MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: 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:

A · B = AiBi ,A × B i = "ijkAjBk ,"ijk"pqk = δipδjq − δiqδjp ··· det A = "i1i2···in A1,i1 A2,i2 An,in Rotation of a Vector:  T Ai = RijAj , Orthogonality: Rij Rik = δjk (R T = I)

j j j  =1 =2 =3  i − =1  cos φ sin φ 0    Rotation about z-axis by φ: Rz(φ)ij = i=2  sin φ cos φ 0  i=3 001 Rotation about axis nˆ by φ:∗∗∗

R(ˆn, φ)ij = δij cos φ +ˆninˆj(1 − cos φ) − "ijknˆk sin φ. Vector Calculus: ∂ Gradient: (∇ ϕ)i = ∂iϕ, ∂i ≡ ∂xi Divergence: ∇·A  ≡ ∂iAi

Curl: (∇×A )i = "ijk∂jAk ∂2ϕ Laplacian: ∇2ϕ = ∇· (∇ ϕ)= ∂xi∂xi Fundamental Theorems of Vector Calculus:

 b Gradient: ∇ ϕ · d, = ϕ(b))− ϕ(a a  Divergence: ∇·A d3x = A  · da V S where S is the boundary of V  Curl: (∇× A) · da = A · d, S 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)δ3(r − r )d3x = 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  r − r  1 ∇· = −∇2 =4πδ3(r − r ) |r − r  |3 |r − r  |       − rˆj xj 1 δij 3ˆrirˆj 4π 3 ∂i ≡ ∂i = −∂i∂j = + δij δ (r) r2 r3 r r3 3 3(d · rˆ)ˆr − d 8π ∇· = − (d · ∇ )δ3(r) r3 3 3(d · rˆ)ˆr − d 4π ∇× = − d × ∇ δ3(r) r3 3 Electrostatics:   F = qE,where  −  −   1 (r r ) qi 1 (r r )  3  E(r)=  3 =  3 ρ(r )d x 4π"0 i |r − r | 4π"0 |r − r | −12 2 2 "0 = of free space = 8.854 × 10 C /(N·m ) 1 =8.988× 109 N · m2/C2 4π"0  r   −   ·  1 ρ(r ) 3  V (r)=V (r 0) E(r ) d, = | − |d x r 0 4π"0 r r ρ ∇·E  = ,E∇× =0,E = −∇  V "0 ρ ∇2V = − (Poisson’s Eq.) ,ρ=0= ⇒∇2V = 0(Laplace’s Eq.) "0 Laplacian Mean Value Theorem (no generally accepted name): If ∇2V =0,then the average value of V on a spherical surface equals its value at the center. Energy:   1 1 qiqj 1 1 3 3  ρ(r)ρ(r ) W = = d x d x  ij | − | 2 4π"0 ij r 2 4π"0 r r  i= j  1 1 2 W = d3xρ(r)V (r )= " E d3x 2 2 0 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 3

Conductors: σ Just outside, E = nˆ "0 1 |  | Pressure on surface: 2 σ E outside − 1 2 Two-conductor system with charges Q and Q: Q = CV , W = 2 CV N isolated conductors:

Vi = Pij Qj ,Pij = matrix, or reciprocal matrix j

Qi = Cij Vj ,Cij = capacitance matrix j

a a2 Image charge in sphere of radius a: Image of Q at R is q = − Q, r = RR

Separation of Variables for Laplace’s Equation in Cartesian Coordinates: cos αx cos βy cosh γz V = where γ2 = α2 + β2 sin αx sin βy sinh γz

Separation of Variables for Laplace’s Equation in Spherical Coordinates: Traceless Symmetric Tensor expansion:   1 ∂ ∂ϕ 1 ∇2ϕ(r, θ, φ)= r2 + ∇2 ϕ =0, r2 ∂r ∂r r2 θ where the angular part is given by   2 2 1 ∂ ∂ϕ 1 ∂ ϕ ∇θ ϕ ≡ sin θ + sin θ ∂θ ∂θ sin2 θ ∂φ2 ∇2 () − () θ Ci1i2...i nˆi1 n ˆi2 ...nˆi = ,(, +1)Ci1i2...i nˆi1 nˆi2 ...nˆi , () where Ci1i2...i is a symmetric traceless tensor and

nˆ =sinθ cos φ eˆ1 +sinθ sin φ eˆ2 +cosθ eˆ3 .

