Re: Coulomb scattering cross-section.

On Tue, 31 Jan 2006, Gregory L. Hansen wrote:

> But if the probability that a particle is scattered in any direction is 1,
> and the cross-section for angle = 0 is infinite, how can that mean
> anything but that there's zero probability that the particle will be
> scattered in any other direction?

Consider a plane electromagnetic wave incident on a spherical particle.
The extinction cross-section is equal to the power removed from the
incident plane wave, P_gone = I_0 * cross-section. What fraction of the
incident plane wave remains? Since the plane wave has infinite total
power, and finite power is removed ...

Along those lines, if you have a collision between a particle with a known
initial momentum and _completely arbitrary_ location and your nucleus, the
deflection is almost certain to be very small, since the probability of
the colliding particle being anywhere near the nucleus is infinitesimally
small. Yes, the probability that a particle will be significantly
deflected is infinitesimally small. But that's because of the way that
cross-sections are defined: for scattering of waves, in terms of an
infinite incident plane wave, and for scattering of particles, in terms of
an infinite random distribution of particles with fixed initial momentum.

The area through which a collider from such an infinite distribution can
pass without being significantly deflected is infinite, since it extends
from a large distance away from the nucleus, to an infinite distance away.
In practice, you have a finite beam, so the probability of an individual
collider being deflected significantly is no longer essentially zero, so
there's no problem.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.

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