PROBLEM SET 1

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
September 3, 2014
Prof. Alan Guth
PROBLEM SET 1
DUE DATE: Friday, September 12, 2014. Due at 5:00 pm in the 8.07 homework box.
The homework boxes are at the intersection of buildings 8 and 16, on the third floor of
bldg. 8 and the 4th floor of bldg. 16.
READING ASSIGNMENT: Chapter 1 of Griffiths: Vector Analysis.
PROBLEM 1: VECTOR IDENTITIES INVOLVING CROSS PRODUCTS
(20 points)
In manipulating cross products, it is useful to define ijk (the Levi-Civita antisymmetric symbol) to be:

 +1 if ijk = (123, 231, 312)
ijk = −1 if ijk = (213, 321, 132)
(1.1.1)

0 otherwise .
That is, ijk is nonzero only when all three indices are different; it is then equal to +1
if ijk is a cyclic permutation of 123, and -1 if ijk is an anti-cyclic permutation. Note
that ijk is totally antisymmetric, in the sense that it changes sign if any two indices are
interchanged:
ijk = −ikj = kij .
(1.1.2)
~ and B
~ can
With this definition, the ith component of the cross product of two vectors A
be written as
~
~
A × B = ijk Aj Bk ,
(1.1.3)
i
where we have used the summation convention that repeated indices are summed over
3 P
3
P
(that is, ijk Ajl Bkm ≡
ijk Ajl Bkm ). For the rest of this problem set, we will
j=1 k=1
always assume that this summation convention is implied, unless explicitly stated otherwise.
(a) From the definition in Eq. (1.1.1), show that
ijk inm = δjn δkm − δjm δkn ,
(1.1.4)
where of course there is an implied sum over the i index in Eq. (1.1.4), but the
indices j, k, n, and m are free.
~ B,
~ and C,
~
(b) Using Eqs. (1.1.3) and (1.1.4), show that for any vectors A,
~ × (B
~ × C)
~ =B
~ (A
~ · C)
~ −C
~ (A
~ · B)
~ .
A
(1.1.5)
8.07 PROBLEM SET 1, FALL 2012
p. 2
~ r ) and B(~
~ r ),
(c) Using Eqs. (1.1.3) and (1.1.4), show that for any vector fields A(~
~ · (A
~ × B)
~ =B
~ · (∇
~ × A)
~ −A
~ · (∇
~ × B)
~ .
∇
~
(d) Using Eqs. (1.1.3) and (1.1.4), show that for any vector A,
1~ 2 ~ ~ ~
~
~
~
A × (∇ × A) = ∇A − A · ∇ A .
2
(1.1.6)
(1.1.7)
Note that, in Cartesian coordinates,
~·∇
~ = Ax ∂ + Ay ∂ + Az ∂ .
A
∂x
∂y
∂z
~ and B,
~
(e) Using Eqs. (1.1.3) and (1.1.4), show that for any vectors A
~ × A
~×B
~ = B
~ ·∇
~ A
~− A
~·∇
~ B
~ +A
~ ∇
~ ·B
~ −B
~ ∇
~ ·A
~ .
∇
(1.1.8)
PROBLEM 2: TRIPLE CROSS PRODUCTS (10 points)
(a) Griffiths Problem 1.2 (p. 4): Is the cross product associative?
? ~
~ × B)
~ ×C =
~ × C)
~ .
(A
A × (B
If so, prove it; if not, provide a counterexample (the simpler the better).
(b) Griffiths Problem 1.6 (p. 8): Prove that
~ × (B
~ × C)]
~ + [B
~ × (C
~ × A)]
~ + [C
~ × (A
~ × B)]
~ =0.
[A
~ × (B
~ × C)
~ = (A
~ × B)
~ × C?
~
Under what conditions does A
PROBLEM 3: PROPERTIES OF THE ROTATION MATRIX R (15 points)
Griffiths Eq. (1.31), p. 11, is
A¯ i =
3
X
Rij Aj ,
j=1
where Rij are components of the rotation matrix that transforms from the unbarred to
~ If we
the barred coordinate system, and Ai are components of some arbitrary vector A.
use the convention that repeated indices are summed over, then this can be written as
A¯ i = Rij Aj .
(1.2.1)
8.07 PROBLEM SET 1, FALL 2012
p. 3
(a) Show that the elements (Rij ) of the three-dimensional rotation matrix must satisfy
the constraint
Rij Rik = δjk
(1.2.2)
~ for all A.
~ Matrices satisfying Eq. (1.2.2) are
in order to preserve the length of A
called orthogonal. Here δjk is the Kronecker delta (δjk is 1 if j = k and 0 otherwise),
and we use the summation convention described above. (Note that Eq. (1.2.2) can
be written simply using matrix notation. If Rij is the (i, j) component of the matrix
R, then Eq. (1.2.2) can be written as RT R = I, where I is the identity matrix.)
(b) Using the orthogonality constraint (1.2.2), show that
