Talk:Nuclear fusion

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[edit] Wikipedia Simple

I've added an entry for Nuclear_Fusion to wikipedia simple, it would be good if someone could check it out and alter any wording that’s a bit too ambiguous or just down right incorrect. Cheers MattOates 16:03, 4 June 2006 (UTC)

I want to compliment you on a short article done well. You have hit the high points accurately, without getting distracted by peripherals. There are a couple of places where someone might quibble, and I'll say more about those, but I want to emphasize that these are complaints "at the margin"; that you have made a simplification that is quite reasonable, and I would not ask for any correction at all if it made the result too complex. That said....

First: Possibly you oversell the role of iron as "most stable nucleus" and "fusion stops here". If you look at the actual data you'll see that the curve is very flat in the middle; there is no single isotope that stands out as a dominant maximum, and in fact the article says that an isotope of nickel is the "most stable", followed by two isotopes of iron. I'm fairly sure that you could find a hypothetical reaction that resulted in something heavier than iron that was still (mildly) energetically favorable. The big point to get right is that the curve has a maximum in the middle. Hence fusion of small nuclei and fission of big nuclei are the energy-releasing reactions for the most part.

Second: Possibly you give the impression that every fusion of light nuclei is energetically favorable. Again I refer to the actual data and point out that the curve is not smooth at the low end. In particular, note the big spike that is 4He. Again I would not recommend pointing out details on the curve; that's way too much detail. JohnAspinall 20:14, 29 August 2007 (UTC)JohnAspinall

[edit] Criticisms

Please attribute the criticisms of fusion power, or provide better arguments for them. At the moment I'd strongly disagree with them. --Robert Merkel

Here's one link

[edit] Strong force vs. Electromagnetic force

Right at the beginning of the article there's a really bad mistake: it says that "strong force only operates over short distances, unlike the electromagnetic force". It is in fact the other way around: electric force gets weaker and weaker at large distances, while the strong one is overwhelming at large distances (because of confinement), getting weaker and weaker at short distances (because of asymptotic freedom). I lack the style and experience to edit the article, but wanted to make this small suggestion...(neko)

i am afraid you are mistaken. the strong force acts only between 0-3 fm (thats 10-15 metres), which is a very small distance. it peaks at around 2 fm, where it reaches forces of up to 30000 N (very strong - hence the name). the electromagnetic force on the other hand, acts over every distance, and follows an inverse-square law, meaning that it is strong at small distances but weaker as distance increases. i hope this clears things up for you -mastodon 15:16, 19 March 2006 (UTC)

The confusion here is between the strong force between quarks (what used to be called the colour force), which neko is talking about, and the nuclear force between hadrons (which used to be called the strong interaction or strong force). I never noticed the definition change either, which is confusing. It might be better to use nuclear force throughout.--Syd Henderson 06:34, 23 September 2006 (UTC)

nuclear force would be appropriate since it takes into account both the weak and strong force. Even with this confusion, EM force should be active at larger distances. QCD, even though mediated by the massless gluons, have many additional factors that keep it from being a "long-range" force such as EM or gravity --Blckavnger 23:11, 17 November 2006 (UTC)
Apparently in modern usage "nuclear force" does not refer to the weak interaction. I guess it's been too many years since I went to graduate school. --Art Carlson 16:10, 14 March 2007 (UTC)

I put nuclear force (so called "residual strong force") instead of strong force and strong nuclear force. Kopovoi 15:11, 14 March 2007 (UTC)

Actually the strong and weak forces only act over a very small distance it IS the electromagnetic force that acts over the large distances (jon)

[edit] 24.6 Watt / cubic meter of nucleary active sun ?

Could someone explain why there is any hope in terestrial nuclear fusion considering the following facts :

(1) the sun is huge R = radius = 0.696×10^9 m

(2) considering the ratio (surface of the sun [m2])/ (volume of the sun [m3]) we get (4piR3/3)/(4piR2)= R/3, so each square meter got the energy of 2.32*10^8 cubic meters of sun to evacuate!

(3) knowing that a square meter of sun produces a mean Intensity of 2.009×10^7 W/m2 we got each cubic meter of the sun producing about : 2.009×10^7 / 2.32*10^8 = 0.087 Watt / cubic meter of sun ?

(4) so even if we suppose that only about one tenth of the cubic meter of the sun are the real productor of the global energy, we still got only 1 watt per cubic meter

! ---> "The core is the only part of the Sun where an appreciable amount of heat is produced by fusion.... The core extends from the center to about 0.2 solar radius"

Ok, so number (2) and further need to be corrected, so we got :


R_active = radius = 0.696×10^9 /5 m = 1.4 * 10^8 R_tot = radius = 0.696×10^9 m = 6.96* 10^8

(2) considering the ratio (surface of the sun [m2])/ (volume of the sun [m3]) we get (4pi*R_active^3/3)/(4pi*R_tot^22)= (R_active^3/3)/(R_tot^2)= 1.4^3 * 10^8 / 6.96^3 = 8.14 * 10^5 , so each square meter got the energy of 8.14*10^5 cubic meters of sun to evacuate!

(3) knowing that a square meter of sun produces a mean Intensity of 2.009×10^7 W/m2 we got each cubic meter of the sun producing about : 2.009×10^7 / 8.14*10^5 = 24.6 Watt / cubic meter of nucleary active sun ? So how can we have hope that the fusion will be one day interesting in earth conditions where the number of cubic meter to harness the power from is very limited ?

Can anybody explain that ?

Yes. The sun uses the Proton-proton chain reaction It also is self regulating. When the sun heats up, it expands, which lowers density and slows the reaction. Fusion reactors will probably use D + T reactions, which go much faster. pstudier 2005 July 3 22:50 (UTC)
The Sun is also stone cold, 13.6 MK at the core, compared to 300MK in a fusion reactor. A medium-sized experimental tokamak running on D-D already has a power density close to that of the Sun. Art Carlson 2005 July 4 06:52 (UTC)
I would go so far as to call 13.6 MK stone-cold, but also, in point two , you're dividing volume by  : surface, as opposed to the other way around (which you say you do)
Also, the volume in which actual fusion takes place is confined to the core of the sun. The rest of the gaseous mass acts as a "dampener", reducing the rock hard X-rays and Gamma rays into soft visible light that we enjoy. One of the main problems with fusion power is finding a way to deal wtih this kind of radiation. Sikkema 08:49, 5 March 2007 (UTC)
When looking extra-terrestrially for a high fusion-power/volume example, try another star other than our own. A white dwarf going Type 1A supernova with carbon and oxygen fusion will do. A white dwarf has a radius on the order of Earth's, and probably less than that, shrinking as it picks up mass and approaches Type 1A conditions of 1.4 Solar mass. The Type 1A fusion event releases ~10^46 joules in a few seconds. The power density then is ~ 10^24 Watts/M^3. Unfortunately this particular reactor design unbinds itself in the same time frame, and has density 10^9kg-m^3 at startup. ;-) 192.160.51.70 23:08, 5 April 2007 (UTC)
Then a supernova doesn't have gravitational confinement but inertial confinement. I don't know the real numbers, but I imagine a (terrestrial) ICF plant operating a 1 Hz with a compressed pellet of 0.1 mm diameter and 1 GW power. That comes out to a power density about equal to that of your supernova. --Art Carlson 09:43, 6 April 2007 (UTC)
Thats still quite a bit short of ET White Dwarf Reactors, Inc. NIF has plans for densities of 10^6kg/m^3 . Impressive, yes, but WDs are electron degenerate matter w/ density at least 10^9kg/m^3 as above, and will burn most of the mass. Sorry ICF, can't get there from here. 192.160.51.70 22:17, 6 April 2007 (UTC)

[edit] Belated Move from Talk:Fusion

The following two 'graphs apparently created Talk:Fusion in the edit "M 15:43, 2002 Feb 25 . . Conversion script (Automated conversion)":
could somebody add something about tokamaks laser fusion and any other current research here please. Also, please check my physics as I am not that sure of myself here. Thanks
What is the energy for DT reactions?

