# Difference between revisions of "Fusion"

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Note: missing an order of magnitude preamble lecture \\

### Nuclear Reaction Rates

\The cross section for nuclear reaction depends upon the strong force, and it also depends upon the quantum mechanical probability for tunneling through the Coulomb barrier. These two dependences are independent of one another. The quantum mechanical part also happens to be the part that primarily sets the temperature sensitivity of fusion, and is also the part that is easier to understand. \\Tunneling \To understand tunneling, we start by writing down the Schrodinger equation.

${\displaystyle \left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V\right)\psi =E\psi .\,\!}$

For a simple step like potential, outside of the step, the potential is V = 0, and the solution is a plane wave, and

${\displaystyle E={\frac {k^{2}\hbar ^{2}}{2m}}.\,\!}$

Here, ${\displaystyle k}$ is the wavenumber. In the step, the potential is just some constant ${\displaystyle V=V_{0}}$, and

${\displaystyle E-V_{0}={\frac {k^{2}\hbar ^{2}}{2m}}.\,\!}$

\Large portion missing here (notetaker had to leave lecture)

### Fusion Reaction Rates

\We ended last time by writing down the cross section for fusion in terms of the probability of tunneling and an intrinsic strong nuclear physics part that we will use without trying to understand the detailed physics. This cross section is

${\displaystyle \sigma (E)={\frac {S(E)}{E}}e^{-(E_{G}/E)^{1/2}}.\,\!}$

Here, ${\displaystyle S(E)}$ is the nuclear physics part, ${\displaystyle E}$ is the energy of both particles, and the expontential term is the probability of tunneling through the Coulomb barrier. We wrote in the last class ${\displaystyle E_{G}}$ as

${\displaystyle E_{G}=(1{\rm {\;MeV}})Z_{1}^{2}Z_{2}^{2}{\frac {m_{r}}{m_{p}}}.\,\!}$

This expression can be used to find the probability of a single tunneling event.We want to write down the rate of fusion for a large system of particles. We can start by writing down the rate for lots of particles that all have the same velocity. For distinguishable fusing particles that we will label 1 and 2 with number densities ${\displaystyle n_{1}}$ and ${\displaystyle n_{2}}$, we can write down a mean free path

${\displaystyle \ell _{2}={\frac {1}{n_{1}\sigma }}\,\!}$

and a mean free time

${\displaystyle \tau _{2}={\frac {1}{n_{1}\sigma v}}.\,\!}$

The number of reactions per unit volume per unit time is then

${\displaystyle r_{12}={\frac {n_{2}}{\tau _{2}}}=n_{1}n_{2}\sigma v.\,\!}$

If we average over all possible velocities, the result is

${\displaystyle r_{12}=n_{1}n_{2}<\sigma (E)v>.\,\!}$

To find ${\displaystyle <\sigma (E)v>}$, we need to integrate over the Maxwell-Boltzmann distribution

${\displaystyle <\sigma (E)v>=\int d^{3}v\;prob(v)\sigma (E)v.\,\!}$

Plugging in (with normalization) the MB distribution, the result is

${\displaystyle r_{12}=n_{1}n_{2}\int d^{3}v\sigma (E)v\left({\frac {m_{r}}{2\pi kT}}\right)^{3/2}e^{-{\frac {{\frac {1}{2}}m_{r}v^{2}}{kT}}}.\,\!}$

We would like to put this entirely in terms of the energy, substituting out the velocity. We’ll use

${\displaystyle E={\frac {1}{2}}m_{r}v^{2},\,\!}$
${\displaystyle dE=m_{r}vdv,\,\!}$
${\displaystyle d^{3}v=4\pi v^{2}dv=4\pi {\frac {v^{2}}{v}}{\frac {dE}{m_{r}}},\,\!}$
${\displaystyle vd^{3}v={\frac {8\pi E}{m_{r}}}{\frac {dE}{m_{r}}}.\,\!}$

Now we can rewrite our velocity averaged cross section as

${\displaystyle <\sigma (E)v>=\left({\frac {2}{kT}}\right)^{3/2}{\frac {1}{\sqrt {\pi m_{r}}}}\int dEE\sigma (E)e^{-E/kT}.\,\!}$

Now, plug in our original expression for ${\displaystyle \sigma (E)}$ to find

${\displaystyle <\sigma (E)v>=\left({\frac {2}{kT}}\right)^{3/2}{\frac {1}{\sqrt {\pi m_{r}}}}\int dES(E)e^{-(E_{G}/E)^{1/2}\;-\;E/kT}.\,\!}$

