Gamow peak and nuclear reaction rate

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SUMMARY

The discussion centers on the Gamow peak and its relation to nuclear reaction rates in stars, specifically defined by the equation $$R_{ab}=n_a n_b\left<\sigma v\right>$$. The formula incorporates factors such as the Maxwell velocity distribution and tunneling probability, highlighting that the reaction rate approaches zero as temperature increases, contradicting typical fusion behavior. Participants clarify that the Gamow peak represents a balance between particle energy distribution and fusion rates, akin to the Planck function in quantum mechanics.

PREREQUISITES
  • Understanding of nuclear physics concepts, particularly fusion processes.
  • Familiarity with the Maxwell velocity distribution and its implications in particle interactions.
  • Knowledge of the Boltzmann constant and its role in thermodynamic equations.
  • Basic grasp of quantum tunneling and its significance in nuclear reactions.
NEXT STEPS
  • Study the derivation and implications of the Gamow peak in stellar nucleosynthesis.
  • Explore the Maxwell-Boltzmann distribution and its application in particle physics.
  • Investigate quantum tunneling effects in nuclear fusion processes.
  • Examine the Planck function and its relevance to energy distribution in thermodynamic systems.
USEFUL FOR

Astrophysicists, nuclear physicists, and students studying stellar dynamics and nuclear fusion processes will benefit from this discussion.

wieblinger
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It's known that the nuclear reaction rate (inside a Star) can be determined with

$$R_{ab}=n_a n_b\left<\sigma v\right> \, \approx \, n_a n_b \Big(\frac{8}{\pi m_e}\Big)^{1/2} \frac{S(E_0)}{(k_BT)^{3/2}} \Delta \frac{\sqrt{\pi}}{2} e^{-3E_0/k_BT},$$

where ##k_B## is Boltzmann constant, ##T## the temperature, ##v## the velocity, ##\sigma## the cross section, ##E## the energy, and
$$\Delta=\frac{4}{\sqrt{3}}\Big(\frac{b}{2}\Big)^{1/3}(k_B T)^{5/6}.$$

The above formula is found by using Maxwell velocity distribution and tunneling probability, since
$$\left<\sigma v\right>=\int_0^{\infty} \sigma(E)v(E)f(E)dE.$$

The maximum of the reaction rate is called Gamow peak and I guess that is achieved with a proper trade-off of the Maxwell velocity distribution and the tunneling probability.

As the temperature tends to infinity, the reaction rate approaches to 0. Is it due to the fact that according to Maxwell velocity distribution there are less particles with higher temperature and, thus, there will be less probability that two atoms collide?
 
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wieblinger said:
As the temperature tends to infinity, the reaction rate approaches to 0. Is it due to the fact that according to Maxwell velocity distribution there are less particles with higher temperature and, thus, there will be less probability that two atoms collide?

If the temperature is approaching infinity, then the number of particles with high temperatures should increase, not decrease, right?
 
There must be something wrong with that formula, because fusion rates generally rise, and quite drastically, with temperature. That formula would give a peak in the fusion rate as a function of temperature, but there is not such a peak. The "Gamow peak" is a peak in what you said, the product (at a given temperature) of the number of particles with a given energy, times the fusion rate for that energy per particle. It's analogous to the Planck function, which is like the excitation number of the states at a given energy, times the number of states at that energy per photon. The Planck function also peaks with energy for a given T, but it is a monotonically increasing function of T at any energy.
 

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