Chapter 4 tutorial

<< Miscellaneous`Units`
<< Miscellaneous`PhysicalConstants`
Needs["Miscellaneous`RealOnly`"]
ℏ = PlanckConstant
c = SpeedOfLight
k = BoltzmannConstant

Problem 6

The pressure is given by equation 3.26:

p = -(Subscript[((∂ (U/∂) V)), σ])

\(where\ \((∂\ U\/∂\ V)\)\_σ\ represents\ the\ partial\ derivative\ with\ respect\ to\ the\ volume\ V\ while\ keeping\ the\ entropy\ constant . \ \ The\ average\ energy\ U\ \(\(is\)\(:\)\)\)

U = (∑_j) (Subscript[s, j]) (Subscript[E, j])
=(∑_j) (Subscript[s, j]) ℏ (Subscript[ω, j])

\(where\ \(s\_j\) 
        is\ the\ number\ of\ photons\ in\ the\ mode\ j\ and\ E\_j\  = \ ℏ\ ω\_j\ is\ the\ energy\ of\ the\ photon\ mode\ j\  . \ \ \ \(Note : \ 
            we\ do\ not\ know\ the\ distribution\ s\_j . \ \ It\ could\ be\ thermal\ equilibrium\ but\ also\ any\ other\ \(\(distirbution\)\(.\)\(
\)\)\)\)

Z = (∑_j) (E^(-(Subscript[E, j])/τ ))

By definition, we get:

p = -(Subscript[((∂ (U/∂) V)), σ])
=-(Subscript[((∂ (((∑_j) (Subscript[s, j]) ℏ (Subscript[ω, j]))/∂) V)), σ])
=-(∑_j)(ℏ (Subscript[ω, j]) (Subscript[((((∂(Subscript[s, j])  )/∂) V)), σ]) + ℏ (Subscript[s, j]) (Subscript[((∂ ((Subscript[ω, j])/∂) V)), σ]))
=-(∑_j) ℏ (Subscript[s, j]) (Subscript[((∂ ((Subscript[ω, j])/∂) V)), σ])

\(ℏ\ \(ω\_j\) \((\(\(∂s\_j\)\(\ \ \)\)\/∂\ V)\)\_σ\) is zero, but I am not entirely sure why unless we are at equilibrium (and therefore (Subscript[s, j]) does not change).  In general, I am not sure why it would be zero.

We know from Eq. 4.15 that:

(Subscript[ω, j]) = ((j π c)/L)

therefore, using chain rules, we can calculate:

(Subscript[((∂ ((Subscript[ω, j])/∂) V)), σ]) = ∂ ((Subscript[ω, j])/∂) 
              L ∂ (L/∂) V
=∂ ((Subscript[ω, j])/∂) L/∂ 
              (V/∂) L
=-((j π c)/(L^2))/∂ ((((L^3)))/∂) L
=-((j π c)/(L^2))/3 
         (L^2)
=-(Subscript[ω, j])/3  (L^3)
=- (Subscript[ω, j])/3  V

We substitute and obtain:

p = (∑_j) ℏ (Subscript[s, j]) (Subscript[ω, j])/3  V = U/3  V

Now, we are asked to compare the  kinetic pressure with the thermal radiation pressure.  Because we now deal with thermal radiation pressure,(i. e.  pressure of radiation at thermal equilibirum), we can use the results of p. 94 in the page, which are valid for thermal equilibrium.  We know that at thermal equilibrium, the energy density of radiation is:

ThermalRadiationEnergyDensityUoverV[
        T_] = ((π^2)/(15 (ℏ^3) (c^3)  )) (((k T))^4)

And therefore, using the derived result p=U/3V, we can say that the thermal radiation pressure is:

(Subscript[p, tr]) = 
      ThermalRadiationEnergyDensityUoverV[2 × (10^7) Kelvin]/
        3

The pressure of an ideal gas at that temperature is given by Eq. (4.74):

(Subscript[p, ig]) = ((N/V)) k T  =  C k T

where C is the concentration of molecules per volume. With C= \(\(\(1\)\(\ \)\(Mole\)\(\ \)\)\/\(\(\((0.01\ \ Meter)\)\^3\)\(\ \)\)\), the pressures are equal when:

C k T == (1/3) ((π^2)/(15 (ℏ^3) (c^3)  )) (((k T))^4)

and therefore

T == ((((45 (ℏ^3) (c^3)  )/((π^2) (k^3))) ((1 Mole )/((((0.01  Meter))^3) )))^(1/3))

In the  sun, the kinetic pressure is roughly:

(Subscript[p, ig]) == ((1 Mole )/((((0.01  Meter))^3) ))  2 × (10^7) Kelvin ×k

and in the center it is a hundred times more:

(Subscript[p, ig])  == ((100 Mole )/((((0.01  Meter))^3) ))  2 × 
        (10^7) Kelvin ×k

Note :   A Joule is a Newton × 
          Meter .   We therefore have the units of Newton × 
          Meter/(Meter^3) = 
      Newton /(Meter^2) which is a Pascal.

