§16. Disintegration of particles
IN many cases the laws of conservation of momentum and energy alone can
be used to obtain important results concerning the properties of various mech-
anical processes. It should be noted that these properties are independent of
the particular type of interaction between the particles involved.
Let us consider a "spontaneous" disintegration (that is, one not due to
external forces) of a particle into two "constituent parts", i.e. into two other
particles which move independently after the disintegration.
This process is most simply described in a frame of reference in which the
particle is at rest before the disintegration. The law of conservation of momen-
tum shows that the sum of the momenta of the two particles formed in the
disintegration is then zero; that is, the particles move apart with equal and
opposite momenta. The magnitude Po of either momentum is given by the
law of conservation of energy:
here M1 and m2 are the masses of the particles, E1t and E2i their internal
energies, and E the internal energy of the original particle. If € is the "dis-
integration energy", i.e. the difference
E= E:-E11-E2i,
(16.1)
which must obviously be positive, then
(16.2)
which determines Po; here m is the reduced mass of the two particles. The
velocities are V10 = Po/m1, V20 = Po/m2.
Let us now change to a frame of reference in which the primary particle
moves with velocity V before the break-up. This frame is usually called the
laboratory system, or L system, in contradistinction to the centre-of-mass
system, or C system, in which the total momentum is zero. Let us consider
one of the resulting particles, and let V and V0 be its velocities in the L and
the C system respectively. Evidently V = V+vo, or V -V = V0, and SO
(16.3)
where 0 is the angle at which this particle moves relative to the direction of
the velocity V. This equation gives the velocity of the particle as a function
41
42
Collisions Between Particles
§16
of its direction of motion in the L system. In Fig. 14 the velocity V is repre-
sented by a vector drawn to any point on a circle+ of radius vo from a point
A at a distance V from the centre. The cases V < vo and V>00 are shown
in Figs. 14a, b respectively. In the former case 0 can have any value, but in
the latter case the particle can move only forwards, at an angle 0 which does
not exceed Omax, given by
(16.4)
this is the direction of the tangent from the point A to the circle.
C
V
V
VO
VO
max
oo
oo
A
V
A
V
(a) V<VO
)V>Vo
FIG. 14
The relation between the angles 0 and Oo in the L and C systems is evi-
dently (Fig. 14)
tan
(16.5)
If this equation is solved for cos Oo, we obtain
V
0
(16.6)
For vo > V the relation between Oo and 0 is one-to-one (Fig. 14a). The plus
sign must be taken in (16.6), so that Oo = 0 when 0 = 0. If vo < V, however,
the relation is not one-to-one: for each value of 0 there are two values of Oo,
which correspond to vectors V0 drawn from the centre of the circle to the
points B and C (Fig. 14b), and are given by the two signs in (16.6).
In physical applications we are usually concerned with the disintegration
of not one but many similar particles, and this raises the problem of the
distribution of the resulting particles in direction, energy, etc. We shall
assume that the primary particles are randomly oriented in space, i.e. iso-
tropically on average.
t More precisely, to any point on a sphere of radius vo, of which Fig. 14 shows a diametral
section.
§16
Disintegration of particles
43
In the C system, this problem is very easily solved: every resulting particle
(of a given kind) has the same energy, and their directions of motion are
isotropically distributed. The latter fact depends on the assumption that the
primary particles are randomly oriented, and can be expressed by saying
that the fraction of particles entering a solid angle element doo is proportional
to doo, i.e. equal to doo/4rr. The distribution with respect to the angle Oo is
obtained by putting doo = 2m sin Oo doo, i.e. the corresponding fraction is
(16.7)
The corresponding distributions in the L system are obtained by an
appropriate transformation. For example, let us calculate the kinetic energy
distribution in the L system. Squaring the equation V = V0 + V, we have
2.2 = V02 + +200 cos Oo, whence d(cos 00) = d(v2)/200V. Using the
kinetic energy T = 1mv2, where m is m1 or M2 depending on which kind of
particle is under consideration, and substituting in (16.7), we find the re-
quired distribution:
(1/2mvov) dT.
(16.8)
The kinetic energy can take values between Tmin = 3m(e0-V)2 and
Tmax = 1m(vo+V)2. The particles are, according to (16.8), distributed
uniformly over this range.
