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:
·B
= Ai Bi ,
A
×B
i = "ijk Aj Bk ,
A
"ijk "pqk = δip δjq − δiq δjp
det A = "i1 i2 ···in A1,i1 A2,i2 · · · An,in
Rotation of a Vector:
Ai = Rij Aj ,
Orthogonality: Rij Rik = δjk
j=1
Rotation about z-axis by φ: Rz (φ)ij
Rotation about axis n̂ by φ:∗∗∗
j=2

i=1  cos φ − sin φ


= i=2 

 sin φ cos φ
i=3
0
0
(RT T = I)
j=3
0
0
1







R(n̂, φ)ij = δij cos φ + n̂i n̂j (1 − cos φ) − "ijk n
ˆ k sin φ .
Vector Calculus:
∂i ≡
Gradient:
ϕ)i = ∂i ϕ ,
(∇
Divergence:
·A
≡ ∂i A i
∇
Curl:
× A)
i = "ijk ∂j Ak
(∇
Laplacian:
· (∇
ϕ) =
∇2 ϕ = ∇
∂
∂xi
∂ 2ϕ
∂xi ∂xi
Fundamental Theorems of Vector Calculus:
b
Gradient:
ϕ · d, = ϕ(b) − ϕ(a)
∇
a
Divergence:
V
Curl:
S
· da
A
·A
d3 x =
∇
S
where S is the boundary of V
· d,
(∇ × A) · da =
A
P
where P is the boundary of S
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 2
Delta Functions:
ϕ(r )δ 3 (r − r ) d3 x = ϕ(r )
ϕ(x)δ(x − x ) dx = ϕ(x ) ,
d
dϕ ϕ(x) δ(x − x ) dx = −
dx
dx x=x
δ(x − xi )
, g(xi ) = 0
δ(g(x)) =
|g (xi )|
i
1
r
−
r
·
= 4πδ 3 (r − r )
∇
= −∇2
3
|r − r |
|r − r |
x r̂j
1
δij − 3r̂i r̂j
4π
j
δij δ 3 (r)
≡ ∂i 3 = −∂i ∂j
+
=
∂i
2
3
r
r
3
r
r
· 3(d · r̂)r̂ − d = − 8π (d · ∇
)δ 3 (r )
∇
3
r3
δ 3 (r )
× 3(d · r̂)r̂ − d = − 4π d × ∇
∇
3
3
r
Electrostatics:
, where
F = qE
1
1 (r − r ) qi
(r − r )
=
E(r ) =
r ) d3 x
3 ρ(
4π"0 i |r − r |3
4π"0
|r − r |
"0 =permittivity of free space = 8.854 × 10−12 C2 /(N·m2 )
1
= 8.988 × 109 N·m2 /C2
4π"0
r
1
ρ(r ) 3 E(r ) · d, =
d x
V (r ) = V (r 0 ) −
4π"0
|r − r |
r0
·E
= ρ ,
×E
= 0,
= −∇V
∇
E
∇
"0
ρ
(Poisson’s Eq.) ,
ρ = 0 =⇒ ∇2 V = 0 (Laplace’s Eq.)
∇2 V = −
"0
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
ρ(r )ρ(r )
W =
=
d3 x d3 x
2 4π"0
rij
2 4π"0
|r − r |
1
W =
2
ij
i=j
1
d xρ(r )V (r ) = "0
2
3
2 3
E
d x
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 3
Conductors:
= σ n̂
Just outside, E
"0
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
a
a2
Image charge in sphere of radius a: Image of Q at R is q = − 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
r
+ 2 ∇2θ ϕ = 0 ,
∇ ϕ(r, θ, φ) = 2
r ∂r
r
∂r
where the angular part is given by
1 ∂
∂ϕ
1 ∂ 2ϕ
2
sin θ
+
∇θ ϕ ≡
sin θ ∂θ
∂θ
sin2 θ ∂φ2
2
()
()
∇2θ Ci1 i2 ...i n
ˆ i1 n
ˆ i2 . . . n̂i = −,(, + 1)Ci1 i2 ...i n̂i1 n̂i2 . . . n̂i ,
()
where Ci1 i2 ...i is a symmetric traceless tensor and
n̂ = sin θ cos φ ê1 + sin θ sin φ ê2 + cos θ ê3 .