Problem 7

\(We\ rewrite\ the\ sum\ as\ an\ integral\ in\ spherical\ coordinates . \ \ A\ sum\ over\ the\ states\ labelled\ by\ \(n\& ⇀ \)\  = \ \((n\_x, n\_y, 
            n\_z)\)\  . \ \ \ The\ sum\ over\ all\ elements\ is\ the\ same\ as\ as\ sum\ over\ the\ elementary\ volume\ element\ dn\_x dn\_y dn\_z\  . \ \ However\ , \ 
    it\ is\ not\ convenient\ to\ use\ Cartesian\ coordinates\ because\ the\ energy\ is\ given\ as\ a\ function\ of\ n, \ 
    not\ \((n\_x, n\_y, 
          n\_z)\) . \ \ Therefore\ we\ go\ to\ Soehrical\ coordinates\ where\ the\ elementary\ volume\ is\ element\ \(n\^2\) 
      Sin[θ] dn\ dθ\ dϕ . \ \ The\ integrals\ over\ Sin[θ]\ dθ\ dϕ\ are\ trivial . \ \ Finally, \ we\ use\ the\ following\ Taylor\ \(series : 
Log[
            1 - \(e\^-\)\^\(\(k\ n\ ℏ\ ϖ\)\/τ\)]\) = \(-\(∑\+\(k = 1\)\%∞\( e\^-\)\^\(\(k\ n\ ℏ\ ϖ\)\/τ\)\/k\)\)\)
\(Finally, \ we\ note\ that\ there\ are\ two\ polarizations\ per\ mode, \ 
    so\ we\ must\ multiply\ by\ 2\ to\ obtain\ the\ number\ of\ \(\(states\)\(.\)\)\)

\(\(=\)\(2\ ×τ \(∫\_0\%\(\(\(π\)\(\ \)\)\/2\)\(∫\_0\%\(\(\(π\)\(\ \)\)\/2\)\(∫\_0\%∞\ \ Log[
                  1 - \(E\^-\)\^\(\(k\ n\ ℏ\ ϖ\)\/τ\)]\ \ \(n\^2\) 
                Sin[θ] \[DifferentialD]n\ \[DifferentialD]θ\ \[DifferentialD]ϕ\)\)\)\)\)
=π τ(∫_0^∞) (n^2)(\(-\(∑\+\(k = 1\)\%∞\( E\^-\)\^\(\(k\ n\ ℏ\ ϖ\)\/τ\)\/k\)\)) \[DifferentialD]n

We substitute for clarity:

x ≡ (( n ℏ ϖ)/τ)
\[DifferentialD]x = ((  ℏ ϖ)/τ) \[DifferentialD]n

to obtain:

=-(( π (τ^4))/(((ℏ )^3) (ϖ^3)))(∫_0^∞) (x^2)\(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\)  \[DifferentialD]x

\(We\ integrate\ by\ parts\ with\ \(∫u dv\) = 
      uv - ∫v du, \ \ and\ we\ use\ u = 
      x\^2, \ dv = ∑\+\(k = 1\)\%∞\( e\^-\)\^\(k\ x\)\/k\ \ dx, \ 
    therefore\ we\ have\ du = \(2  
          x dx\ and\ \ v = \(\(\(\(-∑\)\+\(k = 
                1\)\%∞\)\(\(e\^-\)\^\(k\ x\)\/k\^2\)\)\(\ \)\(:\)\)\)\)

=-(( π (τ^4))/(((ℏ )^3) (ϖ^3)))(\(\((\(x\^2\) \((\(-1\))\)\ \(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\^2\))\)\_\(x = 
                  0\)\%\(x = ∞\)\)-(∫_0^∞) 2 x(-1)\(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\^2\)  \[DifferentialD]x)
=-(( π (τ^4))/(((ℏ )^3) (ϖ^3)))((∫_0^∞) 2x \(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\^2\)  \[DifferentialD]x)

\(We\ integrate\ by\ parts\ with\ u = 
      2  x, \ dv = ∑\+\(k = 1\)\%∞\( e\^-\)\^\(k\ x\)\/k\^2\ \ dx, \ 
    therefore\ we\ have\ du = \(2 dx\ and\ \ v = \(\(\(\(-∑\)\+\(k = 
                1\)\%∞\)\(\(e\^-\)\^\(k\ x\)\/k\^3\)\)\(\ \)\(:\)\)\)\)

=-(( π (τ^4))/(((ℏ )^3) (ϖ^3)))(\(\((2  
                    x\ \(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\^3\))\)\_\(x = 0\)\%\(x = ∞\)\)-(∫_0^∞)2(-1)\(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\^3\)  \[DifferentialD]x)
=-(( 2  π (τ^4))/(((ℏ )^3) (ϖ^3)))(∫_0^∞)\(∑\+\(k = 1\)\%∞\( E\^-\)\^\(k\ x\)\/k\^3\)  \[DifferentialD]x
=-(( 2  π (τ^4))/(((ℏ )^3) (ϖ^3)))(-1)((∑_(k = 1)^∞) (0/(k^4)) - (∑_(k = 1)^∞) (1/(k^4)))

We can actually find the limit of the series:

(∑_(k = 1)^∞) (1/(k^4))

Therefore we obtain the final result:

F = -(( 2  π (τ^4))/(((ℏ )^3) (ϖ^3))) ((π^4)/90)