Statistical Physics: October 9, 2012

Solutions for the Homework 6

Problem 7.20: At the center of the sun, the temperature is approximately 107 K and the concentration of electrons is approximately 1032 per cubic meter. Would it be (approximately) valid to treat these electrons as a “classical ” ideal gas (using Boltzmann statistics), or as a degenerate Fermi gas (with T ≈ 0), or neither? Solution: Calculate the Fermi energy of this system and compare with the real temperature. From the Eq. 7.39, ( ) h2 3N 2/3 ϵF = = 9.1166 × 106 K. (1) TF = k 8mk πV Because of the real temperature is almost same as the Fermi temperature, we can not approximate this system to the classical ideal gas nor the degenerate Fermi gas with T ≈ 0. Problem 7.23: A white dwarf star is essentially a degenerate electron gas, with a bunch of nuclei mixed in to balance the charge and to provide the gravitational attraction that holds the star together. In this problem you will derive a relation between the mass and the radius of a white dwarf star, modeling the star as a uniform-density sphere. White dwarf stars tend to be extremely hot by our standards; nevertheless, it is an excellent approximation in this problem to set T = 0. (a) Use dimensional analysis to argue that the gravitational potential energy of a uniform-density sphere (mass M , radius R) must equal Ugrav = −(constant)

GM 2 , R

(2)

where (constant) is some numerical constant. Be sure to explain the minus sign. The constant turns out to equal 3/5; you can derive it by calculating the (negative) work needed to assemble the sphere, shell by shell, from the inside out. Solution: Let us consider that there is some spherical mass and add spherical shells to that. Then the mass of sphere is m = 4πr3 ρ/3 and the mass of spherical shell is dm = 4πr2 ρdr where the ρ = 3M/4πR3 . Now, integrate the potential energy of shell over r. Then, ( ) ∫ ∫ R m 4πr3 ρ Ugrav = − G 2 dm = − G 4πr2 ρ dr (3) r 3r2 0 ( )( ) ∫ 16Gπ 2 2 R 4 16Gπ 2 9M 2 3 GM 2 5 = − ρ r dr = − R = − . (4) 3 3 16π 2 R6 5 R 0 Because of the work done by gravitational force, sign of gravitational potential is negative.

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Statistical Physics: October 9, 2012

(b) Assuming that the star contains one proton and one neutron for each electron, and that the electrons are nonrelativistic, show that the total (kinetic) energy of the degenerate electrons equals Ukinetic = (0.0088)

h2 M 5/3 5/3

me mp R2

.

(5)

The numerical factor can be expressed exactly in terms of π and cube roots and such, but it’s not worth it. Solution: From the Eq. 7.39 and Eq. 7.42, ϵF =

h2 8me

(

3N πV

)2/3 ⇒

Ukinetic =

3 3 h2 N ϵF = N 5 5 8me

(

3N πV

)2/3 .

(6)

Put the values that N ≈ /(mp + mn ) and V = 4πR3 /3. Thus the energy of the degenerate electrons is ( ) ( ) h2 N 5/3 h2 3N 2/3 37/3 3 N = Ukinetic = (7) 5 8me πV me R 2 40 · 42/3 π 4/3 ( ) 37/3 h2 M 5/3 h2 M 5/3 ≈ (0.00880855) . (8) = 5/3 2 40 · 42/3 · (2.001)5/3 π 4/3 me m5/3 me mp R2 p R (c) The equilibrium radius of the white dwarf is that which minimizes the total energy Ugrav + Ukinetic . Sketch the total energy as a function of R, and find a formula for the equilibrium radius in terms of the mass. As the mass increases, does the radius increase or decrease? Does this make sense? Solution: When I substitute the mass as given by the next part (d), then the graph is like below figure. U 5´10

42

1´107

2´107

3´107

4´107

5´107

R

-5´10 42

-1´10 43

Figure 1: Total energy with the equilibrium radius(dotted line).

