fixed.) Clausius-Mossotti equation: χe =

Magnetostatics: Magnetic Force:  = q( E  + v × B)  = dp , F dt

where p = γm0v ,

γ= 

1 1 − cv2

2

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

 = F



×B  = Idℓ



p. 7

 ×B  d3 x J

Current Density: Current through a surface S: IS =



S

Charge conservation:

J · da

∂ρ  · J = −∇ ∂t

Moving density of charge: J = ρv Biot-Savart Law:  ′   ′ dℓ × (r − r ′ ) µ0 K(r ) × (r − r ′ ) ′ µ0  = B( r) = I da 4π |r − r ′ |3 4π |r − r ′ |3   ′ J(r ) × (r − r ′ ) 3 µ0 d x = 4π |r − r ′ |3 where µ0 = permeability of free space ≡ 4π × 10−7 N/A2 Examples:  = µ 0I φˆ Infinitely long straight wire: B 2πr  = µ0 nI0 zˆ , where n = turns per Infinitely long tightly wound solenoid: B unit length  0, z) = Loop of current on axis: B(0,

µ0 IR2 zˆ 2(z 2 + R2 )3/2

 r ) = 1 µ0 K  ×n Infinite current sheet: B( ˆ, n ˆ = unit normal toward r 2 Vector Potential:   ) coul = µ0 A(r 4π



J(r ′) 3 ′ d x , |r − r ′ |

 =∇  ×A , B

 ·A  coul = 0 ∇

  = 0 (Subject to modification if magnetic monopoles are discovered) ∇ ·B  r ) + ∇Λ(  r ) for any Λ(r ). B  =∇  ×A  is  ′ (r ) = A( Gauge Transformations: A unchanged. Amp`ere’s Law:   = µ0 J , or equivalently ∇ ×B



P

 · dℓ = µ0 Ienc B

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 8

Magnetic Multipole Expansion: Traceless Symmetric Tensor version: ∞

{ rˆi1 . . . rˆiℓ } µ 0  (ℓ ) Aj ( r) = Mj;i1 i2 ...iℓ 4π rℓ+1 ℓ=0  (2ℓ − 1)!! (ℓ) where Mj ;i 1i 2 ...iℓ = d3 xJj(r ){ xi1 . . . xiℓ } ℓ!  Current conservation restriction: d3 x Sym(xi1 . . . x iℓ−1 Jiℓ ) = 0 i1 ...iℓ

where

Sym

i1 ...iℓ

means to symmetrize — i.e. average over all

orderings — in the indices i1 . . . i ℓ Special cases:  ℓ = 1:

ℓ = 2:



d3 x Ji = 0 d3 x (Ji xj + Jj xi ) = 0

 × rˆ  r ) = µ0 m Leading term (dipole): A( , 4π r2 where 1 (1) mi = − ǫijk Mj;k 2   1 1  ℓ = d3 x r × J = Ia , m  = I r × d 2 P 2  where a = da for any surface S spanning P S

 × rˆ µ 0 3(m  · rˆ)ˆ r −m  2µ 0 dip (r ) = µ0 ∇  ×m B = + m  δ 3 (r ) 2 3 4π r 4π r 3  · Bdip (r ) = 0 , ∇ × B  dip( r ) = µ0 Jdip (r ) = −µ0 m  δ3 (r ) ∇  ×∇

Griffiths version:

 ∞  1  r ) = µ0I A( (r′ )ℓ Pℓ (cos θ ′)dℓ′ rℓ+1 4π ℓ=0

Magnetic Fields in Matter: Magnetic Dipoles:   1 1 d3 x r × J = Ia m  = I r × dℓ = 2 P 2

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p. 9

 r δ 3 (r − r d ), where r d = position of dipole Jdip (r ) = −m  ×∇  =∇  (m  F  · B) (force on a dipole)   =m  ×B  U = −m  ·B

(torque on a dipole)

Magnetically Polarizable Materials:  (r ) = magnetization = magnetic dipole moment per unit volume M  ×M  ,  bound = M  ×n K ˆ Jbound = ∇

1  ·B  =0  = Jfree ,  ,  ×H  −M ∇ ∇ B µ0 Boundary conditions: ⊥ ⊥ ⊥ ⊥ ⊥ Bab H⊥ ove − Bbe low = 0 above − Hbe low = −(Mab ove − Mbelow )     ˆ)   B H ˆ above − Bbelow = µ0 (K × n above − H below = Kfree × n

 ≡ H

Linear Magnetic Materials:  = χm H,  M χm = magnetic susceptibility  = µH  µ = µ0 (1 + χm ) = permeability, B