General solution to Laplace’s equation: ∞ () ()  Ci1i2...i V( r)= C r + rˆi rˆi ...rˆi , where r = rrˆ i1i 2...i r+1 1 2  =0 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 4

Azimuthal Symmetry: ∞    B V (r)= A r + { zˆi ...zˆi } rˆi ...rˆi r+1 1  1  =0 where { ...} denotes the traceless symmetric part of ... . Special cases: { 1 } =1

{ zˆi } =ˆzi 1 { zˆizˆ j} =ˆz i zˆ j − δij 3 1 { zˆizˆ jzˆ k} =ˆz i zˆ jzˆ k − zˆiδjk +ˆzj δik +ˆzkδij 5 1 { zˆizˆjzˆkzˆm } =ˆzizˆjzˆkzˆm − zˆizˆjδkm +ˆzizˆkδmj +ˆzizˆmδjk +ˆzj zˆkδim 7 1 +ˆzjzˆmδik +ˆzkzˆmδij + 35 δij δkm + δikδjm + δimδjk Legendre Polynomial / Spherical Harmonic expansion:

General solution to Laplace’s equation: ∞     Bm V (r)= Am r + Ym(θ, φ) r+1 =0 m=−   2π π ∗ Orthonormality: dφ sin θ dθYm ()θ, φ Ym(θ, φ)=δ δ mm 0 0 Azimuthal Symmetry: ∞    B V( r)= A r + P(cos θ) r+1 =0

Electric Multipole Expansion:

First several terms: 1 Q p · rˆ 1 rˆi rˆ j V (r)= + + Qij + ··· ,where 4π" r r2 2 r3  0   3 3 3 2 Q = d xρ(r) ,pi = d xρ(r) xi Qij = d xρ(r)(3xixj −δij |r | ) ,   · · − 1 p rˆ 1 3(p rˆ)ˆr p 1 3  − ∇ − i Edip(r)= 2 = 3 p δ (r) 4π"0 r 4π"0 r 3"0

1 1 3 ∇×E  dip(r)=0 ,∇· E dip(r )= ρdip(r )=− p · ∇ δ (r) "0 "0 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 5

Traceless Symmetric Tensor version: ∞ 1 1 () V( r)= C rˆi ...rˆi , +1 i1...i 1  4π"0 r =0 where  − () (2, 1)!! 3 C = ρ(r) { xi ...xi } d x (r ≡ rrˆ ≡ xieˆi) i1...i ,! 1  ∞  1 (2, − 1)!! r    = { rˆi ...rˆi } rˆ ...rˆ ,forr

Griffiths version: ∞  1 1    3  V (r)= +1 r ρ(r )P (cos θ )d x 4π"0 r =0 where θ = angle between r and r . ∞  ∞ 1 r<  1  = P(cos θ ) , √ = λ P(x) |r − r | +1 − 2 =0 r> 1 2λx + λ =0   1 d 2  P(x)= (x − 1) , (Rodrigues’ formula) 2,! dx  1  2 P(1) = 1 P(−x)=(−1) P(x) dxP (x)P(x)= δ −1 2, +1

Spherical Harmonic version:∗∗∗ ∞  1 4π qm m V (r)= +1 Y (θ, φ) 4π"0 2, +1r =0 m=−  ∗   3  where qm = Ymr ρ(r )d x

∞   1 4π r ∗    = Y (θ ,φ)Ym(θ, φ), forr

Electric Fields in Matter:

