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Statistical Mechanics
White Dwarf Stars
White Dwarf Stars
These are old stars in which most of the Hydrogen is converted into Helium through thermo nuclear reactions. These are the most common end of of evolution of stars ( others are black holes, neutron stars ).The material content is almost Helium. The rate of thermonuclear energy is very slow. Hence the luminosity of the star is very low. The lttle brightness that exists however comes from the gravitaional energy released due to the process of slow contraction .
The mass of the white dwarf is around $10^{33} \mathrm{gm}$ of Helium which is nearly the mass of the sun and the radius is nearly the same as the radius of the earth. Thus the density is roughly $\rho=10^{7} \mathrm{gm} /$ $\mathrm{cm}^{3}$ The temperature of the core is almost $\mathrm{T}=10^{7} \mathrm{~K}$. The high tempereture is due to mean thermal energy of the order of $10^{3} \mathrm{eV}$ whcich is much larger than the ionisation potentials of Helium.
Unlike most other stars that are supported against their own gravitation by normal gas pressure, white dwarf stars are supported by the degeneracy pressure of the electron gas in their interior. Degeneracy pressure is the increased resistance exerted by electrons composing the gas, as a result of stellar contraction. The application of the so-called Fermi-Dirac statistics and of special relativity to the study of the equilibrium structure of white dwarf stars leads to the existence of a mass-radius relationship through which a unique radius is assigned to a white dwarf of a given mass; the larger the mass, the smaller the radius. Furthermore, the existence of a limiting mass is predicted, above which no stable white dwarf star can exist. This limiting mass, known as the Chandrasekhar limit, is on the order of 1.4 solar masses. Both predictions are in excellent agreement with observations of white dwarf stars.
| Property |
Details |
| Formation |
Formed after red giant phase of stars with mass less than about 8 M☉ |
| Supporting Pressure |
Electron degeneracy pressure (quantum mechanical origin) |
| Statistics Obeyed |
Fermi–Dirac statistics |
| Typical Mass |
~ 0.6 M☉ |
| Maximum Mass |
Chandrasekhar limit ≈ 1.4 M☉ |
| Radius |
Approximately equal to Earth's radius (~ 7000 km) |
| Density |
~ 106 to 109 g/cm³ |
| Electron Density |
~ \( 10^{30} \) particles /m³ |
| Temperature (Surface) |
~ 8,000 K to 40,000 K (initially very hot) |
| Composition |
Mostly Carbon and Oxygen (sometimes Helium) |
| Mass–Radius Relation |
R ∝ M−1/3 (Radius decreases as mass increases) |
| Relativistic Effect |
At high density electrons become relativistic leading to instability |
| Fate if Mass > 1.4 M☉ |
Collapse into Neutron Star or trigger Type Ia Supernova |
| Example |
Sirius B |
| Reason of Low Lumninosity |
Energy released in Gravitational Contraction(0.02% times that of sun) |
Important Equation-1 |
\(
P \propto \rho^{5/3} \quad (\text{Non-relativistic})
\) |
Important Equation-2 |
\(
P \propto \rho^{4/3} \quad (\text{Ultra-relativistic})
\)
|
Determination of Chandrasekhar Limit: Method-1
The microscopic contents of a white dwarf may be considered as N electrons each of mass m and $N / 2$ helium nuclei ( each of mass $\approx 4 \mathrm{~m}_{\mathrm{p}}$ )
Thus mass of WHITE DWARF is :
\begin{align*}
M &= N m + \frac{1}{2} N (4 m_p) \approx 2N m_p \\
\Rightarrow \quad n &= \frac{N}{V} = \frac{M / 2m_p}{M / \rho} \\
\Rightarrow \quad n &= \frac{\rho}{2m_p} \\
\Rightarrow \quad n &= \frac{10^{7}}{2 \times 1.6 \times 10^{-24}\ \mathrm{kg}} \\
&= \frac{10^{7}}{3.2 \times 10^{-24}} \\
&= 3.125 \times 10^{30}\ \mathrm{particles/m^3}
\end{align*}
Thus, the white dwarf can be considered as a strongly degenerate electron gas, with \( N \) electrons each having Fermi momentum \( P_{F} \) as:
\[
p_{F} = \left( \frac{3N}{8\pi V} \right)^{1/3} h
= \left( \frac{3n}{8\pi} \right)^{1/3} h
\approx 3 \times 10^{-17} \ \mathrm{g/s}
\]
This momentum is comparable with the charactersitic momentum $\approx m_{0} c \approx 3 \times 10^{-16} \mathrm{gmcm} / \mathrm{s}$ of an electron. Thus the fermi energy of the electron would be comparable with the rest mass energy $m_{0} c^{2}$. It shows that motion of the electron is relativistic and the elctron gas is in completely degenerate state. The contribution of Helium nuclei in comparision to the electrons is insignificant and can be neglected. In addition ,contribution of nuclei in radiation is also neglected.
Thus wite dwarf is left with electron gas only.
