sOF TLCHV0o0
JUL 7 1960
LIBRARY CHANGE OF MAGNETIC MOMENTINSIDE MIRROR CONFIGURATION
by
Shoichi Yoshikawa
B.S., University of Tokyo (1958)
Submitted in Partial Fulfillment of the Requirements for the Degree of Master of Science
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY February, 1960
Signature redacted
Signature of Author: . . . .. . . . .
Deptrtment of Nuci:6ar Ergineering, February 1, 1960
Certified by:
Signature redacted
Dr. David J. Raze, Thesis Supervisor
Accepted by: 0 0 0 0 0 0 0 0 0 0 .
~V? A A
Change of Magnetic Moment Inside Mirror Configuration by
Shoichi Yoshikawa
Submitted in Partial Fulfillment of the requirements for the Degree of Master of Science in Nuclear Engineering, June 1960.
This thesis is concerned with elucidating the practical problem of changes of the magnetic moment inside the collision-less magnetized iasma. In order to do so, the magnetic field is represented as a potential for the 2-dimensional motion of the particle. This is rather a trivial application of the equation of motion, but hitherto has not been presented as an available form.
In an axial magnetic field the Hamiltonian takes a form H 1 2 (P r Pz/2 + U'
U P6 2
U' =(- - eA,)/m + U(r,z)
It can be shown that P is constant and is actually composed of two terms. One is ethe so-called magnetic moment and the other is related to the flux of the magnetic field at the guiding center about which the particle gyrates.
The second part of the thesis deals mainly with the change of the magnetic moment according to time as the
particle makes a motion inside the magnetic field. The con-clusion is that the magnetic moment change (if any) will be the order of (r
/L)
4 where rb is Larmor radius and L is the characteristic Iength of the machine. Therefore, the change strongly depends on mass and energy of the particle. The conclusion is that the smaller the mass, or the smaller the energy, the longer those particles are confined. Thesolutions were obtained by a WKB approximation, and demonstrated by direct numerical computations.
Thesis supervisor: Dr. David J. Rose Title: Professor of Nuclear Engineering
My greatest appreciation and sincerest thanks go to
Dr. David J. Rose, my thesis supervisor, whose guidance and assistance made this thesis possible.
The in valuable suggestions by Dr. Melville Clark, Jr. are gratefully acknowledged. Thanks also go to Mr. Douglas Balcomb who helped me in using Reac analog Computer.
I would like to thank the MIT Computation Center for permission to utilize their IBM 704 Computer to perform the calculations for this thesis.
To Miss Nancy Milner, go my thanks for her patient typing of this thesis.
field is present has been known for several decades. But there has been little use of it for practical applications. This thesis tries to clarify the idea and to apply it to the problem related to confinement of charged particle inside
the magnetic field. This method has a special advantage in application with axial-symmetrical cylindrical coordinate. Then we can reduce the problem into two-dimensional one with a static potential. Devices used in various fields have been using this kind of configuration. For example, in accelerators, mass spectrometers, fusion devices and electronmicroscope.
The generalized Hamiltonian of the system iS (in MKS)9
H= (PrmeAr r A )
/+U(r,z,t)
(1.1)If H has no angular dependence, dP0 H
(1.2)
It means P =constantP, . (1.3)
Then Hamiltonian reduces to
H= ( (PrhAr 2+(Pz Az )2) +Ut(rzt) (14)
where Ut(rz,t) = - eAe)2 +U
This is an ordinary two dimensional motion of charged particles. In particular if Ar z=0, the current which produces magnetic fields is synnetrically encircling the axis. (i.e. jg/O
r
j =0) It is almost the case we see in ordinary machines. For example, in the field of electron-optics, refraction coefficient isIt will be interesting to deal with several examples. (W) The confinement of particles which encircle the axis in the mirror machine.
