Assignment 2
Arfken 1.10.2 Find the work done against the force F~ = (−x̂y + ŷx) / x2 + y 2 going around a unit circle
halfway.
Z
W =−
F~ · d~r
Note that the unit vector here points in a purely angular direction: d~r = θ̂dθ and that the force can be
written F~ = −x̂ sin θ + ŷ cos θ = θ̂. So our integral becomes
Z b
W =−
θ̂ · θ̂ dθ
a
b
= −θ
a
= −(b − a)
where the limits for part (a) are from 0 to π and in (b) are from 0 to −π. Thus the work in the two cases is
−π and π respectively and the work is path dependent.
In section 1.13, there is a discussion on conservative forces and conditions necessary to have them. One
is the vanishing of the line integral of the force around any closed path in our region (in this case, three
dimensional Euclidean space). Said in an equivalent, but slightly different way, the line integral of the force
between two points in the space must not depend on the particular path taken between those two points.
Here it clearly does. So we say that the force is non-conservative. However, note that on taking the curl of
the force field, we get ∇ × F~ = 0 and the force seems to be conservative after all. What’s going on? The
answer lies in a subtle assumption about differentiation (the curl) here. The force field is not differentiable
(well behaved) on the axis, i.e. x = 0, y = 0. To handle this location correctly, we have to use Dirac delta
functions and the curl turns out not to vanish along the axis. It does vanish everywhere else, just not on the
axis. Said another way, any line integral along a closed path that encircles the axis will not vanish. Thus
this force really is nonconservative.
Arfken 1.10.6 Find the curl from the appropriate integral representation:
R
~
d~σ × V
~
=∇×V
lim s R
dτ →0
dτ
Consider the integrand evaluated on the sides of an infinitesimal box of volume dxdydz:
~ = da n̂ × V
~
d~σ × V
~ |x+dx − V
~ |x + dzdx ĵ × V
~ |y+dy − V
~ |y + dxdy k̂ × V
~ |z+dz − V
~ |z
= dydz î × V
~ + · · · + dzdx ĵ × dy ∂y V
~ + · · · + dxdy k̂ × dz ∂z V
~ + ···
= dydz î × dx ∂x V
~ + ĵ × ∂y V
~ + k̂ × ∂z V
~
= dxdydz î × ∂x V
where we have neglected higher order terms in the expectation that we will take the limit in which they go
to zero. Doing that now, we get,
R
~
d~σ × V
~ + ĵ × ∂y V
~ + k̂ × ∂z V
~
= î × ∂x V
lim s R
dτ →0
dτ
= k̂∂x Vy − ĵ∂x Vz + î∂y Vz − k̂∂y Vx + ĵ∂z Vx − î∂z Vy
= î (∂y Vz − ∂z Vy ) + ĵ (∂z Vx − ∂x Vz ) + k̂ (∂x Vy − ∂y Vx )
~
=∇×V
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Arfken 1.11.4 If ψ satisfies Laplace’s equation, show the surface integral of its normal derivative vanishes.
Using the divergence theorem, we get for the vector ∇ψ,
I
Z
∇ψ · n̂ da =
∇ · (∇ψ) dv
S
ZV
=
∇2 ψ dv
V
=0
Arfken 1.11.9 Determine the work needed to assemble a local, steady state system of currents and fields.
Using the vector identitiy
∇ · ~a × ~b = ~b · (∇ × ~a) − ~a · ∇ × ~b
we can write the work as the following
Z
1
W =
2
Z
1
=
2
Z
1
=
2
I
1
=
2
(with dτ the volume element)
~ ·B
~ dτ
H
~ · ∇×A
~ dτ
H
h
i
~×H
~ +A
~·∇×H
~ dτ
∇· A
Z
~· ∇×H
~ dτ
~×H
~ · n̂ da + 1 A
A
2
where we have used the divergence theorem for the first term in the last line. If we argue that our system of
charges and current are localized and that the surface integral is done at very large radius, this surface integral
~ = J~
can be set to zero. Further, using the Maxwell equations with the displacement current zero, ∇ × H
and we have
Z
~ · J~ dτ
W = A
Arfken 1.11.10 Prove the generalization of Green’s theorem.
Z
Z
(vLu − uLv) dτ =
(v∇ · (p∇u) + vqu − u∇ · (p∇v) − uqv) dτ
V
Z
(∇ · [vp∇u] − ∇v · p∇u − ∇ [up∇v] + ∇u · p∇v) dτ
=
Z
∇ · [vp∇u − up∇v] dτ
=
I
p (v∇u − u∇v) · n̂ da
=
S
where in the second line we have canceled two terms and in the last line we have used the divergence theorem.
Arfken 1.12.5 Show the integral form of Ampere’s law from its differential from.
~ = J,
~ over a surface S, we get
Integrating one of Maxwell’s equations, ∇ × H
Z
Z
∇ × Ĥ · d~a =
J~ · d~a
S
S
I
~ · d~r = I
H
C
where we have used Stokes’ theorem and I is the enclosed current.
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Arfken 1.13.3 Examine the force of gravity within a spherical, massive body.
