§18
Scattering
51
For a sphere originally at rest, X = n-202 in all cases, and substitution in (1) gives
do2 = a2|cos 02 do2.
PROBLEM 2. Express the effective cross-section (Problem 1) as a function of the energy E
lost by a scattered particle.
SOLUTION. The energy lost by a particle of mass M1 is equal to that gained by the sphere of
mass M2. From (17.5) and (17.7), € = E2' = [2m22m2/(m1+m2)2] voo2 sin21x = Emax sin21x,
whence de = 1 mex sin X dx; substituting in (1), Problem 1, we have do = na2 de/emax. The
scattered particles are uniformly distributed with respect to € in the range from zero to
Emax.
PROBLEM 3. Find the effective cross-section as a function of the velocity Uoo for particles
scattered in a field U -rrn.
SOLUTION. According to (10.3), if the potential energy is a homogeneous function of order
k = -n, then similar paths are such that p ~v-2/n, or p = Voo-2/nf(x), the angles of deflec-
tion X being equal for similar paths. Substitution in (18.6) gives do ~ Voo-4/n do.
PROBLEM 4. Determine the effective cross-section for a particle to "fall" to the centre of
a field U = -a/r2.
SOLUTION. The particles which "fall" to the centre are those for which 2a > mp20002 (see
(14.11)), i.e. for which the impact parameter does not exceed Pmax = (2a/mvoo). The
effective cross-section is therefore o = Pmax2 = 2na/mvo².
PROBLEM 5. The same as Problem 4, but for a field U = -a/rn (n 2, a > 0).
SOLUTION. The effective potential energy Ueff = depends on r in the
manner shown in Fig. 20. Its maximum value is
Ueff
U0
FIG. 20
The particles which "fall" to the centre are those for which U0 < E. The condition U0 = E
gives Pmax, whence
=
PROBLEM 6. Determine the effective cross-section for particles of mass m1 to strike a sphere
of mass m2 and radius R to which they are attracted in accordance with Newton's law.
SOLUTION. The condition for a particle to reach the sphere is that rmin < R, where r'min
is the point on the path which is nearest to the centre of the sphere. The greatest possible
value of P is given by rmin = R; this is equivalent to Ueff(R) = E or
= , where a = ymim2 (2 being the gravitational constant) and we have put m 22 M1 on
the assumption that m2 > M1. Solving for pmax2, we finally obtain o = mR2(1+2ym2/Rvv3).
52
Collisions Between Particles
§18
When
Voo
8 the effective cross-section tends, of course, to the geometrical cross-section
of the sphere.
PROBLEM 7. Deduce the form of a scattering field U(r), given the effective cross-section
as a function of the angle of scattering for a given energy E. It is assumed that U(r) decreases
monotonically with r (a repulsive field), with U(0) > E and U(00) = 0 (O. B. FIRSOV 1953).
SOLUTION. Integration of do with respect to the scattering angle gives, according to the
formula
(1)
the square of the impact parameter, so that p(x) (and therefore x(p)) is known.
We put
s=1/r,
=1/p2,
[[1-(U|E)]
(2)
Then formulae (18.1), (18.2) become
1/
(3)
where so(x) is the root of the equation xw2(so)-so2 = 0.
Equation (3) is an integral equation for the function w(s), and may be solved by a method
similar to that used in $12. Dividing both sides of (3) by (a-x) and integrating with respect
to x from zero to a, we find
so(a)
dx ds
so(a)
ds
or, integrating by parts on the left-hand side,
This relation is differentiated with respect to a, and then so(a) is replaced by s simply;
accordingly a is replaced by s2/w2, and the result is, in differential form,
=
(11/20)
or
dx
This equation can be integrated immediately if the order of integration on the right-hand
side is inverted. Since for S = 0 (i.e. r oo) we must have W = 1 (i.e. U = 0), we have,
§19
Rutherford's formula
53
on returning to the original variables r and P, the following two equivalent forms of the final
result:
=====
(dx/dp)
(4)
This formula determines implicitly the function w(r) (and therefore U(r)) for all r > rmin,
i.e. in the range of r which can be reached by a scattered particle of given energy E.