considering ground state properties of N relativistic electrons with degeneacy 2 , The number of electrons in the momentum range $p$ and $p+d p$ is :
$$
d N=2 \times \frac{V \times 4 \pi p^{2} d p}{h^{3}}=\frac{8 \pi p^{2} V d p}{h^{3}}
$$
Total Number of electron is:
$N=\int d N=\int_{0}^{P_{F}} \frac{8 \pi p^{2} V d p}{h^{3}}=\frac{8 \pi V}{3 h^{3}} p_{F}^{3}$
Kinetic Energy of the relativistic electron is
$$
\begin{aligned}
& E^{\prime 2}=m_{0}^{2} c^{4}+p^{2} c^{2}=m_{0}^{2} c^{4}\left(1+\left(\frac{p}{m_{0} c}\right)^{2}\right) \\
& \Rightarrow E^{\prime}=m_{0} c^{2}\left(1+\left(\frac{p}{m_{0} c}\right)^{2}\right)^{1 / 2} \\
& \Rightarrow E=E^{\prime}-m_{0} c^{2}=m_{0} c^{2}\left(1+\left(\frac{p}{m_{0} c}\right)^{2}\right)^{1 / 2}-m_{0} c^{2} \\
& \Rightarrow E=m_{0} c^{2}\left(\left\{1+\left(\frac{p}{m_{0} c}\right)^{2}\right\}^{1 / 2}-1\right)
\end{aligned}
$$
Total Kinetic energy of the electron is :
$$
\begin{aligned}
& E_{0}=\int E d N=\int E d N=\int_{0}^{p_{F}} E \frac{8 \pi V}{h^{3}} p^{2} d p \\
& =\frac{8 \pi V m_{0} c^{2}}{h^{3}} \int_{0}^{p_{F}}\left(\left(1+\left(\frac{p}{m_{0} c}\right)\right)^{1 / 2}-1\right) p^{2} d p
\end{aligned}
$$
Total pressure
\begin{align*}
P_{0}&=\int d P_{0}\\
&=\int_{0}^{P_{F}} \frac{p u}{3 V} d N\\
&=\int_{0}^{P_{F}} \frac{p u}{3 V} \frac{8 \pi V}{h^{3}} p^{2} d p\\
& =\int_{0}^{P_{F}} \frac{p}{3 V}\left(\frac{p}{m_{0}\left(1+\left(\frac{p}{m_{0} c^{2}}\right)^{1 / 2}\right)}\right) \frac{8 \pi V}{h^{3}} p^{2} d p \\
& =\int_{0}^{P_{\mathrm{F}}} \frac{8 \pi}{3 h^{3}}\left(\frac{p^{4}}{m_{0}\left(1+\left(\frac{p}{m_{0} c}\right)^{2}\right)^{1 / 2}}\right) d p \\
& =\int_{0}^{\theta_{\mathrm{F}}} \frac{8 \pi}{3 h^{3}}\left(\frac{\left(m_{0} c \sinh \theta\right)^{4}}{m_{0}\left(1+\left(\frac{m_{0} c \sinh \theta}{m_{0} c}\right)^{2}\right)^{1 / 2}}\right)\left(m_{0} c \cosh \theta\right) d \theta \\
& =\int_{0}^{\theta_{F}} \frac{8 \pi}{3 h^{3}}\left(\frac{\left(m_{0} c \sinh \theta\right)^{4}}{m_{0}\left(1+\left(\sin ^{2} h \theta\right)^{2}\right)^{1 / 2}}\right)\left(m_{0} c \cosh \theta\right) d \theta \\
& =\int_{0}^{\theta_{F}} \frac{8 \pi m_{0}^{4} c^{5}}{3 h^{3}}\left(\sinh ^{4} \theta\right) d \theta=\frac{\pi m_{0}^{4} c^{5}}{3 h^{3}} \int_{0} 8\left(\sinh ^{4} \theta\right) d \theta \\
& =\frac{\pi m_{0}^{4} c^{5}}{3 h^{3}} A(x)
\end{align*}
Where $A(x)=\int_{0}^{\theta_{\mathrm{F}}} 8\left(\sinh ^{4} \theta\right) d \theta$ Taking $x=\sinh \theta$
$$
\begin{aligned}
& A(x)=\int_{0}^{\theta_{F}} 2(-2 \sinh \theta)^{2} d \theta \\
& =\int_{0}^{\theta_{F}} 2(\cosh 2 \theta-1)^{2} d \theta \\
& =\int_{0}^{\theta_{F}} 2\left(\cosh ^{2} 2 \theta+1-2 \cosh 2 \theta\right) d \theta \\
& =\int_{0}^{\theta_{F}}\left(2 \cosh ^{2} 2 \theta+2-4 \cosh 2 \theta\right) d \theta\\
& =\int_{0}^{\theta_{F}}(1+\cosh 4 \theta+2-4 \cosh 2 \theta) d \theta \\
& =\int_{0}^{\theta_{F}}(\cosh 4 \theta-4 \cosh 2 \theta+3) d \theta \\
& =\left(\frac{\sinh 4 \theta_{F}}{4}-2 \sinh 2 \theta_{F}+3 \theta_{F}\right) \\
& =\left(\sinh \theta_{F} \cosh \theta_{F} \cosh 2 \theta_{F}-4 \sinh \theta_{F} \cosh \theta_{F}+3 \theta_{F}\right) \\
& =\left(\sinh \theta_{F} \cosh \theta_{F}\left(\cosh 2 \theta_{F}-4\right)+3 \theta_{F}\right) \\
& =\left(\sinh \theta_{F} \cosh \theta_{F}\left(1+2 \sinh ^{2} \theta_{F}-4\right)+3 \theta_{F}\right) \\
& =\left(x \sqrt{1+x^{2}}\left(1+2 x^{2}-4\right)+3 \sinh ^{-1} x\right) \\
& =\left(x \sqrt{1+x^{2}}\left(2 x^{2}-3\right)+3 \sinh ^{-1} x\right) \\
& =\left(\frac{8}{5} x^{5}-\frac{4}{7} x^{7}+\frac{1}{3} x^{9}-\frac{5}{22} x^{11}+\ldots .\right) \quad(\text { for } x<<1) \\
& =\left(\frac{8}{5} x^{5}-\frac{4}{7} x^{7}+\frac{1}{3} x^{9}-\frac{5}{22} x^{11}+\ldots .\right) \quad(\text { for } x<<1) \\
& \text { Where } x=\sinh \theta_{F}=\frac{P_{F}}{m_{0} c}=\left(\frac{3 n}{8 \pi}\right)^{1 / 3}\left(\frac{h}{m_{0} c}\right)
\end{aligned}
$$
As The elctron gas remains stable . The change in energy of the gas due to Adibatic expansion will be balanced by the change in energy due to graviatational force due to change in size considering the shape of white dwarf to be spherical ,Gravitaional Potential energy is :\\
\begin{align*}
\text{GPE} &= -\frac{3}{5} \frac{G M^{2}}{R} \\
&= \alpha \frac{G M^{2}}{R}
\end{align*}
Thus: $d E_{0}+d E_{g}=0$
\[
\begin{align*}
& \Rightarrow -P_{0}(R)\, 4\pi R^{2}\, dR + \alpha \frac{G M^{2}}{R^{2}}\, dR = 0 \\[6pt]
& \Rightarrow P_{0}(R) = \frac{\alpha}{4\pi}\frac{G M^{2}}{R^{4}} \\[6pt]
& \Rightarrow \frac{\pi m_{0}^{4} c^{5}}{3 h^{3}} A(x) = \frac{\alpha}{4\pi}\frac{G M^{2}}{R^{4}}
\end{align*}
\]
$\& A(x)=\dfrac{3 \alpha h^{3} G M^{2}}{4 \pi^{2} m_{0}^{4} c^{5} R^{4}}$.
