Homework 1 - Vector Calculus
Due Tuesday, Sept 8. 1.The gravitational potential is U=Gy J/kg-m. Find the gradient.


g
=
−
∇U
and the gravitational field, which is
.
2. A sticky fluid is moving past a flat horizontal surface
such that the velocity is given by

v = 10yiˆ m/s.
Find both magnitude and direction of the curl,.
3. The electric field inside a uniformly charged dielectric
is

E = 10xiˆ. Find the divergence and therefore the charge
distribution.
I- Class Exercise/Homework- Vector Calculus 

1.The gravitational potential is U=Gy J/kg-m. Find the gradient. and the gravitational field, which is .g = −∇U


∂ ( gy )

g = −∇U = −∇gy = −
ĵ = −g ĵ
∂y
2. A sticky fluid is moving past a flat horizontal surface such that the velocity is given by
Find both magnitude and direction of the curl,.
 

∇ × v = curl( v ) =
iˆ
ĵ
k̂
∂
∂x
∂
∂y
∂
∂z
vx
vy
vz

v = 10 yiˆ
m/s.
( )
⎛ ∂v
⎛ ∂v y ∂v ⎞
∂v y ⎞
∂ 10 y
⎛ ∂vz ∂vx ⎞
z
x
ˆ
=⎜
−
−
−
k̂ = 10 k̂ sec −1
⎟i −⎜
⎟ k̂ =
⎟ ĵ + ⎜
∂z ⎠
∂z ⎠
∂y ⎠
∂x
⎝ ∂x
⎝ ∂y
⎝ ∂x

3. The electric field inside a uniformly charged dielectric is E = 10 xiˆ. Find the divergence and therefore
the charge distribution.
  1 ⎛ ∂E x ∂E y ∂Ez ⎞
∂E x
pC
−12
2
ρ = ε 0∇i E = ⎜
+
+
=
ε
=
9
×
10
F/m
×
10
V/m
=
90
0
ε 0 ⎝ ∂x
∂y
∂z ⎟⎠
∂x
m3
Homework 2- Electromagnetic waves.
Due Tuesday Sept 8
Show that Maxwell's equations in vacuo:

 
 
∂B
1. ε 0 ∇i E = 0
2. ∇ × E +
=0
∂t

 
 
∂E
3. ∇i B = 0
4. ∇ × B − µ 0 ε 0
=0
∂t
can be combined to produce the differential equation for for electromagnetic waves.

2
2
2
2
∂
Ey

∂ Ez
∂
E
∂
E
2
2
2
2
x
∇ E − µ 0 ε 0 2 = 0. ⇒ ∇ E x − µ 0 ε 0
= 0, ∇ E y − µ 0 ε 0
= 0, ∇ E z − µ 0 ε 0
= 0.
2
2
2
∂t
∂t
∂t
∂t



⎛ Combine eqs. 2 and 4 by taking ∇ × (∇ × E)
⎞

 
  
  
⎜
⎟
Hint: Use a vector identity: ∇ × (∇ × E) = ∇ ( ∇ • E ) − (∇ • ∇) E
⎜
 
 ⎟⎟
2
⎜⎝ Use eq.1: ∇ • E = 0, and the definition of the Laplacian: (∇ • ∇)
E ≡ ∇ E⎠
E x = E0 x e
 
i( k i r −ω t+ϕ )
, E y = E0 y e
 
i( k i r −ω t+ϕ )
Repeat the above, but for the magnetic field.
, etc. are solutions.
Electromagnetic waves.
Maxwell’s equations in a vacuum.

 
∂B
2. ∇ × E +
=0
dt

 
∂E
4. ∇ × B − µ 0 ε 0
=0
∂t
 
1. ε 0 ∇i E = 0
 
3. ∇i B = 0
Combine eqs. 2 and 4.
 
2

 
∂∇ × B
∂ E
∇ × (∇ × E) = −
= − µ0 ε 0 2
∂t
∂t
Use a vector identity:

 
  
 
∇ × (∇ × E) = ∇ ∇ • E − ∇ • ∇E

(
)

From eq.1: ∇ • E = 0,
2

∂ E
∇ 2 E − µ0 ε 0 2
∂t
  
( )

and ∇i∇ E ≡ ∇ 2 E
=0
Homework 3. Due Friday Sept 11.