Magnetic Monopoles:   r ) = µ 0 qm rˆ ; B( Force on a static monopole: F = q m B 4 π r2  = µ0qe qm rˆ , where rˆ points Angular momentum of monopole/charge system: L 4π from qe to qm µ 0q eq m 1 Dirac quantization condition: = h ¯ × integer 4π 2 Connection Between Traceless Symmetric Tensors and Legendre Polynomials or Spherical Harmonics: Pℓ (cos θ) =

(2ℓ)! ˆ iℓ { zˆi1 . . . zˆ iℓ } n ˆ i1 . . . n 2ℓ (ℓ!)2

For m ≥ 0,

) Yℓm (θ, φ) = C(i1ℓ,m ˆ i1 . . . n ˆ iℓ , ...iℓ n (ℓ,m)

where Ci1 i2 ...iℓ = dℓm { u ˆ+ ui+mzˆim+1 . . . zˆiℓ } , i1 . . . ˆ  2m (2ℓ + 1) ( −1) m (2ℓ)! , with dℓm = 4π (ℓ + m)! (ℓ − m)! 2ℓ ℓ!

1 ex + ieˆy ) and u ˆ+ = √ (ˆ 2 ∗ Form m < 0, Yℓ,−m (θ, φ) = (−1)m Yℓm (θ, φ)

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 10

More Information about Spherical Harmonics:∗∗∗  2 + 1 ( − m)! m Yℓm (θ, φ) = P (cos θ)eimφ 4π ( + m)! ℓ where Pℓm (cos θ) is the associated Legendre function, which can be defined by P ℓm (x) =

ℓ+m (−1)m 2 m/2 d ) (x2 − 1)ℓ (1 − x 2ℓ ! dx ℓ+m

Legendre Polynomials:

SPHERICAL HARMONICS Ylm(θ , φ)

l=0

Y00 =

1 4π

3 sin θeiφ 8π

Y11 = l=1

3

Y10 =

Y22 =

l=2

4π

1 4

cos θ

15

sin2 θe2iφ

2π

15 sin θ cosθeiφ 8π

Y21 = -

5

Y20 =

4π

( 32 cos2θ

1 ) 2

1 4

35

Y32 =

1 4

105

Y31 =-

1 4

21 sinθ (5cos2θ -1)eiφ 4π

Y33 = -

4π

2π

l=3

Y30 =

7 4π

sin3 θe3iφ

sin2 θ cos θe2iφ

( 52 cos3θ

3 2

cos θ)

Image by MIT OpenCourseWare

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

p. 11

Maxwell’s Equations:  ×E  = − ∂B , (iii)∇ ∂t

 ·E = 1ρ (i) ∇ ǫ0

  ×B  = µ0 J+ 1 ∂E (iv)∇ c2 ∂t

 ·B  =0 (ii) ∇

1 c2  = q(E  + v × B)  Lorentz force law: F where µ0 ǫ0 =

∂ρ  · J = −∇ ∂t Maxwell’s Equations in Matter: Charge conservation:

 and magnetization M : Polarization P  ·P  , ρb = −∇

 ×M  , Jb = ∇

ρ = ρf + ρb ,

f + J b J = J

Auxiliary Fields:   ,  ≡ B −M H µ0

 ≡ ǫ0 E  + P D

Maxwell’s Equations:  ·D  = ρf (i) ∇  ·B  =0 (ii) ∇

  ×E  = − ∂B , (iii)∇ ∂t   ×H  = Jf + ∂D (iv)∇ ∂t

For linear media:  = ǫE  , D

 = 1B  H µ

where ǫ = dielectric constant, µ = relative permeability  ∂D Jd ≡ = displacement current ∂t Maxwell’s Equations with Magnetic Charge:   = −µ0 Jm − ∂B ,  ×E (iii)∇ ∂t   = µ0 Je + 1 ∂E  ·B  = µ0 ρm  ×B (iv) ∇ (ii) ∇ c2 ∂t   1   − v × E  = qm B Magnetic Lorentz force law: F c2  ·E  = 1 ρe (i) ∇ ǫ0

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Current, Resistance, and Ohm’s Law:  + v × B)  , where σ = conductivity. ρ = 1/σ = resistivity J = σ(E Resistors: V = IR ,

P = IV = I 2 R = V 2 /R

Resistance in a wire: R = resistivity

ℓ ρ , where ℓ = length, A = cross-sectional area, and ρ = A

  V 0 −t/RC e , Q = CV0 1 − e−t/RC R   · dℓ , where v is either the velocity EMF (Electromotive force): E ≡ ( E + v × B) Charging an RC circuit: I =

of the wire or the velocity of the charge carriers (the difference points along the wire, and gives no contribution)

