Talk:Neutron temperature

From Wikipedia, the free encyclopedia

WikiProject Physics This article is within the scope of WikiProject Physics, which collaborates on articles related to physics.
Start This article has been rated as Start-Class on the assessment scale.
??? This article has not yet received an importance rating within physics.

Help with this template This article has been rated but has no comments. If appropriate, please review the article and leave comments here to identify the strengths and weaknesses of the article and what work it will need.

Did this replace the article on slow neutrons or something, because I'm getting a lot of pointers to other articles that take me here.theanphibian 16:17, 30 March 2007 (UTC)

[edit] Neutron temperature is wrong by factor of 50%

Actually, folks, the RMS energies of gases are 12.47 J/mole/kelvin, which corresponds to 3404 J/mole at 273 K, which is 0.035 eV/particle, not 0.025 eV. The energy of 0.025 eV corresponds to a temp of only 193 K, which is pretty cold--- certainly not room temp by any criterion. And neutrons at room temp (20 Co) are moving at sqrt(3RT/m) = sqrt (3*8.3*293/.001) = 2.7 km/sec, not 2.2 km/sec as it says in the article (that speed does correspond, of course, with the 193 K temp). So fix this up. If you're going to define the energies this way, you have to give up room temperature. Sorry, but there are no two ways about it. SBHarris 19:20, 23 April 2007 (UTC)