Show that each component of E is a solution to the electromagnetic the wave equation for
A wave traveling in an arbitrary direction 
Write the expression for E in components: i(kx x+ ky y + kz z −ω t+ϕ )
E x = E0 x e
And similarly for E y and Ez
Show that this is a solution provided the speed of the wave is v = fλ =
ω
1
=
= 2.998 × 10 8 m/s = c
k
ε 0 µ0


 
i( k i r −ω t+ϕ ) is a solution to Show
E = E 0e
the electromagnetic the wave equation.

2

∂E
∂2 Ex
2
2
∇ E − µ0ε 2 = 0 ⇒ ∇ Ex − µ0ε
= 0, and similarly for E y and Ez .
∂t
∂t 2
∂2 Ex ∂2 Ex ∂2 Ex
∇ Ex =
+
+
∂x 2
∂y 2
∂z 2
2
 
i( k i r −ω t+ϕ )
and similarly for E y and Ez
i(kx x+ky y+kz z−ω t+ϕ )
E x = E0 x e
= E0 x e
∂2 Ex
∂ ⎛ ∂E x ⎞
=
⎜
⎟
∂x 2
∂x ⎝ ∂x ⎠
∂E x
i(k x+k y+k z−ω t+ϕ )
= ikx E0 x e x y z
∂x
∂2 Ex
i(kx x+ky y+kz z−ω t+ϕ )
2
=
−
ω
E
e
0x
∂t 2
and similarly for E y and Ez
∂2 Ex
i(kx x+ky y+kz z−ω t+ϕ )
2
=
−k
E
e
x 0x
∂x 2
A vector equation implies 3 separate equations-one for each component.
∂2 Ex ∂2 Ex ∂2 Ex
∇ Ex =
+
+
∂x 2
∂y 2
∂z 2
∂2 Ex
∇ Ex − µ0 ε 0
=0
∂t 2
2
2
∂2 Ex
∂ ⎛ ∂E x ⎞
=
⎜
⎟
∂x 2
∂x ⎝ ∂x ⎠
∂E x
i(k x+k y+k z−ω t+ϕ )
= ikx E0 x e x y z
∂x
∂2 Ex
i(kx x+ky y+kz z−ω t+ϕ )
2
=
−k
E
e
and similarly for y and z.
x 0x
∂x 2
i(kx x+ky y+kz z−ω t+ϕ )
∇ 2 E x = −(kx2 + ky2 + kz2 )E0 x e

ˆ + ĵk + k̂k
Define k = ik
x
y
z
 
i( k i r −ω t+ϕ )
∇ E x = −k E0 x e
2
2
k 2 = kx2 + ky2 + kz2

Likewize r = rx iˆ + ry ĵ + rz k̂
= xiˆ + yĵ + zk̂
 
i( k i r −ω t+ϕ )


∂2 Ex
i( k i r −ω t+ϕ )
2
= −ω E0 x e
2
∂t
∇ E x = −k E0 x e
2
2
∂2 Ex
∇ Ex − µ0ε 0 2 = 0
∂t
2
 
i( k i r −ω t+ϕ )
−k E0 x e
2
 
i( k i r −ω t+ϕ )
= − µ 0 ε 0 ω E0 x e
2
k2
This is a solution with 2 = µ 0 ε 0
ω
But k =
⇒
2π
λ
fλ = v =
and ω = 2π f
1
µ0ε 0
=
1
( 4π × 10 )( 8.854 × 10 )
−7
−12
= 2.998 × 10 8 m/s.
Homework 4 DueTues. Sept. 15.
Textbook, Foundations of Special Relativity: P130, Problems 2, 3
Homework 5.
Due. Tues. Sept 18.
Textbook: Space-time. Lorentz Trans: P130-131,
Problems 6,9,11,12
Show that Maxwellʼs wave equation is invariant
under a Lorentz transformation along the x axis for a
wave moving in the x direction .
Use:
∂Ei ∂Ei ∂x ′ ∂Ei ∂t ′
∂E
∂E ∂x ′ ∂Ei ∂t ′
=
+
i = i
+
∂x
∂x ′ ∂x ∂t ′ ∂x
∂t
∂x ′ ∂t
∂t ′ ∂t
⎛
⎝
x ′ = γ (x − vt) t ′ = γ ⎜ t −
∂x ′
=γ
∂x
∂t ′
v
= −γ 2
∂x
c
v
c2
⎞
⎠
x⎟
∂x ′
= −γ v
∂t
∂t ′
=γ
∂t
Homework
6.
• Relativistic Mechanics: P112-121, Problems 13,18,23
• Four Vectors: P121-125, Problems 25,29