This is almost the case of DCX. The explanation of the mirror machine has beep treated elsewhere.(4
In this Ut ( eA6) (1.5a)
While from Hamiltonian,
M = m mr (% s eA) 11
S=mr e + ePrA
e
(1.6)Near the center of the machine, we can roughly see B is parallel to z axis with no variation in r. Then from the equation curl A=B we can get A= rB Therefore,
Pe,=r e+ gr Be (1.7)
The particle has an angular frequency of Wb=( eB 1. As it is essentially diamagnetic in nature, the angular momentum vector must be antiparallel to the magnetic fieldp As the particle encircles the axis, angular momentum seen from the axis has to be anti-parallel too. Which results (rb: Larmour radius)
)BeB
a1
-w2B
P
=(.
r r + - r )=eBr (r -rbeB(.0t max-max o
Where o denotes the value measured at z7O. (Fig. 1.1)
To look at it another way, Ut is proportional to the square a
of velocity in a direction. As far as the particle which encircles the axis can not cross the axis, it always has a velocity different from zero, therefore it never becomes zero.
IL1
AlL
a
.01 4+3
rr-The other way which correlates this quantity to the important physical quantity will be found later.
What is the function Ut?
First we assume the magnetic field is the function of z only. Then we define the corresponding cyclotron frequency as
w(z) = - B(z) & w(0)=w (1.9)
Then 1.5a will be
UI= M(. w s. w(z)r)- (r 2w +r2 )2 (1.10) We want to know the minimum potential with respect to the fixed
z, which will indicate the pass of the potential. The point must have zero value in derivative, so
a
2 -U r w r w 0 = (rw+ r (..2 +w) (1.11) rr r 2 21.o 2 2. r2r ; o r wo = r w (1.12)The condition is satisfied along the magnetic field line. Potential at that point will be,
U -i(r 2w +r 2w) 2 - W.
w
4
0, r ww (1.13)8ri. 1 0 0 0 0 2;-~ wo 0
The minimum potential goes up with w(z). Therefore, the particle whose energy is less than r w w', can not reach that z. Hence absolute confinement. Suppose a particle at z=0 makes an angle in velocity with magnetic line. Then
E= 1/2 mv2, vL = v sin= w0 r0 (1.14$ The absolute confinement criterion says
1/2mw w0 r 2 1 l/2mv
. 1/2mv
<
1/2mv or sin w _ . (1.15)4
Hence if we make the magnetic field R times stronger at the end of the mirror machine, we can confine the paticle which has an inclination with the axis at z=0, which satisfies the condition (1.15). The thorough, though less elegant
discussion of this problem can be found in referencesp (2)(4). (ii) Stability of the cyclotron. The cyclotron can be
considered as a big mirror machine. So it invariably follows that in a certain configuration a beam can be kept forever, because it encircles the axis.
(iii) Stability of the betatron and its frequency. In case (ii) stability for the r direction is not necessary, but
in the betatron and syncrotron, the stability in the r direction is also important. If we inject a beam at very low magnetic
2 field, we may set PQ0. Then U' is proportional to A0 .
A 0 42AO Because of symmetry in the z direction, and are always zero at the midplane. Therefore, the condition for the stable orbit, must satisfy
A 2 2
0, -- ,> 0, -- >0, (1.16)
Y 2y )'z
while condition of no current requires cul. curl A=0O.
Therefore 2A +
'- 2 _
(rA,)=0
---, + -- ,g (rAe) - r,=0(.7
z r r
which can be satisfied if
2 2
(rA (rA ) (-n)l -(rA) (1.18)
The condition that the 2nd derivatives should be positive, requires
0
n1
9)Since A0 (z=0,r) can be represented B rdr where B is 0
parallel to z. If we use that representation for equations (1.16) and (1.18) we have SB 0rdr 0 or B r Brdr (1.20) 0 r 0 0 0 (rA )( B rdr)= (B r) While (1-n) (rA )=(-n)0 (121 Therefore )r90 n (1.21)
So we see that n defined here is nothing but n used in high energy accelerator. Then for the equilibrium point, we have
U'=e 2 /2mo .(B 2 2 2+e (rr )2B (1-n)+e 2 (zz) B n (1.22)
(from 1.%a and by.Taylor expansion) or up to the 2nd term.
t- m W 2 2 02 (1.23
U'- % (r0+(r-r ) (1-n) + (z-o)n2 (1.923)
So the accelerated particle oscillates with frequency (1-n)l/2 and n in r and z directions respectively.