Inductance: Universal flux rule: Whenever the flux through a loop changes, whether due to a  or motion of the loop, E = − dΦ B , where ΦB is the magnetic flux changing B dt through the loop Mutual inductance: Φ2 = M21I1 , M21 = mutual inductance   µ0 dℓ1 · dℓ2 (Franz) Neumann’s formula: M 21 = M12 = 4π P1 P2 |r 1 − r 2 | Self inductance: Φ = LI ,

E = −L

dI ; dt

L = inductance

Self inductance of a solenoid: L = n2 µ0 V , where n = number of turns per length, V = volume  V0  R Rising current in an RL circuit: I = 1 − eL t R

Boundary Conditions: D1⊥ − D2⊥ = σf 1 E1⊥ − E2⊥ = σ ǫ0

  − E  = 0 E 1 2   −D  2 = P 1 − P2 D

B1⊥ − B2⊥ = 0 ⊥ ⊥ ⊥ H⊥ 1 − H2 = M 2 − M 1

 − H   = −n f H ˆ×K 1 2    −B  = −µ0 n  B ˆ×K 1 2

1

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

Conservation Laws: Energy density: uEM

p. 13

  1  |2 + 1 |B|  2 = ǫ 0| E 2 µ0

 = 1 E  ×B  Poynting vector (flow of energy): S µ0 Conservation of energy:  d  · da + U ] = Integral form: [ UEM − S mech dt ∂u  · S , where u = uEM + umech Differential form: = −∇ ∂t 1 1 Momentum density: ℘EM = 2 S ; 2 Si is the density of momentum in the i’th c c direction     1 1 1 2 2   Bi Bj − δij | B| Maxwell stress tensor: Tij = ǫ0 Ei Ej − δij | E| + 2 2 µ0 where −Tij = −Tji = f low in j’th direction of momentum in the i’th direction

Conservation of momentum:     d 1 3 Integral form: Pmech,i + 2 S d x = Tij daj , for a volume V c V i dt S b ounded by a surface S ∂ Differential form: (℘mech,i + ℘EM,i ) = ∂j Tji ∂t Angular momentum:  B)]  Angular momentum density (about the origin): ℓEM = r ×℘EM = ǫ0 [r ×(E×

Wave Equation in 1 Dimension: ∂2f 1 ∂2f − = 0 , where v is the wave velocity ∂z2 v 2 ∂t2 Sinusoidal waves:

f (z, t) = A cos [k(z − vt) + δ] = A cos [kz − ωt + δ] where ω = angular frequency = 2πν ν = frequency ω v = = phase velocity δ = phase (or phase constant) k k = wave number λ = 2π/k = wavelength T = 2π/ω = period A = amplitude Euler identity: eiθ = cos θ + i sin θ ˜ i(kz−ωt) ] , where A˜ = Aeiδ ; “Re” is usually Complex notation: f (z, t) = Re[Ae dropped. ω dω Wave velocities: v = = phase velocity; vgroup = = group velocity k dk

8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012

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Electromagnetic Waves:  1 ∂2E =0, c2 ∂t2 Linea rly Polarized Plane Waves: − Wave Equations: ∇2E

− ∇2 B

 1 ∂2B =0 c2 ∂t2

 r −ωt)  ( r , t) = E ˜ 0 ei(k· ˆ is a unit vector, n ˆ , where E˜0 is a complex amplitude, n E  and ω/| k| = vphase = c.

(transverse wave) n ˆ · k = 0  = 1ˆ  B k×E c Energy and Momentum: u = ǫ 0E02 cos2 (kz − ωt + δ) , (k = k zˆ)   averages to 1/2 ! " 1 1  = ǫ0 E 2 S = E × B = uc zˆ , I (intensity) = | S| 0 µ0 2 1  u ℘EM = 2 S = zˆ c c Electromagnetic Waves in Matter: # µǫ = index of refraction n≡ µ0 ǫ0 c v = phase velocity = n   1 1 2 2   | + | B| u= ǫ| E 2 µ  =nk ˆ ×E B c 1  = uc zˆ S = E ×B µ n Reflection and Transmission at Normal Incidence: Boundary conditions: ǫ1 E1⊥ = ǫ2 E2⊥ B1⊥

=

B⊥ 2

X

 = E , E 1 2 1  1  B = B . µ2 2 µ1 1

Incident wave (z < 0):

V2

V1 Bl