We begin by substituting the U e and U i equations into the equation for B”y”
B”y = 4 p e 2/ m i me [ - By (By 2 + Bz 2 ) /8 p
+ (P - Bo 2 /4 p ) By - B 0 c1 F -( mi 2 - me 2) Bo B’z / 4 p e m t
B”z = 4 p e 2/ m i me [ - Bz (By2 + Bz 2 ) /8 p
+ (P - Bo 2 /4 p ) Bz - B 0 c2 F - ( mi 2 - me 2 ) Bo B’y / 4 p e m t
We can now write the potential part (y) of the force field:
y = 4 p e 2/m i me [(By2 + Bz 2 ) 2 /32 p -
(P - Bo 2 /4 pm )(By2 + Bz 2 )/2 + B 0 F (c1 By + c2 Bz )]
m 2 = e ( mi - me )/m i me
We next write eqns. for B”y
and B”z
in
simpler form:
(IX) B”y = - ¶ y / ¶ By - m 2 Bo B’z
(X) B”z = - ¶ y / ¶ Bz
- m 2 Bo B’y
Since: B = (Bo , By, Bz )
B” ^ = - ÑB y + (m 2 Bo x B ^)
{ }
Velocity-dependent force
Now multiply: B’ y x eqn. (IX) and B’z x eqn. (X) :
è
B’y B” y + B’z B”z = - U dB y /dx (¶ y / ¶ By ) - U dBz /dx (¶ y / ¶ Bz )
And: ½ U d/dx (By 2 + Bz 2 ) = - U d y /dx
->
½ (B’y 2 + B’z 2 ) + y = const. = e
Now take: By x eqn. (X) and: Bz x eqn. (IX)
->
(- By B”z + Bz B”y ) - B y (¶ y / ¶ By ) + Bz (¶ y / ¶ Bz )
+ m 2 Bo ( By B y + Bz B’z ) = 0
= m 2 Bo U/2 d/dx (By 2 + Bz 2 )
Whence:
Bz B”y - By B”z = Bz B”y + B’z B’y - B’z B’y - By B”z
= U d/dx (Bz B’y - By B’z )
Then: Bz (¶ y / ¶ Bz ) - B y (¶ y / ¶ By ) =
- 4 p e 2/ m i me [ - Bz B y (By 2 + Bz 2 ) /8 p + (P - Bo 2 /4 p ) B y Bz -
-
B 0 c1 F
Bz ] + 4 p e 2/ m i me [ - Bz B
y (By 2 + Bz 2 ) /8 p +
(P - Bo 2 /4 p ) B y Bz - B 0 c2 F By] = 4 p e 2 B 0 F / m i me [c1 Bz - c2 By] = 0
So: [c1 - c2 ] = 0
Leading to three new equations:
(XII):
Bz B’y - By B’z + m 2 Bo /2 [(By 2 + Bz 2 )] = const. = L
We then let: B y = B cos q , B z = B sin q
à
(XIII): ½ (B’ 2 + B 2 q’ 2 ) + y (B) = const. = e
(XIV): (m 2 Bo /2 + q’ ) B 2 =
L
Using the above we rewrite the potential field as y (B) :
y (B) = 4 p e 2/ m i me [B 4 /32 p - (P - Bo
2 /4 p )]
= (4 p e 2/ m i me ) B /2 [B 2 /16 p - (P + Bo 2 /4 p )]
Observations:
a) No solution in which q’ , B = const. ( 0) gives linearly polarized Alfven waves.
b) However, solutions exist in which q’ , B = const. ( 0) gives circularly polarized Alfven waves.
c) From this we can deduce a nonlinear Alfven wave velocity:
U = Bo / Ö4 pr
Eliminating q’ from eqns. (XIII) and (XIV):
½ B’ 2 + F 1 = e 1 = const.
F 1 = L 2 / 2 B 2 + m 4 Bo 2 B 2 /8 + y (B)
e 1 = m 2 L Bo 2 /2
d) Then a soliton solution is possible if: e = L = 0, and e 1 = 0
And, finally:
F 1 = 2 p e 2/ m i me [ B 4 /16 p - (P - Bo2 /4 p ) B 2 ] +
m 4 Bo 2 B 2 /8 ) (mi me/ /2 p e 2 ) B 2 ]
~
2 p e 2/ m i me [ B 4 /16 p - (P - Bo2 /4 p - Bo 2/16 p (mi / me/) B 2 ]
And with a certain choice of parameters we find a soliton shock profile, i.e.

For which: ½ B’ 2 + F 1 = 0
If, under certain choice of parameter values, one finds F is possible (with the shape above) then soliton solutions can arise, e.g.

No comments:
Post a Comment