Fusion: what's the diff?

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Topic:	Science > Physics
User:	"Yousuf Khan"
Date:	02 Aug 2005 08:33:32 PM
Object:	Fusion: what's the diff?

What's the difference between core fusion that's supposed to power a
brown dwarf star, and full thermonuclear fusion that's supposed to
power a full star like our Sun? Here's an excerpt from an article
that's got me asking this question:
"Brown dwarfs are the ill-defined middle ground between planets and
stars. A star is a star because it shines on its own, generating light
through thermonuclear reactions in which hydrogen is converted to
helium. Brown dwarfs, though they can burn deuterium in another type of
reaction called "core fusion," fall short of full-blown stellar
thermonuclear fusion."
http://www.space.com/scienceastronomy/solarsystem/planet_confusion_001101-2.html
Yousuf Khan
.

User: "Llanzlan Klazmon"
Title: Re: Fusion: what's the diff?	02 Aug 2005 11:09:19 PM

"Yousuf Khan" <yjkhan@gmail.com> wrote in
news:1123032812.672096.235260@g47g2000cwa.googlegroups.com:

light

through thermonuclear reactions in which hydrogen is converted to
helium. Brown dwarfs, though they can burn deuterium in another type
of reaction called "core fusion," fall short of full-blown stellar
thermonuclear fusion."
http://www.space.com/scienceastronomy/solarsystem/planet_confusion_

0011

01-2.html

Yousuf Khan

Thermonuclear reactions are highly temperature sensitive. The usual
nuclear reactions that a occur in main sequence stars don't occur in a
brown dwarf because the core temperature is not high enough.
Aparrently though, the core temperature of a Brown dwarf does get high
enough for a reaction involving deuterium (heavy hydrogen). This stops
once the deuterium is consumed as no reaction that produces more
deuterium can take place. Similarly the reactions that occur in a main
sequence star break down Lithium 7, so their atmospheres are depleted
of Lithium whereas the primordial Lithium 7 in a Brown dwarf remains
indefinitely.
Klazmon.
.

User: "Llanzlan Klazmon"
Title: Re: Fusion: what's the diff?	02 Aug 2005 11:43:57 PM

Llanzlan Klazmon <Klazmon@llurdiaxorb.govt> wrote in
news:Xns96A7A4575FCADKlazmonllurdiaxorbgo@203.97.37.6:

"Yousuf Khan" <yjkhan@gmail.com> wrote in
news:1123032812.672096.235260@g47g2000cwa.googlegroups.com:

What's the difference between core fusion that's supposed to power

brown dwarf star, and full thermonuclear fusion that's supposed to
power a full star like our Sun? Here's an excerpt from an article
that's got me asking this question:

"Brown dwarfs are the ill-defined middle ground between planets and
stars. A star is a star because it shines on its own, generating

light

through thermonuclear reactions in which hydrogen is converted to
helium. Brown dwarfs, though they can burn deuterium in another

type

of reaction called "core fusion," fall short of full-blown stellar
thermonuclear fusion."
http://www.space.com/scienceastronomy/solarsystem/planet_confusion_

0011

01-2.html

Yousuf Khan

Thermonuclear reactions are highly temperature sensitive. The usual
nuclear reactions that a occur in main sequence stars don't occur in

brown dwarf because the core temperature is not high enough.
Aparrently though, the core temperature of a Brown dwarf does get

high

enough for a reaction involving deuterium (heavy hydrogen). This

stops

once the deuterium is consumed as no reaction that produces more
deuterium can take place. Similarly the reactions that occur in a

main

sequence star break down Lithium 7, so their atmospheres are

depleted

of Lithium whereas the primordial Lithium 7 in a Brown dwarf remains
indefinitely.

Klazmon.

Just another note. The scientist who first figured out the nuclear
reactions that power stars was Hans Bethe who died earlier this year.
http://nobelprize.org/physics/laureates/1967/bethe-bio.html
His proposal led to a puzzle. The reactions he proposed to power the
Sun should emit a certain quantity of sub atomic particles called
neutrinos that can be detected here on Earth. When the first neutrino
detectors were put into operation they detected only about a third of
the expected number of neutrinos based on the energy output of the
Sun. This was known as the solar neutrino puzzle/problem. This was
recently resolved by the discovery that neutrinos flip flop between
three different states. The detectors were only detecting one of the
states.
Klazmon.

.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	03 Aug 2005 02:36:55 PM

Llanzlan Klazmon wrote:

This was
recently resolved by the discovery that neutrinos flip flop between
three different states. The detectors were only detecting one of the
states.

I guess you're talking about the electron, muon, and tau neutrinos,
right? Have they explained what causes each type of neutrino to
flip-flop between each other? Is there a reaction in between (i.e. it
hits something which causes it to flip)? Or is it all just totally
random?
Yousuf Khan
.

User: "Ben Rudiak-Gould"
Title: Re: Fusion: what's the diff?	08 Aug 2005 10:50:40 AM

Yousuf Khan wrote:

Have they explained what causes each type of neutrino to
flip-flop between each other? Is there a reaction in between (i.e. it
hits something which causes it to flip)? Or is it all just totally
random?

In the standard model the different particles are all aspects of the same
thing, and the three neutrinos happen to be more thoroughly mixed up than
other particles. What's emitted as an electron neutrino over there might be
detected as a tau neutrino over here. There's no intermediate interaction
involved. Like everything else in particle physics, it's random, but the
probabilities can be calculated.
-- Ben
.

User: "beavith"
Title: Re: Fusion: what's the diff?	11 Aug 2005 09:00:35 AM

On Mon, 08 Aug 2005 16:50:40 +0100, Ben Rudiak-Gould
<br276deleteme@cam.ac.uk> wrote:

Yousuf Khan wrote:

i was under the impression that neutrino flip flop implied that
neutrinos have mass, which up until proof of flip flopping was found
had neutrinos as massless particles.
.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	03 Aug 2005 02:29:35 PM

Llanzlan Klazmon wrote:

Why? Is it easier to fuse deuterium than regular hydrogen? Is it
because the extra neutron in the deuterium nucleus makes it easier for
it to stick together into helium?
Also is it just deuterium to deuterium reactions that happen here? Or
is it also deuterium to hydrogen (i.e. it works as long as one of them
is a deuterium)?

