# Why is a nucleus isotropic?

I believe in Neutron Scattering the neutrons after hitting a nucleus can bounce in any of 360*3 dimensions -> 1080 degrees?

Why is this so? Shouldn't it only bounce "off" the neutron in approximately the same "direction" that it came in such as when a particle bounces off a mirror -> because of the cross-section ...

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To describe the angular distribution in 3-D space, you should use solid angle, with a maximum of $4\pi$. $360^\circ\times 3$ does not make sense. – C.R. Sep 2 '12 at 2:00
@KarsusRen: can you elaborate? I'm thinking essentially setting the center of the neutron as the origin of a spherical coordinate system. – Eiyrioü von Kauyf Sep 2 '12 at 21:52
Which is exactly what he meant. $360 \equiv 0\ \text{mod}\ 360$. So in your example $1080 \equiv 360\ \text{mod}\ 360$ – CHM Sep 3 '12 at 5:45
Leaving aside the odd misconceptions in the question, not all nuclei are isotropic: polarized observables are big business at transition energies these days. – dmckee♦ Sep 4 '12 at 17:27

## 3 Answers

Physics is not a matter of beliefs, but of measurements with their errors and the analysis of those data according to theoretical( mathematically expressed) models.

In two body scattering, two in two out, the scattering takes place in a plane, because of momentum conservation whether classically or in the quantum mechanical mircrocosm. Thus the angle of scatter is one and goes from 0 to 360 degrees.

If one looks at the data, there is no isotropy. The angular distributions are analyzed (page 6 in the link) in a series of Legendre polynomials, which are a function of the angle theta.

edit after rereading the question:

Each individual scatter of a neutron on a nucleus will have a specific plane in the center of mass, and thus the angle theta will show the functional form of the interaction, and not be isotropic after aligning the scattering planes . The distribution in the angle phi which will define the rotation to align the scattering planes will be isotropic. Because of momentum conservation as mentioned above any phi is equally probable.

There cannot be total isotropy because there are nuclear resonances which are probed when a neutron scatters off a nucleus and the spins involved create in the crossection necessarily a function of theta.

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can you elaborate on which phi and theta  - different conventions.... – Eiyrioü von Kauyf Sep 2 '12 at 21:55
Did you look at the link I gave? It is the normal spherical coordinates in which Legendre polynomials are expressed. In the experiment, which is really the trump card, they have a number of theta angles all around phi, where they measure, find isotropy, and extrapolate from that that they can use the Legendre polynomials for the functional form of the scattering angular dependence. – anna v Sep 3 '12 at 3:37

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Differential cross sections are introduced precisely to quantify the percentage of particles that scatter in a given direction. If all the directions are possibles, they do not have the same differential cross section value. And indeed, the higest value is for scattering directions close to the incoming beam.

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This is misleading--- you are talking about a case (forward scattering) which swamps all off beam scattering in the limit that the neutron wavefunction is spread out. The non-delta-function part of forward scattering is not larger by orders of magnitude than other directions for a localized potential, it is just the limit of off-forward scattering, and this is how the optical theorem works. – Ron Maimon Sep 2 '12 at 22:07

The neutron at low momentum is quantum mechancal during the scattering, and the bouncing off is by wave mechanics, not by bouncing off. The neutron gets reflected into all directions by the small region with a potential, and the amount of scattering is given by the Born approximation to lowest order in the momentum of the neutron:

$$A(k-k') = -i\tilde V(k-k')$$

For scattering from incoming direction k to outgoing direction k'. You decorate this with phase-space factors to take into account the size of the neutron wave. If the neutron wave is large, nearly all the scattering is in the forward direction, and this is expressed by writing the S-matrix as;

$$S = 1 + i A$$

Where the "1" gives a contribution to the scattering which is $\delta(k-k')$ for the case of single-particle scattering. If you smear this with the incoming wavepacket, you get the outgoing wavepacket, which is mostly the incoming wave, plus a spherical outgoing wave. The delta-function guarantees that as you approach a plane-wave limit, there will be no scattering, because the neutron wavefunction area will be much larger than the nuclear area.

This scattering, when the neutron wavefunction is larger wavelength than the nucleus (almost always in real life) leads to a spherical scattering which is roughly isotropic, which you add to the incoming wave to get the full outgoing wave. The imaginary part of the scattering in the forward direction only subtracts some weight from the wavefunction that keeps going forward, and unitarity guarantees that this imaginary part is equal to the scattering probability in all directions added together.

This is covered in scattering theory in most quantum mechanics books, but the treatment in Gribov's Regge theory book "The Theory of Complex Angular Momentum" is most instructive to my mind.

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But Ron, all this is fine and good as an introduction to scattering in the S matrix format, but how about experimental evidence (which I linked in my answer) that shows non isotropy in theta due to nuclear resonances? It would have been isotropic if there were no further interactions open to the scatter which the experiment demonstrates. It is not isotropic in the real world! – anna v Sep 3 '12 at 4:35
@annav: I am talking about energies lower than the ones required for exciting resonances. I get the sense tha OP is asking about slow or thermal neutrons, not KeV/MeV neutrons. – Ron Maimon Sep 3 '12 at 5:50
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