[edit] Fusing elements Iron or heavier

Is it even possible to fuse 2 iron nuclei together? Or an iron and another element's nulcleus to make an element heavier than Iron? It says that fusing elements heavier than iron absorbs energy, but I've been told by a physicist this just simply can't be done, Iron is the limit you can fuse to (check my out my talk page) obviously this page thinks it isn't, it wouldn't say it absorbs energy if it simply can't be done. The snare 05:58, 22 August 2007 (UTC)

Where do you think elements heavier than iron come from in nature? See Nucleosynthesis#Explosive nucleosynthesis and also Transuranic_elements#Super-heavy atoms. --Art Carlson 11:44, 22 August 2007 (UTC)

I thought they come from a process called beta-decay, in which neutrons become protons. The snare 09:38, 24 August 2007 (UTC)

Some come from decay of even heavier elements for example uranium turning to lead, but that still leaves the question of how these heavier than iron elements came from to decay to other slightly less heavy elements. Just because making elements heavier than iron consumes energy rather than releasing it doesn't mean it can't be done, it just means it takes a lot of energy to do it. As linked above the only place with that much energy is a star, and in particular one in its death throws with nothing else left to burn. The rapid collapse of such a star produces unimaginable temperatures and pressures for a brief period before it blows itself apart. The fact that the star explodes has the handy side effect of spreading these newly formed elemets througout space which can then collect to form planets. The upshot of that is every atom in your body apart from most of the hydrogen atoms was once part of a star. We are, in effect, all made of stardust. --LiamE 16:29, 24 August 2007 (UTC)

Ok, but read my talk page, like I was saying. Perhaps you're right but I'm just telling you that's what he told me. The snare 02:18, 12 November 2007 (UTC)

[edit] Misc

Here's another

http://vlt.ucsd.edu/stacey.pdf

I'll rephrase the criticism, but it was a hot topic a few years back when the Department of Energy was deciding funding priorities for future energy research.

Has anybody considered the possible social and environmental costs of obtaining an extremely cheap source of energy? For example, the economic changes that might result, and the environmental damage that could be done if energy costs are no longer a barrier to huge civil engineering projects? -- Heron

Coal plants source of radioactive waste? check change by 66 ...

Yep. If ORNL says it, it must be so: [1] Graft

Can we include something about the possibility of extracting Helium-3 from the lunar surface? This has been bandied about for years by adocates of space exploration and space commercilization, and recently has been advocated by some within the Chinese government. RK

I think there are a few bugs in this page. I corrected one for the D+D reaction, in the 2nd case the new nucleus is a 3He, not a T. Plus, shouldn't the atomic numbers be written top-left instead of top right of the symbol (in the equations)?

[edit] what else is needed for stellar fusion?

In the section of this article discussing gravitational confinement, one finds this assertion:

"Some simple math can demonstrate that the mass of fuel needed to make a star using the D-D reaction is about the size of the Moon."

Certainly Jupiter contains that mass of hydrogen and more. But it isn't a star. So what else is necessary to induce the fusion reaction?

If the "fuel" assumption that yields that result is some typical non-stellar isotopic assortment, the answer may be interesting. If it refers to something else (pure deuterium would be a logical choice for such a calculation), Jupiter should not be expected to qualify. And "and more" is irrelevant since concentration is likely to be crucial.
Also regarding composition, neutron-absorbing substances can poison fusion. (The control rods in fission reactors similarly poison the fission reaction.) This is a plot point in (the novel but IIRC not the film of) The Sum of All Fears.
Clancyphysics, I'm afraid. If you look at the table of reactions, you won't see a neutron as input (left hand side of the arrow) anywhere. How could absorbing the neutrons "poison" the reaction when they are not needed? JohnAspinall 18:01, 29 August 2007 (UTC)JohnAspinall
BTW, "the size of the Moon" is an odd way to put the "mass of fuel": that size under what T & P conditions? It takes no calculation to deduce that a lunar-mass of H2 or D2, compressed to a lunar-diameter sphere, and located near the Earth's or Moon's orbit would dissipate by reaching escape velocity, since we know the moon is far from holding an atmosphere even of heavier gases. (Even Mars is believed to have lost most of its H2O from dissociation and the escape of the H2 that results.) Even if H2 released on the Moon escapes slowly, we know the mass i've specified would expand immediately, producing
Of course i don't mean to assume that the D-D statement is accurate! If it is, it isn't clearly enough stated, IMO.--Jerzy(t) 21:54, 2004 Mar 6 (UTC)

How can nTτ be about 10^14 s/cm³? There is unit inconsistency. It would have to be s cm^-3 K.

10^14 is nτ and actually is s cm^-3. I will fix the reference to make it consistent. Trelvis 19:16, May 17, 2004 (UTC)

[edit] College Research Paper - Peer Review

Has been moved here: /Pompura paper

[edit] Lawson Criterion

As it currently stands, there's a typo in the section on the Lawson criterion. The text states nTτ must be greater than 1014s / cm3. The units don't match the quantity, and - I think - should be s − eV / cm3 or s − keV / cm3. Do any of the current watchers have the correct units on hand?

Also for reference, the textbook Choudhuri. Physics of Fluids and Plasmas: An Introduction for Astrophysicists. Cambridge University Press: Cambridge, 1998. describes the Lawson criterion as the restriction nτ > 1016s / cm3, but I don't know what temperature he assumes.


[edit] Check these numbers?

According to our section "The fusion triple product", :

the nuclei only undergo fusion once in every 1029 seconds. However, the fact that the Sun contains 1059 nuclei means that the net reaction rate is actually quite high, and since the Sun is around for billions of years, eventually the fuel is used up and the total energy released is huge.

This doesn't seem right. If the reaction rate per nucleus is 10-29 s-1, which is order of 10-21 yr-1, then it will take a lot more than mere billions of years to use up the fuel. Indeed the claim that: 1059 nuclei means that the net reaction rate is actually quite high also seems false; our Sun article gives the Sun a mass of 1.9891 × 1030 kg (agreeing with the 4π2r3 = GMT2 law), which (given 99% H, and NA H nuclei to the gram) would result in only about 1057 nuclei. 1057 nuclei × 10-29 s-1 = 1028 fusions per second, and about 40 kg of fuel consumed per second - way too low! It seems likely that the 59 should be 57, but I'm not sure what the 29 should be. I suppose it could be approximated from power output, but if someone already knows, please fix it! Securiger 10:01, 28 Nov 2004 (UTC)