Ai = Rji A¯ j .
(1.2.3)
Note that we can now show that Rji Rki = δjk using this relation, in a manner similar
to the procedure in (a) (you do not have to show this).
(c) Using the chain rule for partial differentiation and the results of (b), show that if f
~ (~r) transforms as a vector; i.e., show
is scalar function of ~r ≡ (x1 , x2 , x3 ), then ∇f
that if
f¯(¯
x1 , x
¯2 , x
¯3 ) = f (x1 , x2 , x3 ) ,
(1.2.4)
where x
¯i = Rij xj , then
∂ f¯
∂f
= Rij
.
∂x
¯i
∂xj
(1.2.5)
PROBLEM 4: USE OF THE GRADIENT (10 points)
Griffiths Problem 1.12 (p.15), Griffiths Problem 1.13 part (a) only (p.15).
(a) Griffiths Problem 1.12 (p. 15): The height of a certain hill (in feet) is given by
h(x, y) = 10(2xy − 3x2 − 4y 2 − 18x + 28y + 12) ,
where y is the distance (in miles) north, x the distance east of South Hadley.
(i) Where is the top of the hill located?
(ii) How high is the hill?
(iii) How steep is the slope (in feet per mile) at a point 1 mile north and one mile
east of South Hadley? In what direction is the slope steepest, at that point?
(b) Griffiths Problem 1.13 (p. 15), part (a) only: Let ~ be the separation vector from a
fixed point (x0 , y 0 , z 0 ) to the point (x, y, z), and let be its length. Show that
~
∇(
2
) = 2~ .
8.07 PROBLEM SET 1, FALL 2012
p. 4
PROBLEM 5: THE DIRAC DELTA FUNCTION AND ∇2 (1/4πr) (20 points)
One of the most used identities in this course is be the relation
1
r
ˆ
2 1
~ · ∇
~
~ ·
= −∇
=∇
= δ 3 (r) = δ(x) δ(y) δ(z) .
−∇
4πr
4πr
4πr2
(1.5.1)
It turns out of course (see Griffiths 1.5.1, p. 45) that
1
−∇2
4πr
is zero everywhere except at the origin, and ill-defined there. To get a better feel for the
fact that
1
−∇2
4πr
is a delta function, let’s look at a different function which approaches −(1/4πr) in some
limit, but which is well-behaved everywhere. The function is
1
1
fa (r) = − √
.
(1.5.2)
2
4π r + a2
For a nonzero, fa (r) is well-behaved everywhere, and
1
lim fa (r) = −
(1.5.3)
a→0
4πr
(a) Calculate ga (r) = ∇2 fa (r) and show that it is also well behaved for all r. Sketch
ga (r) for some value of a as a function of r/a.
(b) Show that
Z
ga (r) d3 x = 1 .
(1.5.4)
all space
(c) Show that
lim ga (r) = 0 if r 6= 0 .
a→0
(1.5.5)
Thus in the limit that a goes to zero, our well-behaved function ga (r) exhibits the
properties we expect of a three-dimensional delta function.
PROBLEM 6: EXERCISES WITH δ-FUNCTIONS (10 points)
(a) A charge Q is spread uniformly over a spherical shell of radius R. Express the volume
charge density using a delta function in spherical coordinates. Repeat for a ring of
radius R with charge Q lying in the xy plane.
(b) In cartesian coordinates, we can write δ 3 (~r − ~r0 ) = δ(x − x0 )δ(y − y 0 )δ(z − z 0 ). How
would one express δ 3 (~r − ~r0 ) in cylindrical coordinates (s, φ, z).
(c) A charge λ per unit length is distributed uniformly over a cylindrical surface of radius
b. Give the volume charge density using a delta function in cylindrical coordinates
p
(d) What is ∇2 ln r in two dimensions? (Here r is the radial coordinate, r = x2 + y 2 .)
8.07 PROBLEM SET 1, FALL 2012
p. 5
PROBLEM 7: COROLLARIES OF THE FUNDAMENTAL INTEGRAL
THEOREMS (15 points)
This problem is closely related to Problem 1.61, p. 56 of Griffiths. You will find
useful hints there — but try without hints first!!. Show that:
R
R
~ d3 x =
ψ d~a, where
S is the surface bounding the volume V . Show that as
(a) V ∇ψ
S
R
a consequence of this, S d~a = 0 for a closed surface S.
R
R
~ ×A
~ d3 x = − A
~ × d~a, where S is the surface bounding the volume V .
(b) V ∇
S
R
H
~ × d~a = − ψ d~l, where Γ is the boundary of the surface S.
(c) S ∇ψ
Γ
R
~ × A)
~ · d~a = 0.
(d) For a closed surface S, one has S (∇

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