(iv) The explanation of Spiral Orbit Spectromter
The basic idea for this can be explained as follows. If certain magnetic field configuration is presnet, particles
which satisfy the cyclotron condition will tend to gather around the orbit, while others tend to move out or spiral
in. Generally the source is put at the center. Then electrons ejected from the source will have no angular momentum therefore the generalized angular momentum also must be zero. (As A
is zero at the axis). So essentially the electrons see 2
potential well Ut proportional to A . Then if it satisfies the betatron condition, for a certain point, -- = 0. So
there is a potential maximum. If the electron has just exactly the energy of Utmax then it will stay around that orbit an
infinite time, and electrons with energy near Utmax stay long time. The electron with less energy will be pushed in because of the potential barrier. And electrons of higher energy
will surpass the potential barrier. (Fig. 1.2)
. I
7
2. Magnetic Moment
It will be worthwhile to deal with the generalized angular momentum. From canonical equation (1.6) we get
2,2 1 e2(21
Pe=mr
e + erA mr + eBr21)This is seen to be a rigorous constant in our case. Next question is how it relates to the so-called magnetic moment.
(a) Particle which does not encircle the axis of syimetry. We take a particle moving in constant magnetic field. It is clear it makes a circular motion in a plane perpendicular
to the magnetic field.line# (cyclotron motion). From Fig. 1.3 it will be seen that a certain point A (and A'), the angular velocity becomes Fero. Let us denote the distance from the center to that point A as R. Then from eq. 2.1 we get P=
1/2eBR2 . Let us define r and rb as shown fig. 1.3. r is called the distance of guiding center and rb is cyclotron radius.
Then from the elementary geometry, R2 2-rb . Therefore P is composed of two terms I and M such that
7 =
/22BR
Ml/2eB2 2 2 - e B2P @ 1/2eBR =1/2e3(r 2 ma n07lem r le m 2)R
M-I > 0
e( .I (2.2)
M
1 e B r 2 9 77 m m b e B r 2 (2.3)We see that difference of M and I is strictly constant. But actually both do not change appreciably. M is prportional to the magnetic flux enclosed by the guiding center. I is the
(b) Particle which encircles the axis. If we notice the point E in fig. 1.2b, as the angular velocity with respec to the guiding center is - , we get - r So e m
Pg =mr
q+
1/2eBr2 l -m(rbmr )rb + 1/2 eB(rbr )2-(M.I M-< 0 (2
t
.4;)
We again see that it leads to the same formula. Needless to say that these arguments do hold for the time dependant magnetic field as long as it changes axial-symietrically.
There are several immediate applications without going into details. One is the precession of the charged particle inside the magnetic field. From (2.1) and(2.2), we get for an off-axis particle, 2- 1 eBR2 2 2 'W
E)
= z B a 1/2 e~'2 =l/2e3(R .r) (2.5) or, 1e=1/2 ( l) m (2.6)We assume that the particle is far off from the axis, it means R ) rb. Therefore, r r-r +rb cos t R+r(cos_' t+ ) or with (2.6) t* 1/24{ rb (Cos t+ + (cos W, i+ 2
= cosa t+ higher terms.) R
If we make a time average over many cyclotron cycles,
(2.7)
a)+
r 2 2 r2
(-.' +=)=R (2*9)
This is the angular velocity of the precession. Of course, this effect has been observed in Project Argus.
So far we are concerned about the axial-symmetric magnetic field. But it is of interest whether there can be any kind of invariance for an arbitrary magnetic field configuration. This problem has been solved by Kruskal. His method is essentially to transform the coordinates by canonical transformation and shows that it can be made constant up to as many powers as possible. It will be of some use to give the qualitative argument for that here. From eq. 1.1, we can make a cononical transformation such that A2 andA equal zero and also = 0. Which
are always possible because we may choose an orthogonal coordinate system such that q, is parallel to A while the direction of q2 and q3 to be cetermined from the condition
-- =0. From which it follws that p
1 is constant. That
is we have an integral of motion which may be said as magnetic moment, with an arbitrary magnetic field.
We want to discuss about the applicability of the preceding argument. If we treat many particles so that we can not neglect the effect due to the induced field (I.e. 1 is not zero), the preceding argument is still valid as long
as we take into account the induced field. (self-consistent field method). Therefore, this method can be applied to Budkar type beam and Astron. The picture can be used to explain
part of the phenomena such as pinch effect and shock wave, if we do not neglect the effect caused by electric force
(as opposed to magnetic field). But generally it is advisable to use the continunus model (magnetohydrodynamics).
