Talk:Neutron temperature

Incipit

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)[reply]

Reading this article I thought it might be a better idea not to connect neutron temperature and neutron energy. The energy of a single neutron is well defined and can be measured pretty precisely. Neutron temperature is a useful term, when we talk about the energy distribution of a large ensemble of neutrons (e.g. like in the moderator vessel of a nuclear reactor) which is (almost) in a thermal equilibrium with the moderator material (water, heavy water, beryllium, graphite or whatsoever). The discussion below has a quite a bit to do with single neutrons vs. "neutron gas". There is definitely nothing magic about any specific neutron energy. We wished we could produce only neutron which have exactely 0.025eV (or any other specifc value for that matter:-) Ulineutron (talk) 20:54, 16 June 2008 (UTC)[reply]

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 C^o) 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. S B H arris 19:20, 23 April 2007 (UTC)[reply]

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! S B H arris 19:36, 23 April 2007 (UTC)[reply]

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)[reply]

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. S B H arris 04:09, 24 April 2007 (UTC)[reply]

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)[reply]

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. S B H arris 18:56, 24 April 2007 (UTC)[reply]

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)[reply]

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. S B H arris 23:26, 24 April 2007 (UTC)[reply]

(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)[reply]

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)[reply]

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)[reply]

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. S B H arris 03:07, 30 April 2007 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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.S B H arris 03:11, 30 April 2007 (UTC)[reply]

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? S B H arris 17:50, 27 April 2007 (UTC)[reply]

??? 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)[reply]

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. S B H arris 00:14, 28 April 2007 (UTC)[reply]

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)[reply]

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. S B H arris 22:55, 6 May 2007 (UTC)[reply]

I guess it is the Maxwell Boltzmann distribution of 290 K folded with the typical 1/v cross section there. Would make sense, as thermal neutrons of 290 K would then have the same cross section as monoenergetic 25 meV neutrons, and you can take the excitation value there. Some student, please check this as an exercise, and correct the article. Peter.steier (talk) 05:06, 5 February 2016 (UTC)[reply]

Pretty much, physicists use (1/40)eV, or 0.025eV in decimal fraction, as kT, ignoring small fractions like 3/2. More specifically, it is usual to round 1/k to 12000K/eV, and use 300K for room temperature. In the case of fission neutrons, they start in the MeV range, so a factor of 2 or so doesn't matter much. I noticed the use of mode in the article and was surprised. It might be true, but I don't believe that is the intention. It is just kT without any additional small constants. Gah4 (talk) 17:52, 29 May 2017 (UTC)[reply]

The 2200m/s is called the most probable velocity (most number of neutrons have this velocity at 20C), not the average velocity. The most probable velocity will give the maxwell equation a maximum value. For most probable velocity E_kinetic = (1/2)mv2 = kT. This will give you 2200 m/s at 20C. I hope this help.

Why moderate neutrons in reactors?

I think the explanation of criticality with thermal neutrons on this page is incorrect. It is true that the neutron capture cross-section of U-238 increases at lower neutron energies -- this is why you breed Pu from U-238 in a thermal reactor. But the fission cross-section of U-235 and Pu-239 increases dramatically for thermal neutrons, which more than compensates for the increased capture by U-238. At 1 eV, the fission cross-section for either fissile isotope is near 100 barns, just about two orders of magnitude greater than for unmoderated fission neutrons, and it keeps going up at lower neutron energies. Operating reactors with natural uranium has nothing to do with the variation in capture cross-section by U-238, but hinges on using a moderator such as heavy water or graphite that absorbs fewer neutrons than the hydrogen in light water.Avm1 (talk) 01:03, 18 January 2009 (UTC)[reply]

You are clearly right about the fissile cross section going up with moderation, so the article is wrong about that and must be changed to mention both effects. But are you sure about which effect (better fission cross section in U-235 vs. poorer capture one in U-238) is the most important neutron temperature effect? S B H arris 00:34, 8 March 2009 (UTC)[reply]