Electric Dipoles:  p = d3xρ(r) r

3 ρdip(r)=−p · ∇ r δ (r − r d),wherer d = position of dipole F =(p · ∇ )E = ∇ (p · E )(forceonadipole)  = p × E ( on a dipole) U = −p · E

Electrically Polarizable Materials: P (r ) = polarization = electric dipole moment per unit volume

ρbound = −∇ · P , σbound = P · nˆ        D ≡ "0E + P , ∇·D = ρfree , ∇×E = 0(for statics) Boundary conditions: ⊥ − ⊥ σ ⊥ − ⊥ Eabove Ebelow = Dabove Dbelow = σfree "0  −   −   −  Eabove Ebelow =0 Dabove Dbelow = Pabove Pbelow Linear Dielectrics:

P = "0χeE , χe = electric susceptibility

" ≡ "0(1 + χe) = permittivity, D = "E " "r = =1+χe = relative permittivity, or dielectric constant "0

Nα/"0 Clausius-Mossotti equation: χe = , where N = number density of atoms 1 − Nα 30 or (nonpolar) molecules, α = atomic/molecular polarizability (P = αE )  1 Energy: W = D · E d3x (linear materials only) 2 on a dielectric: F = −∇ W (Even if one or more potential differences are held fixed, the force can be found by computing the gradient with the total charge on each conductor fixed.)

Magnetostatics: Magnetic Force: dp 1 F = q ( E + v× B )= , where p = γm0v, γ= dt v2 1 − c2 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 7   F = Id, × B= J × B  d3x

Current Density: 

Current through a surface S: IS = J · da S ∂ρ Charge conservation: = −∇·J  ∂t Moving density of charge: J = ρv

Biot-Savart Law:   µ d, × (r − r ) µ K (r )(×r − r ) B ( r)= 0 I = 0 da 4π |r − r |3 4π |r − r |3  µ J(r ) × (r − r ) = 0 d3x 4π |r − r |3

−7 2 where µ0 = permeability of free space ≡ 4π × 10 N/A

Examples: µ I Infinitely long straight wire: B = 0 φˆ 2πr

Infinitely long tightly wound solenoid: B = µ0nI0 zˆ ,wheren = turns per unit length µ IR2 Loop of current on axis: B (0, 0,z)= 0 zˆ 2(z2 + R2)3/2 1 Infinite current sheet: B (r)= µ K × nˆ ,ˆn = unit normal toward r 2 0 Vector Potential:  µ J(r ) A(r ) = 0 d3x , B = ∇× A  , ∇·A  =0 coul 4π |r − r | coul

∇·B  = 0(Subject to modification if magnetic monopoles are discovered)

Gauge Transformations: A(r)=A(r)+∇  Λ(r ) for any Λ(r). B= ∇× A  is unchanged.

Amp`ere’s Law: 

∇×B  = µ0J, or equivalently B · d, = µ0Ienc P 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 8

Magnetic Multipole Expansion:

Traceless Symmetric Tensor version: ∞ { } µ0 () rˆi1 ...rˆi Aj(r )= M 4π j;i1i2...i r+1 =0  − () (2, 1)!! 3 where M = d xJj(r){ xi ...xi } j; i1 i 2...i ,! 1   3 Current conservation restriction: d x Sym(xi1 ...xi−1 Ji )=0 i1...i

where Sym means to symmetrize — i.e. average over all i1...i

orderings — in the indices i1 ...i Special cases: 3 , =1: d xJi =0  3 , =2: d x (Jixj + Jjxi)=0

µ m × rˆ Leading term (dipole): A(r)= 0 , 4π r2 where −1 M(1) mi = "ijk j;k 2  1 1 m = I r × d, = d3xr × J = Ia, 2 P 2  where a = da for any surface S spanning P S