When a particle disintegrates into more than two parts, the laws of con-
servation of energy and momentum naturally allow considerably more free-
dom as regards the velocities and directions of motion of the resulting particles.
In particular, the energies of these particles in the C system do not have
determinate values. There is, however, an upper limit to the kinetic energy
of any one of the resulting particles. To determine the limit, we consider
the system formed by all these particles except the one concerned (whose
mass is M1, say), and denote the "internal energy" of that system by Ei'.
Then the kinetic energy of the particle M1 is, by (16.1) and (16.2),
T10 = 2/2m1 = (M-m1)(E:-E1-EiM where M is the mass of the
primary particle. It is evident that T10 has its greatest possible value
when E/' is least. For this to be so, all the resulting particles except M1
must be moving with the same velocity. Then Ei is simply the sum of their
internal energies, and the difference E;-E-E; is the disintegration
energy E. Thus
T10,max = (M - M1) E M.
(16.9)
PROBLEMS
PROBLEM 1. Find the relation between the angles 01, O2 (in the L system) after a disintegra-
tion into two particles.
SOLUTION. In the C system, the corresponding angles are related by 010 = n-020. Calling
010 simply Oo and using formula (16.5) for each of the two particles, we can put
7+10 cos Oo = V10 sin Oo cot 01, V-V20 cos Oo = V20 sin Oo cot O2. From these two
equations we must eliminate Oo. To do so, we first solve for cos Oo and sin Oo, and then
44
Collisions Between Particles
§17
form the sum of their squares, which is unity. Since V10/2.20 = m2/m1, we have finally, using
(16.2),
(m2/m1) sin202+(m1/m2) sin201-2 sin O1 sin O2 cos(A1+6
PROBLEM 2. Find the angular distribution of the resulting particles in the L system.
SOLUTION. When vo > V, we substitute (16.6), with the plus sign of the radical, in (16.7),
obtaining
(0  .
When vo < V, both possible relations between Oo and 0 must be taken into account. Since,
when 0 increases, one value of Oo increases and the other decreases, the difference (not the
sum) of the expressions d cos Oo with the two signs of the radical in (16.6) must be taken.
The result is
(0    max).
PROBLEM 3. Determine the range of possible values of the angle 0 between the directions
of motion of the two resulting particles in the L system.
SOLUTION. The angle 0 = 01+02, where 01 and O2 are the angles defined by formula (16.5)
(see Problem 1), and it is simplest to calculate the tangent of 0. A consideration of the extrema
of the resulting expression gives the following ranges of 0, depending on the relative magni-
tudes of V, V10 and V20 (for definiteness, we assume V20 > V10): 0 0 77 if V10
< 0 < TT if V V V10, 0 < Oo if V > U20. The value of Oo is given by
sin =
§17. Elastic collisions
A collision between two particles is said to be elastic if it involves no change
in their internal state. Accordingly, when the law of conservation of energy
is applied to such a collision, the internal energy of the particles may be
neglected.
The collision is most simply described in a frame of reference in which the
centre of mass of the two particles is at rest (the C system). As in $16, we
distinguish by the suffix 0 the values of quantities in that system. The velo-
cities of the particles before the collision are related to their velocities V1 and
V2 in the laboratory system by V10 = M2V/(m1+m2), V20 = -m1V/(m1+m2),
where V = V1-V2; see (13.2).
Because of the law of conservation of momentum, the momenta of the two
particles remain equal and opposite after the collision, and are also unchanged
in magnitude, by the law of conservation of energy. Thus, in the C system
the collision simply rotates the velocities, which remain opposite in direction
and unchanged in magnitude. If we denote by no a unit vector in the direc-
tion of the velocity of the particle M1 after the collision, then the velocities
of the two particles after the collision (distinguished by primes) are
V10' m20120/(m1+m2), V20' = -mjono/(m1+m2).
(17.1)
§17
Elastic collisions
45
In order to return to the L system, we must add to these expressions the
velocity V of the centre of mass. The velocities in the L system after the
collision are therefore
V1' =
(17.2)
V2' =
No further information about the collision can be obtained from the laws
of conservation of momentum and energy. The direction of the vector no
depends on the law of interaction of the particles and on their relative position
during the collision.