Case (1) For $R>>10^{8} \mathrm{~cm}$, when $x<<1$
$$
\begin{aligned}
A(x)&=\frac{8}{5} x^{5} \\
\text { Where } x & =\left(\frac{3 n}{8 \pi}\right)^{1 / 3}\left(\frac{h}{m_{0} c}\right)\\
&=\left(\frac{3}{8 \pi} \frac{\rho}{2 m_{p}}\right)^{1 / 3}\left(\frac{h}{m_{0} c}\right) \\
& \Rightarrow x =\left(\frac{9 M}{64 \pi^{2} m_{p}}\right)^{1 / 3}\left(\frac{h / m_{0} c}{R}\right) \\
& \Rightarrow \frac{\pi m_{0}^{4} c^{5}}{3 h^{3}}\left(\frac{8}{5} x^{5}\right)=\frac{\alpha}{4 \pi} \frac{G M^{2}}{R^{4}} \\
& \Rightarrow \frac{\pi m_{0}^{4} c^{5}}{3 h^{3}} \frac{8}{5}\left[\left(\frac{9 M}{64 \pi^{2} m_{p}}\right)^{1 / 3}\left(\frac{h / m_{0} c}{R}\right)\right]^{5} =\frac{\alpha}{4 \pi} \frac{G M^{2}}{R^{4}} \\
\Rightarrow R &=\frac{8}{5} \frac{3\left(9^{2 / 3}\right) h^{2} M^{-1 / 3}}{256 \pi^{4 / 3} m_{p}^{5 / 3} m_{0} G \alpha} \\
&= \frac{3\left(9^{2 / 3}\right) h^{2} M^{-1 / 3}}{160 \alpha \pi^{4 / 3} G m_{p}^{5 / 3} m_{0}} \\
& \Rightarrow R \alpha M^{-1 / 3}
\end{aligned}
$$
Case-(2)
When $x<<1$\\
Then $A(x)=2 x^{4}-2 x^{2}$
$$
\begin{aligned}
\Rightarrow \frac{3 \alpha h^{3} G M^{2}}{4 \pi^{2} m_{0}^{4} c^{5} R^{4}} & =2 x^{4}\left(1-\frac{1}{x^{2}}\right)\\
&=2\left(\frac{9 M}{64 \pi^{2} m_{p}}\right)^{4 / 3}\left(\frac{h / m_{0} c}{R}\right)^{4}\left(1-\frac{1}{\left.\left(\frac{9 M}{64 \pi^{2} m_{p}}\right)^{2 / 3}\left(\frac{h / m_{0} c}{R}\right)^{2}\right)}\right. \\
\Rightarrow \frac{3 \alpha h^{3} G M^{2}}{4 \pi^{2} m_{0}^{4} c^{5} R^{4}} \frac{\left(64 \pi^{2} m_{p}\right)^{4 / 3} R^{4}\left(m_{0} c\right)^{4}}{2(9 M)^{4 / 3} h^{4}}&=\left(1-\frac{\left(64 \pi^{2} m_{p}\right)^{2 / 3} R^{2}\left(m_{0} c\right)^{2}}{(9 M)^{2 / 3}(h)^{2}}\right) \\
\Rightarrow \frac{96 \alpha \pi^{2 / 3} M^{2 / 3} G m_{p}^{4 / 3}}{(9)^{4 / 3} h c}& =\left(1-\frac{\left(64 \pi^{2} m_{p}\right)^{2 / 3} R^{2}\left(m_{0} c\right)^{2}}{(9 M)^{2 / 3}(h)^{2}}\right) \\
\Rightarrow \frac{\left(64 \pi^{2} m_{p}\right)^{2 / 3} R^{2}\left(m_{0} c\right)^{2}}{(9 M)^{2 / 3}(h)^{2}}&=1-\frac{96 \alpha \pi^{2 / 3} M^{2 / 3} G m_{p}^{4 / 3}}{(9)^{4 / 3} h c}\\
\Rightarrow R^{2}&=\frac{(9 M)^{2 / 3}(h)^{2}}{\left(64 \pi^{2} m_{p}\right)^{2 / 3}\left(m_{0} c\right)^{2}}\left(1-\frac{96 \alpha \pi^{2 / 3} M^{2 / 3} m_{p}^{4 / 3}}{(9)^{4 / 3} h c}\right) \\
\Rightarrow R& =\frac{(9 M)^{1 / 3}(h)}{\left(64 \pi^{2} m_{p}\right)^{1 / 3}\left(m_{0} c\right)} \sqrt{1-\frac{96 \alpha \pi^{2 / 3} M^{2 / 3} \mathrm{Gm}_{p}^{4 / 3}}{(9)^{4 / 3} h c}} \\
\Rightarrow R&=\frac{(9 M)^{1 / 3}(h)}{\left(64 \pi^{2} m_{p}\right)^{1 / 3}\left(m_{0} c\right)} \sqrt{1-\frac{M^{2 / 3}}{(9)^{4 / 3} h c /\left(96 \alpha \pi^{2 / 3} G m_{p}^{4 / 3}\right)}} \\
\Rightarrow R&=\frac{(9 M)^{1 / 3}(h)}{\left(64 \pi^{2} m_{p}\right)^{1 / 3}\left(m_{0} c\right)} \sqrt{1-\left(\frac{M}{M_{0}}\right)^{2 / 3}} \\
\quad \text { where } M_{0}^{2 / 3}&=\frac{(9)^{4 / 3} h c}{96 \alpha \pi^{2 / 3} G m_{p}^{4 / 3}} \\
\Rightarrow M_{0}&=\frac{81(h c)^{3 / 2}}{(96 \alpha)^{3 / 2} \pi G^{3 / 2} m_{p}^{2}}=\frac{81(h c / G \alpha)^{3 / 2}}{(96)^{3 / 2} \pi m_{p}^{2}}
\end{aligned}
$$
Hence R adius of the White Dwarf will be real if $ M \ll M_{0}$ i.e mass of the White dwarf must be less than a particular mass $M_{0}$ called as Chandrasekhar Limit . $M_{0}=1.44 M_{\text {sun }}$. This mass is called Chandrasekhar Limit.
ADDITIONAL THEORIES ON WHITE DWARFS
HOW IT IS FORMED:
Every star burns under the nuclear fusion reaction by fusing Hydrogen into Helium, this fusion forces star to outwards (due to internal pressure) but gravity of the star forces it inwards - giving rise to a shape of ball.\\
But less massive stars aren't strong enough to create this pressure, making the star to be crushed down due to its own gravity. This gravity is so strong that even electrons are smashed inwards. When there's no space left for the electrons to be smashed further, it will eventually form a White Dwarf (which is obviously very dense).
