Massachusetts Institute of Technology
Department of Physics
Physics 8.07
Fall 2005
Problem Set 1 Solutions
Problem 1: Cross Products
a) It is sufficient to consider the cross products of all pairs of basis vectors. The
only nonzero cross products are
~ex × ~ey = ~ez ,
~ey × ~ex = −~ez ,
~ey × ~ez = ~ex
~ez × ~ey = −~ex
, ~ez × ~ex = ~ey ,
, ~ex × ~ez = −~ey ,
from which it follows that the only nonzero elements of ijk are
xyz = yzx = zxy = 1 , yxz = zyx = xzy = −1 ,
or, using numerical indices,
123 = 231 = 312 = 1 , 213 = 321 = 132 = −1 .
b) First note that since all ijk = 0 unless all three indices are distinct, ijk lmk =
0 unless i 6= j and l 6= m. Consider now the case where (i, j, l, m) include all
three distinct indices (and one of them appears twice). For example,
12k 13k = 121 131 + 122 132 + 123 133 = 0 × 0 + 0 × (−1) + 1 × 0 = 0 .
Clearly, ijk lmk = 0 unless either i = l and j = m, or i = m and j = l. In
the first case,
ijk ijk = 1 if i 6= j (no sum on i, j)
because only one of the three possible k values contributes to the sum over
k. In the second case,
ijk jik = −ijk ijk = −1 if i 6= j (no sum on i, j) .
These two cases, and all cases of zero, are correctly accounted for by writing
ijk lmk = δil δjm − δim δjl .
~×B
~ = ijk Ai Bj ~ek , we get
c) Using A
~a × (~b × ~c ) = ijk ai (~b × ~c )j ~ek = ijk ai lmj bl cm~ek = −ikj lmj ai bl cm~ek
= (δim δkl − δil δkm )ai bl cm~ek = (~a · ~c )~b − (~a · ~b )~c .
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Problem 2: Spherical Polar Basis Vectors
We start with the expressions given in lecture,
~er = ~ex sin θ cos φ + ~ey sin θ sin φ + ~ez cos θ ,
~eθ = ~ex cos θ cos φ + ~ey cos θ sin φ − ~ez sin θ ,
~eφ = −~ex sin φ + ~ey cos φ .
We see immediately that ∂r~ea = 0, i.e. ωrbc = 0 for all b and c. Next consider ∂θ :
∂θ~er = ~ex cos θ cos φ + ~ey cos θ sin φ − ~ez sin θ = ~eθ ,
∂θ~eθ = −~ex sin θ cos φ − ~ey sin θ sin φ − ~ez cos θ = −~er ,
∂θ~eφ = 0 .
And finally, ∂φ :
∂φ~er = −~ex sin θ sin φ + ~ey sin θ cos φ = ~eφ sin θ ,
∂φ~eθ = −~ex cos θ sin φ + ~ey cos θ cos φ = ~eφ cos θ ,
∂φ~eφ = −~ex cos φ − ~ey sin φ = −~er sin θ − ~eθ cos θ .
The nonzero coefficients are then
ωθrθ = −ωθθr = 1 , ωφrφ = −ωφφr = sin θ , ωφθφ = −ωφφθ = cos θ .
Note that ωabc = −ωacb . This condition holds for any orthonormal basis:
0 = ∂a (~eb · ~ec ) = (ωabd~ed ) · ~ec + ~eb · (ωacd~ed ) = ωabd δcd + ωacd δbd = ωabc + ωacb .
Problem 3: Divergence and Curl in Spherical Coordinates
The solution to this problem is based on writing
~ = ~er ∂r + ~eθ 1 ∂θ + ~eφ 1 ∂φ
∇
r
r sin θ
and applying it to the spherical polar basis vectors.
a) Using the results of Problem 2, we have
1
~ · ~er = ~er · (∂r~er ) + ~eθ · 1 ∂θ~er + ~eφ ·
∂φ~er
∇
r
r sin θ
1
1
2
= ~eθ · ~eθ + ~eφ ·
sin θ~eφ = ,
r
r sin θ
r
2
1
~ · ~eθ = ~er · (∂r~eθ ) + ~eθ · 1 ∂θ~eθ + ~eφ ·
∇
∂φ~eθ
r
r
sin
θ
!
cos θ
cos θ
1
~eφ =
,
= −~eθ · ~er + ~eφ ·
r
r sin θ
r sin θ
1
1
~
∇ · ~eφ = ~er · (∂r~eφ ) + ~eθ ·
∂θ~eφ + ~eφ ·
∂φ~eφ
r
r sin θ
1
(sin θ~er + cos θ~eθ ) = 0 .
