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4.5

Frozen-in field lines

The equation of motion (33) for a species alpha of charged particles is

nαm

α

= n dvα α q α ( E + v α × B ) p α + o t h e r f o r c e d e n s i t i e s . dt

(58)

All terms in this equation except the first on the right hand side contains spatial or temporal deriva- tives. If these derivatives are small, so that we look at almost stationary (slow) and almost spatially

constant (large scale) processes, then the only remaining term is

E + v × B 0.

(59)

Comparing to the Galilean transformation equation (56), we see that this means that the electric field in the plasma frame of reference will be zero. We derived (59) from the equations of motion of the plasma with the assumption of slow and large-scale variations, so the basic physical content of (59) is that given sufficiently long time, the plasma will adjust its flow so as to short-circuit all electric fields in its rest frame.

We will no look at an interesting consequence of equation (59), known as the “frozen-in magnetic field” phenomenon. Consider a closed curve C flowing with the plasma. All parts of the curve are supposed to follow the plasma motion at their particular point, so in general, the shape of the curve as well as its circumference will change with time. The magnetic flux through C is

Φ=

B · dS

(60)

S

where S is a surface whoose boundary is C. The flux Φ will change due to (I) that C will enter regions with different B, and due to (II) that B changes in time. The change in Φ due to (I) is calculated as follows. The area covered by a line element dl of C

due to its motion during a time interval dt is (Figure 4)

dΦ = B · dA = B · (v × dl dt) = (B × v) · dl dt

(62)

where we have used a well known vector relation. Summing up all line elements dl of C and dividing by dt, we get the total flux change due to the motion of C as

(v × B) · dl. dt

=

C

I

(63)

dA = v × dl dt.

The change in flux through C during dt due to the motion of C is

(61)

(64)

(65)

The change in Φ due to (II) is

· dS.

C

E · dl.

Using the Faraday-Henry law (5) and Stokes’ theorem, we get

=

B

S

∂t

(∇ × E) · dS = dt =S II

dt

II

16

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