Another way to put it: if we allow 6.5 MeV (10 − 12J) for one quarter of 4p→4He + 2e (which I think is about right, and is certainly the right order of magnitude), then 10^{-12} J/nucleus \times 10^{57} nuclei \times 10^{-29} s^{-1} = 10^{16} W, which, spread over a sphere of area 4\pi R_e^2 = 2.8 \times 10^{23} m^2 corresponds to a "solar constant" of 35 nanowatts/m2. Our reaction rate seems to be out by 10 orders of magnitude! Securiger 10:28, 28 Nov 2004 (UTC)
There are several different fusion reactions that happen in parallel. See [2] for a discussion of the different rates. Also, the cross-section-velocity product at 20 keV (200,000,000 K) for deuterium-tritum fusion (which I think is the main energy-producing reaction) is 4.2x10 − 22 (from Goldston & Rutherford. Introduction to Plasma Physics. IOP Publishing, Ltd: Philadelphia, 1997.) For a density of 1021m − 3, this gives a reaction rate of 0.42 per second per nucleon... Unfortunately, I don't know the solar core's parameters, although the surface is around 5 eV. A detailed (but technical) treatment of fusion cross-sections in a solar context is given at [3]. Also, [4] gives a power output for a net 4p -> He reaction of 25 MeV/nucleon (one eV = 1.6x10 − 19J) One final reference, [5] is fairly accessible, but I should be getting back to work. Will try to come back here & help distill the numbers into something manageable later.SMesser 17:41, 30 Nov 2004 (UTC)
From the Plasma Formulary: Solar Radius = 6.96x108 m, the upward mass flux is 1.6x10 − 8kg / m2 − s, and mass = 1.99x1030 kg, for a particle confinement time of 2.04x1019 seconds = 6.47x1011 years. An estimate of the energy confinement time can be reached by comparing the sun's luminosity (3.83x1026 J/s) with the energy content of the sun, which is approximately twice its mass times its temperature. (Since temperature is energy / particle, and a zeroth-order model of the sun has only protons and electrons, with the protons contributing nearly all the mass.) To do this calculation properly, you'd need estimates of the densities, temperatures,and compositions of the sun at different depths, but a quick-and-dirty estimate can be reached by using estimates of the dense solar core (below) and of the less-dense surface (above). A geometric mean of the three values gives a mean temperature of 400 eV, for a rough thermal energy of 2x1041 J, for a confinement time of 4x1014 sec = 10 million years. Those are estimates of the τ factor in the triple product, with the energy confinement time being more relevant for the Lawson criterion. Also from the formulary, we can estimate a rough solar density of 1410 kg / m3, although that's certainly lower than the core value (see below). This gives a proton density of 8.42x1029m − 3, which provides a lower limit on n.SMesser 18:04, 3 Jan 2005 (UTC)
From [6], solar fusion reactions typically take place at temperatures of 5 - 30 keV, which should be a decent estimate of T. If we take a proton-proton collision energy of 12 keV, this paper's equations 7 and 8 give a cross section of 2.7x10 − 23 barns or 3.3x10 − 51m2- pretty tiny, which is at the very least consistent with my understanding that the p-p reaction is the rate-limiting part of the process.SMesser 18:04, 3 Jan 2005 (UTC)
[7] quotes a solar hydrogen density of 105kg / m3, which is presumably taken at the core. This translates to a proton density of 6x1031m − 3. Combining this with the estimates of cross section and core density and temperature (above), I get a typical proton-proton reaction time of 6.0x107 seconds = 1.9 years. (i.e. about 37% of the solar core's protons will fuse in 1.9 years. This may or may not be consistent, depending on the size of the core relative to the sun.)SMesser 18:04, 3 Jan 2005 (UTC)
Looking back to the sun's luminosity, if each fusion reaction provides 25 MeV, 3.83x1026J/s implies 9.6x1037 He nuclei produced per second, or 6.4x1011 kg/s of H fusing to form He. This implies a solar lifetime on the order of 3.1x1018 sec = 9.9x1010 years, which is about 20 times too long. To be consistent with the 1.9-year estimate for the fusion core, it also predicts a mass of 3.8x1019 kg of hydrogen in the core.SMesser 18:04, 3 Jan 2005 (UTC)
Can anyone out there check my math, please? Thanks. SMesser 18:04, 3 Jan 2005 (UTC)
Taking another stab at this with numbers from the last paragraph above... I'll estimate the mean lifetime of a proton using the sun's luminosity and density: 9.6x1037 He nuclei produced per second implies 3.8x1038 H nuclei used per second. Since hydrogen makes about 75% of the sun and the mass is 1.99x1030 kg with 1.7x10 − 27 kg per proton, there should be 8.9x1056 protons present. So the typical time-till-fusion of a typical proton is more-or-less the ration of the number of protons to the rate at which they're fusing: 2.4x1018 seconds.
Since it's been a few weeks and I'm the only person posting with hard numbers, I'm going to toss the fallout from this last bit on the page. (Hopefully the lack of posts isn't because I've accidentally marked some of the edits as minor. If it was, I'm sorry.) Anyhow, I'm also going to cross-post on the talk:sun page to ask for more data. SMesser 21:08, 21 Jan 2005 (UTC)
Saw the note on Talk:Sun and have been pondering this. For the estimate of number of particles in the sun, it looks like you're about on track with 1057 - this link has some calculations:[8]. As for the proton-proton reaction rate/lifetime of a proton in the sun, my way of working it is pretty much the same as yours - solar luminosity is 2.9×1026W, and by my calculation each fusion reaction liberates 0.7% of the rest mass of four protons, or 4.64e-29kg. This equals 4.176e-12J. So, to support the solar luminosity, there must be 6.944e37 fusion reactions per second. The number of protons divided by four times the reaction rate (because each reaction requires four protons) gives us a proton lifetime of 4.32e18s, which is near enough similar to your estimate. This is 136 billion years, which means that during the solar lifetime of 10 billion years, rather less than 10% of the protons present would fuse. I believe that is a plausible result. Worldtraveller 18:16, 27 Jan 2005 (UTC)

[edit] Stop! Most of this belongs under Fusion power

The Fusion disambiguation page says

  • For the combining of two atomic nuclei into a single nucleus (with possible emission of radioactivity), see nuclear fusion
    • cold fusion refers to a once hypothetical but now largely discredited form of nuclear fusion
    • See fusion power for a discussion of using nuclear fusion as an energy source

That is, the Nuclear fusion page is only for fusion as a nuclear process. All aspects of fusion as an energy source should be moved to Fusion power. I should say "integrated into Fusion power" since the pages are 80% redundant. I think such a division makes good sense. Art Carlson 17:56, 2005 Jan 30 (UTC)

Yeah, I noticed this, too. I am interested in how the process works (at the moment) and the discussion simply elided into a discussion of the practical ways that this might be used for power generation. Davemenc 04:02, 27 November 2005 (UTC)

There is a mistake in the following calculations:

"solar luminosity is 2.9×1026W, and by my calculation each fusion reaction liberates 0.7% of the rest mass of four protons, or 4.64e-29kg. This equals 4.176e-12J. So, to support the solar luminosity, there must be 6.944e37 fusion "

You did not convert the solar luminosity into J even though the amount of energy per reaction is in J, therefore when you did the division to figure out the number of reactions per seconds your number is slightly off CORECTION: luminosity of sun is approximately = 4e26J =>

  1. of reactions = 4e26 / 4.176e-12 = 9.58e37

[edit] Tunneling

I think the sentence "additional effects, such as quantum tunneling, lower the energy barrier slightly" is incorrect. I don't know about nuclear physics, but in "regular" physical chemistry tunneling is a process by which species "tunnel" through an energy barrier without lowering the barrier. I think it is more correct to say that the "additional effects, such as quantum tunneling, lower the effective activation energy slightly". The difference being that effective activation energy (perhaps there is a more formal term for it) reflects the required amount of energy put into the system and is independent of the actual height of the thermodynamic barrier. Fearofcarpet 17:34, 18 Mar 2005 (UTC)

Your model of tunneling carries over from chemistry to nuclear physics just fine. (Different particles, same wave equation.) Some use of "effective" sounds like the right way to say it. JohnAspinall 13:51, 31 August 2007 (UTC)

[edit] latest condensation

Regarding the latest edit which got rid of most of the explanations of types of fusion confinement: I rather think we should keep this. Yes the individual articles have detailed information on each method but there's no reason we can't have a short synopsis of all the methods here. I welcome Art's edits very much and am glad we have a real plasma physicist here to make these edts, I just get slightly nervous when he gets on with big swaths of deletions. --Deglr6328 07:12, 24 Mar 2005 (UTC)

Addendum- I also have to say I don't really like the whole removal [9] of "fusion triple product" and "breakeven and ignition". Why couldn't these just be improved a bit instead of being deleted? Quell my fears Art! :o) --Deglr6328 07:19, 24 Mar 2005 (UTC)

Thanks for the flowers. I realize I came out redecorating with napalm. I'm surprized I got away with most of it. The issue here is where things belong. I'm not sure having separate articles for nuclear fusion and for fusion power is a good idea, but as long as we do, there should be some system to it. I am trying to draw the line between fusion as a nuclear process and fusion as (potentially) used for electricity production. I discovered that even this line is not so easy to draw. I put the <σv> business in nuclear fusion because it can be discussed (nearly) idependent of any particular confinement scheme, even though it is only of interest if you have power production in mind. That's why I reduced discussions of confinement methods, triple product, breakeven, and ignition to a bear minimum. What do you think is best? One big article for everything? Substantial duplication? Or some line of demarcation with cross references and minimal overlap? Art Carlson 09:29, 2005 Mar 24 (UTC)

I say separate articles are good with some slight duplication necessary for thoroughness of discussion in each article. Wouldn't "breakeven (well... maybe not breakeven) and ignition" be ok to keep here because, for example the term "ignition" can be applied to fusion process in stars? Is the fusion triple product section really THAT inappropriate for this article insted of being in fusion power? Really, I'm asking non-rhetorically 'cause I'm not certain. If it (and breakeven/ignition stuff) does have to go can you incorporate it into fusion power? Also, I would love it if you could provide an explanation/description in the magnetic mirror article of why the "baseball" experiments were considered (largely anyway...?) faiures.--Deglr6328 06:38, 25 Mar 2005 (UTC)

[edit] Fixing the table of reactions

Could someone please fix the table of reactions? The energy released by 3He+3He is missing, and it's not clear what's meant by energy values in parentheses next to the products (eg. D + T -> 4He(3.5MeV) + n (11.4 MeV)). If someone's feeling really ambitious, it would be nice to know what the Lawson criterion is at some "typical" temperature (eg. DT has 10^20 sec/m^3 at 10keV). Thanks much. --NDL


Seconded, 3He should be listed on all the tables ASAP. its all over the @#$@# moon and is one of very few Aneutronic fusion reactions (aka: safe). --Myren

Some of this is on aneutronic fusion (which talks about why aneutronic fusion maybe isn't as safe as it seems). BTW, do you know about the nonequilibrium fusion study (cited there)? It's worth reading as it essentially rules out a lot of the nice-sounding reactions. --Andrew 02:21, Apr 22, 2005 (UTC)

Am I missing something the text talks about neutrons making D+Li6 rxn no good, but the table does not show these neutrons .. (from table)

"(10) D + 6Li → 2 4He + 22.4 MeV"

Is there a missing branch for this rxn?