§19. Rutherford's formula
One of the most important applications of the formulae derived above is
to the scattering of charged particles in a Coulomb field. Putting in (18.4)
U = a/r and effecting the elementary integration, we obtain
whence p2 = (a2/m2voo4) tan2oo, or, putting 00 = 1(-) from (18.1),
p2 =
(19.1)
Differentiating this expression with respect to X and substituting in (18.7)
or (18.8) gives
do = (a/v2cosxdx/sin31
(19.2)
or
do =
(19.3)
This is Rutherford's formula. It may be noted that the effective cross-section
is independent of the sign of a, so that the result is equally valid for repulsive
and attractive Coulomb fields.
Formula (19.3) gives the effective cross-section in the frame of reference
in which the centre of mass of the colliding particles is at rest. The trans-
formation to the laboratory system is effected by means of formulae (17.4).
For particles initially at rest we substitute X = 77 - 202 in (19.2) and obtain
do2 = 2n(a/mvoo2)2 sin de2/cos302
=
(19.4)
The same transformation for the incident particles leads, in general, to a very
complex formula, and we shall merely note two particular cases.
If the mass M2 of the scattering particle is large compared with the mass
M1 of the scattered particle, then X 2 O1 and m 22 M1, SO that
do1 = = (a/4E1)2do1/sin4301
(19.5)
where E1 = 1M1U..02 is the energy of the incident particle.
54
Collisions Between Particles
§ 19
If the masses of the two particles are equal (m1 = M2, m = 1M1), then by
(17.9) X = 201, and substitution in (19.2) gives
do1 = 2(/E1)2 cos 01 d01/sin³01
=
(19.6)
If the particles are entirely identical, that which was initially at rest cannot
be distinguished after the collision. The total effective cross-section for all
particles is obtained by adding do1 and do2, and replacing A1 and O2 by their
common value 0:
do
=
do.
(19.7)
Let us return to the general formula (19.2) and use it to determine the
distribution of the scattered particles with respect to the energy lost in the
collision. When the masses of the scattered (m1) and scattering (m2) particles
are arbitrary, the velocity acquired by the latter is given in terms of the angle
of scattering in the C system by V2' = [2m1/(m1+m2)]%"00 sin 1x; see (17.5).
The energy acquired by M2 and lost by M1 is therefore E = 1M2U2'2
= (2m2/m2)/0002 sin21x. Expressing sin 1x in terms of E and substituting
in (19.2), we obtain
do = de/e2.
(19.8)
This is the required formula: it gives the effective cross-section as a function
of the energy loss E, which takes values from zero to Emax = 2m2voo2/m2.
PROBLEMS
PROBLEM 1. Find the effective cross-section for scattering in a field U = a/r2 (a > 0).
SOLUTION. The angle of deflection is
The effective cross-section is
do
sin
PROBLEM 2. Find the effective cross-section for scattering by a spherical "potential well"
of radius a and "depth" U0 (i.e. a field with U = 0 for r > a and U = - U0 for r < a).
SOLUTION. The particle moves in a straight line which is "refracted" on entering and leav- -
ing the well. According to §7, Problem, the angle of incidence a and the angle of refraction
B (Fig. 21) are such that sin x/sin B = n, where n = W(1+2U0/mVo2). The angle of deflection
is X = 2(a-B). Hence
=
Eliminating a from this equation and the relation a sin a p, which is evident from the
diagram, we find the relation between P and X:
cos
1x
§20
Small-angle scattering
55
Finally, differentiating, we have the effective cross-section
cos
do.
The angle X varies from zero (for p=0) to Xmax (for P = a), where cos 1xmax = 1/n.