3.
Analytical CalculationThe problem arises whether off-axis particles can be confined in the certain domain of the potential well. This is important not only in thermonuclear fusion but also confine-ment of electrons and ions inside a aosmical magnetic
field. From the foregoing discussion we see that A0 and PO have opposite signs. Therefore along a certain line
(which coincides with magnetic line) the potential remains zero. Of course potential can not become negative.
Consequently, the equipotential line (rather surface if we think we eliminated G coordinate) looks like fig. 2.1. This is for a particle which has corresponding P9. If the field is reasonably strong the number of particles which do not encircle the axis would be far much greater than that which do. Therefore, this problem is interesting for low-energy injecting machine like those developed in Livermore. Also it is the case with electrons confined in geomagnetic field, which was demonstrated beautifully in Project Argus. Garren et al3 has computed the confinement of particles inside the potential under no collision. They showed that particles seem to be confined forever. Of course not all but those which satisfy certain conditions. But there
are some points which need to be investigated more throughly. They are (1) How much is the error of computer? Because
we can not integrate the differential equation directly; so we have to use a numerical integration which invariably
3NV~1d 8l31N30
includes some error. (2) The time to make a computation would be enormous even for the high-speed computer. So
the computation is essentially done for a number of transits which is less than that we observe in experiment. So is it possible to say that, after many reflections, particles
are still kept inside the machine from the calculation of small number of reflections?
There have been expressed certain doubts in Gatlinburg Conference in April 1959. From the Ergotic point of view, the orbit of particle must cover a whole E-constant surface of p-q phase space. Therefore the point outside of the
mirror machine must be covered by the particle after a certain time if we admit the Ergodic hypothsis. Of course there
is infinity of which do not satisfy the above hypothsis. But thermodynamics holds that there is a non-denumerable infinity of orbits which do satisfy the above hypothsis. Therfore unless the particles are first given the exact
orientation so that they can stay forever inside the machine, they will finally get out of the system. And to give the exact orientation is completely impossible unless we resort to Maxwell's demon. But Ergodic's hypothesis is based on an assumption. So we can not disprove the result obtained by Garren et al from this standpoint of view. The excellent description of this will be found in reference(5),
configuration is given in several references (2.)4) Here, we consider it from an electron-optical point of view.
As v( ,v0/v=E/ f_); refraction coefficient increases when the particle goes up the potential hill. Therefore particles will be reflected back. (At the line CC' & 001 in fig. 2.1) But because7QAand 00' are not parallel, the particle will
cross the 00' at different angles. If it occurs several times, we can see that finally the particle loses Vz component.
The force is always directing towards Median plane, it, therefore, gets minus velocity and returns to the median plane. The
orbit of such a particle can be seen in Appendix fig. 2.2
which has been done by Reac analog computer. This corresponds to P9=0. Even from these crude pictures, we can find that end point is not constant but subject to som fluctuation the difference of which is order of 1%. The fluctuation of the end-point is what interests us. Because in the case of a mirror machine, if it goes over a bottle neck (EE') then
the force acting on the particle tends to act away from M.P. So it helps for the particle to escape out. (Fig. 3.3)
Consequently there are two questions.
(1) Is there, indeed, the case which seemingly disproves the Ergodic hypothesis? i.e. the absolute confinement of the particle. (2) If not, how long should we wait before a particle escapes out of the bottle neck?
There are some difficulties involved in it. First, the shape of the magnetic field will vary according to each
design. In order to get an applicable result, we can not adhere to a particular type. Secondly, if we want to use a computer, the accuracy required by the problem is so
high that even with present day high speed computers it requires a considerable time. If we want to follow 100 reflections
with the error within 1%, fbr simple particles, it will require about 2 hours.
As for the paper by Garren et al, there are 2 problems. (1) Their result does not include the error accumulated by the computer.
(2) They asserted that, since there is a closed orbit in certain configurations the orbit which lies inside such an orbit can be kept forever. As we must talk about p-q phase space, this conclusion seems to us too rash.
We therefore will treat this problem in simpler form first by an analytical method, second by a numerical method.