If you have a plasma of D and Li-6, you will get a substantial number of D + D -> He-3 + n reactions. --Art Carlson 20:33, 6 March 2006 (UTC)

Could someone comment here on why, for example, the following lines aren't considered practical?

23Na + 35Cl → 54Fe + 4He + 15.3 MeV

23Na + 37Cl → 56Fe + 4He + 16.9 MeV

JohnHarris 20:00, 19 December 2006 (UTC)

Several reasons. It hard to get a high temperature, because heavier ions radiate very quickly (roughly proportional to Z3), there's tons of bremstrahlung due to the mass and charge on the nucleus. And heavier nuclei are harder to fuse (coulomb barrier proportional to ZaZb), so it isn't really in the realm of technological possibility. Even stars can only rarely fuse iron (basically for a few milliseconds before a supernova, as far as I can recall) Danielfong 16:25, 20 December 2006 (UTC)

Thank you Danielfong, that's a helpful start. I brought the question here because I could think of nowhere more likely to get an observation from someone with a suitable background. The problem you describe presumably prevents a sufficient frequency of events rather than making them an impossibility. Banging two heavy ions together is the route taken in producing Super Heavy Elements. If it's a matter of frequency of events, a hot plasma might not be the only way to initiate a power source.

I'm sure this is the wrong place to be discussing the question - if anyone can suggest a more suitable forum I'd be grateful. JohnHarris 13:19, 22 December 2006 (UTC)

[edit] But where does the energy come from?

I've read the article about Nuclear fusion (shouldn't that be either "nuclear fusion" or "Nuclear Fusion") and have faied utterly to understand where the energy actually comes from.

Perhaps a few sentences that make it more explicit would be helpful for those of us who aren't physics experts.

Also, as was mentioned elsewhere in this discussion, much of this is about how to generate power from nuclear fusion when it sould be about how fusion itself works.

Thanks

Dave

Davemenc 04:01, 27 November 2005 (UTC)

Matter can be organized in different ways. Some of those ways are more stable than others. Unstable states are unstable because they have a lot of potential energy. If you take some matter and reorganize it into a more stable state, that potential energy is released. In fusion, deuterium (hydrogen-2) and tritium (hydrogen-3) are not as stable as helium and a neutron, so the reaction D + T -> He + n releases energy (the products come out of the reaction moving much faster than the reactants went in). We capture that energy as heat and use it to generate electricity. Canonymous 18:04, 15 September 2006 (UTC)
Isn't the plain English explanation just that a small part of the reactant mass ceases to exist as matter? It's converted into energy. It's a small portion, to be sure, as opposed to antimatter, where 100% of reactants are converted into some form of energy or neutrinos, with no quarks left to tell the tale. Sacxpert 10:48, 17 November 2006 (UTC)

Smart White Boy 20:16, 20 April 2007 (UTC)Might I just add, the energy release can be explained in e=mc2. Energy equals lost mass times the square of the speed of light. This implies that a large amount of energy is held in a small amount of matter. That's all I wanted to add. Smart White Boy 20:16, 20 April 2007 (UTC)

I would question the stability argument. U235 is relatively stable. The fission products of U235 are both highly unstable, yet the reaction which produces them is exothermic. The paradigm of exothermic reactions leading to stable products belongs to chemistry. It does not necessarily apply to nuclear reactions, fission or fusion.

Perhaps the most lucid explanation (for the layman) is that the mass of an atom is not -quite- equal to the sum of the masses of its constituent parts. Thus, most instances of atoms being combined or split apart result in a small change of total mass. Since the sum of the system's mass and energy must be conserved, this 'mass defect' manifests iself as kinetic energy of the resultant particles. --Anteaus (talk) 22:34, 24 January 2008 (UTC)

The absorption of a neutron places some nuclei into an "invalid" quantum state; the nucleus reacts to this illegal rearrangement by "vibrating" itself into valid smaller nuclei, after throwing away various items, including x-ray photons and more neutrons. Which nuclei? Big ones, where the outer nucleons are "far" from the center of the nucleus, and at these vast distances, the outermost nucleons feel a weaker nuclear force- recall, the nuclear force works only at short distances. -Dawn McGatney 69.139.231.9 (talk) 13:04, 1 April 2008 (UTC)

[edit] reinstating removed link

Some text and citing link was removed because it was dead, but I feel that's a poor way to deal with such a problem, so I found several citations for the paper and restored the link plus a link to the whole paper:

Some ways out of this dilemma are considered — and rejected — in Fundamental limitations on plasma fusion systems not in thermodynamic equilibrium by Todd Rider PDF. -Wikibob 04:05, 12 January 2006 (UTC)

Oops, I linked the PDF to an overview poster referencing Todd, instead here is the link to the 306 page paper: https://dspace.mit.edu/bitstream/1721.1/11412/1/33227017.pdf but I'm getting wikipedia problems when editing large articles. -Wikibob 04:19, 12 January 2006 (UTC)

[edit] Three-body reactions in ICF

Does anyone (meaning Daniel Fong) have the cross sections for any three-body reactions under ICF conditions? Have any such reactions been seriously propsed as candidates for fusion power? --Art Carlson 12:25, 4 March 2006 (UTC)

Sorry this is late. Art, under ICF conditions? I'm not sure if a -serious- proposal has really taken place. But now, come to think of it, light ion beams seem to open up all kinds of possibilities. The only three body reaction that I'm aware of that's really been looked at involves muons (I have given thought to 'seeding' an ICF pulse with Muons, and haven't done enough work on it). I'm not in Princeton right now, so I can't do a serious literature search or ask anyone.

Did you have any particular reactions in mind? I can't think of any that would be useful.

(PS: my gut feeling is that they're unimportant in D-T fusion, and useless in Aneutronic fusion, because the pairs will still fuse and cause havok.)

Note that any reactivity coefficient for a simple linear model of three-body reactions won't even have the familiar dimensions of cross section (i.e. distance squared). For two fluxes and one density on the right hand side, it will have dimensions distance-to-the-fourth times time. All of which is to say that I'm not sure what one would do with such a "cross section" if you had it. How would you compare it to regular cross-sections? JohnAspinall 18:04, 27 August 2007 (UTC)JohnAspinall

[edit] Chain reaction

I had hoped we wouldn't have to revisit this debate but...well. Petri Krohn just added the link to nuclear chain reaction which is used to describe the fusion reaction in the article. Is this right though? Really I don't see it as a chain reaction. A Deuterium nucleus smashes into a tritium nucleus and they combine, a neutron flys off and a high energy He nucleus goes in the other direction. That's it really. Yes the reaction yeilds energy and this energy can go back into heating the unburned nuclei in the plasma but this isn't a particle collision or anything. I think an ignited plasma can be described as being in a kind of "closed state" perhaps (surely there is a more proper term) but I don't think chain reaction is an apt description. thoughts? --Deglr6328 06:28, 13 March 2006 (UTC)

I don't think it is chain reaction either, at least not in the classical sense. --Art Carlson 09:20, 13 March 2006 (UTC)

Perhaps more correctly described as a positive feedback process, in the same sense that chemical combustion is self-augmenting. The heat released by fusion increases the temperature and pressure inside the vessel, and this results in an increasing rate of fusion. There is AFAIK no such thing as a fusion chain-reaction, at least not in the collision-multiplication sense. --Anteaus (talk) 21:51, 24 January 2008 (UTC)

[edit] Is there an exception? Yes.

Opening section: "the fusion of lighter nuclei, which creates a heavier nucleus and a free neutron, will generally release more energy than it took to force them together". Generally? If there is an exception it may be worth actually noting it. Alternatively, I'd tend to generalise and say "fusing light nuclei to produce nuclei no heavier than Iron (56) liberates energy." I'm not sure that the specification that the reaction of unnamed light nuclei to produce a heavier one and a (one) neutron is entirely useful - it looks like a cumulative editing malapropism. Maybe in my version rather than nuclei it should be "products totalling no more than". Midgley 11:26, 19 March 2006 (UTC)

One notable exception is the first step in proton-proton fusion, which already has a full article devoted to it.JohnAspinall 18:46, 27 August 2007 (UTC)JohnAspinall

[edit] The Sun

What kinds of radiant energy are produced by p-p reactions on the sun Where a nhermonuclear reaction occurs deep within the sun how much time will elapse for the products of that explosion to reach the Sun's "surface"? —The preceding unsigned comment was added by 69.230.136.40 (talk • contribs) on 21:16, 17 April 2006.