POSITION OF WHIT DWARFS :
The nearest known white dwarf is Sirius B, at 8.6 light years, the smaller component of the Sirius binary star. There are currently thought to be nine white dwarf stars within 25 light years among the hundred star systems nearest the Sun.
DISCOVERY:
The unusual faintness of white dwarfs was first recognized in 1910.[3] The name white dwarf was coined by Willem Luyten in 1922.
The estimated radii of observed white dwarfs are typically $0.8-2 \%$ the radius of the Sun; this is comparable to the Earth's radius of approximately $0.9 \%$ OF solar radius Although white dwarfs are known with estimated masses as low as 0.17 M and as high as 1.33 M the mass distribution is strongly peaked at 0.6 M ?, and the majority lie between 0.5 and 0.7 M .
It packs mass comparable to the Sun's into a volume that is typically a million times smaller than the Sun's.\\
A typical white dwarf has a density of between $10^{4}$ and $10^{7} \mathrm{~g} / \mathrm{cm}^{3}$\\
A white dwarf, also called a degenerate dwarf, is a stellar core remnant composed mostly of electron-degenerate matter. A white dwarf is very dense: A white dwarf's faint luminos ity comes from the emission of stored thermal energy; no fusion takes place in a white dwarf wherein mass is converted to energy.
HOW STABLE ?
Compression of a white dwarf will increase the number of electrons in a given volume. Applying the Pauli exclusion principle, this will increase the kinetic energy of the electrons, thereby increasing the pressure. This electron degeneracy pressure supports a white dwarf against gravitational collapse. The pressure depends only on density and not on tempera ture.
CHANDRASEKHAR LIMIT :
White dwarfs can attain mass that is smaller than 1.44 times the mass of Sun called as CHANDEASEKHAR LIMIT .
WHAT HAPPENS IF WHITE DWARF CROSSES CHANDRASEKHAR LIMIT ?
If a white dwarf were to exceed the Chandrasekhar limit, and nuclear reactions did not take place, the pressure exerted by electrons would no longer be able to balance the force of gravity, and it would collapse into a denser object called a neutron star or a Black Dwarf
INTERNAL TEMPERATURE :
the interior of the white dwarf maintains a uniform temperature, approximately 107 K . An outer shell of non-degenerate matter cools from approximately 107 K to 104 K
RADIATION :
The visible radiation emitted by white dwarfs varies over a wide color range, from the bluewhite color of an O-type main sequence star to the red of an M-type red dwarf.] White dwarf effective surface temperatures extend from over 150,000 K to barely under 4,000 K.
White dwarfs also radiate neutrinos through the Urca process.
Most observed white dwarfs have relatively high surface temperatures, between 8,000 K and 40,000 A white dwarf, though, spends more of its lifetime at cooler temperatures than at hotter temperatures, so we should expect that there are more cool white dwarfs than hot white dwarfs.
RATE OF COOLING IN WHITE DWARFS :
The rate of cooling has been estimated for a carbon white dwarf of 0.59 M with a hydrogen atmosphere. After initially taking approximately 1.5 billion years to cool to a surface tem perature of 7,140 K, cooling approximately 500 more kelvins to 6,590 K takes around 0.3
billion years, but the next two steps of around 500 kelvins (to 6,030 K and 5,550 K) take first 0.4 and then 1.1 billion years.
White dwarfs are thought to be the final evolutionary state of stars whose mass is not high enough to become a neutron star, that of about 10 solar masses.
Method-2: Derivation of Chandrasekhar Limit
\begin{align}
E&=\int{EdN} \\
& =\int\limits_{0}^{{{p}_{F}}}{E\dfrac{8\pi V}{{{h}^{3}}}{{p}^{2}}dp} \\
& =\dfrac{8\pi V}{{{h}^{3}}}\int\limits_{0}^{{{p}_{F}}}{\sqrt{{{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{4}}}{{p}^{2}}dp} \\
& =\dfrac{8\pi V}{{{h}^{3}}}\int\limits_{0}^{{{p}_{F}}}{pc\sqrt{1+\dfrac{{{m}^{2}}{{c}^{4}}}{{{p}^{2}}{{c}^{2}}}}{{p}^{2}}dp} \\
& =\dfrac{8\pi Vc}{{{h}^{3}}}\int\limits_{0}^{{{p}_{F}}}{\sqrt{1+\dfrac{{{m}^{2}}{{c}^{4}}}{{{p}^{2}}{{c}^{2}}}}{{p}^{3}}dp} \\
& =\dfrac{8\pi Vc}{{{h}^{3}}}\int\limits_{0}^{{{p}_{F}}}{\sqrt{1+\dfrac{{{m}^{2}}{{c}^{2}}}{{{p}^{2}}}}{{p}^{3}}dp} \\