First we assume the current produced by plasma is
negligible compared with the field current. Then inside the mirror machine we get j=0. Or
curl curl A = 0 3.1
The potential curve is shown in fig. 2.1. U' will vary according to the type of device considered as well as the different P corresponding to a different particle. Besides, A. generally can not be represented by analytical form.
15
the higher terms. As U'> 0 and has zero at equilibriumpoint (00' in fig. 2.1) r (z), that is
- eA (r ,z)=0 (3.2)
eq a eq
So with a Taylor expansion
1 ( -eA2 (r-r 2+ 1 e-r e
eq r-r eq
(3.3)
Note constant term and first derivatives are zero
,2 = mw 2(z)
(3.*4)
rreq We may rewrite
U. mw (z)(r-r q)2 (3.5)
Then we assume that the change of magnetic field in the direction of z is very small. Instead of using rectangular
coordinate (r,z coordinates), we may use curvilinear coordinates such that x axis is perpendicular to the magnetic field line. Then we may express x as
x = r-r +x' 3.6
Where is used to show that this correction term is small. If we express every term of Hamiltonian by x,z instead of x,r we get
H 1 (P 2+P )+ (p 2 2 PX + ( )2 2
- )W X ( + )-Z (3+
Sre
If we allow that 3 and neglect compared unity we get
.1 jp2p2 m 2 2
H= 2(P z)+ (z)x + higher terms
(3.8)
The effect from the higher terms is therefore of the order of .O I am not sure whether this assumption is always applicable. But this is applicable in cases where the length of the machine is large compared with the diameter of the machine (say by factor three). Or when we are treating the particles which are close to the axis, this is applicable. If we go to the one higher order of approximation we have to treat the special type of the machine, it will make the calculation applicable only to the particular type of
configuration. By defining w-w0$, where w is the cyclotron angular velocity, we have
>0 for z 0 (3.8)
O(Z)=R)1
Where Z is the point of maximum field strength. R is the mirror ratio. In case of symmetric mirror,
%(Z)=%(-Z)
(3.9)
If we set m, wo, equal unity and, for the unit of length, we set larmour radius as the unit we get,
1 2
2
1
2
2
H =(P 2+Pz2 2+ 2(3.10)
Eq. (3.10), though jUst a classical mechanics problem, can not be solved rigorously. But using WKB method
we can get an approximate solution known as the adiabatic invariance of the magnetic moment.
In order to do that, we first make use of the equation
P $ x2 (3*11)
With the application of WKB method, we set
a cos (a + a.) (3.12)
where
(t
dt (3.13)
a and a are the constants of integration. By using eq. (3.11) repeatedly we get,
d 2P 2 -(z (7 + 0'2)=(" #PX + )+ ( 2O =( X (.-x )+X )+ 0(- )2 2 12 2
X'.
= j
(---
) =
dt)X2
314
Therfore by integrating with respect to time,
( 2 = +HI(t) V: const. (3.15) where t 1 d Ht(t)= - ( X2 )dt (3.16) b dt So (3.10) becomes H= 2 Pz + X(z) +Ht(t)A (3.17)
We can see that, if the quantity HI% is small compared, with total energy H so that it can be neglected, then the
Hamiltonian describes a one dimensional particle inside the potential 1/2%p. This is the rigorous derivation of the socalled adiabatic law of magnetic moment. Now if we use equation 3.12 (so far we have not used it) to
determine the integration constant V at eq. (3.15), we get =a2 by calculating it at t = 0.
If H' were zero, there would be no trouble of confining the particle. Since it is not the case, we are concerned about the behaviour of the reflecting point (z = 0). If we rewrite eq. (3.17)
H= P 2 + (V+2H' (t)) (3.18)
We immediately see that HI takes care of the change of magnetic invariance. One thing is clear that, if H' has an upper limit, then the change of magnetic moment has an upper limit. Therefore, with certain conditions, the particle may be contained inside the machine permanently provided the effects of higher terms do not contribute much.
