Some kinetic problems regarding the motion of neutrons through paraffine

$hysica

IV, No. 6

,Juni’ 193 7

SOME KINETIC PROBLEMS REGARDING THE MOTION OF NEUTRONS THROUGH PARAFFINE by L. S. ORNSTEIN Natuurkundig

Laboratorium

and G. E. UHLENBECK’: der Rijksuniversiteit

Utrecht

Abstract Derivation of a general continuity equation for,the time dependence of the energy distribution of particles going through matter ($ 2;Eq. 7). Application to neutrons going through paraffine when the protons can be considered at rest ($ 3). In § 4 we derive for -this case also the continuity equation for the stationary distribution in space and in velocity (Eq. 13) and compute the mean square displacement for the neutrons with a given, velocity (5 5, Eq. 18).

$ 1. Introdwtiort. In this note we %ill consider the slowing.down and the diffusion of neutrons going through paraffine. Both questions have been treated in detail by F e r m i l), and what we have to add will be only a different and more systematic method of treatment. We will always start from a continuity equation, which is, as always in the kinetic theory, an integro-differential equation and which links the statistical part of the problem with the elementary interaction law. This law has to be found by quantum mechanics, and will be quite different in different cases. The statistical part of the problem though, is always quite similar, and it seems to us therefore that from a systematic point of view it has some advantage to separate as clearly as possible these two aspects of the problem. For our case of the collision between a neutron and a proton the elementary interaction law is wellknown. From the fact that the scattering of the neutron is spherically symmetric in the coordinate system, where the cent,er of gravity is at rest, one obtains easily for the probability W(B)df3 that in a collision of a neutron (velocity V, energy E) against a proton at rest *), the *) In the following we will always suppose that the.velocity of the neutrons is so high that the protons may be considered at rest (“la fase di rallentamento” of Fermi). When the heat motion of the protons begins to play .a role, the problems become much more complicated and are then related with the almost untouched problem of how the M a xw e 11-B o 1 t x m a n n distribution is exactly rea’ched .in time.

- 478 -

SQJfF$&INETIC

neutron

PROBLEMS

is deflected

REGARDING

THE

MOTION

over an angle between w(e)de = 2

~0~

8 sin

ede

,479

OF NEUTRONS

8 and 0 + de: (1) energy of the neutron

When after the collision the velocityyand are resp. v and E, one has: Y = v cos 8 E = E cos2 8

(2) Let in general cp(E, E) dE be the probability that by a collision the energy changes from E to a value between, E and E + do *). Of course: %rp(E, c) &‘= 1 (3) In our case one obtains

immediately

from

(1) and (2):

P)

cp(E, S) dc = $ .

We will further need the probability pat that a neutron with energy- c will make a collision during the time interval At. This probability is immediately related with the total cross section C(E) and with’ the mean free parth A(E), since:

9(E)= VNB(E) =& ifs IV, is the velocity of the particle and N the number pro cc. In our case the theory gives “) :

of protons

(6) when q,, q are the absolute values of the binding 3S and ‘S state of the deuteron.

energies of the

$2. The continzkty equation 107 the distribution in energy. The statistical problem which w&?will consider first consists in the determination of the probability P(E, E, t) that a neutron, starting at t = 0 with energy E, will :have at time t an energy between E and E + dc. The continuity &$ration for P follows immediately. from the relation: P(E, E, t + At) = P(E, E, t) (1 -$(E)

At) + + At/W>

*) rp(E, E) may

be called

the

relative

differential

cross

Y, 4 $4~) CP(Y,4 dr section.

480

L. $. ORNSTEIN

AND

G. E. UHLENBECK

which expresses the fact that the number of particles having energy E at t + At will consist of those particles with energy E at time t which have not collided in At plus those particles which had energy y (between E and E) at time t and which have made one collision in At which has reduced their energy to E. Developing P(E, E, t + At) one obtains the integro-differential equation:

which is the fundamental mediate consequence is :

continuity

equation

required.

An ;

I ;

Pde=

0

s 0

which expresses the conservation We have to find now a solution

of the total number of particles. of (7) with the initial condition

P(E, E, o) = 6(E -

E),

which expresses the fact that at t = 0 all particles have the energy E; 6(E - E) is the well known singular peak function *). With this condition one shows easily (with the help of (8)) that there can be only one solution of (7). 9 3. Application to neutrons. We will apply (7) to the process of the slowing down of neutrons in paraffine, regarding thereby only the dependence of the energy distribution from the time. The great simplification lies in the fact that in this case cp(E, E) is independent of S. With the expression (4) for ‘p, the solution of (7) can be found in the following way. Write: t) P(E, E, t) = e+‘@) 6(E -

then F will fulfill ;

E) + e+‘tE) F(E, E, t)

the inhomogeneous

+ {$(E) -p(E)}

(9)

equation:

F = q

+

;(E,y,t)+y.

