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Subject: Conventional Fusion FAQ Section 1/11 (Fusion Physics)
This article was archived around: 14 Nov 1999 10:27:51 GMT
Disclaimer: While this section is still evolving, it should
be useful to many people, and I encourage you to distribute
it to anyone who might be interested (and willing to help!!!).
1. Fusion as a Physical Phenomenon
Last Revised August 7, 1994
Written by Robert F. Heeter, email@example.com, unless
### Please let me know if anything here is unclear. ###
*** A. What is fusion?
"Fusion" means many things when discussed on the newsgroup.
Technically, "fusion" is short for "Nuclear Fusion," which describes
the class of reactions where two light nuclei fuse together, forming
a heavier nucleus. This heavier nucleus is frequently unstable, and
sometimes splits (fissions) into two or more fragments. "Fusion"
also refers to the type of energy produced, and a "fusion reactor"
describes an energy-producing facility which generates power via
fusion reactors. Finally, "fusion" can also be used to refer to
the scientific program aimed at harnessing fusion for clean,
safe, and hopefully inexpensive energy production - a collaborative
international program which has been carried on for the past 40-some
years. Each of these three uses - the technical, the energy
source, and the scientific research program - is discussed in
a separate section of this FAQ. The technical aspects of
fusion are discussed below in this section.
*** B. How does fusion release energy?
If you add up the masses of the particles which go into a fusion
reaction, and you add up the masses of the particles which come out,
there is frequently a difference. According to Einstein's famous
law relating energy and mass, E=mc^2, the "mass difference" can
take the form of energy. Fusion reactions involving nuclei lighter
than iron typically release energy, but fusion reactions involving
nuclei heavier than iron typically absorb energy. The amount of
energy released depends on the specifics of the reaction; a table
of reactions is given further below to give an idea of the variety
of fusion reactions.
Another way to look at this is to consider the "binding energy"
of the elements in question. If the reactants are bound more
weakly than the products, then energy is released in the reaction.
"Binding energy" is the amount of energy you would have to put
into a system in order to pull its components apart; conversely,
in a system with high binding energy, a lot of energy is released
as the components are allowed to bond together. Suppose you
had two balls connected by a long, thin rubber band, so that they
are not very tightly connected, and the rubber band can be broken
easily. This is a system with low binding energy. Now here's an
analogy to what happens in fusion: imagine the long, thin
rubber band suddenly being replaced by a short, thick one. The
short thick one has to be stretched a lot in order to connect
to the two balls, but it wants to bind them more tightly, so it
pulls them together, and energy is released as they move towards
each other. The low-binding energy, long rubber band system
has been replaced by a high-binding energy, short rubber band
system, and energy is released.
*** C. Where does fusion occur in nature?
The conditions needed to induce fusion reactions are extreme;
so extreme that virtually all natural fusion occurs in stars,
where gravity compresses the gas, until temperature and pressure
forces balance the gravitational compression. If there is enough
material in the star, pressures and temperatures will grow
large enough as the star contracts that fusion will begin to occur
(see below for the explanation why); the energy released will then
sustain the star's temperature against losses from sunlight being
radiated away. The minimum mass needed to induce fusion is roughly
one-tenth the sun's mass; this is why the sun is a star, but
Jupiter is merely a (large) planet. (Jupiter is about 1/1000th
the sun's mass, so if it were roughly 100 times bigger, it
too would generate fusion and be a small, dim star.)
Stellar fusion reactions gradually convert hydrogen into helium.
When a star runs out of hydrogen fuel, it either stops burning
(becoming a dwarf star) or, if it is large enough (so that gravity
compresses the helium strongly) it begins burning the helium into
heavier elements. Because fusion reactions cease to release
energy once elements heavier than iron are involved, the larger
stars also eventually run out of fuel, but this time they
collapse in a supernova. Gravity, no longer opposed by the internal
pressure of fusion-heated gases, crushes the core of the star,
forming things like white dwarfs, neutron stars, and black holes
(the bigger the star, the more extreme the result). (For more
details, try the sci.astro or sci.space.science newsgroups.)