General solution to Laplace’s equation:
()
∞
C
()
i2 ...i
rˆi1 r̂i2 . . . r̂i ,
V (r ) =
Ci1 i2 ...i r + i1+1
r
=0
where r = rr̂
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 4
Azimuthal Symmetry:
∞ B
A r + +1 { ẑi1 . . . ẑi } r̂i1 . . . r̂i
V (r ) =
r
=0
where { . . . } denotes the traceless symmetric part of . . . .
Special cases:
{1} = 1
{ ẑi } = ẑi
{ ẑi ẑj } = ẑi ẑj − 13 δij
ẑi δjk + ẑj δik + ẑk δij
{ ẑi ẑj ẑk ẑm } = zˆi ẑj ẑk ẑm − 71 ẑi ẑj δkm + ẑi ẑk δmj + ẑi ẑm δjk + ẑj zˆk δim
1
δij δkm + δik δjm + δim δjk
+ ẑj ẑm δik + ẑk ẑm δij + 35
{ ẑi ẑj ẑk } = ẑi ẑj ẑk −
1
5
Legendre Polynomial / Spherical Harmonic expansion:
General solution to Laplace’s equation:
∞ Bm
V (r ) =
Am r + +1 Ym (θ, φ)
r
=0 m=−
2π
Orthonormality:
π
dφ
0
0
sin θ dθ Y∗ m (θ, φ) Ym (θ, φ) = δ δm m
Azimuthal Symmetry:
∞ B
V (r ) =
A r + +1 P (cos θ)
r
=0
Electric Multipole Expansion:
First several terms:
1 Q p
· rˆ 1 r̂i r̂j
Q
+
·
·
·
, where
V (r ) =
+ 2 +
ij
r
2 r3
4π"0 r
3
3
Q = d x ρ(r ) , pi = d x ρ(r ) xi Qij = d3 x ρ(r )(3xi xj −δij |r |2 ) ,
dip (r ) = − 1 ∇
E
4π"0
×E
dip (r ) = 0 ,
∇
p · r̂
r2
=
1 3(p · r̂)r̂ − p
1
−
pi δ 3 (r )
3
4π"0
r
3"0
·E
dip (r ) = 1 ρdip (r ) = − 1 p · ∇δ
3 (r )
∇
"0
"0
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 5
Traceless Symmetric Tensor version:
V (r ) =
∞
1 1
()
Ci1 ...i rˆi1 . . . r̂i ,
+1
r
4π"0
=0
where
()
Ci1 ...i
(2, − 1)!!
=
,!
ρ(r ) { xi1 . . . xi } d3 x
(r ≡ rr̂ ≡ xi eˆi )
∞
(2, − 1)!! r 1
=
{ r̂i1 . . . r̂i } r̂i1 . . . r̂i ,
,!
r +1
|r − r |
for r < r
=0
(2, − 1)!! ≡ (2, − 1)(2, − 3)(2, − 5) . . . 1 =
(2,)!
, with (−1)!! ≡ 1 .
2 ,!
Reminder: { . . . } denotes the traceless symmetric part of . . . .
Griffiths version:
∞
1 1
V (r ) =
r ρ(r )P (cos θ ) d3 x
+1
r
4π"0
=0
where θ = angle between r and r .
∞
r
1
<
=
P (cos θ ) ,
+1 |r − r |
r>
∞
√
=0
1
P (x) = 2 ,!