(10)

Es *) With

the

slight h(E

difference -

in definition

E) do =

t) Physically thys substitution which do not collide and which time like exp. (---p(E)).

1

that E /f(~)

is suggesttd keep therefore

6(E

one -E)

by the fact the initial

must

have:

do = f(E)

that the number of parficles energy E, will diminish with

SOME

KINETIC

PROBLEMS

By differentiating

REGARDING

THE

MOTION

OF

NEUTRONS

48 1

after E one obtains:

~+(p(E)--(E)}~+{~/(E)

+V}F=o.

The conditions which have to be imposed besides that at t = 0 F = 0, that further for t and that for E = E and for all times aF/at conditions follow immediately from (IO). Eq. and one finds as a particular solution:

(11)

on the solution are = 0, aF/i3t = $(E)/E = $(E)/E, These last (11) can be separated

where h is the separation parameter. One can then easily verify -that by integrating this solution in the complex A-plane along a closed curve around the origin one is able to fulfill all the boundary conditions. One finds thus: F(E,

E,

t) = dh e’lh 1 +h{j+)-$(E))eXP

k P(5) [Shc r 1 +A{IYE)-$J(E)I

I (12)

By determining the residue one can write this as a power series in t, but only with special. assumptions for $(E) the series can be summed. This is for instance the case when P(E) is a constant, say #,. One obtains then easily:

F(E, E3t) - @ot; (*Z)2n+’ 29otIi(z) z E o n!(lz+ I)! = E when z2 = 4p,t log (El E) and II(z) is the Bessel function of first order and imaginary argument. Unfortunately with the theoretical expression (6) for the collision cross section O(E), fi(~) becomes quite complicated. The power series is now very involved, and since a direct experimental application does not seem to be possible we have refrained from a precise numerical discussion of the resulting distribution function. Let us remark only that the result is again simple, when all the energies involved are either small or very large, because then p( E) can either be approximated by:

Physica

IV

31

482

L. S. ORNSTEIN

AND

G. E. UHLENBECK

or by:

when C is a constant. One finds in these two cases from (12) the , distribution functions :

where in the first case 7 = 4~&/C(3~r

+ EJ and in the second

7 = t/2c. $4. The contilzzcity eqzla&ioB for the stationary

distribution

in @ace

and irt velocity. We will consider now with Fermi the following problem. Suppose that from a point P, which is completely surrounded’by paraffine, Q neutrons with velocity V are emitted pro second continuously. What will be the distribution in space and in velocity when the stationary state is reached? In contrast therefore to the problem in 5 3 the dependence on the time will not be considered. Let F(r, v, cos 0) drdvd (cos 0) be the number of neutrons inside a spherical shell with radii Y and Y + dr around P, which have a velocity between v and v + dv in a direction which makes an angle between I3 and 0 + dfl with the outgoing radius. The continuity equation for F will be again the suitably adapted B o 1 t zm a n n equation. One finds: +r-

sin2 8

aF a(c0se)

=-

Ff A(V)

+ $&~cp

v

F(r, v’, cos fI’)

( 13)

0

At the left hand side are the well known ,,streaming” terms transformed into the polar coordinates, which we use, and divided by v. The first term on the right hand side represents the losses from the group F due to collisions against the protons *), which *) The

collisions

between

the

neutrons

are

of oourse

neglected.

SOME

KINETIC

PROBLEMS

REGARDING

THE

MO’I’ION.OF

483

NEUTRONS

occur with the probability v/A(v) per second (see (5)). The last term represents the gains due to collisions. Here v’ and 8’ are the magniude and direction of the velocity beforet the collision; according to (2) the angle a between v and v’ is given by: cos a

=

y V

.

The probability for such a restituting collision _is v’/?,(v’), and since according to (4) the probability that in a collision the velocity changes from v’ to v is given by:

Fig.

1.

the general structure of the last term in (13) is clear. We will omit the more precise justification, since the consequence which we need can be interpreted very easily. Let us define namely the functions G,(r, v) by: G,,(r, v) =p

sin 0 ~09 0 I;@, v, cos 0)

then one finds from (12) the simultaneous

equations:

v

Go + w

aG1 -=--

al

2

G,(r, v’)

&I’

va(vl> vs V +

2% "

2

+z

GJ-Gl)

=-L

G2 w

2

+/&[(I

+) Mathematically

are,

except

it is better

for

a factor,

G(r,

b

v')

* (15)

+

/I

.represent the continuity to use instead

Gn =,?dO which

&

-$)Go-(l-32j

”

etc. *) These equations

s

of the Gn the functions

sin 0 Pm (cos 8) F(r, the

coefficients

equations

in the

for the

G’e defined

by:

v, cos 0) development

of F in

Legendre

484

L. S. ORNSTEIN

AND

G. E. UHLENBECK

number of particles in the spherical shell dr and with velocities in the range dv, for the radial imp& of these particles (= MvGi), for the kinetic energy of their motion in radial direction (= @!v2G,) etc. One sees this quite simply, not only for the. streaming terms on the left side and for the direct collision terms, but also for the restituting collision terms. Take for instance the second equation The gain in G1 due to a restituting collision will be cos 8 F(r, v’, dos 0’) drdv’. To find the total gain we have to average cos 0 .over all possible collisions. Since (see fig.. 1) : cos 8 = cos 0’ cos cc + sin 8’ sin a cos ‘p’ and 3

= 0, one gets: cose=cose~cosa=~cose~.