*** D. Why doesn't fusion occur anywhere else in nature?
Current scientific knowledge indicates that very little fusion
occurs anywhere else in nature. The reason is because in order
to get two nuclei to fuse, you first have to get them close together.
(This is because the nuclear forces involved in fusion only act
at short range.) However, because the two nuclei are both positively
charged, they repel each other electrically. Nuclei will not fuse
unless either (a) they collide with enough energy to overcome the
electrical repulsion, or (b) they find a "sneaky" way to circumvent
their repulsion (see muon-catalyzed fusion in section 4). The
energy required for fusion is so high that fusion only occurs in
appreciable amounts once the temperature gets over 10 million
degrees Kelvin, so (a) doesn't happen anywhere outside of stars.
Current knowledge suggests that the sort of processes that would
allow sneaky-fusion as in (b) are very rare, so there just isn't
much fusion in the everyday world.
*** E. What are the basic fusion reactions?
While it is possible to take any two nuclei and get them to fuse,
it is easiest to get lighter nuclei to fuse, because they are
less highly charged, and therefore easier to squeeze together.
There are complicated quantum-mechanics rules which determine which
products you will get from a given reaction, and in what amounts
("branching ratios"). The probability that two nuclei fuse is
determined by the physics of the collsion, and a property called
the "cross section" (see glossary) which (roughly speaking)
measures the likelihood of a fusion reaction. (A simple analogy
for cross-section is to consider a blindfolded person throwing
a dart randomly towards a dartboard on a wall. The likelihood
that the dart hits the target depends on the *cross-sectional*
area of the target facing the dart-thrower. (Thanks to Rich
Schroeppel for this analogy.))
Below is an annotated list of many fusion reactions discussed
on the newsgroup. Note: D = deuterium, T = tritium, p = proton,
n = neutron; these and the other elements involved are discussed
in the glossary/FUT. (FUT = list of Frequently Used Terms; section
10 of the FAQ.) The numbers in parentheses are the energies
of the reaction products (in Millions of electron-Volts, see
glossary for details). The percentages indicate the branching
ratios. More information on each of the elements is given below.
Table I: Fusion Reactions Among Various Light Elements
D+D -> T (1.01 MeV) + p (3.02 MeV) (50%)
-> He3 (0.82 MeV) + n (2.45 MeV) (50%) <- most abundant fuel
-> He4 + about 20 MeV of gamma rays (about 0.0001%; depends
somewhat on temperature.)
(most other low-probability branches are omitted below)
D+T -> He4 (3.5 MeV) + n (14.1 MeV) <-easiest to achieve
D+He3 -> He4 (3.6 MeV) + p (14.7 MeV) <-easiest aneutronic reaction
"aneutronic" is explained below.
T+T -> He4 + 2n + 11.3 MeV
He3+T -> He4 + p + n + 12.1 MeV (51%)
-> He4 (4.8) + D (9.5) (43%)
-> He4 (0.5) + n (1.9) + p (11.9) (6%) <- via He5 decay
p+Li6 -> He4 (1.7) + He3 (2.3) <- another aneutronic reaction
p+Li7 -> 2 He4 + 17.3 MeV (20%)
-> Be7 + n -1.6 MeV (80%) <- endothermic, not good.
D+Li6 -> 2He4 + 22.4 MeV <- also aneutronic, but you
get D-D reactions too.
p+B11 -> 3 He4 + 8.7 MeV <- harder to do, but more energy than p+Li6
n+Li6 -> He4 (2.1) + T (2.7) <- this can convert n's to T's
n+Li7 -> He4 + T + n - some energy
From the list, you can see that some reactions release neutrons,
many release helium, and different reactions release different
amounts of energy (some even absorb energy, rather than releasing
it). He-4 is a common product because the nucleus of He-4 is
especially stable, so lots of energy is released in creating it.
(A chemical analogy is the burning of gasoline, which is relatively
unstable, to form water and carbon dioxide, which are more stable.
The energy liberated in this combustion is what powers automobiles.)
The reasons for the stability of He4 involve more physics than I
want to go into here.
Some of the more important fusion reactions will be described below.
These reactions are also described in Section 2 in the context of
their usefulness for energy-producing fusion reactors.
*** F. Could you tell me more about these different elements?
(Note: there's more information in the glossary too.)