The total effective cross-section, obtained by integrating do over all angles within the cone
Xmax, is, of course, equal to the geometrical cross-section 2
a
to
a
FIG. 21
§20. Small-angle scattering
The calculation of the effective cross-section is much simplified if only
those collisions are considered for which the impact parameter is large, so
that the field U is weak and the angles of deflection are small. The calculation
can be carried out in the laboratory system, and the centre-of-mass system
need not be used.
We take the x-axis in the direction of the initial momentum of the scattered
particle M1, and the xy-plane in the plane of scattering. Denoting by P1' the
momentum of the particle after scattering, we evidently have sin 01 = P1y'/P1'.
For small deflections, sin 01 may be approximately replaced by 01, and P1' in
the denominator by the initial momentum P1 = MIUoo:
(20.1)
Next, since Py = Fy, the total increment of momentum in the y-direction is
(20.2)
The
force Fy = - JULY (dU/dr)dr/dy = (dU/dr)y/r.
Since the integral (20.2) already contains the small quantity U, it can be
calculated, in the same approximation, by assuming that the particle is not
deflected at all from its initial path, i.e. that it moves in a straight line y = p
with uniform velocity Voo. Thus we put in (20.2) Fy = - (dU/dr)p/r,
dt = dx/voo. The result is
56
Collisions Between Particles
§20
Finally, we change the integration over x to one over r. Since, for a straight
path, r2 = x2+p2, when x varies from - 00 to + 80, r varies from 8 to P
and back. The integral over x therefore becomes twice the integral over r
from p to 80, and dx = r dr/v(r2-p2). The angle of scattering O1 is thus
given byt
(20.3)
and this is the form of the function 01(p) for small deflections. The effective
cross-section for scattering (in the L system) is obtained from (18.8) with 01
instead of X, where sin 01 may now be replaced by A1:
(20.4)
PROBLEMS
PROBLEM 1. Derive formula (20.3) from (18.4).
SOLUTION. In order to avoid spurious divergences, we write (18.4) in the form
PO
and take as the upper limit some large finite quantity R, afterwards taking the value as R
00.
Since U is small, we expand the square root in powers of U, and approximately replace
rmin by p:
dr
The first integral tends to 1/11 as R 00. The second integral is integrated by parts, giving
=
This is equivalent to (20.3).
PROBLEM 2. Determine the effective cross-section for small-angle scattering in a field
U=a/m(n) 0).
t If the above derivation is applied in the C system, the expression obtained for X is the
same with m in place of M1, in accordance with the fact that the small angles 01 and X are
related by (see (17.4)) 01 = m2x/(m1 +m2).
§20
Small-angle scattering
57
SOLUTION. From (20.3) we have
dr
The substitution p2/r2 = U converts the integral to a beta function, which can be expressed
in terms of gamma functions:
Expressing P in terms of 01 and substituting in (20.4), we obtain
do1.
3
CHAPTER V
SMALL OSCILLATIONS
$21. Free oscillations in one dimension
A VERY common form of motion of mechanical systems is what are called
small oscillations of a system about a position of stable equilibrium. We shall
consider first of all the simplest case, that of a system with only one degree
of freedom.
Stable equilibrium corresponds to a position of the system in which its
potential energy U(q) is a minimum. A movement away from this position
results in the setting up of a force - dU/dq which tends to return the system
to equilibrium. Let the equilibrium value of the generalised co-ordinate
q be 90. For small deviations from the equilibrium position, it is sufficient
to retain the first non-vanishing term in the expansion of the difference
U(q) - U(90) in powers of q-qo. In general this is the second-order term:
U(q) - U(q0) 112 1k(q-90)2, where k is a positive coefficient, the value of the
second derivative U"(q) for q = 90. We shall measure the potential energy
from its minimum value, i.e. put U(qo) = 0, and use the symbol
x = q-90
(21.1)
for the deviation of the co-ordinate from its equilibrium value. Thus
U(x) = .
(21.2)
The kinetic energy of a system with one degree of freedom is in general
of the form 1a(q)q2 = 1a(q)x2. In the same approximation, it is sufficient to
replace the function a(q) by its value at q = qo. Putting for brevity a(go) = m,
we have the following expression for the Lagrangian of a system executing
small oscillations in one dimension:
L = 1mx2-1kx2.