This stops
once the deuterium is consumed as no reaction that produces more
deuterium can take place. Similarly the reactions that occur in a main
sequence star break down Lithium 7, so their atmospheres are depleted
of Lithium whereas the primordial Lithium 7 in a Brown dwarf remains
indefinitely.

Okay, so what's the ratio of deuterium to regular hydrogen in nature?
Something like 99% to 1% in favour of regular hydrogen, last I heard?
So what happens to the brown dwarf after all of the deuterium is gone?
Do you call it just a big planet then?
Yousuf Khan
.

User: ""
Title: Re: Fusion: what's the diff?	03 Aug 2005 08:26:06 PM

In article <1123097375.806850.40160@g43g2000cwa.googlegroups.com>, "Yousuf Khan" <yjkhan@gmail.com> writes:

Llanzlan Klazmon wrote:

Why? Is it easier to fuse deuterium than regular hydrogen?

Yes, many orders of magnitude easier.

Is it because the extra neutron in the deuterium nucleus makes it easier for
it to stick together into helium?

In a way, yes, but there is more to it.
Consider the term "capture". The naive image is of two (or more)
particles coming from infinity to within a range where attraction
forces are active and become captured. This naive image is wrong
since the system has excess energy (enough to send the particles back
to infinity) so no capture is possible, in a two particle interaction.
It is only when there exists some process through which some enrgy can
be given away, that you can have capture.
So, lets look at an interaction of two protons. Coming from,
effectively, infinity they've an excess of energy, so in order for
fusion to occur some energy must be given away, with the two proton
system dropping to a lower energy state. So, what state it can be?
It turns out that there exists no bound state of two protons. On
the other hand, there does exist a bound state of a proton and neutron
(the deuteron, of course). So, a proton-proton system can transform
to a bound state if one of the protons changes to a neutron (with some
of the energy carried away by a positron and neutrino). Possible?
Sure. But that's a weak interaction process and the weak interaction
is, indeed, weak. Small cross section, small probability.
If you collide two deutrons, that's another story. In this case a
bound state of the system (comprising of two protons and two
electrons) exists (alpha particle). So, no "identity change is
required", a transformation to a bound system can be obtained through
an EM process (with a photon being emitted). Now, EM is stronger than
the weak interaction (by orders of magnitude) so the cross section is
much larger.

Also is it just deuterium to deuterium reactions that happen here? Or
is it also deuterium to hydrogen (i.e. it works as long as one of them
is a deuterium)?

This'll work too. Again, an EM transition, producing He3.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: "Sbharris[atsign]ix.netcom.com"
Title: Re: Fusion: what's the diff?	04 Aug 2005 02:59:54 PM

wrote:

If you collide two deutrons, that's another story. In this case a
bound state of the system (comprising of two protons and two
electrons) exists (alpha particle).

Two neutrons, of course.
So, no "identity change is

required", a transformation to a bound system can be obtained through
an EM process (with a photon being emitted).

Well, it's both a nuclear and EM process.

Now, EM is stronger than
the weak interaction (by orders of magnitude) so the cross section is
much larger.

In this case, it's the cross section due to the attractive nuclear
force. If only EM operated, the things would never stick at all. And
yes, the difference is due to nuclear force cross sections vs. weak
force ones.
Nuclear closs sections of course participate even in the proton-proton
interaction, since the bound state of the deuteron is due to them. To
see this purely, look at the cross section of hydrogen for neutrons.

Also is it just deuterium to deuterium reactions that happen here? Or
is it also deuterium to hydrogen (i.e. it works as long as one of them
is a deuterium)?

This'll work too. Again, an EM transition, producing He3.

Also a nuclear interaction participates there also. Deuterium has a
absorption cross section for both protons
and neutrons, and the EM interaction is the difference between one and
other other (neglect spins and a lot of other ugly stuff by insisting
that they are all controlled). The rest is basically the nuclear
process.
SBH
.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	04 Aug 2005 08:39:35 AM

wrote:

So, lets look at an interaction of two protons. Coming from,
effectively, infinity they've an excess of energy, so in order for
fusion to occur some energy must be given away, with the two proton
system dropping to a lower energy state. So, what state it can be?
It turns out that there exists no bound state of two protons. On
the other hand, there does exist a bound state of a proton and neutron
(the deuteron, of course). So, a proton-proton system can transform
to a bound state if one of the protons changes to a neutron (with some
of the energy carried away by a positron and neutrino). Possible?
Sure. But that's a weak interaction process and the weak interaction
is, indeed, weak. Small cross section, small probability.

So what you're saying is that there is no such thing as a He2, only He3
or He4?
So perhaps some portion of the proton-proton interactions results in
the D2 transformation, while others just fly apart again. The portion
that produces D2 then can come back and interact with H1's or other
D2's and produce He3 or He4, respectively?

If you collide two deutrons, that's another story. In this case a
bound state of the system (comprising of two protons and two
electrons) exists (alpha particle).

I assume you mean "two protons and two *neutrons*"?

So, no "identity change is
required", a transformation to a bound system can be obtained through
an EM process (with a photon being emitted). Now, EM is stronger than
the weak interaction (by orders of magnitude) so the cross section is
much larger.

So you're saying we'll find more He4's than He3's in nature?

Also is it just deuterium to deuterium reactions that happen here? Or
is it also deuterium to hydrogen (i.e. it works as long as one of them
is a deuterium)?

This'll work too. Again, an EM transition, producing He3.

Is the same amount of energy released when going to He3 or He4? That
is, is it the same frequency of gamma ray that comes out?
Yousuf Khan
.

User: ""
Title: Re: Fusion: what's the diff?	04 Aug 2005 04:32:53 PM

In article <1123162775.078392.155630@z14g2000cwz.googlegroups.com>, "Yousuf Khan" <yjkhan@gmail.com> writes:

mmeron@cars3.uchicago.edu wrote:

So what you're saying is that there is no such thing as a He2, only He3
or He4?

Yes, right. Would there has been even a very short living He2
isotope, the cross section would've been much, much larger. Mind you,
there are heavier isotopes of He, beyond He4. but that doesn't help
here.