Hydrogen (p): Ordinary hydrogen is everywhere, especially
Deuterium (D): A heavy isotope of hydrogen (has a neutron in
addition to the proton). Occurs naturally at
1 part in 6000; i.e. for every 6000 ordinary
hydrogen atoms in water, etc., there's one D.
Tritium (T): Tritium is another isotope of hydrogen, with two
neutrons and a proton. T is unstable
(radioactive), and decays into Helium-3 with a
half-life of 12.3 years. (Half the T decays
every 12.3 years.) Because of its short
half-life, tritium is almost never found in
nature (natural T is mostly a consequence
of cosmic-ray bombardment). Supplies have been
manufactured using fission reactors; world
tritium reserves are estimated at a few
kilograms, I believe. Tritium can be made by
exposing deuterium or lithium to neutrons.
Helium-3 (He3): Rare light isotope of helium; two protons and a
neutron. Stable. There's roughly 13 He-3 atoms
per 10 million He-4 atoms. He-3 is relatively
abundant on the surface of the moon; this is
believed to be due to particles streaming onto
the moon from the solar wind. He3 can also be
made from decaying tritium.
Helium-4 (He4): Common isotope of helium. Trace component of the
atmosphere (about 1 part per million?); also
found as a component of "natural gas" in gas
Lithium-6 (Li6): Less common isotope of lithium. 3 protons, 3
neutrons. There are 8 Li-6 atoms for every 100
Li-7 atoms. Widely distributed in minerals and
seawater. Very active chemically.
Lithium-7 (Li7): Common isotope of lithium. 3 protons, 4 neutrons.
See above info on abundance.
Boron (B): Common form is B-11 (80%). B-10 20%.
5 protons, 6 neutrons. Also abundant on earth.
Note: Separating isotopes of light elements by mass is not
*** G. Why is the deuterium-tritium (D-T) reaction the easiest?
Basically speaking, the extra neutrons on the D and T nuclei make
them "larger" and less tightly bound, and the result is
that the cross-section for the D-T reaction is the largest.
Also, because they are only singly-charged hydrogen isotopes,
the electrical repulsion between them is relatively small.
So it is relatively easy to throw them at each other, and it
is relatively easy to get them to collide and stick.
Furthermore, the D-T reaction has a relatively high energy yield.
However, the D-T reaction has the disadvantage that it releases
an energetic neutron. Neutrons can be difficult to handle,
because they will "stick" to other nuclei, causing them to
(frequently) become radioactive, or causing new reactions.
Neutron-management is therefore a big problem with the
D-T fuel cycle. (While there is disagreement, most fusion
scientists will take the neutron problem and the D-T fuel,
because it is very difficult just to get D-T reactions to go.)
Another difficulty with the D-T reaction is that the tritium
is (weakly) radioactive, with a half-life of 12.3 years, so
that tritium does not occur naturally. Getting the tritium
for the D-T reaction is therefore another problem.
Fortunately you can kill two birds with one stone, and solve
both the neutron problem and the tritium-supply problem at
the same time, by using the neutron generated in the D-T
fusion in a reaction like n + Li6 -> He4 + T + 4.8 MeV.
This absorbs the neutron, and generates another tritium,
so that you can have basically a D-Li6 fuel cycle, with
the T and n as intermediates. Fusing D and T, and then
using the n to split the Li6, is easier than simply trying
to fuse the D and the Li6, but releases the same amount of
energy. And unlike tritium, there is a lot of lithium
available, particularly dissolved in ocean water.
Unfortunately you can't get every single neutron to stick
to a lithium nucleus, because some neutrons stick to other
things in your reactor. You can still generate as much
T as you use, by using "neutron multipliers" such as
Beryllium, or by getting reactions like
n + Li7 -> He4 + T + n (which propagates the neutron)
to occur. The neutrons that are lost are still a problem,
because they can induce radioactivity in materials that
absorb them. This topic is discussed more in Section 2.
*** H. What is aneutronic fusion?
Some researchers feel the advantages of neutron-free fusion
reactions offset the added difficulties involved in getting
these reactions to occur, and have coined the term
"aneutronic fusion" to describe these reactions.