Using the Poisson equation for a gravitational potential from section 1.14, ∇2 φ = 4πGρ, and the fact
~ we can write down a gravitational analog of Gauss’ law:
that F~ = −∇φ,
Z
Z
F~ · n̂da = −4πGρ dv
Evaluating this on a spherical gaussian surface within a massive, spherical body with constant density, ρ0 ,
and radius a, we get
4
−|F~ | 4πr2 = −4πGρ0 πr3
3
4
F~ = − πGρ0 rr̂
3
~
where we have assumed that the direction of F was opposite that of the normal, n̂, to the gaussian surface
of radius r < a. Finally, we note that this is the force on a unit mass, m0 .
The gravitational potential which corresponds to this force is
Z r
φ=−
F~ · d~r
0
Z r
4
=−
− πGρ0 r0 r̂0 · r̂0 dr0
3
0
4
r2
= πGρ0
3
2
Finally, the equation of motion for a particle traversing the inside of the earth assuming it has constant
density as above is
4πGρ0
~r
F~ = ~¨r = −
3
where, again, we assume that ourqparticle has unit mass, m0 = 1. This, of course, is the equation for simple
0
harmonic motion with frequency 4πGρ
. Putting appropriate numerical values in here, we get 2π over this,
3
the period, to be about 5070 seconds, not quite an hour and a half.
Arfken 1.13.4 Find the potential from the given tidal force:
1
F~ = −GM m 3 (x, y, −2z)
R
The potential is, assuming that R is essentially a constant,
Z
φ = − F~ · d~r
Z
GM m
xdx + ydy − 2zdz
=
R3
GM m 1 2
=
(x + y 2 − 2z 2 )
R3 2
~ for a magnetic induction B given by
Arfken 1.13.5 Find a vector potential, A,
~ = ∇ × A:
~
The equations are for B
µ0 I −y
2π x2 + y 2
µ0 I
x
By = ∂z Ax − ∂x Az =
2π x2 + y 2
Bz = ∂x Ay − ∂y Ax = 0
Bx = ∂y Az − ∂z Ay =
3
µ0 I
2
2π (−y, z, 0)/(x
+ y2 )
As in the text and as we did in class, we will choose Ax to be zero. This seems odd, but we can do it
~ The Bz equation then implies that Ay is a function
because of the arbitrariness of the gauge potential A.
that depends only on y and z. At the same time we can integrate the By equation with respect to x:
µ0 I
2π
µ0 I
Az = −
2π
µ0 I
=−
2π
∂x Az = −
x
x2 + y 2
Z
x
dx
x2 + y 2
1
ln |x2 + y 2 | + f (y, z)
2
where f is an arbitrary function of y and z. Substituting this expression in the Bx equation, we get
y
µ0 I
2π x2 + y 2
µ0 I 1 2y
y
µ0 I
−
+ ∂y f (y, z) − ∂z Ay = −
2
2
2
2π 2 x + y
2π x + y 2
∂y Az − ∂z Ay = −
leaving us with
∂z Ay − ∂y f (y, z) = 0
which can be integrated to give
Z
Ay =
∂y f (y, z)dz + g(y)
where both f and g are arbitrary functions of their arguments. We can witout loss of generality choose
f = g = 0 and our final solution is then Ax = Ay = 0 and
Az = −
µ0 I
ln |x2 + y 2 |
4π
Arfken 1.14.1 Find a two-dimensional Gauss’ law.
Recall that we originally developed the divergence theorem in class in its two-dimensional form and
showed that it extended to three and other dimensions. We will use that here as well as our derivation of
the usual Gauss’ law from class and the text.
~ , on that surface,
Imagine a two-dimensional surface, S, bounded by a contour, C. For any vector field, V
the divergence theorem in two dimensions is
Z
~ da =
∇·V
S
I
~ · n̂ dl
V
C
where n̂ is the outward normal to the contour and dl is the integration variable along the contour, C. Now
~ =E
~ = q ρ̂ , the electric field of a point charge, q, placed at the origin in this two-dimensional space.
let V
2π0 ρ
If our contour, C, does not enclose the origin, the integrals above are zero since we can show that
~ ∝∇·
∇·E
ρ̂
=0
ρ
However, if the contour, C, encloses the origin, we must make an argument similar to the one we made
in class. Namely, let C be composed of two contours, C1 and C2 , with C1 being the “outer” contour and
C2 a small circular contour surrounding the origin and connected to C1 by a thin “passageway” whose
thickness will be considered negligible (and whose contributions to the integral will ultimately cancel). Since
4
~ around it will be zero by our earlier
the contour C does not actually include the origin the integral of E
argument. Thus we have
I
q
ρ̂
0=
· n̂ dl
2π0 C1 +C2 ρ
I
Z 2π
ρ̂
q
ρ̂
q
=
· n̂ dl +
· (−ρ̂) δdθ
2π0 C1 ρ
2π0 0 δ
I
ρ̂
q
q
· n̂ dl −
=
2π0 C1 ρ
0
where we have assumed that the contour C2 around the origin has a radius of δ with the normal, n̂, pointed
inward towards the origin (out of the contour of interest).
Putting all of this together, we see that
I
~ · n̂ dl = q
E
0
C
if C encloses charge and zero otherwise.
Arfken 1.15.9 Evaluate the derivative of the delta function.
∞
Z
Z ∞
0
δ (x)f (x) dx = δ(x)f (x)
−
−∞
−∞
∞
δ(x)f 0 (x) dx
−∞
= −f 0 (0)
where we have used the assumed properties of f (x), namely that it is smooth and drops to zero at ±∞ to
eliminate the first term. Similarly, we can argue that except at x = 0, δ(x) = 0.
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