After a bit of google search, I find that everybody loosely assumes things (including neutrons) have a mean thermal energy of "kT" at room temp, which is where the 0.025 eV number comes from. But classically, the RMS energy of ideal particles at room temp is not kT but (3/2) kT. Which is where my gripe HERE comes from. Sloppy physicists! SBHarris 19:36, 23 April 2007 (UTC)
Ain't my job, bub. Constants for first-order terms are the engineers problem. And they're just going to make up some fudge factors instead of using the theoretical value anyway. Physicsts are a laid-back crowd, man. Off by a factor of c squared and can't find where it went? Here, take mine. Switched an h with an h-bar someewhere? Don't worry about it dude, what's a pi or two beetween friends? KonradG 03:17, 24 April 2007 (UTC)
Yo, KG. Oppenheimer himself was famous for getting the 3/2's and pi's and i's and other dimensionless stuff wrong, or leaving it out or on the wrong side of the divisor. But my problem is that somebody went ahead in this article and calculated everything to 2 places, where 1 place is all you deserve if you're going to sloppily use kT for thermal neutron energy. And the weird thing is that as I see on google, a lot of papers DEFINE thermal neutrons using 0.025 (2 sig digits) as the cutoff, when they started with just order of magnitude kT (leaving out the 3/2) plus room temp, to get THAT. Argghh. SBHarris 04:09, 24 April 2007 (UTC)
As far as cross-section calculations go, any resemblance to the mean KE of a particular Boltzmann distribution is entirely coincidental. Working with a whole spectrum of energies is what you're trying to avoid. Take all the neutrons in the bottom part of the energy spectrum, pretend they're going the same speed, and there's your thermal neutrons. That speed, by the way, isn't so much an average as what is politely referred to as "empirical data". As in, "I correct for empirical data, you apply a fudge factor, he throws in an arbitrary constant".
Like you say, they're defined that way. Just like spherical cows of uniform density, thermal neutrons are perfectly homogenous creatures. They're point particles which travel at exactly 2200 m/s, and avoid each other with perfect precision so they can be scattered and absorbed in an orderly fashion. KonradG 18:11, 24 April 2007 (UTC)
I don't mind that they're defined that way. I just mind when some busybody puts in the article that they have the same energy as ideal particles at room temperature. SBHarris 18:56, 24 April 2007 (UTC)
I agree with you that the article's explanation is wrong. But (I think) the real explanation isn't sloppy physicists droping a factor of 3/2 for order-of-magnitude calculations. That would be too much of a coincidence. Splitting up your neutron spectrum at that point in a two-group model makes it fit the experimental data. KonradG 22:58, 24 April 2007 (UTC)
Really? Is there really an objective inflexion point in the behavior of neutrons in any system, causing any reaction, that honestly occurs at 0.025 eV (and below), but has clearly statistically passed, by the time they have a true room temp energy of 0.0375 eV? I don't believe it-- it would implie some 0.025 eV resonance in the nucleus which is narrow enough to start to go away by the time energy is 50% more. Show me. Yes, there are probably articles where people have arbitrarily looked at neutrons colder than 0.025 (cutoff picked for traditional reasons) and show that they behave differently from hotter ones. But that's not the same as showing that 0.025 is actually the inflexion. People at any age over 14 or so are more likely to die than the sum of people under that given age. But this proves nothing about that age. Only puberty is the true inflexion. The rest is just an exponentially rising curve and you can get a result by dividing the group in half at any point along it. SBHarris 23:26, 24 April 2007 (UTC)
(edit conflicted) No, but there's no such point for defining "fast" neutrons either. The only thing objective about it is that it makes the math easier for specfic reactions in in specific systems. 0.025 eV happens to work well for fission in both U and Pu, and that makes it a pretty neat point, as far as arbitrary values go. KonradG 00:52, 25 April 2007 (UTC)
Yes, there is a big bump right after 0.025 in Pu. Which means any two-group model for has to duck below that, or turn into a big mess. And no, I'm not saying there's something special about that point. Not to nature at least. But it's special for lazy physicists who want realistic predictions without doing anything complicated enough to prevent them from drinking on the job. Allegedly. KonradG 00:58, 25 April 2007 (UTC)
I thought the 2.5 or 2.3 eV number is used in reactors, which doesn't necessarily make it a room temperature neutron because 1.) the moderator in the core is at more like 500 deg F and 2.) with 3 or so collisions before fission, a neutron shouldn't completely thermalize. That could be a part of the article where are more accurate and accepted definition could be applied, but it's certainly not worth this effort. Someone just going through and writing details about uses, creation, and neutronics of all the energies would be more useful. theanphibian 04:47, 29 April 2007 (UTC)
theanphibian, you have a good point that number of hits needed to dethermalize being more than the average neutron gets, in a real reactor. But still, there's something way wierd about that number 0.025 eV. There's no justification for it anywhere, except as the energy of using a bad figure of kT for neutron energy if T is 290 K (room temp). Otherwise, it's not the 0.25 eV that neutrons need for optimal performance, and it's not the thermalized E of neutrons that actually ARE at room temp (0.0375 eV), and it's not even the mean E of neutrons in a real reactor. It's just a garbage number. Which was the point I made trying to start this discussion of why thermal neutrons are defined that way. There isn't a good/practical reason. There just isn't. Nobody wants to admit this. SBHarris 03:07, 30 April 2007 (UTC)
I'm fairly sure it's used in real problems. In fact, I know it is. I don't think this is so much a matter of it being a useless number, but us not really knowing what the significance is. theanphibian 08:38, 30 April 2007 (UTC)
SBH, with regard to Theanphibian's post, don't forget to think of it in the context of modeling a spectrum of energies in a simplified way. The "thermal neutron" energy is not necessarily chosen to give the optimum representation of thermal neutrons themselves, but the flux distribution of the model as a whole. You don't really care how much energy it had at the next fission, since the end result is the same. In a sense, the goal is to figure out how many fission neutrons survive to join the thermal background in the moderator. From there, the problem becomes manageable. KonradG 04:28, 1 May 2007 (UTC)

I took a quick look at some LWR models, where the contribution of Pu fission is less significant, and they do indeed use neutron speeds of ~4 km/s. That would fit with my guess about the reason for a 0.25 eV "magic number", though it's still a post-hoc theoretical justification for something that's likely an empirical fudge. Can't prove it, though. KonradG 17:09, 27 April 2007 (UTC)

