Where does the energy emitted by a star come from?
One source is obviously gravitational potential energy U.
If r decreases, U decreases too, and can be converted into radiated energy. However, because of the Virial Theorem (E = <U>/2) only 1/2 of the change in U can be radiated away (the other half remains as thermal energy to heat the star).
We can calculate U as:
This is done by approximating the density ρ as equal to the average density of the star. R is the radius of the star and M is the mass of the star.
Likewise, the total energy E:
It assumed that the only source of radiation of the star is by loss of gravitational potential energy. This allows to calculate a Kelvin-Helmholtz timescale, i.e. an estimation of the age of the Sun based on constant luminosity and that the initial radius of the Sun was much larger than the current one. The K-H timescale gives an age of ~10e7 years, which clashes with estimations of the age of rocks on the Moon, of 4*10e9 years. Thus another source of energy must exist: it is, as you perhaps already know, nuclear fusion reactions.
The nuclear timescale gives times of ~10e10 years.
Nuclear fusion releases energy as a product of mass loss. For example, a reaction commonly occurring in the center of a star is the fusion of 4 hydrogen atoms into one helium atom, in this process there is necessarily a mass loss of the 0.7%, which is equal (using Einstein’s equation, E = mc2) to 26.71 eV (which is quite a lot). This energy is released due to changes in the configuration of the system, and is called binding energy, Eb.
Factors influencing nuclear reaction rates:
- Energy distribution: We can use Maxwell-Boltzmann to determine the proportion of particles within a given range of energies.
- Interaction probability: Introduction of a cross-section = number of reactions per target nucleus per unit time, divided by the flux of incident particles. It is nothing else that the reaction probability.
- Coulomb Potential Energy Barrier: needs to be overcame for nuclear reactions to take place. It is due to the electrostatic repulsion between nuclei, which at some point is overwhelmed by the attractive strong nuclear force.
- Quantum tunnelling: due to the Uncertainty Principle, knowing the energy of a particle with a certain precision, there is a certain inprecision about its position, and viceversa. This can be translated to the potential energy curve above as: if the particle has energy near the peak, the peak is thin enough so that the particle is not defined to be on one side or the other of the barrier. It jumps between one side and the other, so to say. Not even that, it is at both sides simultaneously. This makes possible for the particle to jump across the barrier, even if it does not have enough energy to do it, classically speaking. Including tunnelling, the temperature estimated to overcome a barrier is:
The “Z” are nuclear charges. e is the charge of an electron. µm is the reduced mass of the two nuclei. k is Boltzmann’s constant, and h is Planck’s constant.
- Electron screening: The free electrons resulting from ionization lower the Coulomb barrier because they screen the positive charge of the nuclei.
The potential added from electron screening is always negative.
We obtain two terms on the energy, that affect the reaction rate for nuclear fusion, a Boltzmann term and a Coulomb term:
The product of these gives a peak (Gamov Peak):
We can write the energy as a power law centered in a particular temperature range.
Once we have an idea of the energy caused from nuclear fusion, we can have the total energy:
Then, the luminosity gradient is:
Lr is the interior luminosity, i.e. the luminosity within a sphere of radius r. Here we use the same idea of spherical shell as in the previous post.