Avm1 is right. The higher fission cross section of U-235 is the reason. Also in a HEU reactor with no U-238 one would use a moderator, as you would need less U-235 to become critical than with pure U-235 (i.e. a U fission bomb). Resonance capture, however is a common problem, not only by U-238, thus you should use a fast moderator like H to quickly pass through the resonance region. Peter.steier (talk) 05:21, 5 February 2016 (UTC)[reply]

The capture cross section of U-238 increases for thermal neutrons, but not as fast as the neutrons slow down. So, the rate of absorption decreases. It takes a longer time. Otherwise, yes, the important thing is the ratio of cross sections. Not mentioned is the distribution, which goes back to Chicago_Pile-1 where the Uranium is in lumps forming a lattice within the graphite. Appropriate spacing makes it more likely that the first collision is with moderator instead of uranium. This is better than a uniform mixture of uranium and moderator, and likely also easier to build. Gah4 (talk) 06:41, 23 March 2020 (UTC)[reply]

Why "which?"

"A thermal neutron is a free neutron with a kinetic energy of about 0.025 eV (approx. 4.0×10−21 J; 2.4 MJ/kg, hence a speed of 2.2 km/s) which is the energy corresponding to most probable velocity at a temperature of 290 K (17°C or 62°F), the mode[which?] of the Maxwell–Boltzmann distribution for this temperature."

Why is "mode" above marked as a 'weasel word'? The Maxwell-Boltzmann distribution has exactly one maximum (= peak = mode of the distribution). Or did I miss something? —Preceding unsigned comment added by 92.201.96.39 (talk) 14:29, 6 December 2010 (UTC)[reply]

The 0.025ev, or (1/40)eV is kT at about room temperature. As far as I know, there is no intent to use mode, but just to ignore small factors like 3/2. One should probably use mean, but that means more unneeded small constants. Gah4 (talk) 18:03, 29 May 2017 (UTC)[reply]

Neutron energy distribution ranges

"Moderated and other, non-thermal neutron energy distributions or ranges are listed below: Fast neutrons have kinetic energies greater than 1 eV, 0.1 MeV or approximately 1 MeV, depending on the definition."

Is the number 0.1 Mev incorrect? — Preceding unsigned comment added by Zedshort (talk • contribs) 15:41, 31 March 2012 (UTC) −[reply]

I don't know. This article now says 1 MeV. — Arthur Rubin (talk) 14:48, 8 November 2012 (UTC)[reply]

Removal of three images from the article

Plane wave

Wave packet

Propagation of de Broglie waves in 1d – real part of the complex amplitude is blue, imaginary part is green. The probability (shown as the colour opacity) of finding the particle at a given point x is spread out like a waveform, there is no definite position of the particle. As the amplitude increases above zero the curvature reverses sign, so the amplitude begins decrease again, and vice versa – the result is an alternating amplitude: a wave.

According to the de Broglie hypothesis, every object in the universe is a wave, a situation which gives rise to this phenomenon through wavefunction collapse.

Three images have been removed from the Neutron temperature article from Arthur Rubin (talk) today, the 7th of November 2012 @ 8:29 AM] with the following motivation:

'"(...) Reverted good faith edits by Maurice Carbonaro (talk):

"The added material is either original research or independent of the concept described here. ([[WP:...) (...)" .

Please participate in commenting if this should be applied or not when {{paradoxical}} judgements are being made:

is it either "original research" or
"independent of the concept described here"?).

I am starting to consider taking a peek in WP:ANI for this administrator seeking opinions from other non involved ones... comments are welcome. Thanks. Maurice Carbonaro (talk) 20:44, 7 November 2012 (UTC)[reply]

Because, quite simply, they have nothing to do with "neutron temperature". As you've made other bad edits at the same time, I just want to point out that that is the reason for removing those edits. — Arthur Rubin (talk) 14:49, 8 November 2012 (UTC)[reply]

I'm surprised that just Dr. Arthur Rubin (talk) and noone else has commented this topic. Thanks for the (only) answer though. Anyway I will come back to this talk page in the future to see if some other editor will post something different from what I honestly consider not being a WP:NPOV (Which stands for "Neutral Point of View" and NOT for "Neutron Point of View"). Arthur please refrain from being to harsh against edits that have been performed in absolutely good faith, because if you keep behaving like that the "Neutrality temperature" could change dramatically...

