Thermalising neutrons

Lbeard

How does thermalising neutrons (slowing them down) in heavy water increase the probability of a nuclear collision with uranium?

Morbert

The slower neutrons would have a longer wavelength, and neutron wavelength is proportional to the interaction cross section as it is assumed the wavelength approximates the "size" of the neutron.

(See equation 1 here)

Lbeard

Morbert wrote: »

The slower neutrons would have a longer wavelength,

Yeah, it was slowly dawning on me that it might have something to do with the DeBroglie wavelength. But I'm still not 100% sure what that means in terms of nuclear decay. If the neutron is going too fast it will just recoil. If it's going slowly, due to its' wavelength getting longer, does it gently drift into the nucleus and how exactly does it cause the decay?

and neutron wavelength is proportional to the interaction cross section as it is assumed the wavelength approximates the "size" of the neutron.

I believe the cross section is a not a physically real area in space. And the size of the cross section is a function of the neutrons velocity. I believe in early simulations (ones used in Los Alamos. ) the nuclei were represented as 2 dimensional shapes (maybe circles, but you could probably do it with any area, like squares) suspended in a volume. The areas are not a physical reality. But i believe the way everything works out you get the correct figures.

Do cross sections exist in reality? Does an electron have a physically real cross section?.........At higher energy levels does this cross section change?....Does a more energetic electron in fact get smaller?

Morbert

Oops! Apologies to Lbead. I accidentally hit the "edit" button instead of the quote button. Newbie mod mistake. I think I've restored your post to the way it was.

Lbeard wrote: »

Yeah, it was slowly dawning on me that it might have something to do with the DeBroglie wavelength. But I'm still not 100% sure what that means in terms of nuclear decay. If the neutron is going too fast it will just recoil. If it's going slowly, due to its' wavelength getting longer, does it gently drift into the nucleus and how exactly does it cause the decay?

A larger wavelength means the area in which a neutron has some non-negligible probability of being found is larger, which means it is more likely to be found near a nucleus, interacting with it.

I believe the cross section is a not a physically real area in space. And the size of the cross section is a function of the neutrons velocity. I believe in early simulations (ones used in Los Alamos. ) the nuclei were represented as 2 dimensional shapes (maybe circles, but you could probably do it with any area, like squares) suspended in a volume. The areas are not a physical reality. But i believe the way everything works out you get the correct figures.

Do cross sections exist in reality? Does an electron have a physically real cross section?.........At higher energy levels does this cross section change?....Does a more energetic electron in fact get smaller?

In the standard model, the electron is a lepton and therefore point like. What gives it a cross section is the wave function, which defines the area in which it can be found, and its surrounding electric field. So this cross section is not a "hard" cross section in the billiard ball sense.

The interaction cross section appears to be 2 * pi * R^2 *D where R is the radius of the neutron plus the wavelength of the neutron, and D is an additional term dependent on the strength of the interaction between the neutron and the nucleus, and the phase shift of a neutron passing through a nucleus. So you can see that, while it is not intuitively real in the classical sense, it is a reasonably simple model that resembles a typical circle.

Lbeard

Morbert wrote: »

A larger wavelength means the area in which a neutron has some non-negligible probability of being found is larger, which means it is more likely to be found near a nucleus, interacting with it.

My assumptions on nuclear stability, was the nuclear particles; neutrons and protons, were, in a stable state - or at least there was a potential barrier high enough to stop them leaving, and the addition of a neutron would unbalance the potential, causing a decay.

I also assumed that the neutron point particle would be in superposition - and it's wave function could overlap and interfere with the other neutrons. So, that the distribution of a nuclear particle would interfere with each other.

My own assumptions on superposition (I'm only hacking away at the maths as I go - I've only been getting a conceptual grip on the maths of quantum mechanics recently). My assumption on superposition is a particle by itself, will have more positions (it will have no spin or pole). A particle in a field or two particles interacting - the spin will become definite. (also I have a rough idea, multiple particles arranged in an atom, will have a polar angle relative to each other).

In the standard model, the electron is a lepton and therefore point like. What gives it a cross section is the wave function, which defines the area in which it can be found, and its surrounding electric field. So this cross section is not a "hard" cross section in the billiard ball sense.

I think the idea of a cross section is quite old. It think it goes back to Beer (of Beer Lambert) in the 18th century. It's probably very useful for measuring a chemical concentration in solution (or volume) - though I'm curious as how it functions when you have to consider spectral absorption. But if you have a table of experimental results - your curve could be quite usable.

There are a few things that confuse me about the point like particle. I can imagine the probability density functioning as the charge density of the electron (who electrons will behave in relation to each other). What really bothers me is the idea that the actual electron is a point somewhere in that density. Because I assumed the electron to be at the centre of its' charge density.

The interaction cross section appears to be 2 * pi * R^2 *D where R is the radius of the neutron plus the wavelength of the neutron, and D is an additional term dependent on the strength of the interaction between the neutron and the nucleus, and the phase shift of a neutron passing through a nucleus. So you can see that, while it is not intuitively real in the classical sense, it is a reasonably simple model that resembles a typical circle.

There's plenty about it that bothers me though. If the point particle is infinitely small, how can there be collisions?...........

I was reading a piece from the Feynman lectures - his description of sunlight passing through the atmosphere an interacting with electrons. And electrons moving in the electric field of the passing photon - from his description, though I'm not 100% sure of my interpretation, that the electron physically presents as a two dimensional cross section and that is how it polarises sun light. I think the description is classical. But what's bothering about it, among other things, is why a photon is not moving all electrons in its' path. Summation of all paths, for the path of a point particle is fine if it only interacts with a few electrons, even without damping - the scattering angle will be small. But the more it interacts with, the greater the angle, until it will exceed 2 pi,

For a Morgan LaFey, a Mirage, or even looking to the horizon, there aren't deflections greater than 2 pi - they're much much smaller. The way mirages appear, it would seem to imply, that light in air is being largely - over a reasonable distance, completely - propogated through a consistent fog of electron charge. The optical effect of an explosive shockwave (they're on youtube) is substantial.

I don't know.....it's very confusing.

Morbert

Sorry for the absence of a reply. I'll need to do a little digging before I can make any more statements with confidence about the specific model of interaction between the nucleus and the neutron. Will get back to this a.s.a.p.

Carawaystick

Lbeard wrote: »

There are a few things that confuse me about the point like particle. I can imagine the probability density functioning as the charge density of the electron (who electrons will behave in relation to each other). What really bothers me is the idea that the actual electron is a point somewhere in that density. Because I assumed the electron to be at the centre of its' charge density.



There's plenty about it that bothers me though. If the point particle is infinitely small, how can there be collisions?...........

I don't think you can say an electron is a point, just very small compared to nucleons.
As an electron has rest mass, if it was a point it'd be a black hole

Morbert

Carawaystick wrote: »

I don't think you can say an electron is a point, just very small compared to nucleons.
As an electron has rest mass, if it was a point it'd be a black hole

In the standard model it can be a point. This is allowed because of renormalisation. Essentially, some of the mass-energy of the particle exists in the surrounding field.

http://en.wikipedia.org/wiki/Renormalisation#Self-interactions_in_classical_physics

Morbert

I haven't really found anything more involved than the Ramsauer model I mentioned previously.

http://www.osti.gov/scitech/servlets/purl/641282