This sounds like a homework question. You might try reading at Sun and proton-proton chain reaction for more information on these topics. --Christopher Thomas 22:43, 2 May 2006 (UTC)

[edit] help me

what if there is one nuclei that is heavier than iron or nickel and one that is lighter than iron and nickel and they both joined together? —The preceding unsigned comment was added by 68.237.238.121 (talk • contribs) on 23:48, 30 April 2006.

They would only fuse at all if there was a net release of energy from the reaction. You might get two nuclei closer to iron coming out, or just a scattering event. --Christopher Thomas 22:46, 2 May 2006 (UTC)
That sounds a bit off. The new super-heavies like Unununium are made by fusing, for instance Bi and Ni. The reaction is "endothermic", but it goes.--Deglr6328 03:17, 3 May 2006 (UTC)


You can find a graph of binding energy per nucleon, against number of nucleons. The gradient relates to which process occurs. The graph has a turning point around Iron. Naturally, reactions give out heat (when talking nuclear physics), e.g. fusion in lighter-than-iron elements is natural, as it gives out heat, and all things go towards iron on the graph, hence fission is natural in heavier-than-iron elements. However, it is possible to cause fission (i think?) in lighter-than-iron elements, but requires you to go against the gradient, e.g. put energy into the system. Same with fusion and heavier-than-iron elements. —The preceding unsigned comment was added by 203.218.230.58 (talk) 13:04, 17 April 2007 (UTC).

[edit] Ionization energy vs. electron affinity

I'm not sure if this line means what im thinking, but isin't ionization energy the energy required to REMOVE an electron from an atom... so In my opinion, it should be the "electron affinity" energy. Not sure.

"the ionization energy gained by adding an electron to a hydrogen nucleus"

If you have a lone proton and you ADD an electron, you get energy out of the system in the form of em radiation....YOU '"gain" that energy'. right? what am I missing?--Deglr6328 19:33, 20 May 2006 (UTC)

[edit] Optimum fusion temperatures

I found what seems to me an inconsistency in the values of the optimum fusion temperatures between :

  1. the curve of the deuterium-tritium fusion reaction rate (and the corresponding data taken from this document);
  2. the data in the table near the end of the "Criteria and candidates for terrestrial reactions" paragraph.

Despite the fact that there is no reference for the second source, my feeling is that these data are more accurate than the first ones, but I am not absolutely sure.

Can anybody fix this inconsistency and add references where needed ?

Croquant 22:07, 5 July 2006 (UTC)

Where exactly do you see a discrepancy? Remember that the table in the article gives the maximum of <σv>/T^2, not the maximum of <σv> itself, and that it uses meters where the NRL Formulary uses centimeters. For D-T, for example, we have
(1.3e-17 cm^3/s)/(5 keV)^2 = 5.2e-25 m^3/s/keV^2
(1.1e-16 cm^3/s)/(10 keV)^2 = 1.1e-24 m^3/s/keV^2
(4.2e-16 cm^3/s)/(20 keV)^2 = 1.05e-24 m^3/s/keV^2
(8.7e-16 cm^3/s)/(50 keV)^2 = 3.48e-25 m^3/s/keV^2
so a maximum of 1.24e-24 m^3/s/keV^2 at 13.6 keV seems about right. --Art Carlson 07:33, 6 July 2006 (UTC)

OK. I didn't see that units and optimum temperature definition were different. What are your references for the table values ? Thanks. Croquant 10:17, 6 July 2006 (UTC)

Now that's a good question. I first wrote this stuff many years ago, back when "internet" and "reference" were never mentioned together. For that kind of info, my first address is always the NRL Formulary, but this table has more information than can be found there. I suppose I got it from a paper, but which one?. Might be in Todd Rider's thesis, but the links seem to be dead. I'll try to figure it out. Wish me luck. --Art Carlson 11:55, 6 July 2006 (UTC)

You may find Todd Rider's thesis here or there. Good luck Art ! Croquant 12:30, 6 July 2006 (UTC)

Nope. Rider doesn't show the cross sections. --Art Carlson 14:22, 6 July 2006 (UTC)

Your source might be this page or that one I just found, or a clone of them. Croquant 22:06, 7 July 2006 (UTC)


I'll be a monkey's uncle! This is my original essay that was the basis for the wiki article, though I haven't the foggiest what obscure paths it has followed. I don't know if I actually differentiated the polynomial approximations, but I could have. In any case we have a short list of general references:

  • Vol. A271 of Nucl.Instr.Meth.Phys.Research (1988--special issue)
  • Bosch&Hale,Nucl.Fusion 32(1992)611)
  • Glasstone and Lovberg, Controlled Thermonuclear Reactions (Van Nostrand, New York, 1960, Chap. 2
  • McGowan, et al., Nucl. Data Tables A6, 353 (1969), and A8, 199 (1970)
  • Miley, Towner and Ivich, Fusion Cross Section and Reactivities, Rept. COO-2218-17 (Univ. Illinois, Urbana, 1974)
  • Cox, Larry T., _Thermonuclear Reaction Bibliography with Cross Section Data for Four Advanced Reactions._, AF-TR-90-053, Edwards Air Force Base: Phillips Laboratory Technical Services Office, 1991

And we have analytical forms for the reaction cross sections as a function of temperature:

Robert Nachtrieb contributed coefficients to curve fits of the reaction rate parameter, sigma-v, for DT, DDn, DDp, DHe3, pLi6, and pB11. The fit is actually to the logarithm (base 10) of the sigma v curves. The coefficients come from:
Cox, Larry T., _Thermonuclear Reaction Bibliography with Cross Section Data for Four Advanced Reactions._, AF-TR-90-053, Edwards Air Force Base: Phillips Laboratory Technical Services Office, 1991

input: t (in keV)
output: sigma-v (in cm^3/s)
range of validity: 1 < t (keV) < 1000

x=alog10(t)
y= a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4
sv=10^y (in cm^3/s)

DT a0=-20.15761 a1=6.318869 a2=-2.421732 a3=.2708006 a4=0.0
DDn a0=-22.08780 a1=5.701349 a2=-2.082958 a3=0.2981119 a4=0.0
DDp a0=-21.97105 a1=5.643589 a2=-2.404634 a3=0.5716992 a4=-0.5730514e-1
DHe3 a0=-25.67344 a1=10.10471 a2=-3.310354 a3=0.3535589 a4=0.0
pLi6 a0=-28.12494 a1=11.87649 a2=-4.265839 a3=0.5712202 a4=0.0
pB11 a0=-33.51636 a1=17.40498 a2=-5.833501 a3=0.6879438 a4=0.0

I also have an /Annotated bibliography of p-B11 fusion. I have created a subpage for it here because it seems very useful but not necessarily in encyclopedic style (or entirely NPOV). It may be more appropriate for Aneutronic fusion. Somebody now needs to figure out the appropriate way to integrate this information into the article. (I renounce any claims to copyright.) The easiest call is referencing Cox as the source for the cross sections. --Art Carlson 09:21, 8 July 2006 (UTC)

That's a lot of references! I am not sure I can find all of them, but nevertheless thanks for your hard work. Croquant 14:22, 8 July 2006 (UTC)

Plugging the above coefficients into a spreadsheet I get the following values for the temperature at which <σv>/T² is maximized, and the value of the maximum:

fuel Topt[keV] (<σv>/T²)max[m³/s/keV²]
D-T 12.4 1.06×10-24
D-Dn 15.6 7.41×10-27
D-Dp 14.8 6.43×10-27
D-3He 47 1.89×10-26
p-6Li 68 1.46×10-27
p-11B 126 3.02×10-27

The current version of the article contains the following table:

fuel T [keV] <σv>/T² [m³/s/keV²]
D-T 13.6 1.24×10-24
D-D 15 1.28×10-26
D-3He 58 2.24×10-26
p-6Li 66 1.46×10-27
p-11B 123 3.01×10-27

I think it is appropriate to add the values for DDp and DDn to get a total value for DD. If I do this, the temperatures and reaction rates are within about 20% for the two tables, which I suppose is reasonable agreement. The excellent match for the last two reactions is probably a result of taking the values from this source to begin with. --Art Carlson 13:48, 15 September 2006 (UTC)

A more recent reference: W.M.Nevins and R.Swain, The thermonuclear fusion rate coefficient for p-11B reactions, Nuclear Fusion, Vol. 40, No. 4, pp. 865-872. If I'm reading it right, their best fit to the coefficient is about twice as high as what I calculate here. Somebody would have to read this paper more carefully and maybe more of the literature too to figure out what is going on here. --Art Carlson 17:11, 15 September 2006 (UTC)

[edit] How many protons in an alpha particle?