I don't know if you ever heard of the WP:DGAF humorous policy. If you will ever decide to take a peek at it and smile (!) ... well... this will make me happy. It just seems to me that you are generally taking Wikipedia a bit too seriously.

Please try to relax and... have a nice week-end.

M aurice Carbonaro 11:25, 30 November 2012 (UTC)[reply]

Removal of explanation?

Why remove the layman's explanation? The article has a 'this is too technical' tag, so why remove an effort to explain things in plain terms, and put back the too technical tag? How is that useful? — Preceding unsigned comment added by 76.100.248.41 (talk • contribs) 01:30, December 18, 2012‎

Because the "explanation" is your creation (see WP:OR), and is simply wrong. — Arthur Rubin (talk) 13:30, 18 December 2012 (UTC)[reply]

...Okay. Then how about a contribution to improve my addition rather than simply reverting without comment? You are familiar with the wikipedia policy to assume good will? if every attempt to add new material (substantive edits) is met with blanket reversions, how will Wikipedia ever grow and become better?

Reverting again, please do not revert. If you think the addition is incorrect, please help us all to improve it. — Preceding unsigned comment added by 76.100.248.41 (talk) 19:34, 18 December 2012 (UTC)[reply]

Woops! Apparently I'm supposed to 'sign' these things with tildes. My bad! :D

76.100.248.41 (talk) 19:51, 18 December 2012 (UTC)[reply]

I don't see what can be done to improve a bad explanation. Temperature is related to mean kinetic energy of particles, not vibrations or speed (neutrons always go faster than normal matter of the same temperature). It is a technical scientific concept about which you don't know the first thing (namely that it IS a scientific concept). So why should we let you explain temperature to laymen? Surely they'll welcome you at the temperature article for that? Okay I admit sarcasm there. See that article's TALK page before splashing in. S B H arris 20:30, 18 December 2012 (UTC)[reply]

Wonderful.

So, 'be bold' means absolutely nothing on todays wikipedia? How remarkably constructive of you. No wonder new editors are hard to find.

How about you make some positive contributions to this article?

(sigh)

Morg00 (talk) 22:17, 30 December 2012 (UTC)[reply]

How about you notice that only one other person has made more? [6]. Then check neutron. Then read this talk page from the top to get some idea of the level of discussion. Be bold all you like, but your edits won't last unless they are also correct. S B H arris 02:02, 31 December 2012 (UTC)[reply]

Right. And the 'level' of that discussion is so high, the article is tagged as 'techno-babble'. I understand the topic just fine, thanks. The point is to make it accessible to bright 7th grader.

Well, as much as you can, this is a bit esoteric.

Morg00 (talk) 02:58, 31 December 2012 (UTC)[reply]

Uploading an image?

Okay, I've found/made a good picture (Well, i think it's good, but I'm no graphics dood) of a cold neutron source.

But when I try to upload it, I get nothing.

I swear, it's not random porn. How the hell do I put up a simple picture? Why the hell is this so hard to use?

Clue-in appreciated! :)

Morg00 (talk) 22:19, 30 December 2012 (UTC)[reply]

What is the URL (http etc) of the image? The address appears whenever you're looking at it, if your upload was successful. S B H arris 02:08, 31 December 2012 (UTC)[reply]

Eureka!

I finally crawled through the eleventy-zillion steps and got the pesky thing up.

Morg00 (talk) 02:55, 31 December 2012 (UTC)[reply]

stability and temperature

I don't know if it goes here or not, but the stability of reactors comes from the sensitivity to neutron temperature, such that high temperatures reduce the fission rate. Well, even more, a reactor has to be stable all throughout, not just on average. Mean-free-path isn't quite long enough to equilibrate though a large reactor. But otherwise, power reactors aren't anywhere close to 290K. Gah4 (talk) 16:06, 13 June 2021 (UTC)[reply]

Math

? 75.143.133.217 (talk) 15:58, 23 April 2022 (UTC)[reply]