d
dx
1
=
λ P (x)
2
1 − 2λx + λ
=0
(x2 − 1) ,
(Rodrigues’ formula)
P (1) = 1
1
P (−x) = (−1) P (x)
−1
dx P (x)P (x) =
2
δ 2, + 1
Spherical Harmonic version:∗∗∗
∞
1 4π qm
V (r ) =
Ym (θ, φ)
4π"0
2, + 1 r +1
=0 m=−
where qm =
∗ Ym
r ρ(r ) d3 x
∞ 1
4π r ∗ =
Y (θ , φ )Ym (θ, φ) ,
2, + 1 r +1 m
|r − r |
=0 m=−
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
r δ 3 (r − r d ) , where r d = position of dipole
ρdip (r ) = −p · ∇
= (p · ∇
)E
=∇
(p · E)
F
(force on a dipole)
= p × E
(torque on a dipole)
U = −p · E
Electrically Polarizable Materials:
(r ) = polarization = electric dipole moment per unit volume
P
· n̂
ρbound = −∇ · P ,
σbound = P
≡ "0 E
+P
,
D
·D
= ρfree ,
∇
×E
= 0 (for statics)
∇
Boundary conditions:
⊥
⊥
Eab
ove − Ebelow =
σ
"0
E
above − Ebelow = 0
⊥
⊥
Dab
ove − Dbelow = σfree
D
above − Dbelow = Pabove − Pbelow
Linear Dielectrics:
= "0 χe E,
P
χe = electric susceptibility
= "E
" ≡ "0 (1 + χe ) = permittivity,
D
"
= 1 + χe = relative permittivity, or dielectric constant
"r =
"0
N α/"0
, where N = number density of atoms
1 − Nα
30
or (nonpolar) molecules, α = atomic/molecular polarizability (P = αE)
1
·E
d3 x
(linear materials only)
Energy: W =
D
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 = γm0v ,
1
γ= 1−
v2
c2
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
=
F
p. 7
=
Id, × B
d3 x
J × B
Current Density:
J · da
Current through a surface S: IS =
S
Charge conservation:
∂ρ
· J
= −∇
∂t
Moving density of charge: J = ρv
Biot-Savart Law:
d, × (r − r )
K(r ) × (r − r ) µ0
µ0
B (r ) =
I
=
da
4π
|r − r |3
4π
|r − r |3
µ0
J(r ) × (r − r ) 3
=
d x
4π
|r − r |3
where µ0 = permeability of free space ≡ 4π × 10−7 N/A2
Examples:
= µ0 I φ̂
Infinitely long straight wire: B
2πr
= µ0 nI0 ẑ , where n = turns per
Infinitely long tightly wound solenoid: B
unit length
0, z) =
Loop of current on axis: B(0,
µ0 IR2
ẑ
2(z 2 + R2 )3/2
r ) = 1 µ0 K
× n̂ , n̂ = unit normal toward r
Infinite current sheet: B(
2
Vector Potential:
)coul = µ0
A(r
4π
J(r ) 3 d x ,
|r − r |
=∇
×A
,
B
·A
coul = 0
∇
·B
= 0 (Subject to modification if magnetic monopoles are discovered)
∇
(r ) = A(
r ) + ∇Λ(
r ) for any Λ(r ). B
=∇
×A
is
Gauge Transformations: A
unchanged.
Ampère’s Law:
· d, = µ0 Ienc
B
×B
= µ0 J , or equivalently
∇
P
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 8
Magnetic Multipole Expansion:
Traceless Symmetric Tensor version:
∞
µ0 ()
{ r̂i1 . . . r̂i }
Mj;i1 i2 ...i
r +1
4π
=0
(2, − 1)!!
()
d3 xJj (r ){ xi1 . . . xi }
where Mj ;i1 i2 ...i =
,!
Current conservation restriction:
d3 x Sym(xi1 . . . xi−1 Ji ) = 0
Aj (r ) =
i1 ...i
where
Sym
i1 ...i
means to symmetrize — i.e. average over all
orderings — in the indices i1 . . . i
Special cases: , = 1:
d3 x Ji = 0
d3 x (Ji xj + Jj xi ) = 0
, = 2:
× r̂
r ) = µ0 m
Leading term (dipole): A(
,
4π r 2
where
1
(1)
mi = − "ijk Mj;k
2
1
1
m
= I
r × d, =
d3 x r × J = Ia ,
2
2 P
where a =
da for any surface S spanning P
S
× r̂
µ0 3(m
· r̂)r̂ − m
2µ0
dip (r ) = µ0 ∇
×m
=
m
δ 3 (r )
B
+
2
3
4π
r
3
4π
r
·B
dip (r ) = 0 ,
×B
dip (r ) = µ0 Jdip (r ) = −µ0 m
δ 3 (r )
×∇
∇
∇
Griffiths version:
∞
µ0 I 1
A(r ) =
(r ) P (cos θ )d,
4π
r +1
=0
Magnetic Fields in Matter:
Magnetic Dipoles:
1
1
d3 x r × J = Ia
m
= I
r × d, =
2 P
2
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 9
r δ 3 (r − r d ), where r d = position of dipole
×∇
Jdip (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
× n̂
K
Jbound = ∇
= Jfree ,
,
×H
·B
=0
≡ 1B
−M
∇
∇
H
µ0
Boundary conditions:
⊥
⊥
⊥
⊥
⊥
⊥
Hab
Bab
ove − Bbelow = 0
ove − Hbelow = −(Mabove − Mbelow )
H
B
above − Bbelow = µ0 (K × n̂)
above − Hbelow = Kfree × n̂
Linear Magnetic Materials:
= χm H,
χm = magnetic susceptibility
M
= µH
B
µ = µ0 (1 + χm ) = permeability,
Magnetic Monopoles:
r ) = µ0 qm r̂ ;
B(
Force on a static monopole: F = qm B
4π r 2
= µ0 qe qm r̂ , where r̂ points
Angular momentum of monopole/charge system: L
4π
from qe to qm
µ0 qe qm
1
= h̄ × integer
Dirac quantization condition:
4π
2
Connection Between Traceless Symmetric Tensors and Legendre Polynomials
or Spherical Harmonics:
(2,)!