Ummm, 0.25 eV (note factor of 10 from 0.025 eV) is more like ~7 km/sec. And it's hot PWR reactor temp (600 C) not room temp.SBHarris 03:11, 30 April 2007 (UTC)
Well, and I can't find any papers online which show (for free) E dependence of slow neutron efficiency well enough to show if there's something magic at 0.025 eV, so I'm still skeptical. I don't believe in nuclear resonances that small, and if they exist, they're going to be swamped by the energy of the incoming neutron which certainly isn't anything like the diffusion energy. The strong force draws the neutron into the nucleus with an energy on the order of a MeV, at the end, and that's all it "sees". It has no idea what the neutron was doing before that. What's 0.025 eV vs. 0.0375 eV, on top of a MeV going to make for any difference? And if it's a wavelength thing, it has to be a smooth curve, and any resonance would have to come from large scale external distance-scale things, like spacing between Pu atoms. Now, a neutron moving a 2.2 km/sec has a wavelength of 1.8 A. which is an interesting coincidence since it's very much atomic scale and might be a crystal latice dimension for Pu (or more likely, half of one). So that's MAYBE plausable. Neutrons of 50% higher energy are 1-(2/3) = 33% "smaller" (1.2 A) and that's getting quite small for big atoms. Anyway, have you got a reference? Can't you ask some bigwig coworker? SBHarris 17:50, 27 April 2007 (UTC)
??? It's kinda hard to miss. [1] [2] [3] [4] Here's one where they use .025eV all over the place, with the peak shown in Figure 12.3 [5]. KonradG 23:34, 27 April 2007 (UTC)
From looking at your last reference only: the graph is at 12.3 and a magnified version at 12.6. For both U and Pu, the resonance for neutron energy is clearly between 10^0 and 10^(-1) eV, which is to say, between 1 and 0.1 eV. It's certainly larger than 0.1 eV. I make it about 10^(.6) eV, which is 0.25 eV, which is exactly 10 times more than your stated max. At 0.24 eV we're into sqrt(10) times the velocities and temps we've been discussing, which puts it rougly 600 K and 600 F both. Or 340 C, which is interestingly enough just about where the max PWR stream reactors opperate (550 F, I read somewhere)-- perhaps to take maximal advantage of this resonance??

Anyway, as for 0.025 eV, You're seeing what you want to see. In science, we're not supposed to do that! Yeah, it's hard to miss, especially if you refuse to really look! You can observe a lot, just by watching (Yogi Bera).

Looking at ALL your other references, they all show the same thing. Strongest resonance between 1 eV and 0.1 eV. Not between 0.1 and 0.01 eV. Okay, over to you. Let's see you get out of this. SBHarris 00:14, 28 April 2007 (UTC)

0.25, 0.025, whatever. You're going to take that tone of voice with me over a single decimal point? Dammit Jim, I'm a physicist, not an accountant! Anyway, I said it was just a post-hoc explanation for one guess at the number's origin. The only thing I can say with any degree of confidence, is that it's an empirical fudge. Maybe this post-hoc explanation is better. (See the end of page 4.) It does relate it to the crystal lattice effects you mentioned.
The operating temperature isn't related to catching resonances, since you've got plenty of reactivity with enriched fuel. The main factor is turbine efficiency, and I think the only reason they don't run hotter is that the fuel would melt. As for asking coworkers, we don't do our own neutronics models here, so I'll have to wait until I get a chance to ask our vendor. KonradG 22:06, 30 April 2007 (UTC)
I think the resonance is what it is: 0.25 eV. However, I think I've resolved the 3/2 kT vs kT issue. The first is the mean (RMS) energy, and the second is the most probable energy (peak of the curve in E distribution). Because the curve is not symmetrical, they aren't the same numbers. So I'm going to have to go back and change it again in the original article. Physicists here haven't been as sloppy as I thought-- it's just that they've been using a different measure or descriptor of an asymmetrical distribution, than is usually used by (say) chemists. SBHarris 22:55, 6 May 2007 (UTC)