# With what probability does nuclear fusion occur at energies far below the Coulomb barrier?

Even at the core of the sun, the temperature of $\sim 10^7$ K only results in $kT\sim1$ keV, which is about a thousand times less than the electrical potential energy of $\sim1$ MeV needed in order to bring two hydrogen nuclei to within the ~1 fm range of the strong nuclear force. Therefore nuclear fusion reactions can only occur inside the sun, or in any other normal star, through the process of quantum-mechanical tunneling. The low probability of this tunneling, along with the need for a weak interaction in order to fuse two protons into a deuterium nucleus, are the two factors that make stars have lifetimes billions of years long.

How is the tunneling probability calculated?

• – Ben Crowell Aug 12 '13 at 16:28

The WKB approximation states that in one dimension, the tunneling probability $P$ can be approximated as

$\ln P=-\frac{2\sqrt{2m}}{\hbar}\int_a^b \sqrt{V-E} dx$ ,

where the limits of integration $a$ and $b$ are the classical turning points, $m$ is the reduced mass, the electrical potential $V$ is a function of $x$, and $E$ is the total energy. Setting $V=kq_1q_2/x$, we have for the integral

$I=\int_a^b \sqrt{V-E} dx$

$=\frac{kq_1k_2}{\sqrt{E}}\int_{A}^1\sqrt{u^{-1}-1} du$ ,

where $A=a/b$. The indefinite integral equals $-u\sqrt{u^{-1}-1}+\tan^{-1}\sqrt{u^{-1}-1}$, and for $A\ll 1$ the definite integral is then $\pi/2$. The result is

$\ln P=-\frac{\pi kq_1q_2}{\hbar}\sqrt{\frac{2m}{E}}$ .

This result was obtained in Gamow 1938, and $G=-(1/2)\ln P$ is referred to as the Gamow factor or Gamow-Sommerfeld factor.

The fact that the integral $\int_A^1\ldots$ can be approximated as $\int_0^1\ldots$ tells us that the right-hand tail of the barrier dominates, i.e., it is hard for the nuclei to travel through the very long stretch of $\sim 1$ nm over which the motion is only mildly classically forbidden, but if they can do that, it's relatively easy for them to penetrate the highly classically forbidden region at $x\sim1$ fm. Surprisingly, the result can be written in a form that depends only on $m$ and $E$, but not on $a$, i.e., we don't even have to know the range of the strong nuclear force in order to calculate the result.

The generic WKB expression depends on $E$ through an expression of the form $V-E$, which might have led us to believe that with a 1 MeV barrier, it would make little difference whether $E$ was 1 eV or 1 keV, and fusion would be just as likely in trees and houses as in the sun. But because the tunneling probability is dominated by the tail of the barrier, not its peak, the final result ends up depending on $1/\sqrt{E}$.

Because $P$ increases extremely rapidly as a function of $E$, fusion is dominated by nuclei whose energies lie in the tails of the Maxwellian distribution. There is a narrow range of energies, known as the Gamow window, in which the product of $P$ and the Maxwell distribution is large enough to contibute significantly to the rate of fusion.

Gamow and Teller, Phys. Rev. 53 (1938) 608

• $P$ is a probability per collision? What constitutes collision in plasma? And why not go a bit further than Gamow and produce reactivity rate $\langle \sigma v \rangle$ averaged over thermal distribution? – user23660 Aug 12 '13 at 16:27
• @user23660: Yes, $P$ is the probability per collision (or per barrier assault). I think it should probably be interpreted as the chance of success when the two nucleons approach each other in an s-wave collision, which I think occurs about as often as you would expect based on the simple geometrical cross-section. (This is all a bit fuzzy because the result is derived in a one-dimensional approximation.) And why not go a bit further than Gamow and produce reactivity rate ⟨σv⟩ averaged over thermal distribution? That would be cool to see -- maybe you could do it as a separate Q&A? – Ben Crowell Aug 12 '13 at 16:34
• I mean in 1D collision (classically) is when E=V and 1D approximation works. But in 3D we have mostly glancing collisions and classically $E > V$ nearly always. – user23660 Aug 12 '13 at 16:36
• @user23660: The standard way of generalizing to three dimensions is to use a radial Schrodinger equation, in which the potential has a centrifugal barrier added to it. That results in a different integral, with a term appearing in it that looks like $l(l+1)/r^2$ and that reduces the probability. But for $l=0$ (s-wave), this term vanishes. The generalization of WKB to the radial wave equation is discussed here: galileo.phys.virginia.edu/classes/752.mf1i.spring03/WKB.pdf – Ben Crowell Aug 12 '13 at 16:43
• Of course! Thanks. Now on to how to obtain the rate of collisions with an impact parameter $b$ and center of mass energy E. – user23660 Aug 12 '13 at 17:01

$$E_{out} = 4 \pi n_1 n_2 (m_0-m_a) c^2{(\frac {m}{2k_B T})}^{3/2} \int_0^ \infty \sigma(v) v^2 e^{\frac {-mv^2}{2k_B T}} dv$$

$$m_0$$ is the sum masses of the reacting nuclei, $$m_a$$ is the sum masses of the product nuclei, $$n_1$$ is the number density of one of the reacting nucleus and $$n_2$$ is the number density of the other reacting nucleus, $$\sigma(v)$$ is the probability of fusion reaction happening, if the energy is less than the coulomb barrier, it is the probability of quantum tunneling through the coulomb barrier. If we could find that probability we could calculate the total energy output of a nuclear fusion reaction.

• As far as I understand, the initial question was about the calculation of $\sigma(v)$. – Gec Dec 25 '18 at 13:27