µ m × rˆ µ 3(m · rˆ)ˆr − m 2 µ B (r)= 0 ∇× = 0 + 0 mδ 3(r ) dip 4π r2 4π r3 3 3 ∇·B  dip(r)=0 ,∇× B  dip(r)= µ0Jdip(r)=− µ0m × ∇ δ (r) Griffiths version: ∞ µ0I 1     A(r )= (r ) P(cos θ )d, 4π r+1 =0

Magnetic Fields in Matter: Magnetic Dipoles:   1 1 m = I r × d, = d3xr × J = Ia 2 P 2 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 9

3 Jdip(r )=−m × ∇ r δ (r − r d), where r d = position of dipole F = ∇(m · B  )(forceonadipole)  = m × B (torque on a dipole) U = −m · B Magnetically Polarizable Materials: M (r) = magnetization = magnetic dipole moment per unit volume

Jbound = ∇×M  , K bound = M × nˆ 1 H ≡ BM −  , ∇× H = Jfree , ∇·B  =0 µ0 Boundary conditions: ⊥ − ⊥ ⊥ − ⊥ − ⊥ − ⊥ Babove Bbelow =0 Habove Hbelow = (Mabove Mbelow)  −   ×  −   × Babove Bbelow = µ0(K nˆ) Habove Hbelow = Kfree nˆ Linear Magnetic Materials:

M = χmH , χm = magnetic susceptibility

µ= µ0(1 + χm) = permeability, B= µH  Magnetic Monopoles: µ0 qm B (r)= rˆ ; Force on a static monopole: F= qmB  4π r2 µ qe qm Angular momentum of monopole/charge system: L = 0 rˆ ,whererˆpoints 4π from qe to qm µ qe q m 1 Dirac quantization condition: 0 = ¯h × integer 4π 2 Connection Between Traceless Symmetric Tensors and Legendre Polynomials or Spherical Harmonics: (2,)! P(cos θ)= { zˆi ...zˆi } nˆi ...nˆi 2(,!)2 1  1  For m ≥ 0, (,m) Ym(θ, φ)=Ci1...i nˆi1 ...nˆi , (,m) { + + } where Ci i ...i = dm uˆi ...uˆi zˆim+1 ...zˆi , 1 2  1 m (−1)m(2,)! 2m (2, +1) with dm = , 2,! 4π (, + m)! (, − m)!

+ 1 and uˆ = √ (ˆex + ieˆy) 2 m ∗ Form m<0, Y,−m(θ, φ)=(−1) Ym(θ, φ) 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 10

More Information about Spherical Harmonics:∗∗∗ − 2 +1( m)! m imφ Y£m(θ, φ)= P£ (cos θ)e 4π ( + m)! m where P£ (cos θ) is the associated Legendre function, which can be defined by (−1)m d£+m P m(x)= (1 − x2 )m/2 (x 2 − 1)£ £ 2£ ! dx£+m

Legendre Polynomials:

SPHERICAL HARMONICS Ylm(θ , φ)

1 l = 0 Y00 = 4π

3 iφ Y11 = - sin θe 8π l = 1 3 Y10 = cos θ 4π

151 2 2iφ Y22 = sin θe 4 2π

15 iφ l = 2 Y21 = - sin θ cosθe 8π

5 3 2θ 1 Y20 = ( 2 cos 2 ) 4π

1 35 3 3iφ Y33 = - sin θe 4 4π

1 105 2 2iφ Y32 = sin θ cos θe 4 2π l = 3

1 21 2 eiφ Y31 = - sinθ (5cos θ -1) 4 4π

7 5 3 3 Y30 = ( cos θ cos θ) 4π 2 2

Image by MIT OpenCourseWare. 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 11

Maxwell’s Equations: 1 ∂B (i) ∇·E  = ρ (iii)∇× E = − , "0 ∂t 1 ∂E (ii) ∇·B  =0(iv) ∇×B  = µ J + 0 c2 ∂t 1 where µ " = 0 0 c2 Lorentz force law: F = q(E + v × B ) ∂ρ Charge conservation: = −∇·J  ∂t Maxwell’s Equations in Matter: Polarization P and magnetization M :