{ ẑi1 . . . ẑi } n̂i1 . . . n̂i
P (cos θ) = 2 (,!)2
For m ≥ 0,
(,m)
Ym (θ, φ) = Ci1 ...i n̂i1 . . . n̂i ,
(,m)
ˆ+
ˆim+1 . . . ẑi } ,
where Ci1 i2 ...i = dm { û+
i1 . . . u
im z
2m (2, + 1)
(−1)m (2,)!
,
with dm =
2 ,!
4π (, + m)! (, − m)!
1
and û+ = √ (êx + iêy )
2
∗
Form m < 0, Y,−m (θ, φ) = (−1)m Ym
(θ, φ)
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 10
More Information about Spherical Harmonics:∗∗∗
2 + 1 ( − m)! m
P (cos θ)eimφ
Y£m (θ, φ) =
4π ( + m)! £
where P£
m (cos θ) is the associated Legendre function, which can be defined by
P£m (x) =
£+m
(−1)m
2 m/2 d
(1
−
x
)
(x2 − 1)£
£
£+m
2 !
dx
Legendre Polynomials:
SPHERICAL HARMONICS Ylm(θ , φ)
l=0
1
Y00 =
4π
3
sin θeiφ
8π
Y11 = l=1
3
cos θ
4π
Y10 =
Y22 =
l=2
1
4
15
sin2 θe2iφ
2π
15
sin θ cosθeiφ
8π
Y21 = -
Y20 =
5
( 32 cos2θ
4π
1
)
2
35
sin3 θe3iφ
4π
Y33 = -
1
4
Y32 =
1
4
105
sin2 θ cos θe2iφ
2π
Y31 = -
1
4
21
sinθ (5cos2θ -1)eiφ
4π
l=3
Y30 =
7
( 5 cos3θ
4π 2
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) ∇
"0
×B
= µ0 J + 1 ∂E
(iv)∇
c2 ∂t
·B
=0
(ii) ∇
1
c2
+ v × B)
Lorentz force law: F = q(E
where µ0 "0 =
∂ρ
· J
= −∇
∂t
Maxwell’s Equations in Matter:
Charge conservation:
:
Polarization P and magnetization M
· P ,
ρb = −∇
×M
,
Jb = ∇
ρ = ρf + ρb ,
J = Jf + Jb
Auxiliary Fields:
≡ B −M
,
H
µ0
≡ "0 E
+ P
D
Maxwell’s Equations:
·D
= ρf
(i) ∇
·B
=0
(ii) ∇
×E
= − ∂B ,
(iii)∇
∂t
×H
= Jf + ∂D
(iv)∇
∂t
For linear media:
= "E
,
D
= 1B
H
µ
where " = dielectric constant, µ = relative permeability
∂D
= displacement current
Jd ≡
∂t
Maxwell’s Equations with Magnetic Charge:
= −µ0 Jm − ∂B ,
×E
(iii)∇
∂t
= µ0 Je + 1 ∂E
·B
= µ0 ρm
×B
(iv)∇
(ii) ∇
c2 ∂t
1
Magnetic Lorentz force law: F = qm B − 2 v × E
c
·E
= 1 ρe
(i) ∇
"0
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 12
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 = ρ , where , = length, A = cross-sectional area, and ρ =
A
resistivity
V0 −t/RC
Charging an RC circuit: I =
,
Q = CV0 1 − e−t/RC
e
R
+ v × B)
· d, , where v is either the velocity
EMF (Electromotive force): E ≡ (E
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 = − dΦB , where ΦB is the magnetic flux
changing B
dt
through the loop
Mutual inductance: Φ2 = M21 I1 , M21 = mutual inductance
µ0
d,1 · d,2
(Franz) Neumann’s formula: M21 = 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
t
L
1−e
Rising current in an RL circuit: I =
R
Boundary Conditions:
D1⊥ − D2⊥ = σf
1
E1⊥ − E2⊥ = σ
"0
− E
= 0
E
