Homework 2 Solutions
Problem 2.1: Tangent Power Series Find the first three terms in the
power series of tan θ. Do so by using the set of power series that you are asked
to memorize.
Solution:
We begin with
1 3
1
x + x5 + · · ·
3!
5!
1
1
cos x = 1 − x2 + x4 + · · ·
2
4!
1
= 1 − x + x2 − x3 + · · ·
1+x
sin x = x −
(1)
(2)
(3)
The third line above can come either from the (1 + z)p power series (with
p = −1), or by taking a derivative of the ln(1 + z) power series. The latter is
actually a little easier.
Given these, we can write
sin θ
cos θ
1 3
1 5
θ − 3!
θ + 5!
θ + ···
=
1 2
1 4
1 − 2 θ + 4! θ + · · ·
tan θ =
(4)
(5)
1
Now the bottom looks a lot like 1+θ
, so we can use that power series:
1
1
1
1
1
1
tan θ = θ − θ3 + θ5 + · · ·
1 − (− θ2 + θ4 + · · · ) + (− θ2 + θ4 + · · · )2 + · · ·
3!
5!
2
4!
2
4!
(6)
We now just need to multiply this all out and collect terms with the same power,
making sure that we have all the terms with that power, and drop the higher
order terms. Since we only want the first three terms (up to θ5 ), we can start
dropping out higher terms than that.
1
1
1
1
1
1
tan θ = θ − θ3 + θ5 + · · ·
1 − (− θ2 + θ4 + · · · ) + (− θ2 + θ4 + · · · )2 + · · ·
3!
5!
2
4!
2
4!
(7)
1
1
1
1
1
1 + θ2 − θ4 + θ4 + · · ·
(8)
= θ − θ3 + θ5 + · · ·
3!
5!
2
4!
4
1
1
1
5
= θ − θ3 + θ5 + · · ·
1 + θ2 + θ4 + · · ·
(9)
3!
5!
2
4!
1
5
1
1
1
= (θ + θ3 + θ5 + · · · ) − (θ3 + θ5 + · · · ) + (θ5 + · · · ) + · · · (10)
2
4!
3!
2
5!
2
16
= θ + θ3 + θ5 + · · ·
(11)
3!
5!
1
1
2
= θ + θ3 + θ5 + · · ·
3
15
(12)
You could have found this power series by taking five derivatives of tan θ (if
the question hadn’t stated otherwise), but that would have been rather more
painful.
Problem 2.2: Plotting a vector relation
a) Make a sketch of the graph
|~r − ~a| = 2
(13)
for each of the following values of ~a:
~a = ~0
~a =
~a =
(14)
2ˆ
x − 3ˆ
y
points due east and is 2 units long
(15)
(16)
Solution: These are the equations for a circle of radius 2 centered at (i)
the origin, (ii) (2, 3) and (iii)(2, 0), respectively.
b) Write a brief description of the geometric meaning of the equation
|~r − ~a| = 2
(17)
Solution: This is the equation for a circle centered at the tip of vector
~a with radius 2 units.
Problem 2.3: Quadrupole
a) A linear quadrupole is a series of three charges in a line, in this case, along
the z-axis. Charges +Q at z = ±D and charge −2Q at z = 0. Find the
electrostatic potential at a point P in the x,y-plane at a distance s from
the center of the quadrupole.
Solution: Choose the point at which you are trying to measure the
potential to lie on the x-axis or choose to work in cylindrical coordinates.
By the principle of superposition, the potential due to the three charges
is just the sum of the potentials due to the individual charges.
V (~r)
=
1
4π0
=
1
4π0
Q
2Q
Q
−
+
0
00
|~r − ~r | |~r − ~r | |~r − ~r000 |
Q
2Q
Q
−
+
|sˆ
x − Dˆ
z|
r
|sˆ
x + Dˆ
z|
2
(18)
(19)
2Q
4π0
=
√
1
1
−
s
s2 + D2
(20)
b) Assume s D. Find the first two non-zero terms of a Laurent series
expansion to the electrostatic potential you found in the first part of this
problem.
Solution: In the first term from your answer to part (a) above, pull
out a factor of s2 from the square root. Then use the memorized series
1
expansion for (1 + )p with = D
s and p = 2 .
2Q
1
1
√
V (~r) =
−
(21)
4π0
s
s2 + D2