The tunneling term here favors reactions at high energies, but the Boltzmann term favors reactions at low energies where there are a large number of available particles. Compared to the energy terms in the exponential, ${\displaystyle S(E)}$ is slowly varying, and we can pull it out of the integral, writing

${\displaystyle <\sigma (E)v>=\left({\frac {2}{kT}}\right)^{3/2}{\frac {1}{\sqrt {\pi m_{r}}}}S(E)I.\,\!}$

with

${\displaystyle I=\int _{0}^{\infty }e^{-(E_{G}/E)^{1/2}\;-\;E/kT}dE.\,\!}$

We can draw a picture to understand roughly what this integral will give us. The MB distribution falls quickly with energy, and the probability of tunneling rises exponentially with energy. The product of the two is at a maximum at some energy ${\displaystyle E_{0}}$ that falls part way along the tail of each. We can find this ${\displaystyle E_{0}}$ by differentiating the integrand above, which we’ll call ${\displaystyle f(E)}$. Then

${\displaystyle {\frac {df}{dE}}=0={\frac {1}{kT}}-{\frac {E_{G}^{1/2}}{2E^{3/2}}}.\,\!}$

Solving we find

${\displaystyle E_{0}=\left({\frac {1}{2}}E_{G}^{1/2}kT\right)^{2/3}.\,\!}$

In terms of the more fundamental quantities that define ${\displaystyle E_{G}}$,

${\displaystyle E_{0}=(5.7\;{\rm {keV}})Z_{1}^{2/3}Z_{2}^{2/3}T_{7}^{2/3}\left({\frac {m_{r}}{m_{p}}}\right)^{1/3}.\,\!}$

This is the energy at which most reactions occur, even though it is far smaller than ${\displaystyle E_{G}}$ and far larger than ${\displaystyle kT}$, thanks to the competition between the two. In order to perform the integral above, we will treat the overlap between the two as a Gaussian (method of steepest descent).

${\displaystyle f(E)=f(E_{0})+{\frac {1}{2}}(E-E_{0})^{2}f^{''}(E_{0}),\,\!}$

where

${\displaystyle f^{''}(E_{0})={\frac {3E_{G}^{1/2}}{4E_{0}^{5/2}}}.\,\!}$

If we put this in our integral ${\displaystyle I}$, we get a Gaussian integral that we can perform. The result is

${\displaystyle I={\frac {e^{-f(E_{0})}{\sqrt {2\pi }}}{\sqrt {f^{''}(E_{0}}}}.\,\!}$

The final results for our velocity averaged cross section is

${\displaystyle <\sigma (E)v>=2.6S(E_{0}){\frac {E_{G}^{1/6}}{(kT)^{2/3}{\sqrt {m_{r}}}}}e^{-3(E_{G}/4kT)^{1/3}}.\,\!}$

\Now, we can start to do something more macroscopic. Let’s define the energy per unit mass per unit time from fusion as ${\displaystyle \epsilon }$. Then the total luminosity is

${\displaystyle L=\int \epsilon dM_{r}=\int \epsilon 4\pi r^{2}\rho dr.\,\!}$

Then the change in luminosity over radius is

${\displaystyle {\frac {dL_{r}}{dr}}=4\pi r^{2}\rho \epsilon .\,\!}$

We can also define an energy released per reaction as ${\displaystyle Q}$. Using our rate ${\displaystyle r_{12}}$ of reactions per volume per time, we can find ${\displaystyle \epsilon }$, with

${\displaystyle \epsilon _{12}={\frac {r_{12}Q}{\rho }}.\,\!}$

We want to put everything in terms of mass density, which we will do with

${\displaystyle n_{1}={\frac {X_{1}\rho }{m_{1}}}.\,\!}$

Here, ${\displaystyle X_{1}}$ is the fraction of the total mass that is in particle 1. Together, these allow us to write the energy generation from fusion.