The best simple answer I've seen so far is this one:
(I've done some proofreading and modified the notation a bit.)
[ Clarifying notes by rfheeter are enclosed in brackets like this.]
>From: firstname.lastname@example.org (John W. Cobb)
>Risto Kaivola <email@example.com> wrote:
[[ Sorry I don't have the date or full reference for this anymore;
this article appeared in sci.physics.fusion a few months ago.]]
>>Basically, what is aneutronic fusion? The term aneutronic
>>confuses me considerably. Could you give me an example of
>>an aneutronic fusion reaction? How could energy be produced
>>using such a reaction? Can there be a fusion reaction in which
>>a neutron is never emitted?
>D + He3 --> He4 + p + 18.1MeV
>(deuteron + helium-3 --> helium-4 + proton + energy)
>p + Li6 --> He4 + He3 + 4.0MeV
>(proton + lithium-6 --> helium-4 + helium-3 + energy)
>D + Li6 --> 2 He4 + 22.4MeV
>(deuteron + lithium-6 --> 2 helium-4's + energy)
>p + B11 --> 3 He4 + 8.7Mev
>(proton + boron-11 --> 3 helium-4's + energy)
>All of these reactions produce no neutrons directly.
[[ Hence "aneutronic." ]]
>There are also other reactions that have multiple branches possible,
>some of which do not produce neutrons and others that do
>(e.g., D + D, p + Li7).
>The question is how do you get a "reactor" going and not get
>any neutrons. There are 2 hurdles here. The first is getting the
>fuel to smack together hard enough and often enough for fusion
>The easiest fusion reaction is D + T --> He4 + n (the D-T fuel
>cycle). A magnetic reactor can initiate fusion in one of these
>things at about a temperature of 10keV.
[1 keV = 1000 eV = 11,000,000 (degrees) kelvin, more or less].
>The other reactions require much higher temperatures (for example
>about 50KeV for the D+He3 reaction). This is a big factor of 5.
>The second hurdle is neutron production via "trash" (secondary)
>reactions. That is, the main reaction may be neutron-free,
>but there will be pollution reactions that may emit neutrons.
[ The products of the main reaction, e.g. He3, can be trapped in
your reactor temporarily, and fuse with other ions in the system
in messy ways. ]
>Even if this is only a few percent, it can lead to big neutron
>emission. For example, the D+He3 reaction will also have some D+D
[ Because in your reactor you will have a lot of Ds and He3s, and
the Ds will collide with each other as well as with the He3s. ]
>At 50Kev temperatures, the reaction
>cross-section for D+D reactions is about 1/2 of the D+He3
>cross-section, so there will be some generation of neutrons from
>the 50% branch reaction of D + D-->He3 + n.
>Also, the other 50% goes to T+p, The triton (T) will then undergo
>a D-T reaction and release another neutron.
[ Because the cross-section for D-T reactions is much higher.]
>If the reactor is optmized (run in a He3 rich mode) the number
>of neutrons can be minimized. The neutron power can be as low
>as about 5% of the total. However, in a 1000 megawatt reactor,
>5% is 50 MW of neutron power. That is [still] a lot of neutron
>irradiation. This lower neutron level helps in designing
>structural elements to withstand neutron bombardment, but it
>still has radiation consequences.
>On the other hand, it is my understanding that the p-B11 reaction
>is completely neutron free, but of course it is much harder
*** I. What sort of fusion reactor is the sun?
Fortunately for life on earth, the sun is an aneutronic fusion
reactor, and we are not continually bombarded by fusion neutrons.
Unfortunately, the aneutronic process which the sun uses is
extremely slow and harder to do on earth than any of the reactions
mentioned above. The sun long ago burned up the "easy" deuterium
fuel, and is now mostly ordinary hydrogen. Now hydrogen has a
mass of one (it's a single proton) and helium has a mass of four
(two protons and two neutrons), so it's not hard to imagine sticking
four hydrogens together to make a helium. There are two major
problems here: the first is getting four hydrogens to collide
simultaneously, and the second is converting two of the four protons
The sun evades the first problem, and solves the second, by using a
catalyzed cycle: rather than fuse 4 protons directly, it fuses a
proton to an atom of carbon-12, creating nitrogen-13; the N-13 emits
a neutrino and a positron (an antielectron, that is an electon with
positive instead of negative charge) and becomes carbon-13.
