Homework 1 solutions

Physics 321, Autumn Quarter 2014
Electrodynamics
Homework Assignment 1
Turn in all problems and clearly note all
constants and assumptions you use.
(1-point penalty otherwise)
Due 9:30am Thursday October 2
Problems 1, 2, 3 and 5 will not be graded.
1. Find the function φ (r) which satisfies the condition ∇ • φ (r)r = 0
2. With ω a constant vector parallel to the z-axis and r the position
vector, find
(i) ∇ × (ω × r)
with C a constant vector and r the position vector, find
(ii) ∇ • r
(iii) ∇ × r
(iv) ∇(C • r)
(v) (C • ∇)r
3. With a and b constant vectors and r the position vector, find the
divergence and curl of the following vectors.
(i) (a • r)b
(ii) (a • r)r
(iii) a × r
4. For incompressible fluids, the flux of the vector ρ v (with ρ the
fluid density and v the fluid velocity vanishes over any closed
surface. Show that in an incompressible fluid ∇ • ρ v is everywhere
zero.
5. Find the general form and the solution of Laplace’s equation
∇ 2 f = 0 for the scalar functions f (ρ ) , f (φ ) and f (z) where ρ , φ
and z are cylindrical coordinates. (This challenge problem will not be
graded. Hint: Bessel Functions. )
6. Show that the general vacuum Maxwell equations contain current
conservation ∇ • j+ ∂ρ / ∂t = 0
7. Show that ∇φ (r) = −∇#φ (r) where φ (r) = φ (r(x − x")) is a function
of the difference of the unprimed and primed position vectors.
∇ operates on unprimed coordinates, and ∇! operates on primed
coordinates. Hence, show that ∇(1 / r) = −∇#(1 / r) . We will use this
relation later in the course.
PHYS 321 - HW 1 Solutions
October 5, 2014
Problem 1
r · ( (r)r) = (r (r)) · r + (r)r · r = 0
(1)
In the second term, r · r = 3 . To compute r (r) it is convenient to write r in spherical coordinates. Then the
ˆ
r component will be the only non vanishing term. So, (1) becomes
@
ˆ
r·r+3 =0
(2)
@r
Let a prime denote di↵erentiation with respect to r. Rearranging terms and using ˆ
r · r = r, we find the first
order ODE
0
=
3
r
(3)
Solving this gives
(r) = e
3ln(r)+C
=
A
r3
(4)
where C and A are constants. Hence the general solution to (1) is the r12 ˆ
r form exhibited by e.g., the electric
field a point particle or the gravitational field of a massive spherical body.
Problem 2
1. r ⇥ (!ˆ
z) ⇥ r = 2!ˆ
z
2. r · r = 3
3. r ⇥ r = 0
4. r(C · r) = C
5. (C · r)r = C
Problem 3
1. r · ((a · r)b) = a · b
r ⇥ ((a · r)b) = a ⇥ b
2. r · ((a · r)r) = 4(a · r)
r ⇥ ((a · r)r) = a ⇥ r
3. r · (a ⇥ r) = 0
r ⇥ (a ⇥ r) = 2a
1
Problem 4
For an incompressible fluid, the flux through a closed surface @ vanishes:
I
dA · (⇢v) = 0
(5)
@
From the divergence theorem, it follows that
I
Z
dA · (⇢v) =
dV (r · (⇢v)) = 0
(6)
@
Since this holds true for any closed surface @ , or equivalently since the interior space over which the volume
integral is taken is arbitrary, the volume integral vanishes only if its argument vanishes:
r · (⇢v) = 0
(7)
Problem 5
For a general function (⇢, , z) that satisfies Laplace’s equation in cylindrical coordinates, the Laplacian can be
written in cylindrical coordinates as follows
@2
1@
1 @2
@2
+
+
+
=0
@⇢2
⇢ @⇢
⇢2 @ 2
@z 2
(8)
We can proceed to find the general form of solutions in the case where depends only on ⇢, , or z respectively
by using separation of variables to break apart the PDE (8). To do this, we assume that can be written in
a separated form (⇢, , z) = R(⇢)Q( )Z(z). Let a prime denote a derivative of a function with respect to its
given argument. Then (8) becomes
✓
◆
R0
R
00
QZ R +
+ 2 ZQ00 + RQZ 00 = 0
(9)
⇢
⇢
Divide by RQZ and rearrange Then
R00 1 R0
1 Q00
+
+ 2
=
R
⇢R
⇢ Q
Z 00
Z
(10)
The RHS depend of (10)only on z, while the LHS is independent of z. Thus for the equality to be satisfied, both
sides must be equal to the same constant, which we will call k 2 . So we end up with two equations:
Z 00
k2 Z = 0
(11)
R00 1 R0
1 Q00
+
+ 2
+ k2 = 0
R
⇢R
⇢ Q
(12)
Multiply (12) be ⇢2 and rearrange. Then
⇢2
R00
R0
+ ⇢ + ⇢2 k 2 =
R
R
Q00
Q
(13)
Now the LHS of (13) depends only on ⇢, while the RHS depends only on . Hence both sides must be equal to
the same constant, which we will call ⌫ 2 . We thus get two equations:
Q00 + ⌫ 2 Q = 0
✓
1 0
R + R + k2
⇢
00
The general solutions to (11), (14), and (15) are
2
⌫2
⇢2
◆
(14)
R=0
(15)
Z(z) = e±z
(16)
Q( ) = e±i⌫
(17)
R(⇢) = J±⌫ (k⇢)
(18)
where J±⌫ (k⇢) are Bessel functions of the first kind. R(⇢), Q( ), and Z(z) correspond to the functions f (⇢),
f ( ), and f (z) referred to in the problem.
Problem 6
To derive the continuity equation, we start with the following Maxwell equations:
r ⇥ B = µ0 j + µ0 ✏ 0
r·E=
@E
@t
(19)
⇢
✏0
(20)
Taking the divergence of (19) and using that the divergence of the curl vanishes identically, we find
r · j + ✏0
@
r·E=0
@t
(21)
which after substituting in (20) for r · E gives
r·j+
@⇢
=0
@t
(22)
Problem 7
We can show that r (r) = r0 (r) where r = r(x x0 ) is a function of the di↵erence between two position
vectors by repeated application of the chain rule in Cartesian coordinates:
r (r) =
X
i
x
ˆi
X @r @
@
(r) =
x
ˆi
(r)
@xi
@xi @r
(23)
i
where i = 1, 2, 3 labels the basis vectors in Cartesian coordinates. Let
component of x x0 . Then
@r
@r @ xi
=
=
@xi
@ xi @xi
xi = xi
@r @ xi
=
@ xi @x0i
x0i be the di↵erence in the ith
@r
@x0i
Since the basis for the position vectors x and x0 must be the same for the expression x
we must have x
ˆi = x
ˆ0i for all i. So it follows that
r (r) =
X
i
x
ˆi
@r @
(r) =
@xi @r
So for the case where (r) = 1/r = 1/|x
x0 |,
X
i
x
ˆ0i
@r @
(r) =
@x0i @r
we have
1
r =
r
3
r0
1
r
X
i
x
ˆ0i
@
(r) =
@x0i
(24)
x0 to be meaningful,
r0 (r)
(25)
(26)

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