= −~eφ ·
r sin θ
b) Similarly,
1
~ × ~er = ~er × (∂r~er ) + ~eθ × 1 ∂θ~er + ~eφ ×
∇
∂φ~er
r
r sin θ
1
1
sin θ~eφ = 0 ,
= ~eθ × ~eθ + ~eφ ×
r
r sin θ
1
1
~
∇ × ~eθ = ~er × (∂r~eθ ) + ~eθ ×
∂θ~eθ + ~eφ ×
∂φ~eθ
r
r sin θ
!
cos θ
1
1
~eφ = ~eφ ,
= −~eθ × ~er + ~eφ ×
r
r sin θ
r
1
~ × ~eφ = ~er × (∂r~eφ ) + ~eθ × 1 ∂θ~eφ + ~eφ ×
∇
∂φ~eφ
r
r sin θ
1
1
(sin θ~er + cos θ~eθ ) =
(cos θ~er − sin θ~eθ ) .
= −~eφ ×
r sin θ
r sin θ
c) Using the results of part (a), the divergence of a vector field ~v = va~ea expanded in a spherical coordinate basis is
~ · (va~ea ) = ~ea · ∇v
~ a + va (∇
~ · ~ea )
∇
1
1
~ · ~er ) + vθ (∇
~ · ~eθ ) + vφ (∇
~ · ~eφ )
= ∂r v r + ∂θ v θ +
∂φ vφ + vr (∇
r
r sin θ
2
1
cos θ
1
= ∂r v r + v r + ∂θ v θ +
vθ +
∂φ v φ
r
r
r sin θ
r sin θ
1 ∂
1 ∂ 2 1 ∂vφ
= 2
r vr +
(sin θvθ ) +
,
r ∂r
r sin θ ∂θ
r sin θ ∂φ
in agreement with Griffiths equation (1.71). Similarly, using the results of
part (b),
3
~ × (va~ea ) = (∇v
~ a ) × ~ea + va (∇
~ × ~ea )
∇
"
#
"
#
1 ∂vr
∂vθ
1 ∂vr
1 ∂vθ
=
~eθ +
~eφ × ~er +
~er +
~eφ × ~eθ
r ∂θ
r sin θ ∂φ
∂r
r sin θ ∂φ
"
#
∂vφ
1 ∂vφ
~ × ~eθ ) + vφ (∇
~ × ~eφ )
~er +
~eθ × ~eφ + vθ (∇
+
∂r
r ∂θ
1 ∂vr
∂vθ
1 ∂vθ
1 ∂vr
~eφ +
~eθ +
~eφ −
~er
= −
r ∂θ
r sin θ ∂φ
∂r
r sin θ ∂φ
1 ∂vφ
vθ
vφ
∂vφ
~eθ +
~er + ~eφ +
(cos θ~er − sin θ~eθ )
−
∂r"
r ∂θ
r#
r sin
#
"θ
1
∂
∂vθ
1 ∂vr
∂
1
=
(sin θvφ ) −
− (rvφ ) ~eθ
~er +
r sin θ ∂θ
∂φ
r sin θ ∂φ
∂r
"
#
∂vr
1 ∂
~eφ ,
(rvθ ) −
+
r ∂r
∂θ
in agreement with Griffiths equation (1.72).
Problem 4: Practice with Vector Calculus
a) Writing ~er = 1r ~r = xr ~ex + yr ~ey + zr ~ez ,
x
y
~u = 3 ~ex − 3 ~ey .
r
r
√ 2
Now using ∂r/∂x = ∂ x + y 2 + z 2 /∂x = x/r and similarly for ∂r/∂y, we
obtain
~ · ~u = ∂ y − ∂ x = − 3y ∂r + 3x ∂r = − 3yx + 3xy = 0 .
