Talk:Nuclear fusion/Archive 2

Archive 1

Archive 2

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)

This does happen, but isn't the reason. Consider someone asking how a flashlight battery works. As with all cases, as the stored energy decreases, the mass decreases, but that doesn't explain the chemical reactions that cause the mass change. In both the battery and fusion reaction, the products have a lower (free) energy, making energy available for our use. Gah4 (talk) 09:04, 5 September 2017 (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)

Fission products have lower potential (nuclear) energy than the input U235. They normally have more neutrons than the more stable nuclide with that number of nucleons. Some neutrons are released in fission, but not enough for the products to be most stable. Gah4 (talk) 09:04, 5 September 2017 (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)

Let me ask the question a different way: at the quantum level, why is fusion exothermic. It's easy to see this in fusion: you start out with a bunch of nucleons held together by strong force against electrostatic repulsion of protons. Add an extra neutron, and the nucleus is too big and splits. This releases some of the potential energy (from the repulsion). In additon, some of the gluons that were holding the nucleus together are released, that's a fair amount of mass available to convert to energy.

But to hold two or more protons together in a nucleus, you need to add gluons. That should require more energy to "generate" them. What mass is given up to make this happen? (Actually, I suspect it's more accurate to say that virtual gluons, spontaneously created and destroyed all the time, become actual when you put together a larger nucleus. Either way, you need energy to account for them. But then, my understanding of quantum chromodynamics is highly deficient. Even my knowledge of quantum electrodynamics is based on Feynman's popularization, Q.E.D.) Bgoldnyxnet (talk) 18:18, 25 November 2008 (UTC)

A large part of the "mass" of a proton already "is" energy from the beginning: The sum of the masses of the three quarks of which the proton consists only make up about 10% of the total mass of a proton, the rest is "field energy" from the strong force's field. So you have a large pool of energy to draw from in fusions. It's similar to an asteroid falling on Earth, which also produces lots of energy without destroying any particles from either the asteroid or the Earth: there the energy comes from the loss of gravitational energy of the asteroid in Earth's gravitational field. Hope this helps.--Roentgenium111 (talk) 16:25, 22 April 2009 (UTC)

Similar to the stability of noble gases, with their full electron shells, the alpha particle (He4 nucleus) has full nuclear shells. The binding is especially strong for alpha. There can be two (opposite spin) protons in the first proton shell, and two neutrons in the first neutron shell. The best model for alpha decay of heavy nuclei is that alpha particles are moving around at high speed (the Fermi velocity) inside, and eventually tunnel out. They tunnel as a bound state, similar to Cooper pairs in superconductors tunneling through a Josephson junction. Gah4 (talk) 09:04, 5 September 2017 (UTC)

BTW: I suppose you mean "fission" in your 2nd sentence? --Roentgenium111 (talk) 16:25, 22 April 2009 (UTC)

The title is a sentence (or something like one). Nuclear fusion. Midgley (talk) 16:54, 22 April 2009 (UTC)

For quantum mechanical reason, the alpha particle (Helium nucleus) is especially stable, for similar reasons to the chemical stability (inertness) of Helium. Two protons (with opposite spin) and two neutrons (also with opposite spin) pack very tightly, minimizing the quantum exchange energy. They are so tightly bound, that larger nuclei can be considered as having alpha particles moving around (at the fermi velocity), and alpha decay occurs when one tunnels out of the nucleus. Fundamentally, it is the result of the quantum mechanical exchange interaction (sometimes called exchange force). There is no classical analog, and no easy way to explain it. It is related to indistinguishability, which also has no classical analog. Gah4 (talk) 23:35, 15 March 2019 (UTC)

Nuclear diameter

Just a minor point. The article states "208 nucleons, corresponding to a diameter of about 6 nucleons" in the 4th paragraph of the requirements section. I may be completely on the wrong track here, but it seems to me that if a nucleus is roughly spherical (which I am pretty sure is true, and if not it's pointless to talk about its diameter) then you should be able to get an approximation of the total number of nucleons with 4π/3 * r^3.

If you have a diameter of 6 nucleons (thus r = 3), that gives roughly 38 nucleons total. Which seems to disagree with the article.

If you take the diameter as 7 nucleons (r = 3.5), then you end up with more like 180, which is significantly closer to 208

Might be splitting hairs, but it was just something I noticed.