So perhaps some portion of the proton-proton interactions results in
the D2 transformation, while others just fly apart again. The portion
that produces D2 then can come back and interact with H1's or other
D2's and produce He3 or He4, respectively?

If you collide two deutrons, that's another story. In this case a
bound state of the system (comprising of two protons and two
electrons) exists (alpha particle).

I assume you mean "two protons and two *neutrons*"?

Oops, yes, sorry.

So you're saying we'll find more He4's than He3's in nature?

As, indeed, we do.

Also is it just deuterium to deuterium reactions that happen here? Or
is it also deuterium to hydrogen (i.e. it works as long as one of them
is a deuterium)?

This'll work too. Again, an EM transition, producing He3.

Is the same amount of energy released when going to He3 or He4? That
is, is it the same frequency of gamma ray that comes out?

Actually, no. He4 is a very tightly bound nucleus (even-even, magical
numbers and so on) so the amount of energy released when it is
produced is significantly larger.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	04 Aug 2005 11:00:23 PM

wrote:

Now what exactly are you talking about when you talk about a "cross
section"? It doesn't exactly sound like my common-usage sense of the
term, cross section.

Actually, no. He4 is a very tightly bound nucleus (even-even, magical
numbers and so on) so the amount of energy released when it is
produced is significantly larger.

Now getting back to the original subject of the topic, the difference
between brown dwarf fusion and main sequence star fusion. What is the
main product produced in a brown dwarf reaction, He3 or He4? My
assumption is that brown dwarves produce mainly He3 as you said it is a
much smaller bang produced than He4 production.
Yousuf Khan
.

User: ""
Title: Re: Fusion: what's the diff?	05 Aug 2005 01:04:14 AM

In article <1123214423.912419.290110@g43g2000cwa.googlegroups.com>, "Yousuf Khan" <yjkhan@gmail.com> writes:

mmeron@cars3.uchicago.edu wrote:

Now what exactly are you talking about when you talk about a "cross
section"? It doesn't exactly sound like my common-usage sense of the
term, cross section.

Well, it is not much different really. Think about a particle as
representing a target for other particles, such that if they had
within a specific part of the target, such and such will happen. The
area of this target is the "cross section". To word it differently,
the cross section for process X occuring between particles A and B
(where A and B may be same or different) is the area, presented by particle
A, such that if particle B hits within this area, process X occurs.
Mind you, this doesn't have to equal the geometrical size of the
particle as probabilities are taken into account as well. For,
example, I noticed that you were asking about collisions where two
deuterons come together then fly apart again with nothing happening.
So, if the cross section for two deuterons to collide (say, come
within range of nuclear forces) is (just for example) 10^(-30) m^2
and, once they collided, the probability for fusion occuring is just
10^(-6) (i.e. one fusion event in million collisions), then the cross
section for the deuterons fusing is 10^(-6)*10^(-30) m^2 = 10^(-36) m^2.
So, what good it is for? Well, if you have a "hot soup" of whatever
particles, then the rate of process X is
density*average_velocity*cross_section(for process X)
i.e. how m many (per unit volume) there are times how fast they move
times what area they've to hit for X to happen.
In standard textbook language this will be written as rho*v*sigma
(rho for density, v for velocity, sigma for cross section).
Note that other than the direct dependence of velocity in the above,
through v, there is also indirect dependence through sigma which, in
general, is a function of the collision velocity.

Actually, no. He4 is a very tightly bound nucleus (even-even, magical
numbers and so on) so the amount of energy released when it is
produced is significantly larger.

The bang is not the only thing that matters, the type of reaction
matters as well.
In D-D fusion you could in principle get one of the following
1) He4 + photon
2) He3 + neutron
3) T (meaning H3) + proton
Now, the emission of a photon is an EM process while the emission of a
proton or neutron is a strong force process which is far more probable
(since the strong force is, well, much stronger). So, initially,
you'll be mostly producing He3 and Tritium with only a small amount of
He4. However, once the process goes for a while, you accumulate
significant amount of tritium and He3 in the system and the cross
section of these to fuse with deuterium is much larger than the D-D
cross section. And these fusion events produce mainly He4.
So, my guess is that you'll still end with more He4 than He3, but I'm
not sure what the ratios will be.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	05 Aug 2005 05:50:49 PM

wrote:

So, what good it is for? Well, if you have a "hot soup" of whatever
particles, then the rate of process X is

density*average_velocity*cross_section(for process X)

i.e. how m many (per unit volume) there are times how fast they move
times what area they've to hit for X to happen.

In standard textbook language this will be written as rho*v*sigma

(rho for density, v for velocity, sigma for cross section).

Note that other than the direct dependence of velocity in the above,
through v, there is also indirect dependence through sigma which, in
general, is a function of the collision velocity.

Okay, I think I got it. The "Cross Section" is a true areal cross
section reduced by a probability fudge-factor. So when you guys compare
/Cross Sections/ you're really making an implicit assumption that the
areal cross-section is the same between all of these reactions, and
you're really just comparing their probabilities.

The bang is not the only thing that matters, the type of reaction
matters as well.

In D-D fusion you could in principle get one of the following

1) He4 + photon
2) He3 + neutron
3) T (meaning H3) + proton

Now, the emission of a photon is an EM process while the emission of a
proton or neutron is a strong force process which is far more probable
(since the strong force is, well, much stronger). So, initially,
you'll be mostly producing He3 and Tritium with only a small amount of
He4. However, once the process goes for a while, you accumulate
significant amount of tritium and He3 in the system and the cross
section of these to fuse with deuterium is much larger than the D-D
cross section. And these fusion events produce mainly He4.

Okay, so which is more likely to happen He3+neutron or T+proton? Is
there a difference in ranking within the Strong family of forces, or
are all Strongs the same?
Also noticed that you guys are giving higher Cross Sections to Strong
interactions, over Electro-magnetic, which in turn are higher than Weak
interactions. I've also heard that these forces meld into one at really
high temperatures. So I assume that means their Cross Sections become
pretty equal at these high temperatures?

So, my guess is that you'll still end with more He4 than He3, but I'm
not sure what the ratios will be.

