Hello guys and welcome to the third installment of our Guide to Thermonuclear Fusion. You can check out the table of content here where you can navigate through different topics on our journey to getting to know Fusion. So far I have talked about "economic aspects" of the issue. How does harvesting fusion energy compare to other clean and renewable energy sources and how viable it is through an engineers point of view. Now we will start with the physics of it, to be more precise what is the catalyst for fusion of two particles in nature.

Before we begin with the physics of it I just want to say that I very much appreciate the response I am getting thus far. To make my contribution even more original I intend to make as many graphics as I can on my own. I have only started learning animation, so bare with me for a few horrendous gifs that I hope will still make my point clear.

Where fusion occurs naturally and what drives it?

I mentioned this already in pretty much every post so you should know the answer already. The answer to where fusion occurs is very simple. In stars. The power of fusion lights up the Sun for example. The light "steals" a part of the Suns energy and brings it to Earth where it enables the prospering of life. That is all very poetic but what actually drives the fusion process.

The center of the Sun is actually very hot as we all know. Later on we will make a simple calculations how hot exactly and where did the heat home from initially, but lets just focus for now on the fact it is hot enough that all atoms are in a state of plasma. This means that center of the Sun consists of nuclei and of electrons. Electrons contribute almost nothing to the overall picture so we will forget them also. All nuclei consist, in general, of protons and neutrons which means they are positively charged. Charges of the same sign are repulsed from each other so how can they fuse. Well the answer is disappointingly unpoetic. It is pure chance. As you can see in the right figure the force between two nuclei is repulsive from about 1 fm up to infinity. But here a surprising fact comes in. In the region of 0 to 1 fm the force is attractive. This is because when particles that form nuclei, meaning neutrons and protons, come very close together they actually like to stick together. When close enough even protons tolerate each others company although for not very long as we all know that a p-p nucleus exists. But wait a minute a p-n nucleus does exist and is stable. I am talking about deuterium of course. One of the three isotopes of hydrogen.

So where does the chance come in, you might ask. Well in a Sun like star the particle density in the center of the star is about n = 10³⁰ m^-3 which we calculate from the mass of the Sun, its radius and the mass of the proton, assuming the entire Sun is made of only protons. From this we can calculate how much space each proton has by taking the inverse cube root. We get that the space between two neighboring protons is about L = 10^-10 m. So this means that particles that are about 0.1 nm apart need to somehow come to a vicinity of 1 fm = 10^-15 m. This can only happen if the protons are headed straight for each other into a head-on collision with a speed fast enough that they dont have the time to "feel" the Coulomb repulsion which would throw them of course. This high speed, when speaking in terms of gas, means very high temperature. For a p-p fusion event to happen with regular frequency the temperature must be in the region of 10 million K.

So how does the Star heat up to such temperatures?

Well once the ignition of H in the star by p-p fusion happens, enough energy is released to keep the star hot enough. But how did it get so hot in the first place? Well this is where my first gif comes into play.

Well all stars at first are just super massive clouds of mainly hydrogen. Because they are so large almost no gravitational force is exerted on it and the temperature of the cloud is that of the surrounding universe which is almost 0 K. For our estimation we will assume that the cloud of H at first has an infinite volume and 0 temperature. Because it is isolated from the universe its mass is constant, therefore the density at first is almost 0. As the infinitesimal gravity force starts to take effect the cloud starts to collapse towards the center of mass. Its density rises with mass being constant of course. At some point H ignition occurs. How would one estimate the radius of the Star when ignition happens by knowing the mass of the beginning cloud?

We can use the conservation of energy and assuming the star is made out of ideal gas! At first with 0 temperature and with infinite size the energy of the cloud is 0. At the end the the Star is a sphere with radius R and mass M. It also has a homogeneous profile T in this assumption which is of course not true, but it will get the job done for this example.

E_start = E_final
0 = n4π/3R³kT - 3GM²/5R

The first part is the internal energy of an ideal gas with particle density n with temperature T trapped in a volume equal to a sphere with radius R. The second is the gravitational energy of a sphere with mass M and radius R. G is the gravitational constant and k is the Boltzmann constant. The gravitational energy is negative because gravitational energy 0 is defined in a way that the infinitely large cloud at the beginning has an energy of 0. By expressing R we get:

So for the Sun with mass equaling M = 2 x 10³⁰kg and a radius of 7 x 10⁸m we get a temperature of around one million K, which is a pretty good estimate.

The p-p cycle

This is the most basic fusion chain reaction that is the dominant one while the core of the star is dominated by H (protons). You can read all this stuff on wikipedia here. I will just quickly summarize and again torture you with my own gif. This one was the second I made, so you can actually notice the improvement already.

The nuclear reaction chain starts off simply enough with two protons (yellow colort) having a head on collision and a fusion reaction occuring turning them into a p-p pair which quickly decays into a p-n pair - a deuteron (blue color)!

p + p → ²₁D + ν_e + e⁺ + 0.42 MeV

The diproton nucleus has a beta decay and emits a neutrino and a positron. The positron usually quickly anihilates with an electron in the plasma and emits two photons with combined energy of 1.02 MeV. The deutoron which is a stable nucleus is then flying around in the plasma and eventually hits another proton and fuses into a isotope of helium (green color):

p + ²₁D → ³₂He + γ + 5.49 MeV

The helium nucleus is left in an excited state at first because it captured another proton. This is not a pun. Its just physical terminology. It soon decays by emitting a gamma photon with the released energy. Again the Helium-3 nucleus is stable which means it again goes flying into the plasma. There it eventually finds another Helium-3 and they fuse into Beryllium-6 (red ball). Because this isotope of Beryllium is unstable it decays by a double beta decay emitting two positrons and neutrinos. The positrons again annihilate with electrons int he plasma which means the final product is a Helium-4 nucleus and a sum of four neutrinos. The energy generated in the entire process is 26.732 MeV.

³₂He + ³₂He → ⁴₂He + 2ν_e + 26.732 MeV

So this is the main energy source of the star as it it dominant through most of the lifetime of the Star. Only when all of the hydrogen in the core is spent does a Helium fusion chain reaction become important coined the CNO cycle. There are of course other fusion chain reactions that fuse heavier elements into elements up to iron but only in the heaviest of stars where temperatures are sufficient. Once iron forms the star is dead. This is because iron is the most stable element in the universe. This means neither fusion nor fusion of iron is energetically favorable. This means the star just keeps pumping energy into trying to fuse iron, but it doesnt work and the star is effectively dead.

Conclusion

Thank you for tuning into my third section in the introduction section in the Guide to Thermonuclear Fusion. I will be back again with the fourth and last section of the introduction next week, as these posts are quite time consuming as you might imagine.

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