2Q 
1
1
q
=
(22)
−
2
4π0 
s
s 1+ D
s
!
)
(
2
4
2Q
1 D
3 D
1
1
=
1−
+
+ ... −
(23)
4π0 s
2 s
8 s
s
(
!)
2
4
2Q
1 D
1
3 D
=
−
+
+ ...
(24)
4π0 s
2 s
8 s
Problem 2.4: Curvilinear distance
a) Find the distance |~r2 − ~r1 | between the point ~r1 = (x1 , y1 , z1 ) and the
point ~r2 = (x2 , y2 , z2 ) in rectangular coordinates.
r2
z
|r 2 - r 1 |
y
r1
x
Solution:
In rectangular coordinates:
|~r2 − ~r1 | =
=
|(x2 x
ˆ + y2 yˆ + z2 zˆ) − (x1 x
ˆ + y1 yˆ + z1 zˆ)|
p
2
2
(x2 − x1 ) + (y2 − y1 ) + (z2 − z1 )2
3
(25)
(26)
b) Show that this same distance written in cylindrical coordinates is:
q
|~r2 − ~r1 | = s22 + s21 − 2s1 s2 cos(φ2 − φ1 ) + (z2 − z1 )2
Solution:
In cylindrical coordinates:
x
= s cos φ
y
= s sin φ
z
= z
Plug these coordinates into the answer to part (a) above. (Note: It is
nontrivial to start this calculation from the position vector. Why? What
is the position vector ~r in cylindrical coordinates?)
p
|~r2 − ~r1 | =
(s2 cos φ2 − s1 cos φ1 )2 + (s2 sin φ2 − s1 sin φ1 )2 + (z2 − z1 )2
=
2
s2 (cos2 φ2 + sin2 φ2 ) + s21 (cos2 φ1 + sin2 φ1 )
−2s1 s2 (cos φ2 cos φ1 + sin φ2 sin φ1 ) + (z2 − z1 )2
=
12
q
s22 + s21 − 2s1 s2 cos(φ2 − φ1 ) + (z2 − z1 )2
c) Show that this same distance written in spherical coordinates is:
q
|~r2 − ~r1 | = r22 + r12 − 2r1 r2 [sin θ2 sin θ1 cos(φ2 − φ1 ) + cos θ2 cos θ1 ]
Solution:
In spherical coordinates:
x
= r sin θ cos φ
y
= r sin θ sin φ
z
= r cos θ
Plug these coordinates into the answer to part (a) above. (Note: It is
nontrivial to start this calculation from the position vector. Why? What
is the position vector ~r in spherical coordinates?)
|~r2 − ~r1 | = (r2 sin θ2 cos φ2 − r1 sin θ1 cos φ1 )2
+(r2 sin θ2 sin φ2 − r1 sin θ1 sin φ1 )2
1
+(r2 cos θ2 − r1 cos θ1 )2 2
=
r22 [sin2 θ2 (cos2 φ2 + sin2 φ2 ) + cos2 θ2 ]
+r12 [sin2 θ1 (cos2 φ1 + sin2 φ1 ) + cos2 θ1 ]
4
1
−2r1 r2 [sin θ2 sin θ1 (cos φ2 cos φ1 + sin φ2 sin φ1 ) + cos θ2 cos θ1 ]} 2
q
=
r22 + r12 − 2r1 r2 [sin θ2 sin θ1 cos(φ2 − φ1 ) + cos θ2 cos θ1 ]
d) Now assume that ~r1 is in the x-y plane. Simplify the previous two formulas.
Solution:
Cylindrical Coordinates at z1 = 0
q
r22 + r12 − 2r1 r2 cos(φ2 − φ1 ) + z22
Spherical Coordinates at θ1 =
π
2
(27)
⇒ cos θ1 = 0 and sin θ1 = 1.
q
r22 + r12 − 2r1 r2 sin θ2 cos(φ2 − φ1 )
(28)
e) Find the distance |~r − ~r0 | between the point ~r and the point ~r0 in terms
of the magnitudes of ~r and ~r0 and γ, the angle between them. (Do not
choose a coordinate system.) Then assuming that r r0 , find a series
expansion for |~r − ~r0 |, correct to second order in r0 . This expansion is the
basis of multipole expansions, used in both electromagnetic theory and
quantum mechanics.
Solution:
|~r − ~r0 | =
=
p
(~r − ~r0 ) · (~r − ~r0 )
(29)
q
r2 + r0 2 − 2rr0 cos γ
(30)
Now we assume r r0 , and factor out the “big” thing.
s
r0 2
r0
|~r − ~r0 | = r 1 + 2 − 2 cos γ
r
r
(31)
At this point we just need to do a series expansion of the square root,
02
0
where the small quantity is rr2 − 2 rr cos γ
2
r0
r0
− 2 cos γ
2
r
r
p
1
1
1 + ξ = 1 + ξ − ξ2 + · · ·
2
8
1
1 2
0
|~r − ~r | = r 1 + ξ − ξ + · · ·
2
8
ξ≡
5
(32)
(33)
(34)

1
= r 1 +
2
2
r0
r0
−
2
cos γ
r2
r
!
2
r0
r0
−
2
cos γ
r2
r
1
+
8
!2

+ ···
(35)
02
=r 1−
1
r0
r
cos γ + (1 + cos2 γ) 2 + · · ·
r
2
r
!
(36)
2
1
r0
= r − r0 cos γ + (1 + cos2 γ)
+ ···
2
r
(37)
In the second-to-last step, we collected terms with equal powers of r0 ,
which allowed us to drop a whole lot of terms that would have contributed
had we gone to higher order.
6