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Statistical Physics: October 9, 2012

The equilibrium radius will be extremum value, so differentiate the total energy with respect to R and set equal to zero. Then the formula for the equilibrium radius in terms of the mass is R → 9.00225 × 1016 M −1/3 .

(9)

So, if the mass increases then the equilibrium radius will be decrease and it make sense because of the gravitational potential is much larger (∼ M 2 ) then the kinetic energy of degenerate electrons (∼ M 5/3 ), in the other words, the white dwarf star will shrink by gravity when the mass increases. (d) Evaluate the equilibrium radius for M = 2 × 1030 kg, the mass of the sun. Also evaluate the density. How does the density compare to that of water? Solution: From the result of part (c), R = 7.14509 × 106 m. Then the mass density of the sun is ρ = M/V = 1.30894 × 109 kg/m3 and it’s 1.30894 × 106 times higher density than the case of water. (e) Calculate the Fermi energy and the Fermi temperature, for the case considered in part (d). Discuss whether the approximation T = 0 is valid. Solution: From the result of part (b) with given values, ϵF = 3.12437 × 10−14 J = 1.95030 × 105 eV

,

TF = 2.26297 × 109 K.

(10)

Real temperature at the center of the sun is near the 108 K which can be occur the nuclear fusion process. So, the Fermi temperature is much more high and then, we can do the approximation T = 0. (f) Suppose instead that the electrons in the white dwarf star are highly relativistic. Using the result of the previous problem, show that the total kinetic energy of the electrons is now proportional to 1/R instead of 1/R2 . Argue that there is no stable equilibrium radius for such a star. Solution: From the result of the problem 7.22, the Fermi energy is changed and it depends on the V −1/3 which is proportional to 1/R. So, now the kinetic energy of degenerate electrons also proportional to 1/R same as the gravitational energy and it makes that there is no extremum value of radius. (g) The transition from the nonrelativistic regime to the ultrarelativistic regime occurs approximately where the average kinetic energy of an electron is equal to its rest energy, mc2 . Is the nonrelativistic approximation valid for a one-solar-mass white dwarf? Above what mass would you expect a white dwarf to become relativistic and hence unstable? Solution: Average kinetic energy of electron is U/N = 0.6ϵF = 1.87462 × 10−14 J and the rest energy is 8.178105 × 10−14 J. So, the nonrelativistic approximation valid for the one-solar-mass case. Now, to find the critical mass of that the white dwarf to become relativistic, set the equation like this: ( ) 3 h2 3N 2/3 = me c2 ⇒ M → 6.0372 × 1030 kg. (11) 5 8me πV The value is almost two times bigger than the Chandrasekhar limit which is well known value of critical mass of white dwarf star. Because of the realistic model is not uniform and used much more complex Fermi model.

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Statistical Physics: October 9, 2012

Problem 7.24: A star that is too heavy to stabilize as a white dwarf can collapse further to form a neutron star: a star made entirely of neutrons, supported against gravitational collapse by degenerate neutron pressure. Repeat the steps of the previous problem for a neutron star, to determine the following: the mass-radius relation; the radius, density, Fermi energy, and Fermi temperature of a one-solar-mass neutron star; and the critical mass above which a neutron star becomes relativistic and hence unstable to further collapse. Solution: The model is same but the electron mass changes to the neutron mass and the way to find the total number of Fermi particles is also changed. Then, the mass-radius relation is R → 1.55212 × 1014 M −1/3 .

(12)

Now, set the mass as a one-solar-mass. Then, R = 1.23192 × 104 m,

(13)

ρ = 2.55384 × 10 kg/m , 17

3

−12

(14)

ϵF

= 9.06858 × 10

J,

(15)

TF

= 6.56834 × 10 K,

(16)

11

and finally the critical mass is → 2.41263 × 1031 kg.