BackStabbath (talk) 01:14, 3 February 2010 (UTC)

If you want the ratio of the radii (or diameters) of two spheres, one of which is 208 times the volume of other, the ratio is 208^(1/3) = 5.92. The sphere with 208 times the volume will have almost 6 times the linear dimensions of the other. The constants in front drop out of these ratios. That factor is what you're adding, but shouldn't. SBHarris 01:56, 3 February 2010 (UT

If you furthermore want the ratio of diameters of a sphere packing process you have to increase the spherical volume size to accommodate the unfilled volume by about 35% and thus the spherical diameter by about 10% which in this case argues in favor of a 7 diameter multiple. But if you want to take into consideration the proposed volumetric independence of each sphere for spin compatibility purposes you also need to conceptualize a larger volume. Also, if you're willing to consider other accumulation configurations, such as cylindrical configurations as a means of accumulation consideration, you might consider the accumulation configurations shown in my nuclear model image in Talk:nuclear model. You might note that in his book "General Chemistry" Dr Linus Pauling showed images of spherical packing atomic models as "hypothetical models" with a degree of skepticism.WFPM (talk) 18:14, 13 March 2010 (UTC)

No. It is usual to picture a nucleus as a ball of protons and neutrons packed together. Note that Pauli exclusion applies to protons and neutrons separately, such that two (opposite spin) protons and two (opposite spin) neutrons can pack as one. They are still quantum objects, and so somewhat fuzzy, but not as fuzzy as electrons in atomic shells. The actual requirement is to fill up shells satisfying Pauli exclusion, with additional nucleons having higher and higher Fermi energy, and so Fermi momentum, which, with Heisenberg uncertainty, determines the size. Consider also the sizes of atoms as atomic number increases. Gah4 (talk) 09:28, 5 September 2017 (UTC)

It is common to describe nuclei, not by diameter but by area, called cross section. If you shoot a neutron, for example, at a nucleus, the cross section is the area that describes the chance of hitting (interacting) with the nucleus, as it might for a neutron to go through a hole with that area. Because of the large de Broglie wavelength for slow neutrons, use the fast neutron cross section. The unit barn is commonly used for nuclei cross section. Gah4 (talk) 00:28, 16 March 2019 (UTC)

Greenhouse Item

Is it appropriate to mention Greenhouse Item in the opening?

It seems to me we should be discussing self-sustaining fusion there, and Greenhouse Item wasn't. It was the fusion of a few grams of D-T driven by several kilograms of fissile material, not self-sustaining fusion. Boosting isn't used to provide extra energy to a nuclear weapon, it's used to provide additional neutrons. I'm not sure it's appropriate to include it. Staged thermonuclear weapons? Sure, mention it because the secondary stages clearly output more energy than goes into them, but I'm not so sure about boosted weapons. Kylesenior (talk) 07:15, 28 May 2020 (UTC)

break-even, boosterism

the overview claims that tokamaks have demonstrated break-even. I don't recall reading this (I didn't think any reactor designs had yet actually accomplished break-even) and a quick googling turns up lots of discussions about potential break-even designs but no break-even.

the overview also makes it sound like ITER is going to be a working model for generating energy, that generating energy from fusion is right around the corner. my understanding is that neither of these things are actually considered to be true except by those who are financially involved in ITER.

it could use a little sourcing. especially given how many contrary opinions there are about whether we should expect fusion power anytime soon. --98.217.14.211 (talk) 01:56, 5 April 2009 (UTC)

Yep. Controlled fusion for commercial power production has been "just 20 years away" for the last 50 years. I'm beginning to wonder if it won't be just 20 years away for the next 50 years, too. ;). SBHarris 01:59, 5 April 2009 (UTC)

It is the energy of the future! Meaning it will always be a future date when we will learn to harness it.72.12.199.20 (talk) 01:34, 17 December 2009 (UTC)

Tokamaks have not demonstrated break-even. ITER is designed to do so. ITER is not designed to be a 'commercial-style' demonstration plant, that is planned for the next generation reactor, DEMO. http://www.iter.org/sci/beyonditer 2001:630:12:10D0:2CB4:9415:E385:C14D (talk) 15:19, 19 July 2013 (UTC)

I haven't followed this for a while, but there are different definitions of breakeven. One counts only the energy in the fusion reactor, but the energy needed to confine the reaction. (That is, the magnets for Tokamaks.) If you use superconducting magnets (low electrical power input, but there is need for coolant). There is also energy needed to make Tritium, which needs to be considered for a production power plant. Gah4 (talk) 09:11, 5 September 2017 (UTC)

There is Fusion_energy_gain_factor which has a good explanation for the different levels of breakeven, such as scientific, engineering, and economic. I suspect it is better there, and to link from here. Gah4 (talk) 03:22, 25 June 2020 (UTC)

Neutrinos??