Would you expect there to be a larger ratio of He4 vs. He3 produced in
a main sequence star than in a brown dwarf? I know that probably would
be a research paper on its own, so I'm only asking you to take a gut
feeling.
Yousuf Khan
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	08 Aug 2005 12:30:10 PM

Yousuf Khan wrote:
|> Okay, I think I got it. The "Cross Section" is a true areal cross
|> section reduced by a probability fudge-factor. So when you guys compare
|> /Cross Sections/ you're really making an implicit assumption that the
|> areal cross-section is the same between all of these reactions, and
|> you're really just comparing their probabilities.
Well, it is true that the object of the calculation or measurement is a
probability (rate of event occurrence per unit volume), but the role of
"cross section" is the other way around. In the 19th century people did
the first kinetic theory calculations for a scattering collision rate
assuming a distribution of hard spheres of a given size. This related a
probability to a cross sectional area. Nowadays, the probability is
calculated and the result expressed in terms of a cross section. But
since the velocities themselves now have a distribution, this cross
section is meaningful only as an averaged quantity. The central
ingredient in the theory is <sigma v>, a probability "rate coefficient"
which is interpreted as an area times a velocity averaged over the
velocity distribution.
|> Okay, so which is more likely to happen He3+neutron or T+proton? Is
|> there a difference in ranking within the Strong family of forces, or
|> are all Strongs the same?
The NRL formulary says (sheesh, 70 pages now!) 50 percent. I think it
is 50 percent within something like 1 in 10^6 or 10^7. The direct to
alpha version, D + D --> alpha + gamma, has a probability less than
1 in 10^7.
I guess this is energy dependent. At low energies (not sure how low is
low, but clearly these are way below 1 GeV, the proton/neutron mass
scale), the strong forces involving up or down quarks are not very
different.
|> Also noticed that you guys are giving higher Cross Sections to Strong
|> interactions, over Electro-magnetic, which in turn are higher than Weak
|> interactions. I've also heard that these forces meld into one at really
|> high temperatures. So I assume that means their Cross Sections become
|> pretty equal at these high temperatures?
Only on the unification scale of everything except gravity. Many orders
of magnitude outside relevance.
|> Would you expect there to be a larger ratio of He4 vs. He3 produced in
|> a main sequence star than in a brown dwarf? I know that probably would
|> be a research paper on its own, so I'm only asking you to take a gut
|> feeling.
Look up the literature on He3 in giant planets. Since the BDs are
unlikely to do much of the He3 + He3, there should be a lot of
accumulation. Is it enough to overcome primordial He4? Probably not.
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: ""
Title: Re: Fusion: what's the diff?	05 Aug 2005 06:44:21 PM

In article <1123282249.723637.318940@g43g2000cwa.googlegroups.com>, "Yousuf Khan" <yjkhan@gmail.com> writes:

mmeron@cars3.uchicago.edu wrote:

No, you really don't have to make any assumptions of the sort, cross
sections can be measured. If you create a controlled situation with a
beam of particles, at particle current (measured in number of
particles per second) I, impinging on a target of thickness d, containing
a known target density rho, then the number of interaction of specific
type X produced within time dt is
dN = I*sigma(X)*rho*d*dt
where sigma(X) is the cross section for process X. The above is a tad
simplified since it doesn't take into account that the beam diminishes
as it goes through the target, so assume that d is very small or, you
can account for this too. Bottom line is, you don't need to assume
anything, you count interactions and measure the cross section.
On the other hand, yes, once you know the cross sections, the
probability of various processes occuring is proportional to the cross
section.
It is worth mentioning here (a bit going off on a tangent but very
important) that if you happen to have a model of what's happening
during the interaction, you can use this model to calculate the cross
section. Then, you can compare the calculated result with the
measured, to check how good your model is. This is the main mechanism
of studying atomic and subatomic interactions, starting with the famed
Rutherford experiment.

The bang is not the only thing that matters, the type of reaction
matters as well.

In D-D fusion you could in principle get one of the following

1) He4 + photon
2) He3 + neutron
3) T (meaning H3) + proton

Now, the emission of a photon is an EM process while the emission of a
proton or neutron is a strong force process which is far more probable
(since the strong force is, well, much stronger). So, initially,
you'll be mostly producing He3 and Tritium with only a small amount of
He4. However, once the process goes for a while, you accumulate
significant amount of tritium and He3 in the system and the cross
section of these to fuse with deuterium is much larger than the D-D
cross section. And these fusion events produce mainly He4.

Okay, so which is more likely to happen He3+neutron or T+proton? Is
there a difference in ranking within the Strong family of forces, or
are all Strongs the same?

Pretty much the same, but there are small differences in masses in
these two branches that influence the probabilities a tad. So, no, I
don't think that the two diviede excatly 50-50 but probably not too
far from this. Don't have numbers but Bruce probably knows.

Also noticed that you guys are giving higher Cross Sections to Strong
interactions, over Electro-magnetic, which in turn are higher than Weak
interactions. I've also heard that these forces meld into one at really
high temperatures. So I assume that means their Cross Sections become
pretty equal at these high temperatures?

Yes, that's true. Mind you, these "high" are really very, very high.
A thermonuclear reaction is freezing cold in comparison.

So, my guess is that you'll still end with more He4 than He3, but I'm
not sure what the ratios will be.

My feeling is that you're right and that relatively speeking there'll
be more He3 in brown dwarves. But, yes, that would be a topic for a
very extensive research paper (so you're right twice in a row:-)) and
I can claim no more than passing acquintance with the topic. Thus,
don't take my gut feeling too seriously, on this.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: ""
Title: Re: Fusion: what's the diff?	05 Aug 2005 04:16:23 PM

In article <200508051254.j75CsvWe027021@ipp.mpg.de>, Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> writes:

Mati Meron wrote:

[cross sections]

This is the reaction rate per unit particle...

Yes, you're right, I should've mentioned it.

where I got the density
squared factor from is in the reaction rate per unit volume, which for
reactants a and b is

<sigma v> n_a n_b

where the angle brackets are the average over the velocity distribution
function (different probabilities to have different relative encounter
velocities) and the two n's are the densities. For pp the incident
particles are identical, so the reaction rate goes like density squared
times <sigma v>, the latter factor which contains the temperature
dependence (temperature is average kinetic energy of random motion,
which translates to an average velocity).