The Wikipedia article for alpha particle says that it "consist of two protons and two neutrons bound together". The current article suggests that an alpha particle contains four protons. Can someone clarify? Adambisset 09:45, 8 September 2006 (UTC)

Two is correct. The current article is also correct, because the "extra" protons are equivalent to neutrons plus positrons. Do you think it is necessary to burden the article with a clarification? --Art Carlson 10:07, 8 September 2006 (UTC)

[edit] SVG diagrams

Is it just me or do the two SVGs of the proton-proton chain and CNO cycle look considerably more aliased than their PNG counterparts when resized to that width? Should the PNGs be used inline purely for cosmetic reasons or the SVGs just because we can? -- Borb 13:25, 4 October 2006 (UTC)

[edit] Hydrogen Fusion

Why does a search for hydrogen fusion redirect to Nuclear_fusion? I think it should at least have a stub of its own. If someone is looking for information specifically about the nuclear fusion of hydrogen, they shouldn't have to sift through everything else in the article. I might do it myself if there are no objections and if I can figure out how to make an article. LOL! Any input would be appreciated. Thx! --Dimblethum 03:06, 19 October 2006 (UTC)

The easiest fusion reactions to do invlove hydrogen. All stars do hydrogen fusion. All other elements come from hydrogen fusion. "hydrogen fusion" is practically the definition of fusion itself. they are virtually inextricable. it doesn't need its own article. --Deglr6328 04:55, 19 October 2006 (UTC)


Hydrogen fusion is the most common, and provides the most energy per reaction...

[edit] History of Nuclear Fusion?

Should we add a History section describing the history of events relating to this?

Sharkface217 22:36, 4 November 2006 (UTC)

there is a sort of history at fusion power and timeline of fusion power ....--Deglr6328 22:17, 5 November 2006 (UTC)

[edit] Question about unit energy release

Okay, I never took particle physics in university, so forgive me, but here's what I don't get. A nuclear fusion reaction is said to be far more powerful than a fission reaction in most discussion of nuclear weapons, hence the development of the hydrogen bomb in the first place. However, the energy release from the fission of U-236 is approximately 200 MeV, whilst the release from a D-T fusion is only 17.6 MeV. The energy release per AMU is higher for fusion, but isn't it still wrong to say that a fusion reaction is more powerful than fission? Isn't it true that you can only get more energy from fusion because there are many more fusion reactions per gram of hydrogen than fission reactions per gram of uranium? I don't know that the article needs to discuss this; it's just a matter of curiosity. Sacxpert 11:08, 17 November 2006 (UTC)

I think the real difference is that fission is a chain reaction, which has a critical mass. If you try to build a bomb with many times the critical mass, it likes to blow up before you get it completely put together. Fusion is a thermal reaction, so there is no critical mass and (theoretically, at least) no limit to the size of a bomb. They stopped making them bigger around 50 MT, when the bomb would blow a hole in the atmosphere, sending most of the additional explosive power into space. --Art Carlson 11:26, 17 November 2006 (UTC)
Almost all of the energy from a thermonuclear device results from fission and not fusion. (Fusion is not a self-sustaining reaction in the same sense as fission.) Fusion is employed to produce fast neutrons, which are absorbed by, say, U-238, vastly increasing the energy release of the device compared to simple fission. -Dawn McGatney 69.139.231.9 (talk) 13:14, 1 April 2008 (UTC)

[edit] Michael Jackson comment?

I've tried to edit out the sentence in the first paragraph of the Overview section, the one that mentions Michael Jackson. However, that text doesn't appear in the Edit page. I'm not sure where else to post this or how to report it as something odd. Please could someone more knowledgeable take the necessary steps to have this comment removed.

[edit] Hydrogen Bomb image -- factual errors?

Seems to me that there might be some problems in the H-bomb picture and caption in the Neutronicity, confinement requirement, and power density subsection of the Important fusion reactions section. Here's the caption:

The only fusion reactions thus far produced by humans to achieve ignition are those which have been created in hydrogen bombs; the first of which, shot Ivy Mike, is shown here.

Fusion ignition has been achieved (by humans) in many settings besides bombs. Labs do it routinely, and the article lists a bunch of examples, in the methods to produce fusion section. I'm also not sure that the caption would be correct if "ignition" were replaced by "break-even" -- fusion has broken even in the lab before, right?

Also, didn't Ivy Mike use liquid deuterium fuel? And so shouldn't it be a thermonuclear weapon instead of a hydrogen bomb? (A tiny point relative to the other one...) Wesino 17:16, 22 November 2006 (UTC)