1
2
−D
= P − P
D
B1⊥ − B2⊥ = 0
H1⊥ − H2⊥ = M2⊥ − M1⊥
−H
= −n̂ × K
f
H
1
2
= −µ0 n̂ × K
−B
B
1
2
1
2
1
2
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 13
Conservation Laws:
1
1 2
2
|B|
Energy density: uEM =
"0 |E | +
2
µ0
= 1 E
×B
Poynting vector (flow of energy): S
µ0
Conservation of energy:
d
· da
Integral form:
[UEM + Umech ] = − S
dt
∂u
, where u = uEM + umech
·S
Differential form:
= −∇
∂t
1 1
; 2 Si is the density of momentum in the i’th
Momentum density: ℘EM = 2 S
c
c
direction
1
1
1
2
2
Maxwell stress tensor: Tij = "0 Ei Ej − δij |E| +
Bi Bj − δij |B|
2
µ0
2
where −Tij = −Tji = flow in j’th direction of momentum in the i’th direction
Conservation of momentum:
d
1
3
Si d x =
Tij daj , for a volume V
Integral form:
Pmech,i + 2
dt
c V
S
bounded by a surface S
∂
Differential form:
(℘mech,i + ℘EM,i ) = ∂j Tji
∂t
Angular momentum:
× B)]
Angular momentum density (about the origin): ,EM = r ×℘EM = "0 [r ×(E
Wave Equation in 1 Dimension:
1 ∂ 2f
∂2f
−
= 0 , where v is the wave velocity
v 2 ∂t2
∂z 2
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 à = Aeiδ ; “Re” is usually
Complex notation: f (z, t) = Re[Ae
dropped.
ω
dω
= group velocity
Wave velocities: v = = phase velocity; vgroup =
k
dk
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 14
Electromagnetic Waves:
1 ∂ 2E
=0,
c2 ∂t2
Linearly Polarized Plane Waves:
−
Wave Equations: ∇2 E
−
∇2 B
1 ∂ 2B
=0
c2 ∂t2
r −ωt)
(r , t) = E
˜0 ei(k·
n
ˆ , where Ẽ0 is a complex amplitude, n̂ is a unit vector,
E
and ω/|k| = vphase = c.
n̂ · k = 0
(transverse wave)
= 1 k̂ × E
B
c
Energy and Momentum:
u = "0 E02 cos2 (kz − ωt + δ) , (k = k ẑ)
averages to 1/2
! " 1
= 1E
×B
= uc zˆ ,
= "0 E02
S
I (intensity) = |S|
µ0
2
1 u
℘EM = 2 S
= ẑ
c
c
Electromagnetic Waves in Matter:
#
µ"
n≡
= index of refraction
µ0 "0
c
v = phase velocity =
n
1
|2 + 1 |B|
2
u=
"|E
2
µ
= n k̂ × E
B
c
= 1E
×B
= uc ẑ
S
µ
n
Reflection and Transmission at Normal Incidence:
Boundary conditions:
"1 E1⊥ = "2 E2⊥
B1⊥
=
B2⊥
i(k1 z−ωt)
ET
El
V2
V1
Bl
Incident wave (z < 0):
I (z, t) = Ẽ0,I e
E
X
,
= E
E
1
2
1 1 B1 =
B .
µ1
µ2 2
BT
Z
êx
I (z, t) = 1 Ẽ0,I ei(k1 z−ωt) êy .
B
v1
ER
BR
V1
Interface
Y
Image by MIT OpenCourseWare.
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 15
Transmitted wave (z > 0):
T (z, t) = Ẽ0,T ei(k2 z−ωt) êx
E
T (z, t) = 1 Ẽ0,T ei(k2 z−ωt) êy .
B
v2
Reflected wave (z < 0):
R (z, t) = Ẽ0,R ei(−k1 z−ωt) êx
E
R (z, t) = − 1 E
˜0,R ei(−k1 z−ωt) êy .