First we are interested in the change of the magnetic moment from the center (t=O, z=O) to the reflecting point
(t=T, Z=0). At z=0, it is reasonable to assume 0. Further this term is generally small, we may use the approximate solution for x, namely eq. (3.12). Then we get,
2 H( T 1 2) dt= 2 2( )
T 2
=p ( 1(1+cos 2(a+a ))dt~= (3.*19)
0
The first term inside the parenthesis is seen to vanish trivially with the boundary condition we imposed on ,t = 0 and T. Therefore it is only necessary to consider 2nd term. By partial integrationa nd application of boundary condition#, we get 1 jT((d I d( cos 2 (a+ai)) dt TT 02 1 d 1( d 1 d21 dd )cos 2 (4 + (- ) - L, 0 2
Xcos 2(o( +k ) dt =- ( )cos 2('+)t(.0
This integral, can be evaluated correctly, if and only if we know the information regarding f and the change of z with
respect to time. But we still get some useful information from the general property of the integral. If we can
d21
approximate 2 g by the- average value (which is possible if the term is slowing varying) then we have
2 T 2
I= - d [. cos 2( +0d t
j=a
os 2a 1cos 2(Xk (T)+t k)) (3.21)
This shows that the change of the magnetic moment will vary as the 2nd harmonics of the initial angle distribution which has been demonstrated by MIT computer. Again if we may replace
the bracket by 1/4 cos2Q ( which is possible if (TI >l) 2
1 . and if we may replace the average value by g
t=o where g is order of unity, we get
I cos 20 (3.22)
t=o
But it is easy to calculate the bracket which gives us
(3.23)
t=0
At (3.18), we may take as a first approximation HI=0. So we get
1P2+ 1 =const. (3.24)
Therefore at t=0 and T quantities must be the same
+ -= RO(T) Z P . )- ) (3.25) Further we may make
2 O(T)-l (3.26)
dZ L
combining together we get
=1t
((
1)2 cos 20( (3.27)Therefore the maximum change of the magnetic moment for the trip from the center to the edge is
LLI 2
2H (t) ((OT)-l) 3.28)
The value of L may be defined as the length in which
the change of magnetic field takes place. ) Equation (3.27) says the importance of the initial phase. With the
this order, I is bounded by equation 3.28 no matter how many traversals the particle under consideration makes
between the reflecting points of mirror machine.
Therefore a particle, put into the Ymchine which is completely symmetrical with respect to axis and z=0 plane, will leak out of the system very slowly, (if it can), due to the higher term effect. But things are completely different in case there is a slight asyumetry of order of larmor radius inside the machine. Then because of the precession of the particle, the particle may forget the
initial angle when he left the center previously. Or higher terms may contribute much, if the particle is situated
far off the axis of -symmetry. It is obvious that theCusp type mirror is just this case. Then we can anticipate that the change of the magnetic moment occurs in random fazhion. Then if we define a reflection as such that the particle leaves the center (z=0) and returns there after being reflected
back from the end. Thai after N reflections,
g (O(T) 1)2 (3.29)
L
where the change of magnetic moment was assumed to be small. While on the other hand from eq. (3.18), in order to
arrive at Z(maximum field strength), the magnetic moment should change such that
holds.
Eliminating ApL, we get,
N- 2 L2 R-O(T)
g(9(T)-l) R (3.31)
With our urit this becomes 2 L 4'd
N= (-7:- -L (l) (3.32)
g rb R (O(T)-1)
# (T) may be considered as the ratio of magnetic field for
the first reflection. If we express the equation by conventinnal units we get
B4
Nor/ (3.33)
Where V stands for energy of the particle. The confinement time 5 will be
B4
L
B ~00 (3.34)
Both clearly indicates the strong dependence on the intensity of magnetic field. And also they show that the confinement will be the better the smaller the particle, which has been inferred with Van Allen radiation belt.
Suppose we have magentic marror with D ions; and let B=
10
gaussV 100 key Ll1m
R = 4, fC')=2
Then r'b -4.5 m (3.35)
N = 3.04 x 104
and if we assume average velocity in z direction of the order of 106m/see, will be -. 100 msec.
The conclusion is, then, that we can not keep the particle forever. The escape of particles is governed by equation (3.32). If it is very small compared with the c6llision time, the leakage will be dominated by collision. But if 1) weak magnetic field, 2) very low
density,
3)
heavy mass or,4)
high energy is the case (or sonm combined), the leakage due to this mechanism may be more important.4.
Numerical CalculationIn order to see the validity of the analytical method, we have done computations using the IBM 704 computer.