Yes, quite right.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	05 Aug 2005 07:54:57 AM

Mati Meron wrote:
[cross sections]

This is the reaction rate per unit particle... where I got the density
squared factor from is in the reaction rate per unit volume, which for
reactants a and b is
<sigma v> n_a n_b
where the angle brackets are the average over the velocity distribution
function (different probabilities to have different relative encounter
velocities) and the two n's are the densities. For pp the incident
particles are identical, so the reaction rate goes like density squared
times <sigma v>, the latter factor which contains the temperature
dependence (temperature is average kinetic energy of random motion,
which translates to an average velocity).
(from Clayton, pp 288-293)
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Y.Porat"
Title: Re: Fusion: what's the diff?	03 Aug 2005 11:44:11 PM

in addition to the above:
the geometric structure of the input particles
*has a lot to do with it *
that is one of the main reasons that actually
the Triton - deuteron is a much essayer process.(yet more expensive)
see my site:
http://www.geocities.com/porat_y/mypage.html
see there the geometric structure of the
Proton (neutron) Deuteron and Triton and ..... the Alpha particle
they are 3D and the Alpha is a tetrahedron
most scientists that come from just 'flat' mathematics are not aware
enough
to that crucial aspect
ATB
Y.Porat
-------------
.

User: "Martin Brown"
Title: Re: Fusion: what's the diff?	03 Aug 2005 04:08:00 PM

Yousuf Khan wrote:

Llanzlan Klazmon wrote:

Why? Is it easier to fuse deuterium than regular hydrogen? Is it
because the extra neutron in the deuterium nucleus makes it easier for
it to stick together into helium?

Also is it just deuterium to deuterium reactions that happen here? Or
is it also deuterium to hydrogen (i.e. it works as long as one of them
is a deuterium)?

The reaction with the lowest activation energy is something like:
D2 + H1 -> He3 + 5.5MeV half life 50000y at 10^6K
Once the stellar core reaches something more than 10^7K the full range
of nuclear reactions can begin to run, but until then it has to rely on
this and a few other minor isotopes from the primordial mix for energy.
If it doesn't reach 10^7 it doesn't get to fledge as a true main
sequence star.

I suppose you call it a brown dwarf and watch it gradually cool down a
bit like a super Jupiter with an unusual abundance of He3 in it.
Regards,
Martin Brown
.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	04 Aug 2005 08:17:30 AM

Martin Brown wrote:

The reaction with the lowest activation energy is something like:

D2 + H1 -> He3 + 5.5MeV half life 50000y at 10^6K

Is that the half-life of the starting reactants (D2 & H1) or the end
product (He3)?
If it's the half-life of the end product, He3, then what does it
degenerate into? Back to D2 & H1?

I suppose you call it a brown dwarf and watch it gradually cool down a
bit like a super Jupiter with an unusual abundance of He3 in it.

What if we only discover the object long after all of its reactants
have depleted? There's no fusion happening anymore, and all of its
stored fusion heat is gone, and it produces no more heat than any
standard gas-giant planet?
As for unusual abundance of He3, how unusual is unusual? If there is
only 1% D2 on average to every H1, then likely if all of that D2 is
converted to He3, it would only be 1% extra He3 more than in the
standard environment, right?
Yousuf Khan
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	03 Aug 2005 03:10:08 PM

Yousuf Khan wrote:

What's the difference between core fusion that's supposed to power a
brown dwarf star, and full thermonuclear fusion that's supposed to
power a full star like our Sun? [...]

The reactions are very different. The brown dwarf (BD) chain,
D + D --> p + T
--> n + He3
plus about 3.5 MeV, with 50-50 branch ratio, followed by the
secondaries,
D + T --> n + alpha
D + He3 --> p + alpha
plus about 18 MeV, is all strong interactions, so the cross section is,
in stellar evolution terms, relatively high. This can occur in the
cores of BDs which reach about 10^5 or 10^6 degrees (look up the actual
numbers).
For a sunlike star, the above reactions occur in the very early stages
and wipe out the original complement of deuterium. Then the star
contracts to the main sequence. The core, at temperatures between 10
and 15 million degrees K, is now hot and dense enough to use a much
slower reaction to produce fusion power and thereby maintain its
stability on the main sequence for several billion years. This new
reaction chain is called PP, and the lowest energy version of it is PPI:
p + p --> D + e+ + nu (W)
p + D --> He3
He3 + He3 --> alpha + 2p
where (W) denotes a weak interaction. This reaction is weak because it
has to convert the charge on one of the protons to a neutron and creates
the lepton-antilepton pair (it actually occurs in one of the quarks).
When small quantities of Li are present you get additional side chains
that produce Li and Be neutrinos which were the first ones detected.
The pp neutrino wasn't detected until 1992.
Basically, the star owes its long life to the weakness of this one
pp interaction. If not for that, stars would burn themselves out very
quickly and evolved life would not be possible.
--
cu,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Steve Willner"
Title: Re: Fusion: what's the diff?	05 Aug 2005 05:24:34 PM

In article <200508032010.j73KA8No007634@ipp.mpg.de>,
Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> writes:

The reactions are very different. The brown dwarf (BD) chain,
D + D --> p + T
--> n + He3

Not p + D?

Basically, the star owes its long life to the weakness of this one
pp interaction. If not for that, stars would burn themselves out very
quickly and evolved life would not be possible.

Are you sure about this? If the pp reaction were stronger, I'd have
expected stars to have about the same luminosity they do now but much
lower core temperatures to give about the same reaction rate.
--
Steve Willner Phone 617-495-7123

Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	08 Aug 2005 12:44:50 PM

Steve Willner wrote:
|> In article <200508032010.j73KA8No007634@ipp.mpg.de>,
|> Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> writes:
|> > The reactions are very different. The brown dwarf (BD) chain,
|> > D + D --> p + T
|> > --> n + He3
|>
|> Not p + D?
Oops... in a star you might be right. I am too used to fusion plasmas
where D is a main constituent. (looks it up... Clayton p 365) Yes, it
is p + D --> He3, since the primordial concentration is D/H < 10^{-4}.
|> > Basically, the star owes its long life to the weakness of this one
|> > pp interaction. If not for that, stars would burn themselves out very
|> > quickly and evolved life would not be possible.
|>
|> Are you sure about this? If the pp reaction were stronger, I'd have
|> expected stars to have about the same luminosity they do now but much
|> lower core temperatures to give about the same reaction rate.
If p+p were actually a strong interaction, not merely "stronger"?
If you have a flexible code handy, try it: multiply rate(p,p) by 10^7 or
so. Would you merely get a larger body with lower T and the same
luminosity? Really? :-)
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Steve Willner"
Title: Re: Fusion: what's the diff?	08 Aug 2005 05:59:12 PM

SW> If the pp reaction were stronger, I'd have
SW> expected stars to have about the same luminosity they do now but much
SW> lower core temperatures to give about the same reaction rate.
In article <200508081744.j78HioUK002673@ipp.mpg.de>,
Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> writes:

If p+p were actually a strong interaction, not merely "stronger"?