& =\dfrac{8\pi Vc}{{{h}^{3}}}\int\limits_{0}^{{{p}_{F}}}{\left( 1+\dfrac{1}{2}\dfrac{{{m}^{2}}{{c}^{2}}}{{{p}^{2}}}+... \right){{p}^{3}}dp} \\
& \approx \dfrac{8\pi Vc}{{{h}^{3}}}\int\limits_{0}^{{{p}_{F}}}{\left( 1+\dfrac{1}{2}\dfrac{{{m}^{2}}{{c}^{2}}}{{{p}^{2}}} \right){{p}^{3}}dp} \\
& =\dfrac{8\pi Vc}{{{h}^{3}}}\left( \dfrac{{{p}^{4}}}{4}+\dfrac{{{m}^{2}}{{c}^{2}}{{p}^{2}}}{4} \right) \\
& \text{Let A=}\dfrac{p}{mc} \\
\Rightarrow {{A}^{2}}
&=\frac{{{p}^{2}}}{{{m}^{2}}{{c}^{2}}} \\
& =\frac{\left( 2m{{E}_{F}} \right)}{{{m}^{2}}{{c}^{2}}} \\
& =2\frac{{{E}_{F}}}{m{{c}^{2}}} \\
& =2\frac{1}{m{{c}^{2}}}\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{3}{8\pi }\frac{N}{V} \right)}^{2/3}} \right) \\
& =2\frac{1}{m{{c}^{2}}}\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{3}{8\pi }\frac{\frac{M}{2{{m}_{p}}}}{\frac{4}{3}\pi {{R}^{3}}} \right)}^{2/3}} \right) \\
& =\frac{2}{m{{c}^{2}}}\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{9}{32{{\pi }^{2}}}\frac{M}{{{m}_{p}}{{R}^{3}}} \right)}^{2/3}} \right) \\
& =\frac{{{h}^{2}}}{m{{c}^{2}}}{{\left( \frac{9}{64{{\pi }^{2}}}\frac{M}{{{m}_{p}}{{R}^{3}}} \right)}^{2/3}} \\
& =\frac{{{h}^{2}}}{m{{c}^{2}}}{{\left( \frac{9}{64{{\pi }^{2}}}\frac{M}{{{m}_{p}}} \right)}^{2/3}}\frac{1}{{{R}^{2}}} \\
& \Rightarrow A=\frac{{{h}^{2}}}{m{{c}^{2}}}{{\left( \frac{9}{64{{\pi }^{2}}}\frac{M}{{{m}_{p}}} \right)}^{1/3}}\frac{1}{R} \\
& =\dfrac{{{\left( \dfrac{9}{64{{\pi }^{2}}}\dfrac{M}{{{m}_{p}}} \right)}^{1/3}}}{R/{{\left( h/mc \right)}^{2}}} \\
& =\dfrac{{{{\bar{M}}}^{1/3}}}{{\bar{R}^{2}}} \\
\end{align}
\[\begin{align}
\Rightarrow E &=\frac{8\pi Vc}{{{h}^{3}}}\left( \frac{{{m}^{4}}{{c}^{4}}{{A}^{4}}}{4}+\frac{{{m}^{4}}{{c}^{4}}{{A}^{2}}}{4} \right) \\
& =\frac{2\pi V{{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( {{A}^{4}}+{{A}^{2}} \right) \\
& \Rightarrow P=-\frac{\partial E}{\partial V}=-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( V\left( \frac{\partial }{\partial V}{{A}^{4}}+\frac{\partial }{\partial V}{{A}^{2}} \right)+\left( {{A}^{4}}+{{A}^{2}} \right) \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( 4{{A}^{3}}V\frac{\partial A}{\partial V}+2AV\frac{\partial A}{\partial V}+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( 2AV\left( 2{{A}^{2}}+1 \right)\frac{\partial A}{\partial V}+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( 2AV\left( 2{{A}^{2}}+1 \right)\frac{\partial }{\partial V}\frac{2{{E}_{F}}}{c}+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( \frac{4}{c}AV\left( 2{{A}^{2}}+1 \right)\frac{\partial {{E}_{F}}}{\partial V}+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( \frac{4}{c}AV\left( 2{{A}^{2}}+1 \right)\frac{\partial }{\partial V}\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{3N}{8\pi V} \right)}^{2/3}} \right)+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( \frac{4}{c}AV\left( 2{{A}^{2}}+1 \right)\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{3N}{8\pi } \right)}^{2/3}}\left( -\frac{2}{3}{{V}^{-5/3}} \right) \right)+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( -\frac{8}{3c}VA\left( 2{{A}^{2}}+1 \right)\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{3N}{8\pi V} \right)}^{2/3}}\left( {{V}^{-1}} \right) \right)+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( -\frac{8}{3c}A\left( 2{{A}^{2}}+1 \right)\left( \frac{{{h}^{2}}}{2m}{{\left( \frac{3N}{8\pi V} \right)}^{2/3}} \right)+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( -\frac{8}{3c}A\left( 2{{A}^{2}}+1 \right){{E}_{F}}+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( -\frac{8}{3c}A\left( 2{{A}^{2}}+1 \right)\frac{Ac}{2}+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( -\frac{4}{3}{{A}^{2}}\left( 2{{A}^{2}}+1 \right)+{{A}^{4}}+{{A}^{2}} \right) \\
& =-\frac{2\pi {{c}^{5}}{{m}^{4}}}{{{h}^{3}}}\left( -\frac{8}{3}{{A}^{4}}+{{A}^{4}}-\frac{4}{3}{{A}^{2}}+{{A}^{2}} \right) \\
& =\frac{2\pi {{c}^{5}}{{m}^{4}}}{3{{h}^{3}}}\left( {{A}^{2}}-{{A}^{4}} \right) \\
\Rightarrow P&=\frac{2\pi m{{c}^{2}}}{3}{{\left( \frac{mc}{h} \right)}^{3}}\left( {{A}^{2}}-{{A}^{4}} \right) \\
\end{align}\]
\[\begin{align}
P&=\frac{2\pi m{{c}^{2}}}{3}{{\left( \frac{mc}{h} \right)}^{3}}\left( {{A}^{2}}-{{A}^{4}} \right) \\
