Subfield: Modern Physics

WHY ARE STARS SO LARGE?

The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen.

<img_4491>

Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts.

$$
\begin{aligned}
& \text { USEFUL CONSTANTS } \\
& \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\
& \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\
& \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\
& \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\
& \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\
& \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\
& \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\
& \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\
& \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg}
\end{aligned}
$$



1. A classical estimate of the temperature at the center of the stars. 

Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision.
Context question:
1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures.
Context answer:
\boxed{$5.5 \times 10^{9}$}


Extra Supplementary Reading Materials:

2. Finding that the previous temperature estimate is wrong.

To check if the previous temperature estimate is reasonable, one needs an independent way of estimating the central temperature of a star. The structure of the stars is very complicated, but we can gain significant understanding making some assumptions. Stars are in equilibrium, that is, they do not expand or contract because the inward force of gravity is balanced by the outward force of pressure (see Figure 2). For a slab of gas the equation of hydrostatic equilibrium at a given distance $r$ from the center of the star, is given by

$$
\frac{\Delta P}{\Delta r}=-\frac{G M_{r} \rho_{r}}{r^{2}}
$$

where $P$ is the pressure of the gas, $G$ the gravitational constant, $M_{r}$ the mass of the star within a sphere of radius $r$, and $\rho_{r}$ is the density of the gas in the slab.



<img_4408>

Figure 2. The stars are in hydrostatic equilibrium, with the pressure difference balancing gravity.

An order of magnitude estimate of the central temperature of the star can be obtained with values of the parameters at the center and at the surface of the star, making the following approximations:

$\Delta P=P_{o}-P_{c}$,

where $P_{c}$ and $P_{o}$ are the pressures at the center and surface of the star, respectively.

Since $P_{c} \gg P_{o}$, we can assume that

$\Delta P=-P_{c}$.

Within the same approximation, we can write

$\Delta r \approx R$,

where $R$ is the total radius of the star, and

$M_{r} \approx M_{R}=M$,

with $M$ the total mass of the star.

The density may be approximated by its value at the center,

$\rho_{r}=\rho_{c}$.

You can assume that the pressure is that of an ideal gas.
Context question:
2a Find an equation for the temperature at the center of the star, $T_{c}$, in terms of the radius and mass of the star and of physical constants only.
Context answer:
\boxed{$T_{c}=\frac{G M m_{p}}{2 k R}$}


Extra Supplementary Reading Materials:

We can use now the following prediction of this model as a criterion for its validity:
Context question:
2b Using the equation found in (2a) write down the ratio $M / R$ expected for a star in terms of physical constants and $T_{c}$ only.
Context answer:
\boxed{$\frac{M}{R}=\frac{2 k T_{c}}{G m_{p}}$}


Context question:
2c Use the value of $T_{c}$ derived in section (1a) and find the numerical value of the ratio $M / R$ expected for a star.
Context answer:
\boxed{$1.4 \times 10^{24} $}


Context question:
2d Now, calculate the ratio $M($ Sun $) / R($ Sun $)$, and verify that this value is much smaller than the one found in $(2 \mathrm{c})$.
Context answer:
\boxed{$2.9 \times 10^{21}$}


Extra Supplementary Reading Materials:

3. A quantum mechanical estimate of the temperature at the center of the stars 

The large discrepancy found in (2d) suggests that the classical estimate for $T_{c}$ obtained in (1a) is not correct. The solution to this discrepancy is found when we consider quantum mechanical effects, that tell us that the protons behave as waves and that a single proton is smeared on a size of the order of $\lambda_{p}$, the de Broglie wavelength. This implies that if $d_{c}$, the distance of closest approach of the protons is of the order of $\lambda_{p}$, the protons in a quantum mechanical sense overlap and can fuse.

Question: 3a Assuming that $d_{c}=\frac{\lambda_{p}}{2^{1 / 2}}$ is the condition that allows fusion, for a proton <br> with velocity $v_{r m s}$, find an equation for $T_{c}$ in terms of physical constants <br> only.

Solution:
We have that 

$\lambda_{p}=\frac{h}{m_{p} v_{r m s}}$, and since

$\frac{3}{2} k T_{c}=\frac{1}{2} m_{p} v_{r m s}^{2}$, and

$T_{c}=\frac{q^{2}}{12 \pi \varepsilon_{0} d_{c} k}$, we obtain:

$T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}} \cdot$

Final answer: $T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$