The results obtained above may be interpreted geometrically. Here it is
more convenient to use momenta instead of velocities. Multiplying equations
(17.2) by M1 and M2 respectively, we obtain
(17.3)
P2' muno+m2(p1+p2)/(m1+m2)
where m = m1m2/(m1+m2) is the reduced mass. We draw a circle of radius
mv and use the construction shown in Fig. 15. If the unit vector no is along
OC, the vectors AC and CB give the momenta P1' and P2' respectively.
When p1 and P2 are given, the radius of the circle and the points A and B
are fixed, but the point C may be anywhere on the circle.
C
p'
no
P'2
B
A
FIG. 15
Let us consider in more detail the case where one of the particles (m2, say) is
at rest before the collision. In that case the distance OB = m2p1/(m1+m2) = mv
is equal to the radius, i.e. B lies on the circle. The vector AB is equal to the
momentum P1 of the particle M1 before the collision. The point A lies inside
or outside the circle, according as M1 < M2 or M1 > M2. The corresponding
46
Collisions Between Particles
§17
diagrams are shown in Figs. 16a, b. The angles 01 and O2 in these diagrams
are the angles between the directions of motion after the collision and the
direction of impact (i.e. of P1). The angle at the centre, denoted by X, which
gives the direction of no, is the angle through which the direction of motion
of m1 is turned in the C system. It is evident from the figure that 01 and O2
can be expressed in terms of X by
(17.4)
C
p'
P2
pi
P2
0
max
10,
X
O2
O2
B
B
A
0
A
Q
0
(a) m < m2
(b) m, m m m
AB=p : AO/OB= m/m2
FIG. 16
We may give also the formulae for the magnitudes of the velocities of the
two particles after the collision, likewise expressed in terms of X:
ib
(17.5)
The sum A1 + O2 is the angle between the directions of motion of the
particles after the collision. Evidently 01 + O2 > 1/1 if M1 < M2, and 01+O2 < 1st
if M1 > M2.
When the two particles are moving afterwards in the same or in opposite
directions (head-on collision), we have X=TT, i.e. the point C lies on the
diameter through A, and is on OA (Fig. 16b ; P1' and P2' in the same direc-
tion) or on OA produced (Fig. 16a; P1' and P2' in opposite directions).
In this case the velocities after the collision are
(17.6)
This value of V2' has the greatest possible magnitude, and the maximum
§17
Elastic collisions
47
energy which can be acquired in the collision by a particle originally at rest
is therefore
(17.7)
where E1 = 1M1U12 is the initial energy of the incident particle.
If M1 < M2, the velocity of M1 after the collision can have any direction.
If M1 > M2, however, this particle can be deflected only through an angle
not exceeding Omax from its original direction; this maximum value of A1
corresponds to the position of C for which AC is a tangent to the circle
(Fig. 16b). Evidently
sin Omax = OC|OA = M2/M1.
(17.8)
The collision of two particles of equal mass, of which one is initially at
rest, is especially simple. In this case both B and A lie on the circle (Fig. 17).
C
p'
P2
Q2
B
A
0
FIG. 17
Then
01=1x,
A2 = 1(-x),
(17.9)
12
=
(17.10)
After the collision the particles move at right angles to each other.
PROBLEM
Express the velocity of each particle after a collision between a moving particle (m1) and
another at rest (m2) in terms of their directions of motion in the L system.
SOLUTION. From Fig. 16 we have P2' = 20B cos O2 or V2' = 2v(m/m2) cos O2. The momen-
tum P1' = AC is given by OC2 = AO2tp12-2AO.p cos or
Hence
for m1 > M2 the radical may have either sign, but for M2 > M1 it must be taken positive.
48
Collisions Between Particles
§18
§18. Scattering
As already mentioned in §17, a complete calculation of the result of a
collision between two particles (i.e. the determination of the angle x) requires
the solution of the equations of motion for the particular law of interaction
involved.
We shall first consider the equivalent problem of the deflection of a single
particle of mass m moving in a field U(r) whose centre is at rest (and is at
the centre of mass of the two particles in the original problem).
As has been shown in $14, the path of a particle in a central field is sym-
metrical about a line from the centre to the nearest point in the orbit (OA
in Fig. 18). Hence the two asymptotes to the orbit make equal angles (o,
say) with this line. The angle X through which the particle is deflected as it
passes the centre is seen from Fig. 18 to be
X = -200.
(18.1)
A
X
to
FIG. 18
The angle do itself is given, according to (14.7), by
(M/r2) dr
(18.2)
taken between the nearest approach to the centre and infinity. It should be
recalled that rmin is a zero of the radicand.