& =\frac{2\pi m{{c}^{2}}}{3}{{\left( \frac{mc}{h} \right)}^{3}}\left( {{\left( {{\frac{\overline{M}}{\overline{R}}}^{1/3}} \right)}^{2}}-{{\left( {{\frac{\overline{M}}{\overline{R}}}^{1/3}} \right)}^{4}} \right) \\
& \Rightarrow P=\frac{2\pi m{{c}^{2}}}{3}{{\left( \frac{mc}{h} \right)}^{3}}\left( {{\frac{\overline{M}}{{{\overline{R}}^{2}}}}^{2/3}}-{{\frac{\overline{M}}{{{\overline{R}}^{4}}}}^{4/3}} \right) \\
\end{align}\]
\begin{align}
\Rightarrow W &=U \\
& \Rightarrow W=\int\limits_{\infty }^{R}{P\left( 4\pi {{r}^{2}} \right)dr} \\
& =4\pi {{\left( \frac{h}{mc} \right)}^{3}}\int\limits_{\infty }^{{\bar{R}}}{P\left( \overline{R} \right){{{\bar{r}}}^{2}}d\bar{r}} \\
& U=-\alpha \frac{G{{M}^{2}}}{R} \\
& =-\alpha G{{\left( \frac{8m}{9\pi } \right)}^{2}}\left( \frac{mc}{h} \right)\frac{{{{\bar{M}}}^{2}}}{{\bar{R}}} \\
& \Rightarrow 4\pi {{\left( \frac{h}{mc} \right)}^{3}}\int\limits_{\infty }^{{\bar{R}}}{P\left( \overline{R} \right){{{\bar{r}}}^{2}}d\bar{r}}=-\alpha G{{\left( \frac{8m}{9\pi } \right)}^{2}}\left( \frac{mc}{h} \right)\frac{{{{\bar{M}}}^{2}}}{{\bar{R}}} \\
& \Rightarrow \int\limits_{\infty }^{{\bar{R}}}{P\left( \overline{R} \right){{{\bar{r}}}^{2}}d\bar{r}}=-\alpha G{{\left( \frac{8m}{9\pi } \right)}^{2}}\frac{1}{4\pi }{{\left( \frac{mc}{h} \right)}^{4}}\frac{{{{\bar{M}}}^{2}}}{{\bar{R}}} \\
& \Rightarrow \frac{d}{d\overline{R}}\left( \int\limits_{\infty }^{{\bar{R}}}{P\left( \overline{R} \right){{{\bar{r}}}^{2}}d\bar{r}} \right)=-\alpha G{{\left( \frac{8m}{9\pi } \right)}^{2}}\frac{1}{4\pi }{{\left( \frac{mc}{h} \right)}^{4}}{{{\bar{M}}}^{2}}\frac{d}{d\overline{R}}\left( \frac{1}{{\bar{R}}} \right) \\
& \Rightarrow P\left( \overline{R} \right){{{\bar{R}}}^{2}}=-\alpha G{{\left( \frac{8m}{9\pi } \right)}^{2}}\frac{1}{4\pi }{{\left( \frac{mc}{h} \right)}^{4}}\frac{{{{\bar{M}}}^{2}}}{{{{\bar{R}}}^{2}}} \\
& \Rightarrow P\left( \overline{R} \right)=-G{{\left( \frac{8m}{9\pi } \right)}^{2}}\frac{1}{4\pi }{{\left( \frac{mc}{h} \right)}^{4}}\frac{{{{\bar{M}}}^{2}}}{{{{\bar{R}}}^{4}}} \\
& \Rightarrow \frac{2\pi m{{c}^{2}}}{3}{{\left( \frac{mc}{h} \right)}^{3}}\left( {{\frac{\overline{M}}{{{\overline{R}}^{2}}}}^{2/3}}-{{\frac{\overline{M}}{{{\overline{R}}^{4}}}}^{4/3}} \right)=-G{{\left( \frac{8m}{9\pi } \right)}^{2}}\frac{1}{4\pi }{{\left( \frac{mc}{h} \right)}^{4}}\frac{{{{\bar{M}}}^{2}}}{{{{\bar{R}}}^{4}}} \\
& \Rightarrow {{\frac{\overline{M}}{{{\overline{R}}^{2}}}}^{2/3}}-{{\frac{\overline{M}}{{{\overline{R}}^{4}}}}^{4/3}}=-\frac{G{{\left( \frac{8m}{9\pi } \right)}^{2}}\frac{1}{4\pi }{{\left( \frac{mc}{h} \right)}^{4}}}{\frac{2\pi m{{c}^{2}}}{3}{{\left( \frac{mc}{h} \right)}^{3}}}\frac{{{{\bar{M}}}^{2}}}{{{{\bar{R}}}^{4}}} \\
& \Rightarrow {{{\bar{M}}}^{2/3}}-{{\frac{\overline{M}}{{{\overline{R}}^{2}}}}^{4/3}}=-\frac{3G{{\left( \frac{8m}{9\pi } \right)}^{2}}}{8\pi hc}\frac{{{{\bar{M}}}^{2}}}{{{{\bar{R}}}^{2}}} \\
& \Rightarrow 1-{{\frac{\overline{M}}{{{\overline{R}}^{2}}}}^{2/3}}=-k\frac{{{{\bar{M}}}^{4/3}}}{{{{\bar{R}}}^{2}}} \\
& \Rightarrow \frac{1}{{{\overline{R}}^{2}}}\left( k{{{\bar{M}}}^{4/3}}-{{{\bar{M}}}^{2/3}} \right)=1 \\
& \Rightarrow \overline{R}={{{\bar{M}}}^{1/3}}\sqrt{k{{{\bar{M}}}^{2/3}}-1} \\
& \Rightarrow k{{{\bar{M}}}^{2/3}}-1>0 \\
& \Rightarrow {{{\bar{M}}}^{2/3}}>\frac{1}{k} \\
& \Rightarrow {{{\bar{M}}}^{2/3}}>\frac{8\pi hc}{3G{{\left( \frac{8m}{9\pi } \right)}^{2}}} \\
& \Rightarrow {{\left( \frac{9}{64{{\pi }^{2}}}\frac{M}{{{m}_{p}}} \right)}^{2/3}}>\frac{8\pi hc}{3G{{\left( \frac{8m}{9\pi } \right)}^{2}}} \\
& \Rightarrow M>1.44{{M}_{sun}} \\
\end{align}
Derivation of Chandra sekhar Limit : Alterate Approach
In white dwarf star all hydrogen have been converted into Helium.The temperature of the core is \( 10^7 K \) the thermal energy is about 100ev which is much larger than the energy required to ionize heium atom , so all the He atoms are assumed to have been ionised and hence , white dwarf star contianing N electrons can be assumed to be composed of N/2 helium nuclei ,Thus the mass of the star is=mass of electrons +mass of Helium nuclei so :
\[
M=N(m)+\dfrac{N}{2}(4m_p) \approx 2N m_p
\]
The electron density of the star is :
\[
n=\dfrac{N}{V}=\dfrac{M/2m_p}{V/\rho}=\dfrac{\rho}{2m_p}=10^{36}
\]
electrons per unt volume.
Consider a sphere of uniform mass density $\rho$ and radius $R$.