∇
∂x r3
∂y r3
r4 ∂x r4 ∂y
r5
r5
b) We have ~er · ~ey = sin θ sin φ, so
~v · w
~ = x sin θ sin φ =
xy
= r sin2 θ cos φ sin φ .
r
c) Using the results of Problem 3c),
~ × ~v = ∇
~ × (vr~er ) = 0 .
∇
2
2
d) We have ~u·w
~ = −(x2 /r3 )e−ar = −r−1 sin2 θ cos2 φ e−ar and d3 x = r2 sin θ dr dθ dφ,
hence
Z
Z ∞
Z π
Z 2π
2π
4
1
3
−ar 2
3
2
~u·w
~d x=−
dr
(π) = − .
re
sin θ dθ
cos φ dφ = −
2a
3
3a
0
0
0
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Problem 5: Griffiths Problem 1.56 (p. 55)
We write the line integral as a sum over the four curve segments:
Z
C
~v · d~l =
Z
C1
~v · d~l +
Z
C2
Z
~v · d~l +
C3
~v · d~l +
Z
C4
~v · d~l .
Along these four line segements (starting from the origin), the infinitesimal displacement
vector is

~er dr = ~ex dx , C1 (θ = π/2, φ = 0)



~
e rdφ ,
C2 (θ = π/2, r = 1)
.
d~l =  φ
ez dz ,
C3 (φ = π/2, y = r sin θ = 1)

~

−~er dr ,
C4 (φ = π/2, y/z = tan θ = 1/2) .
Now we evaluate each integral in turn, using the appropriate (r, θ, φ) for each curve
segment, with ~v = r cos2 θ ~er − r cos θ sin θ ~eθ + 3r~eφ = z~ez + 3r~eφ :
C1 :
~v = 3r~ey , ~v · d~l = 0 ,
C2 :
~v = 3~eφ , ~v · d~l = 3dφ ,
C3 :
C4 :
Z
C1
~v · d~l = 0
Z
C2
~v · d~l = 3
~v = z~ez − 3r~ex , ~v · d~l = z dz ,
Z
C3
π/2
Z
dφ = 3π/2 ,
0
2
Z
~v · d~l =
z dz = 2 ,
0
4
~v = z~ez − 3r~ex , ~v · d~l = −r cos2 θ dr = − r dr ,
5
Z
C4
~v · d~l = −
Z
0
√
5
4
r dr = −2 .
5
H
Adding these four contributions gives ~v · d~l = 3π/2. We check Stokes’s Theorem by
~ × ~v through the two surfaces:
evaluating the flux of ∇
Z
S
Z
~ × ~v ) · d~a =
(∇
S1
~ × ~v ) · d~a +
(∇
Z
S2
~ × ~v ) · d~a .
(∇
~1 lies in the x-y plane with d~a ∝ ~ez while S2 lies in the y-z plane with d~a ∝ ~ex .
Here, S
~ × ~v in spherical coordinates obtained in Problem 3, we find
Using the expression for ∇
~ × ~v = 3 cot θ~er − 6~eθ .
∇
~ × ~v ) · d~a = 0 for S2 . For S1 , d~a = r drdφ ~ez and cos θ = 0, so (∇
~ × ~v ) · d~a =
Thus, (∇
6r drdφ. The flux is therefore
Z
S
~ × ~v ) · d~a = 6
(∇
Z
S1
r drdφ = 6
Z
0
verifying Stokes’s Theorem.
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1
d dr
Z
0
π/2
dφ = 3π/2 ,
Problem 6: Griffiths Problem 1.47 (p. 52)
a)
Z
(r2 + ~r · ~a + a2 )δ 3 (~r − ~a ) d3 r = 3a2 .
b)
Z
V
[(y − 4)2 + (z − 3)2 ]5−3 δ 3 (~r ) d3 r =
1
42 + 32
= .
3
5
5
√
c) Because |c| = 52 + 32 + 22 > 6, δ 3 (~r − ~c ) = 0 everywhere inside V , and the
integral is zero.
√
d) Here, |~e − (2, 2, 2)| = |(1, 0, −1)| = 2 < 1.5, so ~e lies inside V . Therefore,
Z
V
~r · (d~ − ~r )δ 3 (~e − ~r ) d3 r = ~e · (d~ − ~e ) = (3, 2, 1) · (−2, 0, 2) = −4 .
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