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Random Mix-06 An example of Quantum Effects in Stars

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An example of Quantum Effects in Stars

Sarita Vig
Department of Earth and Space Sciences
Indian Institute of Space Science and Technology (IIST)
Valiamala, Tiruvananthpuram 695547, INDIA


We have come across a variety of problems that we solve in order to understand the concepts of Quantum Mechanics. Here, I give an example to illustrate how quantum effects come to our rescue when we face hurdles related to nuclear fusion in stellar interiors. However, at the outset I would like to make it clear that this doesnt involve rigorous calculation, which people (eg. Gamow) have addressed and others are still trying in numerous ways. What I will be describing is the problem and a heuristic approach to overcome it. An important point to bear in mind is that Astrophysics is a subject that deals with numerical quantities that are many orders of magnitude larger than what we encounter in daily life. Hence, sensible approximations are second nature to astrophysicists.

We are aware that the fusion of protons to helium nucleus generates energy in most stellar interiors. The phase of a star when energy from the hydrogen fusion contributes maximum to the energy generation is called main-sequence phase and a star spends most of its lifetime in this phase. The Sun, for example, is roughly midway in this phase. For fusion to occur, the energy of the two interacting nuclei of charges \(Z_1\) and \(Z_2\) must be high enough to overcome the Coulomb potential barrier given by

\begin{equation}  U_c= \frac{Z_1Z_2}{4\pi\epsilon_o} \frac{e^2}{r} \end{equation}

Let us consider energy required  for two protons \((Z_1=Z_2=1)\) to overcome the Coulomb barrier potential and to come together. In this initiating step  deuterium nucleus in formed. The following  steps involve fusion of a deuterium nucleus with a proton to form $^3$He. Thereafter, $^3$He forms $^4$He through a number of plausible fusion chain reactions.

We can estimate the temperature $T$ of the ionised plasma, where most of the particles have sufficient energy to overcome the barrier. This is given by
\begin{equation}
 \frac{1}{2} \mu_{r} v^2 = \frac{3}{2} k_B T
\end{equation}
On the left hand side of the above equation, we have the proton (of mass $m_p$) kinetic energy in the reduced mass frame where the
\begin{equation}
 \mu_{r} = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_p}{2}
\end{equation}
Considering Eqns. 1 and 2, we get
\begin{equation}
 T \ge \frac{2}{3}\frac{1}{4\pi\epsilon_o} \frac{e^2}{r\,k_B}
\end{equation}
At an interparticle distance of roughly 1 fm$=10^{-15}$m, the attractive strong force comes into play. If we calculate the above expression using $r = 10^{-15}$m, we get $T\sim10^{10}$ K. This would mean that classically, a temperature of nearly 10 billion kelvin is needed to initiate nuclear fusion. This is orders of magnitude higher than the central temperature of the Sun (16 million kelvin) that is obtained on the basis of simulated models or simple arguments using the virial theorem. Even if we consider the high energy tail of the Maxwell Boltzmann distribution, the number of particles is so minute that energy production by this method can be ruled out.

Let us next consider the effect of quantum mechanics by considering the wave nature of proton. The particles are able to approach close to each other by tunneling through the Coulomb barrier. Let us estimate the de-Broglie wavelength $\lambda$ where the kinetic energy of particle is the Coulomb barrier energy at this separation. With de Broglie wavelength \( \lambda = \frac{h}{p}\), we have
\begin{eqnarray}
\frac{1}{4\pi\epsilon_o} \frac{e^2}{\lambda} &=& \frac{1}{2}\mu_{r} v^2\ \
\text{and  } \frac{p^2}{2\mu_{r}}  &= &\frac{h^2}{2\mu_{r} \lambda^2}
\end{eqnarray}
Therefore, we get the expression for de-Broglie wavelength as
\begin{equation}
 \lambda = \frac{4\pi\epsilon_o}{e^2}\frac{h^2}{2\mu_{r}}
\end{equation}
Substituting for $r$ in Eqn 4, we have
\begin{equation}
 T \ge \frac{4}{3}\frac{1}{(4\pi\epsilon_o)^2} \frac{e^4\,\mu_r}{h^2\,k_B}
\end{equation}
Substituting the values of various constants in the above expression, we obtain the temperature for proton-proton fusion as $T\sim10^7$ K. This is a more realistic estimate and the values obtained from models show that the central temperature in the Sun is $1.6\times10^7$ K, very close to the value obtained above!

However, this is not the full story. Only a rigorous calculation of tunneling probability can tell us what fraction can tunnel through at a given temperature. In the case of proton-proton fusion to deuterium, the tunneling probability is $\sim10^{-18}$ per proton at $10^6 - 10^7$ K implying that only under high density conditions will a sizeable fraction of protons fuse together to release energy. This first step is the bottle neck as it has the lowest reaction rate among fusion reactions leading to formation of helium. Considering the enormous proton densities in the Sun, this reaction becomes feasible. To see how large the proton densities can be in stellar interiors, a simple exercise would be to calculate the average proton density in the Sun by considering the mass (M$_\odot=2\times10^{30}$ kg) and radius (R$_\odot=7\times10^8$ m) of Sun and assuming that all the hydrogen is ionised.

References

  1.  An Introduction to Modern Astrophysics, B. W. Caroll & D. A. Ostlie, Pearson International (Second Edition), p300
  2.  Solar Fusion Cross Sections, E. G. Adelberger et al. 1998, Reviews of Modern Physics, 70, 1265, and references therein.

 

 

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