The mass of the sphere is
\[
M = \frac{4}{3}\pi R^3 \rho
\]
The work done in bringing an extra shell of thickness $dr$ is
\begin{align}
W &= - \int_0^R \frac{G \left(\frac{4}{3}\pi r^3 \rho\right) 4\pi r^2 \rho \, dr}{r} \\
&= -\frac{16}{3} G \pi^2 \rho^2 \int_0^R r^4 dr \\
&= -\frac{16}{3} G \pi^2 \rho^2 \frac{R^5}{5}
\end{align}
Using $M = \frac{4}{3}\pi R^3 \rho$, we obtain
\begin{equation}
W = -\frac{3GM^2}{5R}
\end{equation}
If the configuration is in equilibrium, the change in its energy for an infinitesimal change in its size must be zero:
\[
- P_0(r) 4\pi r^2 dr + \frac{3GM^2}{R^2} dr = 0
\]
or
\begin{equation}
P_0(r) = \frac{3GM^2}{20\pi R^4}
\end{equation}
The Ground state energy of fermi gas is: (Consdering the factor 2 for the spin degeneracy for fermions and \( x=p/mc \quad \& x_F=\dfrac{p_F}{mc}\) )
\begin{align}
{{E}_{0}} &=2\sum{E}dN \\
& =2\int{E\frac{4\pi V{{p}^{2}}dp}{{{h}^{3}}}} \\
& =\frac{8\pi V}{{{h}^{3}}}\int{E{{p}^{2}}dp} \\
& =\frac{8\pi V}{{{h}^{3}}}\int{\sqrt{{{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{4}}}{{p}^{2}}dp} \\
& =\frac{8\pi V}{{{\left( 2\pi \hbar \right)}^{3}}}\int{\sqrt{{{p}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{4}}}{{p}^{2}}dp} \\
& =\frac{V}{{{\pi }^{2}}{{\hbar }^{3}}}\int\limits_{0}^{{{x}_{F}}}{\sqrt{{{\left( xmc \right)}^{2}}{{c}^{2}}+{{m}^{2}}{{c}^{4}}}}\left( {{x}^{2}}{{m}^{2}}{{c}^{2}}mcdx \right) \\
& =\frac{V{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\sqrt{{{x}^{2}}+1}}\,dx \\
& =\frac{V{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\sqrt{{{x}^{2}}+1}}\,dx \\
& =\frac{V{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}f\left( {{x}_{F}} \right) \\
& \Rightarrow {{P}_{0}}=-\frac{\partial E}{\partial V}=-\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\frac{\partial }{\partial V}\left[ Vf\left( {{x}_{F}} \right) \right] \\
& =-\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left( f\left( {{x}_{F}} \right)+V\frac{\partial f}{\partial {{x}_{F}}}\frac{\partial {{x}_{F}}}{\partial V} \right) \\
& =-\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left( f\left( {{x}_{F}} \right)+V\left[ \frac{\partial }{\partial {{x}_{F}}}\int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\sqrt{{{x}^{2}}+1}}\,dx \right]\frac{\partial }{\partial V}\left( \frac{\hbar }{mc}{{\left( 3{{\pi }^{2}}\frac{N}{V} \right)}^{1/3}} \right) \right) \\
& =-\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left( f\left( {{x}_{F}} \right)+V\left[ x_{F}^{2}\sqrt{x_{F}^{2}+1} \right]\left( \frac{\hbar }{mc}{{\left( 3{{\pi }^{2}}N \right)}^{1/3}}\left( -\frac{1}{3} \right){{V}^{-4/3}} \right) \right) \\
& =-\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left( f\left( {{x}_{F}} \right)+V\left[ x_{F}^{2}\sqrt{x_{F}^{2}+1} \right]\left( \frac{\hbar }{mc}{{\left( 3{{\pi }^{2}}\frac{N}{V} \right)}^{1/3}}\left( -\frac{1}{3} \right){{V}^{-1}} \right) \right) \\
& =-\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left( f\left( {{x}_{F}} \right)-\frac{1}{3}\left[ x_{F}^{2}\sqrt{x_{F}^{2}+1} \right]{{x}_{F}} \right) \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left( \frac{1}{3}\left[ x_{F}^{3}\sqrt{x_{F}^{2}+1} \right]-f\left( {{x}_{F}} \right) \right) \\
\end{align}
For non relativistic case: \( p_F << mc \)
\begin{align}
& \Rightarrow {{x}_{F}}<<1 \\
& \int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\sqrt{1+{{x}^{2}}}}dx \\
& =\int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\left( 1+\frac{1}{2}{{x}^{2}} \right)}dx \\
& =\int\limits_{0}^{{{x}_{F}}}{\left( {{x}^{2}}+\frac{{{x}^{4}}}{2} \right)}dx \\
& =\frac{x_{F}^{3}}{3}+\frac{x_{F}^{5}}{10} \\
& \frac{1}{3}x_{F}^{3}\sqrt{1+x_{F}^{2}}=\frac{1}{3}x_{F}^{3}\left( 1+\frac{1}{2}x_{F}^{2} \right) \\
& =\frac{x_{F}^{3}}{3}+\frac{x_{F}^{5}}{6} \\
& \Rightarrow {{P}_{0}}=\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{1}{3}x_{F}^{3}\sqrt{1+x_{F}^{2}}-f\left( {{x}_{F}} \right) \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{1}{3}x_{F}^{3}\sqrt{1+x_{F}^{2}}-\int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\sqrt{1+{{x}^{2}}}dx} \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{x_{F}^{3}}{3}+\frac{x_{F}^{5}}{6}-\left( \frac{x_{F}^{3}}{3}+\frac{x_{F}^{5}}{10} \right) \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{x_{F}^{5}}{15} \right]=\frac{{{m}^{4}}{{c}^{5}}}{15{{\pi }^{2}}{{\hbar }^{3}}}{{\left( \frac{9\pi M{{\hbar }^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}{{R}^{3}}} \right)}^{5/3}} \\
& \Rightarrow {{P}_{0}}=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow \frac{{{m}^{4}}{{c}^{5}}}{15{{\pi }^{2}}{{\hbar }^{3}}}{{\left( \frac{9\pi M{{\hbar }^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}{{R}^{3}}} \right)}^{5/3}}=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow {{M}^{1/3}}R=\frac{4{{\hbar }^{2}}}{9mG\pi }{{\left( \frac{9\pi }{8{{m}_{p}}} \right)}^{5/3}} \\