yes i was wondering about that caption too. can we edit this please for accuracy. --Blckavnger 18:02, 22 November 2006 (UTC)
Ignition is, in some sense, a self-sustaining reaction. That has never ever been achieved in the lab. And as far as I know there is no distinction between a thermonuclear weapon and a hydrogen bomb. --Art Carlson 21:35, 22 November 2006 (UTC)
I wrote that caption, I am fully confident of its accuracy. Ignition is the fusion plasma state where heating of the plasma is achieved by the energy deposition of the fusion products alone and not by an external source and thus capable of continuing indefinitely (until the fuel runs out or the confinement structure disassembles). This state has never been achieved anywhere by us except in an H-bomb. Also I wonder where you get the idea that there is any difference between thermonuclear bomb/h-bomb/fusion bomb/etc. they're all the same. --Deglr6328 21:42, 22 November 2006 (UTC)
Thanks for the comments. First let me respond to the ignition question. In a nutshell, if they're not wrong I'm concerned that the definition(s) given above might be either confusing or overly restrictive. If they're generally accepted in the fusion community, then I think they should be clearly stated somewhere. They are certainly not mentioned in the article (or clearly anywhere else that I can find on Wikipedia) and I think I can make a case that even a scientifically literate reader is likely to be confused without this clarification. Here's why:
  • I think I'm safe in saying that ignition, as a word in common usage, just refers to getting some kind of "combustion" started. Its conventional use says nothing about the "combustion" being able to continue or continuing indefinitely. For example, I can ignite a match which then sputters out because of a gust of wind. For the common usage of "ignition," the fact that it goes out without consuming all its fuel doesn't mean I can't say I "ignited" it initially. So there is a great potential for confusion with the definition(s) given above.
  • In the bomb example, the heating of the plasma is not achieved by fusion alone -- at least initially it all comes from the fission bit of the bomb. So it seems that the bomb should fail this definition as well. In the case of a supernova, the energy that ignites the final catastrophic fusion explosion comes from the gravitational potential energy of the gas in the star -- see the supernova#current models section, under core collapse. But surely we wouldn't say a supernova is not a fusion reaction ignition because the energy comes from gravitational infall? And just talking about bombs, how can we qualify a thermonuclear bomb as igniting if the initial heating came from an "outside source"?
  • I haven't been following progress in starting fusion in the laboratory, but poking around Wikipedia, it seems that Q=(energy released from fusion)/(energy put in) is defined in here, and here is a mention that a Japanese team achieved a Q>1, which combined with the definition seems to imply that fusion occurred in the laboratory. Also, in the article under Methods to Produce Fusion, there are definitely entries that describe fusion occurring in certain devices. If these claims are wrong, an edit is in order. And coming back to my main point, under the everyday usage of "ignition," it seems that I could say that the Japanese team or an operator of one of these devices "ignited nuclear fusion in the lab."
  • Now perhaps there is a more technical definition of "ignition" which is applied by fusion researchers. (The closest thing I can find is also on the page that defines Q, which says that "ignition corresponds to infinite Q." Confusingly, this does not seem to apply to either the supernova, or the thermonuclear bomb examples -- both have finite Q.) Then I think that this should be somewhere on the page, especially if one wants to draw fine distinctions between fusion reactions occurring and being considered "ignition," and fusion reactions occurring and not being considered "ignition."
So, given these points I think it's quite reasonable that a scientifically literate reader, though one that is not a specialist in nuclear fusion, could easily be confused by the usage of the word "ignition" in the caption (and potentially in the rest of the article as well). So if there's a narrow definition of a word in common usage that one is using, I think it really needs to be defined clearly in the article, with appropriate citations if possible.
As for the "H-bomb" nomenclature issue, as I said, it's a minor point. But since someone has said they're wondering about it, I'll elaborate. First, I did "get the idea" from anywhere, nor do I have any idea (I didn't claim I did) that there was a difference between the weapons mentioned. Certainly as they appear in common usage they all tend to refer to the same thing. However, sometimes a thing has many names, but one is clearly superior -- what I am arguing is that these other usages lack precision and are therefore are inferior. Here's why I say this:
  • If there's no hydrogen perhaps one should consider "hydrogen bomb" is a misnomer. (I know deuterium is an isotope of hydrogen -- but it does have its own distinctive name, unlike other isotopes),
  • Thermonuclear bombs fuse other elements besides isotopes of hydrogen -- eg lithium deuteride designs (Castle Bravo) which also involved lithium, so calling these "hydrogen bombs" is even more misleading,
  • A large fraction of the energy release comes from fission made possible by neutrons released from the fusion reaction -- as explained in the opening paragraphs of Nuclear weapons design. So no light elements (and certainly not hydrogen) are involved in that.
All of these problems are resolved by referring to these devices as "thermonuclear bombs" instead of "hydrogen bombs." It is analogous to the imprecision of "atom bomb" relative to the superior term "nuclear bomb." A similar point is made in the beginning nuclear weapons design article. I'll grant that calling something a "hydrogen bomb" is not terribly uncommon, but after all, shouldn't we be aspiring to precision in an encyclopedia article? Wesino 00:25, 23 November 2006 (UTC)
Well I'd take the nomenclature thing up on the thermonuclear weapons page but a few of your points are off. Just because a thermonuclear weapon "uses lithium" doesn't mean its fusing lithium. The lithium is bred into hydrogen via neutron absorption which is then fused. Its still only fusing hydrogen. Also, your third point about a large fraction of some fusion weapons yield coming from fission produced by neutron irradiation from the secondary is correct but when you then say "no light elements (and certainly not hydrogen) are involved in that".....where do you think those neutrons came from?? They came from fusion of hydrogen isotopes of course.
As for the ignition thing, I think you are confusing breakeven w/ignition. Breakeven is merely a state of Q=1 where the energy being released by the fusion plasma is equal to the energy being used to heat it. Breakeven has also not been achieved yet in fusion reactors, the Japanese case you are referring to is JT-60 which only uses DD fusion and HAS achieved a theoretical equivalency breakeven state with that fusion mode. Meaning if they went to DT (50-50 mix of deut. and tritium) then they WOULD have a breakeven condition on the device but since the DD rxn. doesn't go nearly as easily as DT they haven't strictly achieved breakeven, only a theoretical equivalency. This is all on the talk of the fusion power page. Now ignition is a state where not only is the plasma producing more energy (through fusion reactions) than it is taking to heat it but is creating enough energy to heat itself to the fusion state. ie. it is capable of burning indefinitely if given sufficient fuel. Your match analogy is perfectly apt. If you try to light a wet match and get a few sparks but can't light it you say you cant get it to IGNITE. But if its dry and you do get it to ignite, you have 'achieved an ignition state'. Suppose you had the match head resting on a large block of "match head material" and you lit the matchead by dragging the striking paper across it. You would reach the ignition state in that material and the burn would continue until the entire amount of fuel is consumed. It doesn't HAVE to continue forever, if you throw a bucket of water on it you will QUENCH that reaction and analogously if you breech the vacuum vessel of a fusion reactor you will let in (relatively) high-Z gas and quench the plasma due to excessive energy loss and rapid cooling. The initial heat source that creates the ignition state is irrelevant (at least wrt the definition of ignition, its very relevant if you're trying to build a power plant). I guess I suck at explaining this and maybe we should add something to the fusion gain factor article to prevent confusion on the issue.--Deglr6328 01:59, 23 November 2006 (UTC)
Thank you for your detailed explanation, Wesino. Unfortunately I don't follow you and/or agree with you. I think these terms have a clear technical definition that is generally in line with common usage:
  • fusion took place or fusion reactions occured: a measureable number of reactions, but not necessarily energetically interesting
  • breakeven: a somewhat arbitrary milestone
  • ignition: the fusion reactions deposit enough energy in the fuel to keep it hot enough that the reaction continues; in practice Q = infinity can't be reached, but Q = 100 or possibly even Q = 20 (5% power from external sources) is close enough to use the term "ignition"
On the other hand, I have the eyes of an expert and maybe it can be made clearer for the average reader. Rather than going into more semantics, why don't you suggest a change and we'll talk about that. --Art Carlson 08:45, 23 November 2006 (UTC)


Glad you appreciated the detailed explanation, Art. Since it might have been difficult to follow, I'll reiterate my point more concisely: I think readers are likely to be confused by the usage of "ignition" in the article. This is because 1. it is a word in everyday use, and 2. the reader is never told that "ignition" is a word with a special definition in the context of fusion research, and is being used specifically in that sense throughout the article. I was trying to explain with the bullets how a reader carrying over the everyday definition of the word would find the article confusing.
I that this is something that is likely to cause confusion for the average reader, for the scientifically literate reader, and even for readers that are experts in other subfields of physics -- though as you say, perhaps not for experts in fusion specifically. Since the "intro" article on nuclear fusion is where all of these people will turn if they want to learn more about this topic, this seems like a potential problem.
I agree that arguing more about semantics will be counterproductive. I actually did suggest a solution (define the terms in the article!) in my previous post, but maybe it was concealed in all the other text... Anyway, I think the three bullets defining the terms from your last post are exactly what the article needs. It seems to me that if these important terms have "clear technical definitions," which are used in the body of the article, then the article won't suffer from having the defintions in it. A note to that effect would have saved me, for one, a lot of confusion.
More concretely, as one possibility, the bullets from your post could probably be lifted more-or-less directly into (say) the Requirements for Fusion section, perhaps with a little surrounding text explaining that "fusion researchers are careful to distinguish between several types of fusion reactions" or something along those lines. So in short, I suggest the article needs someone to combine the "eyes of an expert" and the "fingers of a Wikipedia contributor" and define the terms in the article!
...now, I was asked where I thought the neutrons came from for the final stage of a fission-fusion-fission reaction. To respond, I can only assume that the poster has heard of my work on the magical neutron fairies that live in thermonuclear bombs and wants to learn more :) I'm fully aware of where the neturons come from. My point was merely that since a signifcant fraction of the energy comes from fission, perhaps calling it a "hydrogen bomb" is a misnomer. Nonetheless, I can grant that this nomenclature does seem to be a common usage: so despite how strong I think the arguments against it are, one could view it as POV pushing at worst, or tilting at windmills at best, to try and change other people's usage of it.
That all being said, I agree that the fusion gain factor article would be improved by a little elaboration on those points regarding Q. Wesino 10:32, 23 November 2006 (UTC)
Why not link ignition in this page to the fusion energy gain factor article and work on that article. It DOES have a 'this page is confusing' tag. --Deglr6328 10:41, 23 November 2006 (UTC)
Sounds like a good idea. Not sure I'm the one to do it though -- obviously I'm not up on the proper use of the terms and I wouldn't know where to link to citations. By the way, I was poking around to see if there was some policy page on these "jargon" issues, and found it -- Wikipedia:Explain_jargon. Cheers, Wesino 11:48, 23 November 2006 (UTC)
I am a scientist and i was confused by the definition of ignition, thus my beginning comment. I did not know there was a Q limit. While fusion isnt my research area i do study some plasma physics and had not come accross the terminology for that. Are there references for this categorization? i was just wondering for my own knowledge (is this use in theory or by applied science/engineering?). Also maybe we should at least add a lil blurb just to clarify the word ignition when it first used. --Blckavnger 17:17, 23 November 2006 (UTC)
Just noticed the addition of an "ignition" def. from a few days ago and a link from the word in the picture caption. (Apparently I only noticed the vandalism on my watchlist but not the actual additions to the article). That would have certainly made things clearer for me on my first read through... Looks good. Wesino 16:45, 27 November 2006 (UTC)

[edit] Vandalism

Ok, I read over the article, and someone named JuanDC added some things, such as "...possible, unless with juanDC's unhumanlike powers."

Can anyone please remove this vandal?

