Description
1. Consider the parameterization of a position vector in 3D:
~r(η, φ, z) = cosh η cos φˆi + sinh η sin φ ˆj + z
ˆk
(a) Demonstrate that η, φ, z constitute a set of orthogonal coordinates.
(b) What is the metric tensor?
(c) What is ∇~ f in these coordinates?
(d) What is ∇ · E ~ in this coordinates, where E~ = Eη ηˆ + Eφ φˆ + Ez
ˆk?
(e) What is ∇2
f in this coordinate system? Where possible, express your answer
only in terms of cosines and hyperbolic cosines.
In parts (b)-(e) Express your answer in terms of cosines and hyperbolic cosines, where
possible.
2. Consider the function
φ(r) = r
r
2 +
2
in three dimensions. Calculate ∇2φ.
(a) Show that for any fixed value of r = r0 6= 0,
limr0→0
lim→0
∇2φ(r0)
= 0.
(b) Show that for a fixed value of 6= 0,
lim→0
limr0→0
∇2φ(r0)
= ∞
(c) Using (a) and (b), construct an argument that ∇2 1
r
is zero at all points save at
the origin, where it diverges. If f(r) is a smooth, well behaved function, calculate
an estimate for
Z
f(r) ∇2
1
r
d
3
r
where the integration range is over all space.
3. Your textbook (and many others) state that the curl of the gradient of a scalar function
is zero. Consider the function f(r, ϕ) = ϕ where r and ϕ are the standard polar
coordinates in two dimensions.
(a) Calculate ~g = ∇~ f in polar coordinates.
(b) Calculate ∇ × ~ ∇~ f = ∇ × ~ ~g in polar coordinates.
(c) From Stoke’s Theorem, We know
Z
∇ × ~ ~g
· n dA ˆ =
I
~g · d
~`
Calculate the last line integral in polar coordinates.
(d) Do your answers agree? Can you explain?
4. This is one long problem, designed to give you experience in working with tensors, and
exposure to one of the fundamental tensors in field theory, that of the electromagnetic
field. The questions below basically encompass questions 11-13 in the text, with some
extensions. Those of you interested in other approaches might look at The Feynman
Lectures, II-27.
The questions are a bit tedious, but should not be too difficult. If you are having a lot
of trouble, then I have probably made a mistake with the question. In that case come
see me for clarification.
Maxwell’s equations are one of the crowning achievements of classical physics. If we
ignore the presence of any polarizing medium, they are:
∇ · E = 4πρ
∇ · B = 0
∇ × E = −
1
c
∂B
∂t
∇ × B =
4π
c
J +
1
c
∂E
∂t
We will develop a more compact notation for expressing these equations.
(a) Derive the law of conservation of charge from Maxwell’s equations:
∂ρ
∂t + ∇ · J = 0
(b) Express the electric field in terms of the scalar potential φ and the vector potential
A where B = ∇ × A. (Note, that in the general time dependent case the answer
is not simply E = −∇φ).
(c) A second-rank antisymmetric tensor in 3 dimensions has only 3 independent elements. We can make a link between it and a vector in 3 dimensions via:
Fij = ijkBk
Prove that if this is the case, then Bk =
1
2
ijkFij .
(d) Prove that for Fij as defined above,
Fij =
∂Aj
∂xi
−
∂Ai
∂xj
≡ ∂iAj − ∂jAi
where A is the vector potential mentioned above.
This tensor is obviously antisymmetric – that is, Fij = −Fji. It has 32 = 9
components. We will now extend it to four dimensions, so that it has 42 = 16
components. (Note that it is still a second rank tensor. The number of dimensions
is not the same as the rank of the tensor.) We will do so by setting x4 = ict, and
A4 = iφ, where i =
√
−1 and φ is the electrostatic scalar potential. The indices
will now in general run from 1 to 4, rather from 1 to 3.
(e) Prove F4j = −Fj4 = iEj where Fij defined as in part (d), and j = 1, 2, 3. (F44 is
zero.) Don’t forget that the electric field is not simply −∇φ.