Well, that's what I remember from my stellar interiors courses, but
admittedly they were a long time ago. As I understand it, though,
luminosity is controlled mostly by opacity. In effect, stellar mass
sets the _pressure_ at the core (which is what you need to hold the
star up), opacity sets the rate at which energy leaks out, and the
nuclear reactions adjust to maintain the pressure at the required
luminosity. Of course everything is coupled to everything else, but
the very steep dependence of reaction rate on temperature means that
temperature doesn't change much no matter what else you do to the
star (within limits, of course).

If you have a flexible code handy, try it: multiply rate(p,p) by 10^7 or
so. Would you merely get a larger body with lower T and the same
luminosity? Really? :-)

No, I don't have any code handy nor even a copy of Clayton's book.
The answer can probably be found analytically with reasonable
approximations... but not by me. :-)
--
Steve Willner Phone 617-495-7123

Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	09 Aug 2005 11:00:51 AM

Steve Willner wrote:

SW> If the pp reaction were stronger, I'd have
SW> expected stars to have about the same luminosity they do now but much
SW> lower core temperatures to give about the same reaction rate.

In article <200508081744.j78HioUK002673@ipp.mpg.de>,
Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> writes:

If p+p were actually a strong interaction, not merely "stronger"?

Well, that's what I remember from my stellar interiors courses, but
admittedly they were a long time ago. As I understand it, though,
luminosity is controlled mostly by opacity.

Density gets into that as well, and if we're holding the mass fixed and
have to expand the star then the density will go down. So will (likely)
the opacity.

In effect, stellar mass
sets the _pressure_ at the core (which is what you need to hold the
star up),

In a constant density model this is an elementary result (the one our
courses started us with), but...

opacity sets the rate at which energy leaks out, and the
nuclear reactions adjust to maintain the pressure at the required
luminosity.

.. I think that is because of the T scaling being so much steeper than
anything else...

Of course everything is coupled to everything else, but
the very steep dependence of reaction rate on temperature means that
temperature doesn't change much no matter what else you do to the
star (within limits, of course).

If you linearise about a given solution, I think all this is correct,
but with really large changes it might not be. What we are
contemplating is different though; increasing the power by a uniform
(large) factor, not by increasing the temperature.

If you have a flexible code handy, try it: multiply rate(p,p) by 10^7 or
so. Would you merely get a larger body with lower T and the same
luminosity? Really? :-)

No, I don't have any code handy nor even a copy of Clayton's book.
The answer can probably be found analytically with reasonable
approximations... but not by me. :-)

If a zero-D model can get the basic gist I might try it in the next
days...
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	19 Aug 2005 06:09:39 AM

For purposes of clarity, here are three responses to the thread in
stellar structure under this subject. Steve raised a number of points
which require a calculation. So the three responses are:
1) responses to Steve's points in light of the calculations I did
2) an outline of the calculations
3) a separate explanation of why Steve's point suggesting opacity
controls the luminosity actually works for these cases
a) hydrostatic equilibrium, such that T in inverse to R for ideal gases
b) opacity independent of T or density for solar core parameters
This is post 1)...
I wrote:

Steve Willner wrote:

SW> If the pp reaction were stronger, I'd have
SW> expected stars to have about the same luminosity they do now but much
SW> lower core temperatures to give about the same reaction rate.

In article <200508081744.j78HioUK002673@ipp.mpg.de>,
Bruce Scott TOK <Use-Author-Supplied-Address-Header@[127.1]> writes:

If p+p were actually a strong interaction, not merely "stronger"?

Well, that's what I remember from my stellar interiors courses, but
admittedly they were a long time ago. As I understand it, though,
luminosity is controlled mostly by opacity.

I did actually find this, but...

Density gets into that as well, and if we're holding the mass fixed and
have to expand the star then the density will go down. So will (likely)
the opacity.

...density did not affect the opacity, because I kept electron
scattering. The Kramer's opacity model, with the coefficient kappa
proportional to density times (T to the -7/2 power), has very
small values for density between 1 and 10 gm/cm^3 (solar average is
about 1 plus plus even if the core is about 100) and temperature in the
range 10 to 100 million degrees Kelvin. So what remains is the electron
scattering value, which is independent of density or temperature.
For beginners, the opacity coefficient is a multiplier of the density,
so that the mean free path of a photon is given by the inverse of
(kappa rho), where rho is the density and kappa is a coefficient. For
electron scattering, the probability of a scattering event scales with
the electron density, so that kappa itself is a constant if the
composition is uniform.

In effect, stellar mass
sets the _pressure_ at the core (which is what you need to hold the
star up),

In a constant density model this is an elementary result (the one our
courses started us with), but...

opacity sets the rate at which energy leaks out, and the
nuclear reactions adjust to maintain the pressure at the required
luminosity.

This turns out to be correct for constant kappa.

. I think that is because of the T scaling being so much steeper than
anything else...

And this turns out to be largely irrelevant :-)

And this is largely irrelevant, because the change in T is relatively
small even for a very large change in the fusion rate.

If you have a flexible code handy, try it: multiply rate(p,p) by 10^7 or
so. Would you merely get a larger body with lower T and the same
luminosity? Really? :-)

No, I don't have any code handy nor even a copy of Clayton's book.
The answer can probably be found analytically with reasonable
approximations... but not by me. :-)

If a zero-D model can get the basic gist I might try it in the next
days...

This is what I did (next post)...
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	19 Aug 2005 06:10:05 AM

This is post 2)...
I wrote:

If a zero-D model can get the basic gist I might try it in the next
days...