B
v1
ω must be the same on both sides, so
ω
c
ω
c
= v1 =
,
= v2 =
n1
k2
n2
k1
Applying boundary conditions and solving, approximating µ1 = µ2 = µ0 ,
n1 − n2 ˜
2n1
˜
˜0,I
E0,R =
E0,I
E
E0,T =
n1 + n2
n1 + n2
Electromagnetic Potentials:
=∇
×A
,
The fields: B
= −∇
V − ∂A
E
∂t
+∇
Λ ,
= A
Gauge transformations: A
·A
=0
Coulomb gauge: ∇
=⇒
·A
= − 1 ∂V
Lorentz gauge: ∇
c2 ∂t
2
2
V =−
1
ρ,
"0
2
V =V −
∇2 V = −
1
ρ
"0
∂Λ
∂t
is complicated)
(but A
=⇒
= −µ0 J ,
A
where
2
≡ ∇2 −
1 ∂2
c2 ∂t2
= D’Alembertian
Retarded time solutions (Lorentz gauge):
r , tr )
r , tr )
1
1
3 ρ(
3 J(
V (r , t) =
,
A(
r
,
t)
=
d
x
d x
4π"0
|r − r |
4π"0
|r − r |
where
tr = t −
|r − r |
c
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 16
Liénard-Wiechert Potentials (potentials of a point charge):
V (r , t) =
1
q
4π"0 |r − r p | 1 −
qvp
(r , t) = µ0
A
4π |r − r p | 1 −
vp
c
vp
c
·ˆ
·ˆ
=
vp
V (r , t)
c2
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 | ,
ˆ =
r − r p
|r − r p |
Fields of a point charge (from the Liénard-Wiechert potentials):
r , t) =
E(
2
q
|r − r p |
2
(
)
+
(
)
v
u
r
−
r
)
×
(
u
×
a
−
c
p
p
p
4π"0 (u · (r − r p ))3
r , t) = 1 ˆ × E(
r , t)
B(
c
where u = c ˆ − vp
Radiation:
Radiation from an oscillating electric dipole along the z axis:
p(t) = p0 cos(ωt) ,
p0 = q0 d
Approximations: d λ r,
cos θ
p0 ω
V (r, θ, t) = −
sin[ω(t − r/c)]
4π"0 c
r
r , t) = − µ0 p0 ω sin[ω(t − r/c)] ẑ
A(
4πr
2
µ
p
ω
sin
θ
0
0
r , t) = 1 r̂ × E(
r , t)
=−
E
cos[ω (t − r/c)] θˆ ,
B(
c
4π
r
2
1 µ0 p0 ω 2 sin θ
Poynting vector: S =
(E × B ) =
r̂
cos[ω(t − r/c)]
µ0
c
4π
r
! " µ p2 ω 4 sin2 θ
$ 2% 1
0 0
=
cos =
Intensity: I = S
r
ˆ
,
using
r2
2
32π 2 c
! "
2 4
· da = µ0 p0 ω
Total power: P =
S
12πc
8.07 FORMULA SHEET FOR FINAL EXAM, FALL 2012
p. 17
Magnetic Dipole Radiation:
Dipole moment: m
(t) = m0 cos(ωt) ẑ , at the origin
µ0 m0 ω 2 sin θ
r , t) = 1 r̂ × E(
r , t)
E=−
cos[ω(t − r/c)] φˆ ,
B(
4πc
r
c
m0
,
θ̂ → −φ̂
Compared to the electric dipole radiation, p0 →
c
General Electric Dipole Radiation:
¨
r , t) = 1 r̂ × E(
r , t) = − µ0 [r̂ ×p]
(r , t) = µ0 [(ˆ
E
r · p¨ )ˆ
r − p¨ ] ,
B(
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
rad =
E
[ ˆ × ( ˆ × ap )]
2
4π"0 c |r − r |
2 2
rad = 1 |E
rad |2 ˆ = µ0 q a
Poynting vector: S
16π 2 c
µ0 c
where θ is the angle between ap and ˆ .
Total power (Larmor formula): P =
sin2 θ
2
ˆ
µ0 q 2 a 2
6πc
(valid for vp = 0 or |vp | c)
Liénard’s Generalization if vp = 0:
2 µ0 q 2 γ 6
v
×
a
P =
a2 − =
6πc
c µ0 q 2 dpµ dpµ
6πm20 c dτ dτ
For relativists only
Radiation Reaction:
Abraham-Lorentz formula:
2
rad = µ0 q ȧ
F
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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∆ ∆
∆
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