${\displaystyle \epsilon _{12}={\frac {2.6QS(E_{0})X_{1}X_{2}}{m_{1}m_{2}{\sqrt {m_{r}}}(kT)^{2/3}}}\rho E_{G}^{1/6}e^{-3(E_{G}/4kT)^{1/3}}.\,\!}$

This expression has units ergs per second per gram, and gives the energy from reactions between particles 1 and 2. It is a function of density, temperature, and composition. Since we do not want to carry this whole expression around most of the time, we would like to just write down a simple proportionality in terms of these fundamental variables. In general, we’ll approximate

${\displaystyle \epsilon \propto \rho ^{\alpha }T^{\beta }.\,\!}$

Depending upon the reaction, ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ can take different values. For any process involving collisions between two particles, ${\displaystyle \alpha =1}$. Finding ${\displaystyle \beta }$ is a bit more complicated. Our expression for ${\displaystyle \epsilon }$ is not a simple power law in temperature. We can approximate it though by noticing that

${\displaystyle \beta ={\frac {d\ln \epsilon }{d\ln T}}.\,\!}$

Writing ${\displaystyle \epsilon }$ in terms of only the temperature dependent quantities and then performing the above differentiation, we find

${\displaystyle \beta =-{\frac {2}{3}}+\left({\frac {E_{G}}{4kT}}\right)^{1/3}.\,\!}$

Note that this is a function of temperature. This is only valid then when we are considering a small range of temperature. Fortunately, this is generally a fine approximation when considering fusion in stars. For the proton-proton chain at the center of a sun like star, we find ${\displaystyle \beta \approx 4.3}$. Then

${\displaystyle \epsilon _{pp}\propto \rho T^{4.3}\,\!}$

at temperatures near ${\displaystyle 10^{7}}$ Kelvin. We can make a bit more detailed an estimate of energy from fusing protons to Helium. Again using ${\displaystyle T_{c}\sim 10^{7}}$ K, ${\displaystyle \rho \sim 1}$ g cm${\displaystyle ^{-3}}$, the cross section ${\displaystyle S(E)}$ is of order 1 keV barn, and ${\displaystyle Q}$ is order 10 MeV. If we put this in our expression for ${\displaystyle \epsilon }$, we find

${\displaystyle \epsilon \sim 10^{20}{\rm {\;erg/s/g}}.\,\!}$

The luminosity of the Sun is then

${\displaystyle L=\int dM_{r}\epsilon \sim \epsilon M_{\odot }.\,\!}$

This is

${\displaystyle L\sim 10^{54}{\rm {\;erg/s}}\sim 10^{20}L_{\odot }.\,\!}$

What happened in our estimate? This is WAY too big. We can think of a couple of resolutions that would give us a more reasonable estimate. One is that the weak interaction is involved somewhere, as it has a cross section that is ${\displaystyle 10^{20}}$ times smaller than the strong interaction. The other possibility is that a proton is fusion with a particle that has a much larger charge, which would suppress the ${\displaystyle E_{G}}$ term and give much lowre energy generation rates. As it turns out, both of these solutions exist in nature. The proton-proton chain involves conversion of protons to neutrons, which requires the weak interaction. The CNO cycle produces Helium by fusion of protons with C, N, and O. Using either, we would find an energy generation rate that is in much better agreement with the observed luminosity of the Sun. \\Let’s look at the proton proton chain in more detail. The net reaction is

${\displaystyle 4p\rightarrow {}^{4}{\rm {He}}+{\rm {energy}}.\,\!}$

This energy comes out as photons, kinetic energy of particles, and a small amount as neutrinos. The kinetic energy and photon energy is all shared with the star through collisions, and produces the heat inside of stars. The energy in neutrinos is not useful, as it just leaves the star without providing any heat (and, by extension, prssure support against gravity). \\As implied already, the key step in the proton proton chain is the step where a proton turns in to a neutron (necessary since there are two neutrons in Helium, and we are starting with only protons). This step gives its identity as a weak interaction away by the presence of a neutrino.

${\displaystyle p+p\rightarrow {}^{2}{\rm {H}}+e^{+}+\nu _{e}.\,\!}$

The cross section for this is ${\displaystyle S(keV)\approx 3.78\times 10^{-22}}$ keV barn. This is, as promised, much weaker than the typical strong reaction, which has a cross section of order of 1 keV barn. The rest of the reactions needed to produce Helium are easy, involving the strong reaction. First,

${\displaystyle {}^{2}{\rm {H}}+p\rightarrow {}^{3}{\rm {He}}+\gamma ,\,\!}$

which has a cross section of 2.5 ${\displaystyle \times 10^{-4}}$ keV barn. and then