[edit] Physics or chemistry

I don't think "chemistry" or "nuclear chemistry" applies to nuclear fusion and reference to these disciplines should be removed from the introduction of this article. Comments? ClaudeSB 20:31, 3 December 2006 (UTC)

Already removed it once. Took it out again.--Deglr6328 21:08, 3 December 2006 (UTC)

[edit] Gamma radiation

For completeness sake, could the respective average fusion energy release of gamma radiation as well as that of neutrino radiation be added to the fusion reactions?

Cheers! Dio1982 13:10, 2 April 2007 (UTC)

Are you referring to the "neutronicity" in the first table in Nuclear fusion#Neutronicity, confinement requirement, and power density? First, that is neutron radiation, not neutrino radiation. Second, the gamma radiation, while it may be important for some special considerations, is not energetically important. For reactions that are fast enough to be of interest for fusion power, there are two (or sometimes three) fusion products that carry the reaction energy away as kinetic energy. Reactions that need to wait for the emission of a gamma are too slow. --Art Carlson 14:14, 2 April 2007 (UTC)
Sorry, I am coming from a nuclear fission background, and as far as I am aware of the physics (I am just a mech. eng., not a high energy physicist) every nuclear reaction also emits gammas, and many reactions also emit neutrinos. Yes, I am aware that the energy contained in this sort of radiation is of no concern for power generation, but the gammas may be of importance for things such as health and safety and engineering issues. I am just curious and asking for completeness sake... --Dio1982 15:40, 29 May 2007 (UTC)
I'm pretty sure that gamma radiation from a fusion power plant is a minor concern - compared to the neutrons. Not only is a much larger proportion of the power released as neutrons, neutrons also have a much higher biological effectiveness. Furthermore, I suspect that more gammas will be produced by interactions of the neutrons with the walls than directly in the fusion reactions, making gamma levels highly dependent on engineering details. I agree that reactor designers should worry about gammas anyway, and nuclear physicists may have their own reasons to be interested in the numbers. The bottom line is that I don't know what the numbers are, so I can neither tell you nor put them in the article. Sorry, too. --Art Carlson 11:09, 31 May 2007 (UTC)

[edit] Introduction

The introduction to this article doesn't seem to know what it wants to talk about. It starts off ok talking about the general principles of nuclear fusion, but then goes off on a tangent about nuclear fissions and it's use in weapons and then talks about the a Uranium-fueled bomb. I think we should cut out the parts about fission and Uranium so it becomes:

In physics and nuclear chemistry, nuclear fusion is the process by which multiple atomic particles join together to form a heavier nucleus. It is accompanied by the release or absorption of energy. Iron and nickel nuclei have the largest binding energies per nucleon of all nuclei and therefore are the most stable. The fusion of two nuclei lighter than iron or nickel generally releases energy while the fusion of nuclei heavier than iron or nickel absorbs energy; vice-versa for the reverse process, nuclear fission.

Although perhaps it should be a bit longer? zzymyn 09:50, 8 June 2007 (UTC)


Agree. The addition about fission ( 22:09, 31 May 2007 66.177.55.17 (I am studying this in class and thought it would be good to add.)) does not fit. Especially in Introduction section. The nuclear fission is already wikilink. I delete the addition. Kopovoi 11:01, 2 July 2007 (UTC)


Talking briefly about fission in the intro may be good, to help put fusion in a broader context. Many non-scientists get the two confused. But, use your judgement. Dstrozzi 16:16, 5 July 2007 (UTC)

Yes, I am not against mentioning the nuclear fission or fusion power in the intro, as it is now. But for more explanation you can just click on the link.Kopovoi 15:41, 17 July 2007 (UTC)

I would keep the intro as short as possible, to let people quickly know what fusion is. I will delete also the new addition about fusion power from the intro. Kopovoi 15:41, 17 July 2007 (UTC)

There is no reason to have a really short LEAD for the sake of one. 3 and 4 paragraph LEADS are recommended in the style manual. In this case, we were short. And we only mentioned that fusion happens in stars. Well, it also happens in bombs and in semi-controlled experiments for physics and power generation. All that, plus some intro history, was already available in a nice paragraph ending the OVERVIEW. So I moved it up, and voila! A two paragraph lead that has just about everything. See what you think. SBHarris 02:04, 12 November 2007 (UTC)

[edit] Coulomb barrier

How does one arrive at the cited figure:

"Using deuterium-tritium fuel, the resulting energy barrier is about 0.01 MeV".

Asuming that the distance between nuclei should not exceed the range of the strong nuclear force (~10-15 m), the nuclei will have to "nearly touch". Inserting a minimum separation of twice the radius of a nucleus (~10-14 m) to the Coulomb potential formula E = -k_C \cdot \frac{e^2}{r} gives af Coulomb barrier of 0.072 MeV?? —Preceding unsigned comment added by 217.116.235.64 (talk) 14:38, 11 November 2007 (UTC)

[edit] Fusion torch

I have come across the article Fusion torch, which seems to be a 1960s idea for recovering elements using fusion power. The article itself strikes me, a non-scientist, as being rather insubstantial, and probably out of date. Is it worth while for someone who knows the subject to take a quick look at it? Patche99z (talk) 17:24, 14 December 2007 (UTC)

There seems to have been a concept around of a device using a plasma-jet to separate materials by differences in atomic weight, mainly for nuclear fuel reprocessing. Not a nuclear fusion application as such, though. To me this Fusion torch page looks extremely questionable science. Its content is mirrored on a number of other sites, some of a rather pseudoscientific style. Hmmm :-/ Inclined to agree. --Anteaus (talk) 23:08, 24 January 2008 (UTC)

[edit] Self-Redirect

The linked word "Thermonucelear" in the "Overview" section directs me back to this article. I clicked it trying to find out what temperatures count as thermonuclear. Would someone please point it to something that answers the question? --Brilliand (talk) —Preceding comment was added at 19:49, 26 January 2008 (UTC)

[edit] Assessment

This article is rated as an A-class article. I think that it is worth to be nominated to be FAC, if it will be better referenced. However, without additional references it should be downgraded to the B-class. Please help with adding missing references.Beagel (talk) 07:14, 1 March 2008 (UTC)

[edit] Reaction equation in introductory section

I am going to delete the reaction equation in introductory section. The picture subtitle should be enough. There is also list of reactions in "Important reactions" section. Kopovoi (talk) 13:44, 10 March 2008 (UTC)

The reaction is:\,^{2}_{1}\mathrm{H} + \,^{3}_{1}\mathrm{H} \, \to \,^{4}_{2}{He} + \,^{1}_{0}\mathrm{n} \; Done. Kopovoi (talk) 13:26, 12 March 2008 (UTC)

[edit] Important reactions

Why do some of the reactions have energies broken out by particle, and others only have the net energy? It makes for a rather confusing read. --Belg4mit (talk) 02:03, 15 April 2008 (UTC)

Generally speaking, if there are only two reaction products, conservation of momentum can be used to figure out how the energy splits up. There are, however, several entries in the table which give a lump energy even though there are only two products. With three products, this is usually not possible, but in the table reaction 6iii does give such an energy breakdown for three products. I suspect that is a result of the particular reaction chain, i.e., He-3 + T → He-5 + p and then He-5 decays to He-4 plus n. In any case, you're right that the table is not entirely consistent. --Art Carlson (talk) 06:50, 15 April 2008 (UTC)

[edit] Philo T. Farnsworth's Fusor...where is it in this article?

Why is there no mention of Philo T. Farnsworth's Fusor in this article? This is a legitimate device that has shown the possibility of negatively charged electrostatic confinement of protons to overcome their electrostatic repulsion and create fusion. There is a great wealth of information out there on this but this article mentions none of it. It is scientifically verified and repeatable, using the same technology as a CRT television with a magnetic containment field, in a vaccuum, creating a negatively charged electrostatic "well" consisting of a "cloud" of free, yet highly confined electrons which attracts the protons etc. and voila! fusion. I know this is not in the mainstream of current science, but was one of the earliest provable experiments on fusion. To leave it out of this article is a disservice. 71.35.224.232 (talk) 16:02, 22 May 2008 (UTC)

Look again. The fusor is mentioned just as often as the tokamak (despite its relative unimportance). Once under Nuclear fusion#Inertial confinement (which may not be entirely appropriate), once under Nuclear fusion#Generally cold, locally hot fusion, and once at the bottom in the template on "Nuclear fusion methods". --Art Carlson (talk) 16:22, 22 May 2008 (UTC)