Prove Alfven’s theorem: In a perfectly conducting fluid (say, a gas of free electrons), the magnetic flux through any closed loop moving with the fluid is constant in time. (The magnetic field lines are, as it were, “frozen” into the fluid.)
(a) Use Ohm’s law, in the form of Eq. 7.2, together with Faraday’s law, to prove that if σ = ∞ and J is finite, then
J = σ(E + v × B) (7.2)
\frac{\partial B }{\partial t}=\nabla \times( v \times B ) .
(b) Let S be the surface bounded by the loop (P) at time t, and S ^{\prime} a surface bounded by the loop in its new position (P ^{\prime}) at time t + dt (see Fig. 7.58). The change in flux is
d \Phi=\int_{ S ^{\prime}} B (t+d t) \cdot d a -\int_{ S } B (t) \cdot d a .
Use ∇ · B = 0 to show that
\int_{ S ^{\prime}} B (t+d t) \cdot d a +\int_{ R } B (t+d t) \cdot d a =\int_{ S } B (t+d t) \cdot d a(where R is the “ribbon” joining P and P ^{\prime}), and hence that
d \Phi=d t \int_{ S } \frac{\partial B }{\partial t} \cdot d a -\int_{ R } B (t+d t) \cdot d a(for infinitesimal dt). Use the method of Sect. 7.1.3 to rewrite the second integral as
d t \oint_{ P }( B \times v ) \cdot d l ,
and invoke Stokes’ theorem to conclude that
\frac{d \Phi}{d t}=\int_{ S }\left(\frac{\partial B }{\partial t}-\nabla \times( v \times B )\right) \cdot d a .
Together with the result in (a), this proves the theorem.
