26
CHAPTER 1 Introduction: Maxwell Equations
1.10. The simplest way is probably to work in Cartesian coordinates,
starting from
(S D V X
rot u) i--
~j SD V j (~i uj-
~jui)~, i = 1, 2, 3.
= __ 1 2
Then the last term is
- fD VvUJ'
and ~D vj 3i uj
SD C~ U j ~i uj ~- fD O~ ~i l ujl =
1 uj 2) 1 uj 2
2 YD~)i (0~] I ---~SD I I ~i0¢.
1.11. It doesn't move, but
they
do: Charge carriers may very well pass
through the region of charge imbalance, .being accelerated by the electric
field and slowed down by the invoked "friction" along the way, and
leave the apparent net charge constant. But how does the charge dynamics
account for this behavior? Imagine two kinds of carriers, positive and
negative but identical in all other respects, and argue against the logical
consistency of the myth we used to justify Ohm's law. (This is more than
a mere exercise, rather a theme for reflection. See the Int. Compumag
Society Newsletter, 3, 3 (1996), p. 14.)
SOLUTIONS
1.1. Eliminate h and d: Then
3tb +
rot e = 0, unchanged, and
-
E 0 3te + rot(~t0 -1 b) = j + 3tp + rot m,
so j can "absorb" p and m at leisure. Alternatively, p can assume the
totality of charge fluxes (integrate j + rot m in t). But one can't put all
of them in rot m, since j + 3tp may not be divergence-free. One calls
rot m the density of
Amperian currents.
1.2. Consider a domain D in configuration space (Fig. 1.5). The decrease
of the mass it contains, which is -JD 3tf, equals outgoing mass. The latter
is the flux through the boundary S of the vector field {v, 7} f, which is
the speed, not of a particle in physical space, but of the representative
point {x, v} in configuration space. By Ostrogradskii, 3tf + div({v, 7} f)=
0. Since 7 does not depend on v, div({v, 7}) = 0. So div({v, 7} f) =
{v, 7} • Vf = v. Vxf + 7. Vf. (Be wary of the wavering meaning of the dot,
which stands for the dot-product in V x V left to the - sign, but for the
one in V right to it.) If 7 depends on v, an additional term f div 7
will appear on the left-hand side of (34). (Here, divv7 is the divergence
of 7 considered as a field on V, the x-coordinates being mere parameters.)
SOLUTIONS 27
n S
i .... I
D
t
X
X
w-
FIGURE
1.5. Notations for Exer. 1.2. The open curve is the trajectory of {x, v} in
configuration space.
1.4. Let X be
E3,
V the associated vector space (denoted V 3 in A.2.2).
With v ~ e + v x b, we are dealing with a V-valued function, the
domain 41 of which is all or part of the vector space V, considered with
its affine structure, and position x and time t (which are what e and b
may depend on) are parameters. (This is an illustration of the notion of
section, cf. A.1.1: section by {x, t} of the function {t, x, v} --~ e(t, x) +
v x b(t, x).) Now, e does not depend on v, and since this is also the case
for b, one has div(v -~ v x b) = 0, after the result of Exercise 1.3.
iv
1.6. Last term in (11) is div(v --~ Qc (e + v x b) c~), the integral of which
over V (with t and x as parameters) is zero if q vanishes fast enough.
And by (8), 3tq + div j= ~v [3tq + div(v q)] = ~v
(~t ~] q- V. V x
q), thus 0
after (11 ).
1.7. The
density
q does not transform like a
function
in the change of
reference frame de~ned by x ~ x + u(x)/2, because the volume element
also changes, unless div u = 0, which characterizes volume-preserving
deformations. A correct computation must therefore explicitly take into
account the Jacobian of the mapping x ~ x + u(x)/2. Hence a more
involved computation in the case when div u ~ 0, for of course the same
final result.
1.11. Let p = 1/a be the conductivity. Assume steady currents. Then
div j = 0 by (1), e = pj if Ohm's law is valid, and q = div(¢ 0 e) = ¢0 e. Vp,
nonzero if p varies with position. This result clashes with the predictions
of the simple-minded model in which there would be two symmetrical,
41Be aware that "domain" has a dual meaning, open connected set as in Note 7, or
domain of definition of a map, as here. Cf. Appendix A for precise definitions.
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