(Effectively, the Carbon-12 converted the proton to a
neutron + positron + neutrino, kept the neutron, and became C-13).
The C-13 eventually fuses with another proton to become N-14.
N-14 then fuses with a proton to become oxygen-15. Oxygen-15 decays
to N-15 (emitting another positron), and N-15 plus another proton
yields carbon-12 plus a helium-4 nucleus, (aka an alpha particle).
Thus 4 protons are tacked one by one onto heavier elements, two of
the protons are converted to neutrons, and the result is production
of helium and two positrons. (The positrons will undergo
matter-antimatter annihilation with two electrons, and the result
of the whole process is formation of a helium, two neutrinos, and
a bunch of gamma rays. The gamma rays get absorbed in the solar
interior and heat it up, and eventually the energy from all this
fusion gets emitted as sunlight from the surface of the sun.)
The whole process is known as the carbon cycle; it's catalyzed
because you start with carbon and still have carbon at the end.
The presence of the carbon merely makes it possible to convert
protons to helium. The process is slow because it's difficult
to fuse protons with carbon and nitrogen, and the positron-emitting
nuclear decays are also slow processes, because they're moderated
by the weak nuclear force.
*** J. Why is it so hard to create controlled man-made fusion
In order to get two nuclei to fuse, you basically have to get
them to collide energetically. It turns out that colliding two
beams of particles yields mostly scattering collisions, and few
fusion reactions. Similarly, blasting a stationary target with
a beam of energetic ions also yields too little fusion.
The upshot is that one must find some way to confine hot,
energetic particles so that they can collide many many times,
and finally collide in just the right way, so that fusion occurs.
The temperatures required are upwards of 100 million degrees
(Kelvin - it would be about 200 million Fahrenheit!). At these
temperatures, your fusion fuel will melt/evaporate any material
wall. So the big difficulties in fusion are (a) getting
the particles hot enough to fuse, and (b) confining them long
enough so that they do fuse.
*** K. What is plasma physics, and how is it related to fusion?
Plasma physics is the area of physics which studies ionized
gases and their properties. In most conventional types of fusion
(muon-catalyzed fusion being the major exception), one must heat
the fusion fuel to extremely high temperatures. At these
temperatures, the fuel atoms collide so much and so hard that
many electrons are knocked loose from their atoms. The result
is a soup of ionized atoms and free electrons: a plasma.
In order to achieve the conditions required for controlled
fusion, an understanding of how plasmas behave (and particularly
how to confine and heat them) is often essential.
*** L. Just how hot and confined do these plasmas need to be?
(Or, what conditions are needed for controlled fusion?)
Basically, the hotter your plasma, the more fusion you will have,
because the more ions will be flying around fast enough to stick
together. (Although actually you can go *too* fast, and the atoms
then start to whiz by too quickly, and don't stick together long
enough to fuse properly. This limit is not usually achieved in
practice.) The more dense your plasma is, the more ions there are
in a small space, and the more collisions you are likely to have.
Finally, the longer you can keep your plasma hot, the more likely
it is that something will fuse, so duration is important too. More
importantly, the slower your plasma loses energy, the more likely
it is that it will be able to sustain its temperature from internal
fusion reactions, and "ignite." The ratio of fusion energy
production to plasma energy loss is what really counts here.
Hotness is measured by temperature, and as explained above, the
D-T fuel cycle (the easiest) requires temperatures of about 10 keV,
or 100,000,000 degrees kelvin. Density is typically measured in
particles-per-cubic centimeter or particles-per-cubic meter.
The required density depends on the confinement duration.
The Lawson product, defined as (density)*(confinement time) is a
key measure of plasma confinement, and determines what
combinations of density and energy confinement will give you
fusion at a given temperature. It is important to note that
what you must confine is the *energy* (thermal energy) stored
in the plasma, and not necessarily the plasma particles.
There's a lot of subtlety here; for instance, you want to
confine your fuel ions as well as their energy, so that they
stick around and fuse, but you *don't* want to confine the
"ash" from the reactions, because the ash needs to get out
of the reactor... But you'd like to get the *energy*
out of the ash to keep your fuel hot so it will fuse better!