For an infinite motion, such as that considered here, it is convenient to
use instead of the constants E and M the velocity Voo of the particle at infinity
and the impact parameter p. The latter is the length of the perpendicular
from the centre O to the direction of Voo, i.e. the distance at which the particle
would pass the centre if there were no field of force (Fig. 18). The energy
and the angular momentum are given in terms of these quantities by
E = 1mvoo²,
M = mpVoo,
(18.3)
§18
Scattering
49
and formula (18.2) becomes
dr
(18.4)
Together with (18.1), this gives X as a function of p.
In physical applications we are usually concerned not with the deflection
of a single particle but with the scattering of a beam of identical particles
incident with uniform velocity Voo on the scattering centre. The different
particles in the beam have different impact parameters and are therefore
scattered through different angles X. Let dN be the number of particles
scattered per unit time through angles between X and X + dx. This number
itself is not suitable for describing the scattering process, since it is propor-
tional to the density of the incident beam. We therefore use the ratio
do = dN/n,
(18.5)
where n is the number of particles passing in unit time through unit area of
the beam cross-section (the beam being assumed uniform over its cross-
section). This ratio has the dimensions of area and is called the effective
scattering cross-section. It is entirely determined by the form of the scattering
field and is the most important characteristic of the scattering process.
We shall suppose that the relation between X and P is one-to-one; this is
so if the angle of scattering is a monotonically decreasing function of the
impact parameter. In that case, only those particles whose impact parameters
lie between p(x) and p(x) + dp(x) are scattered at angles between X and
+ dx. The number of such particles is equal to the product of n and the
area between two circles of radii P and p+dp, i.e. dN = 2mp dp n. The
effective cross-section is therefore
do = 2mp dp.
(18.6)
In order to find the dependence of do on the angle of scattering, we need
only rewrite (18.6) as
do = 2(x)|dp(x)/dx|dx
(18.7)
Here we use the modulus of the derivative dp/dx, since the derivative may
be (and usually is) negative. t Often do is referred to the solid angle element
do instead of the plane angle element dx. The solid angle between cones
with vertical angles X and x+dx is do = 2n sin x dx. Hence we have from
(18.7)
do.
(18.8)
t If the function p(x) is many-valued, we must obviously take the sum of such expressions
as (18.7) over all the branches of this function.
50
Collisions Between Particles
§18
Returning now to the problem of the scattering of a beam of particles, not
by a fixed centre of force, but by other particles initially at rest, we can say
that (18.7) gives the effective cross-section as a function of the angle of
scattering in the centre-of-mass system. To find the corresponding expression
as a function of the scattering angle 0 in the laboratory system, we must
express X in (18.7) in terms of 0 by means of formulae (17.4). This gives
expressions for both the scattering cross-section for the incident beam of
particles (x in terms of 01) and that for the particles initially at rest (x in terms
of O2).
PROBLEMS
PROBLEM 1. Determine the effective cross-section for scattering of particles from a perfectly
rigid sphere of radius a (i.e. when the interaction is such that U = 8 for r < a and U = 0
for r>a).
SOLUTION. Since a particle moves freely outside the sphere and cannot penetrate into it,
the path consists of two straight lines symmetrical about the radius to the point where the
particle strikes the sphere (Fig. 19). It is evident from Fig. 19 that
a sin to = a sin 1(-x) = a cos 1x.
A
to
p
&
FIG. 19
Substituting in (18.7) or (18.8), we have
do = 1ma2 sin X do,
(1)
i.e. the scattering is isotropic in the C system. On integrating do over all angles, we find that
the total cross-section o = na2, in accordance with the fact that the "impact area" which the
particle must strike in order to be scattered is simply the cross-sectional area of the sphere.
In order to change to the L system, X must be expressed in terms of 01 by (17.4). The
calculations are entirely similar to those of $16, Problem 2, on account of the formal resemb-
lance between formulae (17.4) and (16.5). For M1 < m2 (where M1 is the mass of the particle
and m2 that of the sphere) we have
do1,
where do1 = 2n sin 01 d01. If, on the other hand, M2 < M1, then
For m1 = M2, we have do = a²|cos 01| do1, which can also be obtained directly by sub-
stituting X = 201 from (17.9) in (1).