M

(17)

This is also not agree with the more complex model. It may came from that the real temperature of neutron star is very high and that is comparable with the Fermi temperature. So, for this case, T ≈ 0 is not valid. Problem 7.28: Consider a free Fermi gas in two dimensions, confined to a square area A = L2 . (a) Find the Fermi energy (in terms of N and A), and show that the average energy of the particles is ϵF /2. Solution: In two dimensional case, ∫ N

∫

nmax

π/2

π 2 n . 2 max

(18)

h2 2N h2 h2 n2max = = N. 2 8mL 8mA π 4πmA

(19)

n dn dθ =

= 2 0

0

From the Eq. 7.37, ϵF

=

Then, we can find the total energy U : ∫ U

nmax

∫

∫

π/2

= 2

ϵ(n)n dn dθ =

∫ =

0 ϵF

[ ϵ

0

0

4πmA h2

]

∫ dϵ =

ϵ 0

[ √ ϵ π

0 ϵF

ϵF

8mA √ ϵ h2

√

8mA 1 √ h2 2 ϵ

N 1 dϵ = N ϵF . ϵF 2

So, we can easily get the average energy of the particles is U/N = ϵF /2.

4

] dϵ

(20) (21)

Statistical Physics: October 9, 2012

(b) Derive a formula for the density of states. You should find that it is a constant, independent of ϵ. Solution: From the part (a), g(ϵ) = N/ϵF , that is independent of ϵ. (c) Explain how the chemical potential of this system should behave as a function of temperature, both when kT ≪ ϵF and when T is much higher. Solution: Since the density of state is a constant, the behavior of the chemical potential is same as the Fermi-Dirac distribution. As we see when the previous homework, problem 7.12, Fermi-Dirac distribution has some symmetrical property but when the energy difference from µ is higher than the chemical potential, it’s broken because of the occupancy is zero when ϵ < 0. So, the electrons with energy less than the Fermi energy excited to higher than the Fermi energy with broken symmetry, it means that the number of electron loss with energy ϵ < ϵF is lower than the number of electron gain with energy ϵ > ϵF . To make sensible with the number difference of electrons, chemical potential should move, it means that the Fermi-Dirac distribution have to move totally. Since the number of electrons with energy ϵ > ϵF is larger than ϵ < ϵF , Fermi-Dirac distribution have to move left side with fixed ϵF , which means that the chemical potential decrease. When the temperature is slightly grow up, then the difference of the number of electrons also grow up. So, chemical potential will decrease continuously and at some point, it will be negative value. Μ 1.0

0.5

0.5

1.0

1.5

2.0

2.5

T

-0.5

-1.0

-1.5

Figure 2: Result of (d) with k = 1 and ϵF = 1. (d) Because g(ϵ) is a constant for this system, it is possible to carry out the integral 7.53 for the number of particles analytically. Do so, and solve for µ as a function of N . Show that the resulting formula has the expected qualitative behavior. Solution: Since g(ϵ) is a constant, we can analytically integrate the Eq. 7.53: ∫ ∞ ∫ 1 N ∞ e−(ϵ−µ)/kT N = g(ϵ) (ϵ−µ)/kT dϵ = dϵ ϵF 0 1 + e−(ϵ−µ)/kT e +1 0 N N = (−kT ) ln(1 + e−(ϵ−µ)/kT )|ϵ=∞ kT ln(1 + eµ/kT ). ϵ=0 = ϵF ϵF

5

(22) (23)

Statistical Physics: October 9, 2012

From this, { µ = kT ln(e

ϵF /kT

− 1) =

ϵF where kT ≪ ϵF . kT ln(ϵF /kT ) where kT ≫ ϵF

(24)

The second case is approximated and it will be negative value. (e) Show that in the high-temperature limit, kT ≫ ϵF , the chemical potential of this system is the same as that of an ordinary ideal gas. Solution: From the answer of part (d), ( µ = kT ln(ϵF /kT ) = −kT ln

A 4πmkT N h2

(

) = −kT ln

A 2 2 N lQ

) .

(25)

It seems to be the same form with the Eq. 6.93, which is the chemical potential of an ordinary ideal gas with Zint = 2 came from the electron spin.

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