Well, I did this mostly over the weekend and a bit more this week. I
used a set of equations which does not assume anything about the
equilibrium but gets it as a result. Basically the rules are those for
homologous changes, so there are factors of 3 in divergences and in the
gravitational potential. I see now I missed a factor of 2 in the mean
mass per particle by forgetting the electrons but that won't matter to
the scaling. The equations are
(1/rho) d rho/dt + 3 V/R = 0
alpha d V/dt = [(p/rho) - (GM/3R)] / R
(3/2) (k/mu) d T/dt + (p/rho) 3 V/R + 3 (q/rho) / R = S
and
R = (3 M/4 pi rho)^(1/3)
where alpha is a factor to slow down the hydrodynamic scale,
rho, T, and p are mass density, temperature, and pressure,
M and R are the mass and radius, V is the expansion velocity,
q is the outward heat flux, S is the energy generation rate per mass,
k is Boltzmann constant, and mu is the mean mass per particle.
For mu I used the proton mass.
For the equation of state I used p = (k/mu) rho T. The equation for T
above shows I used an thermal energy per unit mass of (3/2) kT. That
is, I ignored electron degeneracy and radiation pressure (for a sunlike
star, degeneracy is negligible and radiation pressure is a percent-level
correction).
For the timestep I used a simple second order ``Runge-Kutta'' scheme
(the names are two German 19th-Century mathematicians). For a system of
equations
du/dt = f(u)
you create a guess
u^* = u + theta tau f(u)
where tau is the timestep and theta is a multiplier, 1/2 in the exact
form but something slightly larger if you want to stabilise nonlinear
oscillations. I used theta = 0.55. Then you correct by using
u^(new) = u + tau f(u^*)
This gave very accurate nonlinear oscillations in the case you start
_out_ of hydrostatic equilibrium. I usually started with a pressure of
1.01 times the equilibrium value, in order to see the oscillations in V
and especially how fast they damp away. This told me how to set the
parameter alpha to keep the hydrostatic time scale somewhat shorter than
the thermal time scale but still long enough so I could afford the
calculation.
For the thermal flux I used a radiative opacity model,
q = (4/3) (c/kappa rho) (a T^4)/R
where c is the speed of light and a is the Stefan-Boltzmann coefficient
for photon energy (a T^4) as a function of temperature. The factor of 4
is for the gradient of T^4. The factor of 3 is a geometric coefficient.
For the opacity I used
kappa = MIN( 0.4 + k0 rho T^(-7/2) , 10^4 ) cm^2/gm
where 0.4 is the electron scattering value for pure hydrogen and the
Kramer's coefficient was k0 = 7.6 x 10^22 with rho in units of gm/cm^3
and T in degrees K. This also assumes pure hydrogen. The cap at 10^4
to avoid potential numerical problems turned out to be unnecessary.
For the fusion energy generation rate per unit mass, I used a simple
formula for proton-proton fusion, assuming pure hydrogen, and neglecting
the parts of the PP chains involving He4 (PPII and PPIII). So the
fusion energy rate per unit mass is given by
S = s_0 rho T6^(-2/3) exp[-e_0 T6^(-1/3)]
where s_0 = 2.38 x 10^6 and e_0 = 33.8, and T6 = T/(10^6 degrees K),
with rho in gm/cm^3 and S in units of erg/(gm sec). In general I am
interested in the result of a very large variation of s_0, modelled by
the parameter s = s_0/(2.38 x 10^6).
The initial parameters were the mass M and the radius R. Then rho is
found from 3M/4 pi R^3. Then the equilibrium value of p_eq is found
from the Virial theorem p_eq/rho = GM/3R, for which the acceleration
d V/dt would be zero. Then p is set equal to 1.01 p_eq and T is found
from the equation of state. Then we begin. The timestep is continually
adjusted to be equal to 0.03 times the inverse of the larger of the
freefall rate GM/3 R^2 and the expansion rate p/(rho R). The slowdown
parameter alpha is mainly chosen by trial and error, so that the initial
oscillations die out slightly faster than the gravitational contraction
proceeds towards thermal fusion/flux equilibrium (``main sequence'').
The result is found once the equilibrium is established and the error
bars over the last 1/4 of the run are small enough that the 3rd decimal
place in any value is no longer uncertain. Typical run time is about 30
seconds on my 600 MHz laptop.
To answer the question ``what if p-p were a strong interaction'' I
varied the coefficient s_0 over 7 orders of magnitude. The results are
presented in terms of R10 = R/(10^10 cm) in which the solar value is 7,
and T6 = T/(10^6 degrees K) in which the solar core value is about 15,
and rho in gm/cm^3 in which the solar average is 1 plus plus but the
core value is about 100, and the luminosity per unit mass L/M in which
the solar value is 2. I got very large values of L/M mostly because of
the low values of the opacity (kappa) and rho.
Here are the results in terms of the hydrostatic and fusion
coefficients, whose physical values are alpha = s = 1.
log alpha log s R10 T6 rho L/M
-6 0 4.52 47.5 5.17 2.11 (x 10^5)
-4 0 4.52 47.5 5.17 2.11
-4 1 6.62 32.5 1.64 2.11
-4 2 9.46 22.7 0.564 2.11
-4 3 13.2 16.3 0.208 2.11
-2 4 18.0 11.9 0.0817 2.11
-2 5 24.1 8.91 0.0340 2.11
-2 6 31.7 6.77 0.0149 2.11
-1 6 31.7 6.77 0.0149 2.11
-1 7 41.1 5.23 0.00690 2.11
0 7 41.1 5.24 0.00690 2.11
You can see that over this wide range of parameters that
1) the equilibrium values are independent of alpha, so that the plan of
slowing down the hydrodynamics works
2) the luminosity is independent of the fusion rate, of T, and of rho.
This is in accordance with Steve's expectations. However, a caveat:
Comments in the next post.
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	19 Aug 2005 06:10:20 AM

This is post 3)...
Steve wrote:

Well, that's what I remember from my stellar interiors courses, but
admittedly they were a long time ago. As I understand it, though,
luminosity is controlled mostly by opacity.

and

In effect, stellar mass
sets the _pressure_ at the core (which is what you need to hold the
star up),
opacity sets the rate at which energy leaks out, and the
nuclear reactions adjust to maintain the pressure at the required
luminosity.