ρb = −∇·P, J b = ∇× M,  ρ= ρf + ρb ,J= J f + J b Auxiliary Fields: B H ≡ − M,D  ≡ "0E + P µ0 Maxwell’s Equations: ∂B (i) ∇·D  = ρf (iii)∇×E  = − , ∂t ∂D (ii) ∇·B  =0(iv) ∇×HJ  = f + ∂t For linear media: 1 D = "E,H  = B µ where " = dielectric constant, µ = relative permeability ∂D Jd ≡ = ∂t Maxwell’s Equations with Magnetic Charge: 1 ∂B (i) ∇·E  = ρe (iii)∇× E = −µ0Jm − , "0 ∂t 1 ∂E (ii) ∇·B  = µ ρm (iv)∇× B = µ Je + 0 0 c2 ∂t   1 Magnetic Lorentz force law: F = qm B − v × E c2 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 12

Current, Resistance, and ’s Law:

J = σ(E + v × B ) , where σ = conductivity. ρ =1/σ =resistivity

Resistors: V = IR , P = IV = I2R = V 2/R , Resistance in a wire: R = ρ,where , =length,A = cross-sectional area, and ρ = A resistivity V Charging an RC circuit: I = 0 e−t/RC ,Q= CV 1 − e−t/RC R 0

EMF (): E≡ (E + v × B ) · d, ,wherev is either the 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 dΦB changing B or motion of the loop, E = − , where ΦB is the magnetic flux dt through the loop

Mutual : Φ2 = M21I1 ,M21 = mutual inductance

µ0 d,1 · d,2 (Franz) Neumann’s formula: M21 = M12 = | − | 4π P1 P2 r 1 r 2 dI Self inductance: Φ = LI , E = −L ; L = inductance dt 2 Self inductance of a solenoid: L = n µ0V , where n = number of turns per length, V =volume V0 R t Rising current in an RL circuit: I = 1 − e L R Boundary Conditions:

⊥ − ⊥  −  D1 D2 = σf E1 E2 =0 ⊥ − ⊥ 1  −   −  E1 E2 = σ D1 D2= P 1 P2 "0

⊥ − ⊥  −  − ×  B1 B2 =0 H1 H2 = nˆ Kf ⊥ − ⊥ ⊥ − ⊥  −  − ×  H1 H2 = M2 M1 B1 B2 = µ0nˆ K 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 13

Conservation Laws: 1 2 1 2 Energy density: uEM = "0|E | + |B  | 2 µ0 1 Poynting vector (flow of energy): S = EB×  µ0 Conservation of energy:  d Integral form: [U + U ]=− S · da dt EM mech ∂u Differential form: = −∇· S, where u = u + u ∂t EM mech 1 1 Momentum density: ℘ = S ; Si is the density of momentum in the i’th EM c2 c2 direction     1 2 1 1 2 Maxwell stress tensor: Tij = "0 EEi j − δij |E | + BiBj − δij |B | 2 µ0 2 where −Tij = −Tji =flowinj’th direction of momentum in the i’th direction Conservation of momentum:   d 1 3 ,i i ij j V Integral form: Pmech + 2 S d x = T da , for av olume dt c V S bounded by a surface S ∂ Differential form: (℘ ,i + ℘ ,i)=∂jTji ∂t mech EM Angular momentum:

Angular momentum density (about the origin): ,EM = r ×℘EM = "0[r ×(E ×B )] Wave Equation in 1 Dimension: ∂2f 1 ∂2f − =0, where v is the wave velocity ∂z2 v2 ∂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 =wavenumber λ =2π/k = wavelength T =2π/ω =period A = amplitude Euler identity: eiθ =cosθ + i sin θ Complex notation: f(z, t)=Re[Ae˜ i(kz−ωt)] , where A˜ = Aeiδ; “Re” is usually dropped. ω dω Wave : v = = phase velocity; v = =groupvelocity k group dk 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 14