(And it gets even more complicated than that!)
Regardless, it's true that for a special value of the Lawson
product, the fusion power produced in your plasma will just
balance the energy losses as energy in the plasma becomes
unconfined, and *ignition* occurs. That is, as long as
the plasma fuel stays around, the plasma will keep itself
hot enough to keep fusing.
A simple analogy here is to an ordinary fire. The fire won't
burn unless the fuel is hot enough, and it won't keep burning
unless the heat released by burning the fuel is enough to keep
the fuel hot enough. The flame continually loses heat, but
usually this loss is slow enough that the fire sustains itself.
You can accelerate the heat loss, however, by pouring water
on the fire to cool it quickly; this puts the fire out.
In fusion, the plasma continually loses heat, much as a fire
gives off heat, and if the plasma loses heat faster than heat
is produced by fusion, it won't stay hot enough to keep burning.
In fusion reactors today, the plasmas aren't quite confined well
enough to sustain burning on their own (ignition), so we get
them to burn by pumping in energy to keep them hot. This is sort
of like getting wet wood to burn with a blowtorch (this last analogy
is usually credited to Harold Furth of PPPL).
For the D-T fuel cycle, the Lawson ignition value for a temperature
of about 200,000,000 Kelvin is roughly 5E20 seconds-particles/m^3.
Current fusion reactors such as TFTR have achieved about 1/10th of
this - but 20 years ago they had only achieved 1/100,000th of this!
How can we improve the Lawson value of a plasma further, so we get
even closer to fusion ignition? The trick is to keep the heat in the
plasma for as long as possible. As an analogy to this problem,
suppose we had a thermos of coffee which we want to keep hot. We can
keep the thermos hotter longer by (a) using a better type of
insulation, so that the heat flows out more slowly, or (b) using
thicker insulation, so the heat has farther to go to escape, and
therefore takes longer to get out.
Going back to the fusion reactor, the insulation can be improved by
studying plasmas and improving their insulating properties by
reducing heat transport through them. And the other way to boost
the Lawson value is simply to make larger plasmas, so the energy
takes longer to flow out. Scientists believe it's technically
feasible to build a power-producing fusion reactor with high
Lawson value *Right Now*, but it would have to be large, so large
in fact that it would cost too much to be able to make electricity
economically. So we're studying plasmas and trying to figure out
how to make them trap energy more efficiently.
*** M. What are the basic approaches used to heat and confine
the plasma? (Or, what is magnetic confinement?
There are three basic ways to confine a plasma. The first is
the method the sun uses: gravity. If you have a big enough
ball of plasma, it will stick together by gravity, and be
Unfortunately for fusion researchers, that doesn't work here on
earth. The second method is that used in nuclear fusion bombs:
you implode a small pellet of fusion fuel. If you do it quickly
enough, and compress it hard enough, the temperature will go way
up, and so will the density, and you can exceed the Lawson
ignition value despite the fact that you are only confining your
pellet for nanoseconds. Because the inertia of the imploding
pellet keeps it momentarily confined, this method is known as
The third method uses the fact that charged particles placed in
a magnetic field will gyrate in circles. If you can arrange the
magnetic field carefully, the particles will be trapped by it.
If you can trap them well enough, the plasma energy will be
confined. Then you can heat the plasma, and achieve fusion with
more modest particle densities. This method is known as
magnetic confinement. Initial heating is achieved by a
combination of microwaves, energetic/accelerated particle beams,
and resistive heating from currents driven through the plasma.
(Once the Lawson ignition value is achieved, the plasma becomes
more-or-less self-heating.) In magnetic confinement, the plasma
density is typically about 1E20 particles per cubic meter, and with
a temperature of about 1E8 kelvin, we see that ignition could be
achieved with a confinement time of about 4 seconds. (All these
numbers in reality vary by factors of 2 or 3 from the rough values
I've given.) Currently, magnetic-confinement reactors are about
a factor of ten short of the ignition value. (TFTR has an
energy confinement time of 0.25 seconds during its best shots.)
More information on these different approaches is given in the
sections that follow.