(21.3)
The corresponding equation of motion is
m+kx=0,
(21.4)
or
w2x=0,
(21.5)
where
w= ((k/m).
(21.6)
+ It should be noticed that m is the mass only if x is the Cartesian co-ordinate.
+ Such a system is often called a one-dimensional oscillator.
58
§21
Free oscillations in one dimension
59
Two independent solutions of the linear differential equation (21.5) are
cos wt and sin wt, and its general solution is therefore
COS wt +C2 sin wt.
(21.7)
This expression can also be written
x = a cos(wt + a).
(21.8)
Since cos(wt+a) = cos wt cos a - sin wt sin a, a comparison with (21.7)
shows that the arbitrary constants a and a are related to C1 and C2 by
tan a = - C2/C1.
(21.9)
Thus, near a position of stable equilibrium, a system executes harmonic
oscillations. The coefficient a of the periodic factor in (21.8) is called the
amplitude of the oscillations, and the argument of the cosine is their phase;
a is the initial value of the phase, and evidently depends on the choice of
the origin of time. The quantity w is called the angular frequency of the oscil-
lations; in theoretical physics, however, it is usually called simply the fre-
quency, and we shall use this name henceforward.
The frequency is a fundamental characteristic of the oscillations, and is
independent of the initial conditions of the motion. According to formula
(21.6) it is entirely determined by the properties of the mechanical system
itself. It should be emphasised, however, that this property of the frequency
depends on the assumption that the oscillations are small, and ceases to hold
in higher approximations. Mathematically, it depends on the fact that the
potential energy is a quadratic function of the co-ordinate.
The energy of a system executing small oscillations is E =
= 1m(x2+w2x2) or, substituting (21.8),
E =
(21.10)
It is proportional to the square of the amplitude.
The time dependence of the co-ordinate of an oscillating system is often
conveniently represented as the real part of a complex expression:
x = re[A exp(iwt)],
(21.11)
where A is a complex constant; putting
A = a exp(ix),
(21.12)
we return to the expression (21.8). The constant A is called the complex
amplitude; its modulus is the ordinary amplitude, and its argument is the
initial phase.
The use of exponential factors is mathematically simpler than that of
trigonometrical ones because they are unchanged in form by differentiation.
t It therefore does not hold good if the function U(x) has at x = 0 a minimum of
higher order, i.e. U ~ xn with n > 2; see §11, Problem 2(a).
60
Small Oscillations
§21
So long as all the operations concerned are linear (addition, multiplication
by constants, differentiation, integration), we may omit the sign re through-
out and take the real part of the final result.
PROBLEMS
PROBLEM 1. Express the amplitude and initial phase of the oscillations in terms of the
initial co-ordinate xo and velocity vo.
SOLUTION. a = (xx2+002/w2), tan a = -vo/wxo.
PROBLEM 2. Find the ratio of frequencies w and w' of the oscillations of two diatomic
molecules consisting of atoms of different isotopes, the masses of the atoms being M1, m2 and
'M1', m2'.
SOLUTION. Since the atoms of the isotopes interact in the same way, we have k = k'.
The coefficients m in the kinetic energies of the molecules are their reduced masses. Accord-
ing to (21.6) we therefore have
PROBLEM 3. Find the frequency of oscillations of a particle of mass m which is free to
move along a line and is attached to a spring whose other end is fixed at a point A (Fig. 22)
at a distance l from the line. A force F is required to extend the spring to length l.
A
X
FIG. 22
SOLUTION. The potential energy of the spring is (to within higher-order terms) equal to
the force F multiplied by the extension Sl of the spring. For x < l we have 81 = (12++2) -
=
x2/21, so that U = Fx2/21. Since the kinetic energy is 1mx2, we have = V(F/ml).
PROBLEM 4. The same as Problem 3, but for a particle of mass m moving on a circle of
radius r (Fig. 23).
m
&
FIG. 23