${\displaystyle {}^{3}{\rm {He}}+{}^{3}{\rm {He}}\rightarrow {}^{4}{\rm {He}}+2p,\,\!}$

with cross section 5000 keV barn. Note that the first two steps of the reaction must happen twice for the last step to happen once. That this whole process requires multiple steps is a result of the exceedlingly low probability of four protons all happening to be in the same place at the same time. Even so, determining the overall rate is made simpler by the fact that the slowest step in a cycle sets the rate for the entire cycle. Here, our rate limiting step is the first step, when two protons form a deuterium atom. Then we can calculate the energy generation rate for the entire chain with

${\displaystyle \epsilon _{cycle}=r_{p-p\;step}Q_{cycle}/\rho .\,\!}$

This fact will be useful in the CNO cycle as well, which has many steps, but is also rate limited by a single step, allowing us to easily write down the energy generation rate. Using our previous (complicated) expression for energy generation and applying it to the proton-proton step, we find the energy for the entire chain to be

${\displaystyle \epsilon _{pp}\propto \rho T^{-2/3}e^{-15.7T_{7}^{-1/3}}.\,\!}$

With the constants,

${\displaystyle \epsilon _{pp}=(5\times 10^{5}){\frac {\rho X^{2}}{T^{2/3}}}e^{-15.7T_{7}^{-1/3}}{\rm {erg/s/g}}.\,\!}$

We can use this to estimate the central temperature of the Sun, using the known luminosity.

${\displaystyle L=\int \epsilon dM\sim \epsilon (center)M_{\odot },\,\!}$
${\displaystyle L_{\odot }\sim 10^{7}{\frac {M_{\odot }}{T_{7}^{2/3}}}e^{-15.7T_{7}^{-1/3}},\,\!}$
${\displaystyle T_{c}\approx 10^{7}K.\,\!}$

This is pretty good, and is (fortunately) self consistent.

### Hydrogen Fusion in Stars

\One method of fusing hydrogen in the centers of stars is called the proton-proton chain.

${\displaystyle p+p\rightarrow {}^{2}H+e^{+}+\nu _{e}\,\!}$
${\displaystyle {}^{2}H+p\rightarrow {}^{3}He+\gamma \,\!}$
${\displaystyle {}^{3}He+{}^{3}He\rightarrow {}^{4}He+2p\,\!}$

This process is important in lower mass stars. For higher mass stars, the CNO cycle becomes important. Carbon, nitrogen, and oxygen are not actually fused in this process. Instead, they are used as catalysts, and the reaction has the exact same input (four protons) and output (one helium atom) as the proton-proton chain does. The steps of the CNO cycle are

${\displaystyle {}^{12}C+p\rightarrow {}^{13}N+\gamma \,\!}$
${\displaystyle {}^{13}N\rightarrow {}^{13}C+e^{+}+\nu _{e}\,\!}$
${\displaystyle {}^{13}C+p\rightarrow {}^{14}N+\gamma \,\!}$
${\displaystyle {}^{14}N+p\rightarrow {}^{15}O+\gamma \,\!}$
${\displaystyle {}^{15}O\rightarrow {}^{15}N+e^{+}+\nu _{e}\,\!}$
${\displaystyle {}^{15}N+p\rightarrow {}^{12}C+{}^{4}He.\,\!}$

Considering our results for the Gamow Energy, which sets the probability of tunneling, how can the CNO cycle be competitive with the proton-proton chain? The charge involved is much larger, so the Gamow Energy of any of the steps are of order 50 MeV. This is five orders of magnitude larger than the typical energy at the center of a star, which is of order 1 keV. At ${\displaystyle 10^{7}}$ K, the tunneling probability is ${\displaystyle 10^{-7}}$ for the proton-proton chain, and it is a staggering ${\displaystyle 10^{-31}}$ for the CNO cycle. The reason that CNO can be competitive turns out to be because the cross section in every step of the CNO cycle that requires fusing is the strong force, whereas the first step in the proton-proton chain is a weak interaction, with a significantly lower cross section. They happen to be different by roughly ${\displaystyle 10^{24}}$, which roughly cancels the difference in tunneling probability. (Question: what about abundance?) The energy generation rate for the CNO cycle is set by the slowest step, as is the case for any reaction. The slowest step is the fourth step, and gives

${\displaystyle \epsilon _{CNO}\approx (4\times 10^{2}7){\frac {\rho }{T_{7}^{2/3}}}XZe^{-70.7T_{7}^{-1/3}}{\rm {\;erg/g/s}}.\,\!}$