Why is the mention of the neutrinos produced almost entirely missing (except in the CNO cycle section) and neutrinos are missing from all reaction equations? --vuo (talk) 11:32, 3 August 2020 (UTC)

The "Human body produces more heat than the sun"-falacy

This infamous falacy is at work in the following 2 sentences:

"At the temperatures and densities in stellar cores, the rates of fusion reactions are notoriously slow. For example, at solar core temperature (T ≈ 15 MK) and density (160 g/cm3), the energy release rate is only 276 μW/cm3—about a quarter of the volumetric rate at which a resting human body generates heat."

The number 276 μW/cm^3 results from dividing the total solar power output (~4*10^26 W) by the total volume of the sun (~4/3π*696,000,000 m^3) and is fairly irrelevant, yet fun to know. But in this case the message is really quite wrong because a point is made regarding the energy release rate at core-temperature (15 MK) and -density (150 t/m^3). So the total energy certainly needs to be divided by the volume in which fusion occurs at all, i.e. the core. Taking the number of 24% of the solar radius, found with a quick search on Wikipedia itself, one arrives at ${\frac {3.8*10^{26}W}{4/3\pi \cdot 0.24\cdot 696,000,000m^{3}}}\approx 19.4mW/cm^{3}(=19.4kW/m^{3}$in SI units) which is about 70x the number currently stated. Can I correct this number? This is obviously original research and I don't have a source for it, but since it is so simple to verify, is a source really necessary? --Felix Tritschler (talk) 12:37, 8 March 2022 (UTC)

Possible merge and redirect

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.

To merge Thermonuclear fusion into a distinct section in Nuclear fusion. Klbrain (talk) 10:06, 2 April 2023 (UTC)

---

This article seems to be duplicated by thermonuclear fusion, which should I think be merged into this one.

They seem to be exactly the same topic, and this article name is the more concise.

Discuss here I think. I have placed a heads-up at Talk:Thermonuclear fusion#Possible merge and redirect. Andrewa (talk) 13:37, 1 May 2021 (UTC)

i think that they should not as in the future they might have a breakthrough in technology might sepreate them both 82.113.249.177 (talk) 17:42, 26 January 2023 (UTC)

Merge. "Thermonuclear fusion" is the one driven by high temperature, so it has a bit narrower scope. However, only the Nuclear fusion's subsections Beam…, Muon…, and Other principles fall outside of that scope, so it does not make sense to duplicate all the rest. Petr Matas 18:26, 18 June 2022 (UTC)

I agree with your point, but only if thermonuclear fusion becomes its own subsection, with other content that has already been mentioned in nuclear fusion being placed to appropriate sections. Nuclear fusion does not only figurate thermonuclear fusion, and other types of fusion, such as pycnonuclear, should be distinguished too (and merged too). The article also seems to disregard the various different types that authors have incorporated in their calculations of nuclear fusion rates - [1], [2] (thermonuclear w/ weak and strong screening, thermopycnonuclear, pycnonuclear), as examples, which further supports my point that distinction is still required. Mk0uQ (talk) 18:24, 9 August 2022 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

  Y Merger complete. Klbrain (talk) 11:28, 2 April 2023 (UTC)

Reaction rate - Units do not add up

Nuclear fusion#Requirements:

The reaction rate (fusions per volume per time) is ⟨σv⟩ times the product of the reactant number densities

Unit is given, in the diagram, as m³/s for the reaction rate. This does not add up. number density is mol/m³, σ (cross section, area) is m² and v is velocity is m/s. In total we get: mol/m³ * mol/m³ * m² * m/s = mol²/(m³*s). We have mol² left over to get to m³/s, missing somewhere, potentially in the cross section?

But it does not end here, this paper[1] states:

The volumetric reaction rate, that is, the number of reactions per unit time and per unit volume

Per time per volume, that would be 1/(s*m³) for the reaction rate. Not m³/s as per wiki. It makes more sense this way, since we want to multiply this by some reactive plasma volume and some amount of time to get the conversion. The formula given is (ignoring the integers, since they have no units):

R = n*n*⟨σv⟩

As we already established, that adds up to mol²/(m³*s), with factors on the right missing the mole(s) (or they need to be included on the left).

1. We have some missing units

2. Reaction rate has a different unit, either mol²/(m³*s) or 1/(m³*s), but not m³/s

I come from the chemical side of things, I am not going to edit this without feedback on how Physicists handle units and what they think is missing. Eheran (talk) 13:17, 17 May 2023 (UTC)