The average gain by a collision

from v’ to v will therefore

be

; Gl(r, v’) drdv’ . Multiplied with v’/h(v’) and (p(v’, v) as given by (14), and integrated over all possible values of v’ one obtains just the last term of (15b). To derive (14~) one has to compute s. One finds :

ca = +(1-$)-j.(l

-$)cosw

so that the average gain in G2 by a collision from v’ to v will be: 4 (1 -

$)

G,(r, v’) drdv’ -

-which gives immediately go on.

4 (1 -

g)

G,(r, v’) drdv’

the last term of (15~). In this way one can

$5. The calculation of certain average val,ues. One has to solve (13) with the boundary condition that at I = 0 F(0, v, cos 0) = (Q/V) 6(V - v) 6(O) and that at Y = 00 for all v + 0 F goes to polynomials

in cos 8. One

obtains

then

from

(13) the simultaneous

equations:

V

av?s+ ar However, the first

1

+

;

(G’,,+l

the equations two of these.

-

G’,,-1)

we actually

= -

use,

;$

namely

$ 2

/

” (15a)

-

dv’

v?k(v’) and

(156)

are

identical

with

SOME

KINETIC

PROBLEMS

REGARDING

THE

MOTION

.OF

485

NEUTRONS

zero sufficiently fast. Consider instead of ( 12) the equations By writing analogously to (9): G,,(r, v) = $ 6(V -

(14).

v) c-+(~) + H&> v)

one takes care of the condition at I = 0, if one still supposes 23, = 0 for I = 0. For H,, one obtains then a set of simultaneous inhomogeneous equations, which are similar to (15) and which we will not write down. The general solution seems quite difficult and we succeeded only in calculating certain average values over Y. Let us take first: /&Y,

v) dr = $ A(V) 6(V -

v) +j+kH”

0

(Y, v)

0

and call the last integral H,, one obtains:

N,,(V). From the first of the equations

By differentiating after v one gets a simple differential in ~V,/A with the solution:, . N,(v) = g

for

equation

A(v)

(16)

which is a special case of a result first obtained by F e r m i “). Unfortunately it does not seem possible to calculate Nr, IV, etc. in this way. Let us introduce next the abbreviations: ~H,,(Y,

v) dr = R,,(v)

j?H&,

v) dr = P,(v)

The average distance velocity v, will then be:

from

the source

of the neutrons

with

486

L.

S. ORNSTEIN

AND

G. E. UHLENBECK

the average square distance : fi) = PO/N0 and so on *). For’ Ii0 and R, one obtains the equations: v -NN,(v)

=-$+2

vs

+

ROW + $$ A(p)

v

RI R,(v’) + 3 W’). A(v) + 2vs & ” Since N1 is unknown, one cannot compute R, and therefore is easily soluble. One finds: neither pi. But the equation for -NN,(v)

= --

RI v. JG(4=212 2QW+-&(V) a? + f$ +. A(v) vs

One needs this for the calculation equation :

(‘7)

of P,, for which one gets the

V.

-

2R,(v) = -3

A(v) + 2s ” This can now be solved, and ‘one obtains then for 72: v

m

= 2A2(v) + 2A2(v) + $

A(

+ 4vh(v)

sv

I!@ de + E2

V

a result, which is in accordance with F e r m i, who has derived it by a completely different method. In this way one can go on. The calculation of the averages of the even powers of I can be done successively, but the result for y4 is already quite complicated. Received

April

22,

1937. REFERENCES

1) E. Fermi, Comp. also:

Ric. Scient., VII 2, 1, 1936. E. Condon and G. Brei t, Phys. Rev. 49, 229, 1936. S. Goudsmi t, Phys. Rev. 49, 406, 1936. L.S. Ornstein, Proc.Amst.Acad.39,810;904; 1049; 1166; 2) B e t h e and B a c h e r, Rev. Mod. Phys. 8, 82, 1936. 3) E. Fermi, Zeeman Feestbundel, p. 128 (The Hague, Nyhoff 1935). one

*) This will hold for vel6cities v # V. For the neutrons obtains of course r(V) = h(V); YP(V) = us(V).

with

the

initial

1936.

velocity