There are sixteen elements of Fij . The four diagonal ones must be zero, since antisymmetry requires Fii = −Fii. Of the 12 left, only half of them are independent,
because of the antisymmetry requirement. Of those six, we have just shown that
three are the magnetic field, and three are the electric field. If we write out the
results of parts (d) and (e) we find:
Fij =
0 Bz −By −iEx
−Bz 0 Bx −iEy
By −Bx 0 −iEz
iEx iEy iEz 0
Define the current as a four dimensional vector: J = (Jx, Jy, Jz, icρ). Our goal
will be to show that all of Maxwell’s equations are subsumed in:
[1] ∂iFjk + ∂jFki + ∂kFij = 0
[2] ∂kF`k = (4π/c)J`
.
Eq.(1) is in fact 43 = 64 equations, so it will help to break this down
(f) Show that eq.(1) holds from the definition of Fij given in (d), that is Fij =
∂iAj − ∂jAi
. (While this is nice, we haven’t really made a link yet to Maxwell’s
equations!)
(g) Show that eq.(1) trivially holds when any pair of indices is the same (i.e. if i = j)
or if all are the same.
(h) We have only cases where i 6= j 6= k, or 4 × 3 × 2 = 24 cases left. Prove (1)
holds when i = 4, when j and k are different spatial indices (1,2,3). This covers
3 × 2 = 6 cases. However, our choice of starting with i = 4 was arbitrary – we
could have started with j. This gives us a factor of 3, so we have covered 18 cases.
(i) Prove (1) holds when i 6= j 6= k 6= 4. (Note that these all involve just the magnetic
field). There are 3 × 2 × 1 = 6 equations. This covers the remaining cases.
(j) Prove eq.(2) gives the rest of Maxwell’s equations.
Thus we have shown that eqs (1) and (2) reproduce Maxwell’s equations and give
nothing more.
For your edification, it is interesting to note that the Lorentz force law is:
fi =
1
c
FijJk,
for i = 1, 2, 3. (Convince yourself of this, but you don’t have to calculate it for
the homework.) The quantity icf4 gives the work done by the electric field.
What have we gained by writing E and B in terms of Fij? Well, we’ve saved a
bit of typing. It is very suggestive that we can express everything so simply. But
recall that Fij is not some arbitrary matrix, it is a tensor.
That means if our
transformation matrix for coordinates is Lij , then when we change to a different
coordinate system,
F
0
ij = LimLjnFmn
Let’s see if this shows up in our new formulation.
Consider an infinite line of charge along the z-axis. You will recall that it produces an electric field: E = λ log(r) ˆr = λ log(r) (cos θxˆ + sin θyˆ) where λ is
a constant involving the charge per unit length. The charge distribution is
J = (0, 0, 0, icρ0(~r)), where ρ0(~r) is zero everywhere but on the z-axis, where
it is a constant.
The transformation matrix we will use will put us in a moving
frame, which is sometimes called a “boost” transformation. In this case we will
“boost” along the z-axis. The matrix is
Lij =
1 0 0 0
0 1 0 0
0 0 cosh α isinh α
0 0 −isinh α cosh α
It can be thought of as a rotation, by an angle iα, where cosh α = 1/
q
1 − (v
2/c2
).
This transformation puts us in a frame moving at velocity v in the positive ˆz
direction. The quantity α is sometimes called the “rapidity”. It is useful because
if we make two successive boosts in the same direction, the final velocity may not
be the sum of the two boost velocities, but the final rapidity is the sum of the
two boost rapidities. (That is, rapidities are additive in relativistic mechanics.)
(k) Calculate J
0
. (It transforms as a vector.) Approximate it for v << c. In this
frame we should now see a current. Can you give a meaning to the change in J4?
(l) Calculate B0
x
and B0
y
, by calculating F
0
23 and F
0
31. Can you explain why in this
frame we see a magnetic field?
Thus when we make a transform to a new coordinate system, we see different
electric and magnetic fields. The value of the fields is consistent with what we
would have gotten if we had started our calculations in the moving frame. All
of this information is built in to the tensor Fij , suggesting that this is more than
just a convenient notation.
Those of you who have a lot of time on your hands
might wonder how all of the above changes if we allow for magnetic charges (i.e.
monopoles).