This turned out to be what I found. However, it depends on the opacity
coefficient kappa being constant over the parameter range I used. The
Kramer's opacity is k0 rho T^(-7/2), with k0 = 7.6 x 10^22. With rho
between 1 and 10 and T larger than 10^7, this is very small, smaller
than the electron scattering value of 0.4 cm^2/gm. So kappa was
constant over the range of parameters used.
Now, hydrostatic equilibrium said that p/rho = GM/3R, so that with the
ideal gas law p = (k/mu) rho T this means that T is inverse to R,
nothing more (i.e., the sound speed scales with the grav potential
energy, which is what the Virial theorem says). Now, the thermal
balance says that
S = 3 q/(rho R)
or that luminosity (rho S) is balanced by the heat flux divergence (3q/R).
Now put in the formula for radiative transport, q goes like
T^4/(kappa rho R). So S times M (note M is a constant here) goes like
T^4/(rho^2 R^2). But we already found that T goes like 1/R and the
definition of rho gives (rho R^3) = constant. Hence, S is constant if
kappa is constant, and hence the luminosity is indeed given by kappa.
But only if kappa is independent of density and temperature.
I hope all this is found intresting... I certainly had fun doing it!
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

User: "Yousuf Khan"
Title: Re: Fusion: what's the diff?	04 Aug 2005 09:06:05 AM

Okay, you're going to have to take it easy on me about these
abbreviations, I'm just a layman.
Bruce Scott TOK wrote:

The reactions are very different. The brown dwarf (BD) chain,

D + D --> p + T
--> n + He3

So here you're saying two Deuteriums will produce either a Tritium and
a proton (or a positron?), or a Helium-3 and a neutron (or a
neutrino?)?

plus about 3.5 MeV, with 50-50 branch ratio, followed by the
secondaries,

D + T --> n + alpha
D + He3 --> p + alpha

plus about 18 MeV, is all strong interactions, so the cross section is,
in stellar evolution terms, relatively high. This can occur in the
cores of BDs which reach about 10^5 or 10^6 degrees (look up the actual
numbers).

For a sunlike star, the above reactions occur in the very early stages
and wipe out the original complement of deuterium. Then the star
contracts to the main sequence. The core, at temperatures between 10
and 15 million degrees K, is now hot and dense enough to use a much
slower reaction to produce fusion power and thereby maintain its
stability on the main sequence for several billion years. This new
reaction chain is called PP, and the lowest energy version of it is PPI:

p + p --> D + e+ + nu (W)
p + D --> He3
He3 + He3 --> alpha + 2p

How much energy is liberated in each of these above reactions?

where (W) denotes a weak interaction. This reaction is weak because it
has to convert the charge on one of the protons to a neutron and creates
the lepton-antilepton pair (it actually occurs in one of the quarks).

When small quantities of Li are present you get additional side chains
that produce Li and Be neutrinos which were the first ones detected.
The pp neutrino wasn't detected until 1992.

So the lithium stays the same after the reaction, but we get a
different kind of neutrino come out of the proton-proton reaction if
lithium is nearby?

Basically, the star owes its long life to the weakness of this one
pp interaction. If not for that, stars would burn themselves out very
quickly and evolved life would not be possible.

What's the ratio of proton-proton reactions that result in a deuterium
being produced vs. those which just fly apart again? I assume the slow
rate of deuterium production is what slows it down?
Yousuf Khan
.

User: "Bruce Scott TOK"
Title: Re: Fusion: what's the diff?	04 Aug 2005 12:33:20 PM

Yusuf Khan wrote:

Okay, you're going to have to take it easy on me about these
abbreviations, I'm just a layman.

Yeah I get lazy...

Bruce Scott TOK wrote:

The reactions are very different. The brown dwarf (BD) chain,

D + D --> p + T
--> n + He3

So here you're saying two Deuteriums will produce either a Tritium and
a proton (or a positron?), or a Helium-3 and a neutron (or a
neutrino?)?

Yes, with about equal probability (tritium and proton, or He3 and a
neutron). These reactions do not involve neutrinos because they are
("conventional") strong nuclear reactions. Look at the details of the
above reactions and note that no proton gets converted into a neutron or
vice versa.

plus about 3.5 MeV, with 50-50 branch ratio, followed by the
secondaries,

D + T --> n + alpha
D + He3 --> p + alpha

[...]

p + p --> D + e+ + nu (W)
p + D --> He3
He3 + He3 --> alpha + 2p

How much energy is liberated in each of these above reactions?

Details for laboratory fusion reactions including tables here:
http://www.rzg.mpg.de/~bds/phys/fusion-energies.html
(from the NRL plasma formulary):
http://wwwppd.nrl.navy.mil/nrlformulary/
I don't remember all the PPI chain numbers. Go to a library and look
them up in:
Clayton, D, Principles of Stellar Evolution and Nucleosynthesis,
McGraw-Hill (1968)

So the lithium stays the same after the reaction, but we get a
different kind of neutrino come out of the proton-proton reaction if
lithium is nearby?

The neutrino from pp is different from the one coming from Li or Be
reactions (element 3: lithium, element 4: beryllium).
For more on this and a much better summary than I could write, look at
Prof. John Bahcall's web site:
http://www.sns.ias.edu/~jnb/

What's the ratio of proton-proton reactions that result in a deuterium
being produced vs. those which just fly apart again? I assume the slow
rate of deuterium production is what slows it down?

Basically, if they fly apart and do nothing then it is not a reaction,
just a scattering event. The cross section for pp fusion is many orders
of magnitude smaller than the one for scattering.
The probability of a reaction per unit volume depends on the density
(amount of stuff) and temperature (how fast it is flying about and
potentially running into each other). Higher density --> more
encounters within a given distance. Higher temperature --> more kinetic
energy per encounter on average.
The cross section basically scales as density squared (things running
into the same things) and is a very steep function of temperature since
the actual reactions at these low relative energies depend on quantum
tunneling (to overcome the electric repulsion between two nuclei).
--
ciao,
Bruce
drift wave turbulence: http://www.rzg.mpg.de/~bds/
.

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