114 lines
4.8 KiB
Text
114 lines
4.8 KiB
Text
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
|