We can do the same trick as with the proton-proton chain to get a simple power law scaling, and find that

${\displaystyle \beta ={\frac {-2}{3}}+{\frac {23.6}{T_{7}^{1/3}}},\,\!}$

with

${\displaystyle \epsilon \propto \rho T^{\beta }.\,\!}$

This is phenomenally sensitive to energy, due to being way out on the tail of the Maxwell-Boltzmann distribution. Raising the temperature just a little bit will cause the number of particles at that energy to rise significantly. While the proton-proton chain was also sensitive to temperature, it is not nearly as sensitive as the CNO cycle. This means that at high temperatures, CNO dominates, and at low temperatures, the proton-proton chain dominates. It turns out that they are roughly equal for stars that are a bit more massive than the Sun. In the Sun, 99% of the luminosity is produced by the proton-proton chain. \\

### Observational Evidence for Fusion in the Sun

\Detections of neutrinos from the Sun provide evidence that the center of the Sun is undergoing fusion. Neutrinos are very light particles, and for a long time were believed to be massless. Now, they are thought to have a mass of order (or smaller than) 0.1 eV, which is six orders of magnitude lighter than an electron. They have no charge (or color), meaning they only are affected by the weak force and gravity. There are three types of neutrinos, corresponding to the three leptons: the electron, muon, and tau particle each has a corresponding neutrino. We are only really concerned with electron neutrinos though, since that is what stars provide in fusion (which can be confirmed by looking back at the reactions we wrote down). There are multiple branches in the proton-proton chain. We talked about the dominant branch, but the other branches produce different energy neutrinos. \The typical cross section for an interaction between neutrinos and matter is

${\displaystyle \sigma \sim 10^{-44}\left({\frac {E_{\nu }}{m_{e}c^{2}}}\right)^{2}{\rm {\;cm^{2}}}.\,\!}$

Note again the twenty orders of magnitude difference between this cross section and something representative of the electromagnetic force, like the Thomson cross section. We can use this to calculate the mean free path of neutrinos in the Sun.

${\displaystyle \ell ={\frac {1}{n\sigma }}.\,\!}$

Using ${\displaystyle E_{\nu }\sim }$ MeV, and the average number density in the Sun, we find

${\displaystyle \ell \sim 10^{9}R_{\odot }.\,\!}$

This means that a neutrino typically leaves the Sun without interacting at all. A neutrino that we can detect on Earth thus tells us about the conditions at the center of the Sun, where the neutrino was produced. Most importantly, the energy of neutrinos depends on the reaction that produced them, so the energy of neutrinos that we detect tells us about fusion in the center of the Sun. Of course, since a neutrino gets out of the Sun unimpeded, detecting a neutrino on Earth is challenging. The first experiment to do this used cleaning fluid in a huge tank underground, and resulted in a Nobel Prize for Ray Davis in 2002. This cleaning fluid had chlorine, and the reaction of interest was

${\displaystyle {}^{37}Cl+\nu _{e}\rightarrow {}^{37}Ar+e^{-}.\,\!}$

Every day, ${\displaystyle 10^{22}}$ neutrinos go through the tank, but on average one neutrino is detected per day. This reaction only operates for neutrinos more energetic than 0.814 MeV, so this experiment was only able to detect neutrinos from some of the less common fusion reactions, as those from the proton-proton chain were of lower energy. \This experiment began in the 1960s, and wound up producing what is called the Solar Neutrino problem. The number of neutrinos detected was one third of the number expected. There was debate for many years between astrophysicists and particle physicists. The resolution was finally provided by Sudbury Neutrino Observatory. The SNO used heavy water, which has deuterium rather than lone protons in the water molecule. An incoming neutrino can break up the deuterium in to two protons and a positron:

${\displaystyle \nu _{e}+D\rightarrow p+p+e^{-}.\,\!}$

This experiment could also detect mu and tau neutrinos, with

${\displaystyle \nu +D\rightarrow p+n+\nu .\,\!}$

This measurement confirmed that the total number of neutrinos produced is the total number expected. The total was expected to be made up entirely of electron neutrinos though, and this experiment showed that only one third of them were electron neutrinos. It turns out that since neutrinos have mass, they can oscillate between different neutrino types. That means that despite the Sun producing only electron neutrinos, we can observe mu and tau neutrinos from the Sun. Thus the neutrino observation experiments provided direct confirmation of solar models, and also proved that neutrinos have mass. \\