Equations of motion, though simple in form, can not be integrated so easily because we need high precision for a long period. The energy integral might have been used as a means of reducing the four variables (x, z, p , Pz)
into three. But here we have used it as a means to calculate the error. The equations of motion
62
PX x x p
PZ dz z
may be integrated if we assume a function for
0.
We used forX
the form of 1 + k2z2As for the initial condition, we started the particle from the center. That is, at
t
=
0, z = 0, and z = Q. With our units=l.
Therefore the first end point will be very close to z =Q/k.
So0 (T) will be 1 + Q 2 The time required for one reflection will be "'/k (With the error of order of HI). And pb/L ratio may be set l/Zm =k/Q.
This problem may be solved streight-forwardly. But since we have' an approximate solution, namely, the adiabatic invariance (2.12). If we set
t
x = u+h
(4.3)
and calculate h which is a very small quantity. In thisway we can. economize the steps required to compute the function with a given accuracy.
But if we do not change the expression of u, the magnitude of the deviation h becomes so large after several reflections that no longer can we say that h is much smaller compared with u. Then the calculations may require the same number of steps as without this device. In order to circumvent this difficulty, we changed u at certain points so that we can keep h always
small. Here we must stress that the above method still gives a correct solution (subject to truncation error, of course) and not an approximate one. But if we treat something which is only a small deviation of the true solution, the accuracy of it need not be so high. This procedure seems to give a new method for numerical analysis of certain differential
equations. The physical interpretation of this will be as follows: We first calculate the orbit of the particle
being started at a certain place and follow to such an extent that the approximate solution is still a good solution. Then we annihilate the particle and create a new particle whose initial conditions are exactly the same as the old particle. And start the computation again with a new approxirmte solution.
The aquare of a, though a little different from conventional magnetic moment, still gives a good measure of the magnetic
moment. It coincides with the conventional magnetic moment exactly at the center or at the reflecting point.
It has a more stable property than the conventional one, because the latter is subject to fluctuation.(7) However the difference is an order of k. The equation used in the computation is rather tedi.ous. (See appendix).
Initial condition for x will be,
x Cos Q x -sin a
(44)
has a bearing with vr and v at t=0 by V
= - tan a
(4.5)
We did several calculations varying Q, k, and a.
The typical result is tabulated in table 4.1. Figure
4.1 shows the variation of the change of
sL
vs. the function of the initial phase. Though we could not find a spectacular escape of the particle, fig. 4.2 actually shows a systematic decreasing of a, hence increasing of the reflecting-pont.Maximum change of p vs. k is shown in fig. 4.3. It
shows k2 dependence clearly. Besides the curve by different
Q shows the slope which is proportional to Q2 which is also predicted by equation (3.28). The solid line represents eq. (3.28) with g = 1. It-can be noticed that if the k becomes large, derivation also becomes significant. This
Conclusion and discussion
It is necessary to review the assumptions we have made( (1) We have neglected effects of order k compared with unity, so as to get the general characteristics.
(2) In order to derive eq. (3.32), we assumed the randomness in the phase of the particle which is valid if there is a slight asymmetry in the machine.
(3) The effectobthe actual machine magnetic field, especially if it has a bottle-neck, has been omitted completely. But it may contribute for higher terms.
Then with results from the computer, we may say the following things:
(i) The particle can not be kept forever inside the machine. (ii) The rate of the escaping, however, is quite low. And it will be given by eq. (3.32).
(iii) Ions have more probability of getting out from the system, which offers a new possibility for confining
particles.
Namely, using the idea of the Homopolar device (8) we may impose an axial electric field making the core positive. Then it can form a potential well for ions. Therefore by that device, we can keep ions inside the machine. Of course this electric field tends to make electrons get out from the system. But since electrons are better confined by the adiabacy of magnetic moments, there may be an optimum electric field
for which the confinement becomes best.
Next step will be to calculate the orbit by using the double precision method. And also to check the dependence of the potential field will be interesting.
At least, I hope, this settles the dispute about the constancy of the magnetic moment.
vs Initial Phase 0.025 error rb 1 L 10-0.3 0.2 0.1 0 0 0 0 0.* -0.2-~10 1 3 Phase
* .