Electromagnetic Waves: 1 ∂2E 1 ∂2B Wave Equations: ∇2E − =0, ∇2B − =0 c2 ∂t2 c2 ∂t2 Linearly Polarized Plane Waves: i(k·r −ωt) E ( r,t)=E˜0 e n,ˆ where E˜0 is a complex amplitude, nˆ is a unit vector,  and ω/|k| = vphase = c. nˆ · k = 0(transverse wave) 1 B = kˆ × E c Energy and Momentum: 2 2 −  u = "0E0 cos (kz ωt + δ) , (k = k zˆ) averages to 1/2  1 ×  || 1 2 S = EB = uc z,ˆ I(intensity) = S = "0E0 µ0 2 1 u ℘ = S = zˆ EM c2 c Electromagnetic Waves in Matter: µ" n ≡ = index of refraction µ0"0 c v = phase velocity = n 1 1 u = "|E |2 + |B |2 2 µ n B = kˆ × E c 1 uc S = E× B  = zˆ µ n Reflection and Transmission at Normal Incidence: Boundary conditions: ⊥ ⊥ X "1E = "2E E = E , 1 2 1 2 E El T 1 1 B⊥ = B⊥ B = B . 1 2 1 2 V µ1 µ2 V1 2 B Incident wave (z<0): l BT Z  ˜ i(k1z−ωt) EI (z, t)=E0,I e eˆx ER BR

1 i(k1z−ωt) V1 Interface B I (z, t)= E˜0,I e eˆy . v1 Y

Image by MIT OpenCourseWare. 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 15

Transmitted wave (z>0):

i(k2z−ωt) ET (z, t)=E˜0,T e eˆx

1 i(k2z−ωt) B T (z, t)= E˜0,T e eˆy . v2 Reflected wave (z<0):

i(−k1 z−ωt) ER(z, t)=E˜0,R e eˆx

1 i(−k1z−ωt) B R(z, t)=− E˜0,R e eˆy . v1 ω must be the same on both sides, so ω c ω c = v1 = , = v2 = k1 n1 k2 n2

Applying boundary conditions and solving, approximating µ1 = µ2 = µ0 ,   n1 − n2 2n1 E˜0,R = E˜0,I E0,T = E˜0,I n1 + n2 n1 + n2

Electromagnetic Potentials:

∂A The fields: B = ∇×A, E  = −∇ V − ∂t ∂Λ Gauge transformations: A = A + ∇ Λ ,V = V − ∂t 1 Coulomb gauge: ∇·A  =0= ⇒∇2V = − ρ (but A is complicated) "0 1 ∂V Lorentz gauge: ∇·A  = − =⇒ c2 ∂t 2 2 − 1 2  −  2≡∇ 2 − 1 ∂ V = ρ, A = µ0J ,where 2 2 "0 c ∂t 2 = D’Alembertian

Retarded time solutions (Lorentz gauge):      1 3  ρ(r ,tr)  1 3  J(r ,tr) V (r,t)= d x , A(r,t)= d x  4π"0 |r − r | 4π"0 |r − r | where |r − r | tr = t − c 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 16

Li´enard-Wiechert Potentials (potentials of a point charge): 1 q V (r,t)= vp 4π"0 |r − r p| 1 − c · ˆ