\end{align}
For Relatisvistic case : \( x_F >>1 \because p_F >>mc \)
\begin{align}
& \Rightarrow {{P}_{0}}=\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{1}{3}x_{F}^{3}\sqrt{1+x_{F}^{2}}-\int\limits_{0}^{{{x}_{F}}}{{{x}^{2}}\sqrt{1+{{x}^{2}}}dx} \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{1}{3}x_{F}^{4}\sqrt{\frac{1}{x_{F}^{2}}+1}-\int\limits_{0}^{{{x}_{F}}}{{{x}^{3}}\sqrt{1+\frac{1}{{{x}^{2}}}}dx} \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{1}{3}x_{F}^{4}\left( \frac{1}{2x_{F}^{2}}+1 \right)-\int\limits_{0}^{{{x}_{F}}}{{{x}^{3}}\left( 1+\frac{1}{2{{x}^{2}}} \right)dx} \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{x_{F}^{2}}{6}+\frac{x_{F}^{4}}{3}-\int\limits_{0}^{{{x}_{F}}}{\left( {{x}^{3}}+\frac{1}{2}x \right)dx} \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{x_{F}^{2}}{6}+\frac{x_{F}^{4}}{3}-\left( \frac{x_{F}^{4}}{4}+\frac{x_{F}^{2}}{4} \right) \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{{{\pi }^{2}}{{\hbar }^{3}}}\left[ \frac{x_{F}^{4}}{12}-\frac{x_{F}^{2}}{12} \right] \\
& =\frac{{{m}^{4}}{{c}^{5}}}{12{{\pi }^{2}}{{\hbar }^{3}}}\left( x_{F}^{4}-x_{F}^{2} \right) \\
& \Rightarrow {{P}_{0}}=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow \frac{{{m}^{4}}{{c}^{5}}}{12{{\pi }^{2}}{{\hbar }^{3}}}x_{F}^{2}\left( x_{F}^{2}-1 \right)=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
\end{align}
\begin{align}
{{x}_{F}}&=\hbar {{\left( \frac{3{{\pi }^{2}}N}{V} \right)}^{1/3}}={{\left( \frac{9\pi M}{8{{m}^{4}}{{c}^{3}}{{R}^{3}}} \right)}^{1/3}} \\
& \Rightarrow {{P}_{0}}\left( r \right)=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow \frac{{{m}^{4}}{{c}^{5}}}{12{{\pi }^{2}}{{\hbar }^{3}}}{{\left( \frac{9\pi M{{h}^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}{{R}^{3}}} \right)}^{2/3}}\left( {{\left( \frac{9\pi M{{h}^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}{{R}^{3}}} \right)}^{2/3}}-1 \right)=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow \frac{{{m}^{4}}{{c}^{5}}}{12{{\pi }^{2}}{{\hbar }^{3}}}{{\left( \frac{9\pi {{h}^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}{{R}^{3}}} \right)}^{2/3}}\frac{{{M}^{2/3}}}{{{R}^{2}}}\left( {{\left( \frac{9\pi {{h}^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}} \right)}^{2/3}}\frac{{{M}^{2/3}}}{{{R}^{2}}}-1 \right)=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow ab\frac{{{M}^{2/3}}}{{{R}^{2}}}\left( b\frac{{{M}^{2/3}}}{{{R}^{2}}}-1 \right)=\frac{3G{{M}^{2}}}{20\pi {{R}^{4}}} \\
& \Rightarrow ab\left( b\frac{{{M}^{2/3}}}{{{R}^{2}}}-1 \right)=\frac{3G{{M}^{4/3}}}{20\pi {{R}^{2}}} \\
& \Rightarrow ab\left( b\frac{{{M}^{2/3}}}{{{R}^{2}}}-1 \right)=\frac{3G{{M}^{4/3}}}{20\pi {{R}^{2}}} \\
& \Rightarrow a{{b}^{2}}\frac{{{M}^{2/3}}}{{{R}^{2}}}-ab=\frac{3G{{M}^{4/3}}}{20\pi {{R}^{2}}} \\
& \Rightarrow \frac{{{M}^{2/3}}}{{{R}^{2}}}\left( a{{b}^{2}}-\frac{3G{{M}^{2/3}}}{20\pi } \right)=ab \\
& \Rightarrow {{R}^{2}}={{M}^{2/3}}\left( b-\frac{3G{{M}^{2/3}}}{20\pi ab} \right) \\
& \Rightarrow R={{M}^{1/2}}\sqrt{b-\frac{3G{{M}^{2/3}}}{20\pi ab}} \\
& \Rightarrow b>\frac{3G{{M}^{2/3}}}{20\pi ab} \\
& \Rightarrow {{M}^{2/3}}<\frac{20\pi a{{b}^{2}}}{3G} \\
& \because a=\frac{{{m}^{4}}{{c}^{5}}}{12{{\pi }^{2}}{{\hbar }^{3}}},b={{\left( \frac{9\pi {{\hbar }^{3}}}{8{{m}^{3}}{{c}^{3}}{{m}_{p}}} \right)}^{2/3}} \\
& a=\frac{{{\left( 1.67\times {{10}^{-27}} \right)}^{4}}{{\left( 3\times {{10}^{8}} \right)}^{5}}}{12{{\pi }^{2}}{{\left( 6.62\times {{10}^{-34}}/2\pi \right)}^{3}}} \\
& =\frac{{{\left( 1.67\times {{10}^{-27}} \right)}^{4}}{{\left( 3\times {{10}^{8}} \right)}^{5}}8\pi }{12{{\left( 6.62\times {{10}^{-34}} \right)}^{3}}} \\
& =\frac{{{1.67}^{4}}\times 243\times 8\times 3.142\times {{10}^{-108+40}}}{12\times 290.11\times {{10}^{-102}}} \\
& =1.364\times {{10}^{35}} \\
& b= \\
& \Rightarrow a=1.364\times {{10}^{35}},b=1.14\times {{10}^{-6}}\text{ }{{\text{m}}^{2}},1.02\times {{10}^{-31}} \\
& \Rightarrow a{{b}^{2}}=1.364\times {{10}^{35}}\times 1.2996\times {{10}^{-12}} \\
& =1.77265\times {{10}^{23}} \\
& =1.425\times {{10}^{-27}} \\
& \Rightarrow M<{{\left( \frac{20\pi a{{b}^{2}}}{3G} \right)}^{3/2}} \\
& \Rightarrow M<{{\left( \frac{20\times 3.142\times 1.77\times {{10}^{23}}}{3\times 6.67\times {{10}^{-11}}} \right)}^{3/2}} \\
& \Rightarrow M<{{\left( \frac{20\times 3.142\times 1.77\times {{10}^{34}}}{3\times 6.67} \right)}^{3/2}} \\
& \Rightarrow M<3.40\times {{10}^{30}} \\
& \Rightarrow M<1.7\times 1.99\times {{10}^{30}} \\
& \Rightarrow M<1.7{{M}_{0}} \\
& \Rightarrow M<1.4{{M}_{0}} \\
\end{align}
White Dwarf Theory – MCQ Quiz