22
e
.A. ,b W ITH g =1 m 2 L 0.01-100 200 300 40e 4 rb @6
TIME 0. 0. 1.5820 3.1360 4.7180 6.2720
7.8540
9.4080 11.004012.5440
14.1399 15.6939 17.2759 18.8299 20.4119 21.9659 23.5199 25.1019 26.6559 28.2519 29.7919 31.3879 32.9419 (Z) 0. -0,0000 0.2422 0.0146 0.2165 0.0757 0.2013 0.1658 0.1699 0.2541 0.1566 0.2684 0.1462 0*3177 0.1250 0.3248 0.0974 0.2879 0.0645 0.2510 0.0293 0.1673 0,.0046 PHASE 1.5708 0.1616 4.4'84 3.0654 1.0686 -0.3168 3.9680 2.8599 0,5906 -0.5296 3.6416 2.3605 002762 -1.0363 3.2003 1.2905 -0.1521 4.1720 2.9239 0.7692-0.4159
3.9273 FINAL POINT (Z) 0. 0.9951 -0.0015 -0.9951 0.0023 0.9950 -0.0022 -0.9948 0.0013 009946-0.0133
-0-9944
0.0105 0.9942 -0.0068 -0.9943 0.0021 0-9946
-0.0106 -0.9949 0-0043 0.9951 0. 0. 0. VELOCITY 1.0000 -0.0229 -0.9994 0.0124 0.9983 -0.0018 -0.9968 0.0168 0.9954 -0.0068 -0.9940 0.0012 0.9933 -0.0009 -0.9929 0*0039 0.9929 -0.0075 -0.9931 0.0045 0.9935 -0.0118 ERROR (%) 0.-0.0004
-0.0374
-0.0418
-0.0876 -0.0903-0.1423
-0.1447
-0*2033-0.2054
-0.2599 -O .2615 -0.3188 -0.3229 -0.3720 -0.3790-044180
-0.4277
-0.4558
-0.4651
-0.4942
-0.5062 -0A5260 0.Appendix
Fow diagram of the computer program is given in fig. A-i, We have used the Runge-Kutta method for computing programmed by Lockeed Aircraft Company. With substitution of x=u+h. The equations of motion are rewritten,
h ph p= uq-2h -2k f 0x (A-1) 1+Q 2 r2
T
k 2 q = ( 2 -3Q i 2 2 -2k2S x 2 )After the particle hits the reflection point (of course within the error of the step of the Runge-Kutta method), new us is calculated from the knowledge of x and px. (Which is possible by definition). The new parameters of u are stored to be
printed after the end of the run. Along with it, the error of the computation which is defined as
v-v
(A-2) V 0
is stored. Where V is the energy calculated at that point, while V0 is the energy of the particle at t=0. Strict
calculations should be V = V at any time. But because of the error we get this quantity which will be the measure of the
error. Also time, position, and speed (in z direction) are stored.
One step of Runge-Kutta mthod takes roughly 150
msec. And compuation of one unit (center to reflecting point or vice-versa) takes about 70 steps. The nuinber of steps of course depends upon k.
00 00 H 0 a.a W 0 +3 *4 +) 0 -0 44 0 0 *
. . . . . . ... . . . . . . . . . .. ... . ... 9TV*o vcij is4ndmon OVU Aq pa4ndmoc) wlaqc)
1. Livingston, M. S., "High Energy Accelerators, Interscience.
2. SpitzerJr. L., "Physics of Fully Ionized Gases,
Interscience.
3.
Garren, A.; Riddell, R.J.; Smith, L.; Bing, G.; Henrich, L.R.; Northrop, T.G.; and Roberts, J.E.; "Individual ParticleMotion and the Effect of Scattering in an Axially Synmetric Magnetic Field", Second United Nations
International Conference on the Peaceful Uses of Atomic Energy.
4.
Rose, D.J., "22.61 Introduc tion to Thermonuclear Processes" Class Notes.5.
Becker, Richard, "Theorie der WVarme", p. 97. 6. Schiff, L.I., "Quantum Mechanics", pp. 178-187.7. See last paragraph of 2nd page of reference (3). 8. Wilcox, J.M., "Review of High-Temperature Rotating
Plasma Experiments", Rev. of Mod. Phys. 31, pp.
1045-1051 (1959 ).Cl
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