µ qvp vp A( r,t)= 0 = V (r,t) vp 2 4π |r − r p| 1 − c · ˆ c

where r p and vp are the position and velocity of the particle at the retarded time tr,and

r− r p  = r − r p , = |r − r p| , ˆ = |r − r p| Fields of a point charge (from the Li´enard-Wiechert potentials): | − |  q r r p 2 − 2 − × × E(r,t)= 3 ()c vp u+( r r p) (u ap) 4π"0 ((u· r− r p)) 1 B (r,t)= ˆ × E (r,t) c

where u = c ˆ − vp

Radiation: Radiation from an oscillating electric dipole along the z axis:

p(t)=p0 cos(ωt) ,p0 = q0d Approximations: d  λ  r,   p ω cos θ V (r, θ, t)=− 0 sin[ω(t − r/c)] 4π"0c r µ p ω A(r,t)=− 0 0 sin[ω(t − r/c)] zˆ 4πr   µ p ω2 sin θ 1 E = − 0 0 cos[ω( t − r/c)] θ,ˆ B (r,t)= rˆ × E (r,t) 4π r c   1 µ p ω2 sin θ 2 Poynting vector: S = (E× B  )= 0 0 cos[ω(t − r/c)] rˆ µ0 c 4π r   µ p2ω4 sin2 θ 1 Intensity: I = S = 0 0 r,ˆ using cos2 = 32π2c r2 2  µ p2ω4 Total power: P  = S · da = 0 0 12πc 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 17

Magnetic Dipole Radiation: Dipole moment: m (t)=m cos(ωt)ˆz,at the origin   0 µm ω2 sin θ 1 E = − 0 0 cos[ω(t − r/c)] φ,ˆ B (r,t)= rˆ × E (r,t) 4πc r c m Compared to the electric dipole radiation, p → 0 , θˆ →−φˆ 0 c General Electric Dipole Radiation: µ 1 µ E (r,t)= 0 [(ˆr· p¨)ˆ r− p¨ ] , B (r,t)= rˆ × E (r,t)=− 0 [ˆr ×p¨] 4πr c 4πrc Multipole Expansion for Radiation: The electric dipole radiation formula is really the first term in a doubly infinite series. There is electric dipole, quadrupole, ... radiation, and also magnetic dipole, quadrupole, ... radiation. Radiation from a Point Particle: When the particle is at rest at the retarded time, q E = [ ˆ ×( ˆ × ap)] rad 4π" c2|r − r | 0   1 µ q2a2 sin2 θ |  |2 ˆ 0 ˆ Poynting vector: Srad = Erad = 2 2 µ0c 16π c where θ is the angle between ap and ˆ . µ q2a2 Total power (Larmor formula): P = 0 6πc (valid for vp =0or|vp|c)

Li´enard’s Generalization if vp =0: 2 6 × 2 2 µ µ0 q γ 2 − v a µ0q dpµ dp P = a = 2 6πc c 6πm0cdτ dτ For relativists only Radiation Reaction: Abraham-Lorentz formula: µ q2 F = 0 a˙ rad 6πc The Abraham-Lorentz formula is guaranteed to give the correct average energy loss for periodic or nearly periodic motion, but one would like a formula that works under general circumstances. The Abraham-Lorentz formula leads to runaway solutions which are clearly unphysical. The problem of radiation reaction for point particles in classical electrodynamics apparently remains unsolved. 8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012 p. 18

Vector Identities:

Triple Products

A . (B x C) = B . (C x A) = C . (A x B)

A x (B x C) = B(A . C) - C(A . B)

Products Rules

∆ ∆

∆ (f g) = f ( g) + g ( f)

∆ ∆ ∆ ∆

∆ (A . B) = A x ( x B) + B x ( x A) + (A . )B + (B . )A

∆ ∆

∆ (f A) = f ( . A) + A . ( f)

∆ ∆

∆ (A x B) = B . ( x A) - A . ( x B)

∆ ∆

∆ x (f A) = f ( x A) - A x ( f)

∆ ∆ ∆ ∆

∆ (A x B) = (B . )A - (A . )B + A ( . B) - B( . A)

Second Derivatives ∆

∆ . ( x A) = 0 ∆

∆ x ( f) = 0

∆ ∆ ∆ ∆

∆ x ( x A) = ( . A) - 2A

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8.07 Electromagnetism II Fall 2012

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