Quantum Mechanics and Thermodynamics: Linking Thermal Radiation and Plank Distribution, Study notes of Engineering Physics

Download Quantum Mechanics and Thermodynamics: Linking Thermal Radiation and Plank Distribution and more Study notes Engineering Physics in PDF only on Docsity! Thermal Radiation and the Plank Distribution Adrian Down September 28, 2005 1 Chapter 3 Summary 1.1 Boltzmann factor Applicable to the case of a small system coupled to a very large system in thermal equilibrium. The state of the small system is specified by the quantum number l. The probability for the small system to be in state with energy l is given by P (l) = e− l τ Z where Z is the partition function Z = ∑ l e− l τ 1.2 Pressure Pressure is different than other thermodynamic properties in that it must be consistent with Newton’s Laws. p = − ( ∂U ∂V ) σ = τ ( ∂σ ∂V ) U Note. Be careful of which variablse are held constant when taking a deriva- tive. 1 1.3 Thermodynamic identity dU = τdσ︸︷︷︸ heat added to system − pdV︸︷︷︸ work done on system 1.4 Helmholtz Free energy F = U − τσ Key importance is that F is minimized for a system in thermal equilibrium at constant τ and V . F led to two new expressions for previously-defined quantities, σ = − ( ∂F ∂τ ) V p = − ( ∂F ∂V ) τ F is also related to the partition function, F = −τ ln Z ⇒ P (l) = e (F−l) τ By replacing the partition function with the macroscopic quantity F , we no longer need the micrscopcic partition funcction to calculate probalities. This expression for P (l) is a step towards a more macroscopic formulation of thermodynamics. 1.5 Classical ideal gas For one particle, Z1 = nQV where nQ is the quantum concentration. nQ = ( mτ 2π~2 ) 3 2 The energy of the gas depends only on the temperature, and U = 3 2 τ = 3 2 kbT 2 3.2.3 Average energy per mode Consider a particular mode with angular frequency ω. The mode is charac- terized by the number of photons m which occupying it. Each photon has energy ~ω. Because photons are bosons, there can be an infinite number in the same energy level.  = m~ω,m ∈ Z+ Example. Suppose a given energy level is 3~ω. Then there are three photons in the ω mode. Note. We will see that the average occupancy is not an integer. 3.3 3 dimensional generalization 3.3.1 Setup Take a cubical box of side length L. Consider the z component of a section of EM radiation traveling in the x̂ direction. Note. We could have taken any other direction or the ~B component of the EM radiation. We take a particular direction only to apply the wave equation more easily. 3.3.2 Wave equation This section of EM radiation must obey the wave equation, ∇2Ez = ( ∂2 ∂x2 + ∂2 ∂y2 + ∂2 ∂z2 ) Ez = 1 c2 ∂2Ez ∂t2 We assume a solution for the ~E field as Ez(x, y, z, t) = Ez0 sin (ωt) sin (nxπx L ) sin (nyπy L ) sin (nzπz L ) where ni ∈ N. 5 3.3.3 Substitute the assumed solution Substituting the assumed solution into the wave equation gives the desired relation between n and ω. Each second derivative will pull factors out of the sin terms. All of the sin terms cancel out from both sides of the equation to give, −π 2 L2 ( n2x + n 2 y + n 2 z ) = 1 c2 ( −ω2 ) . Thus our assumeded solution really satisfies the wave equation provided the above equation is satisfied. Let n2x + n 2 y + n 2 z = n 2 Since each ni term is an integer, n 2 is also an integer. With this substi- tituon, n2 = L2ω2 π2c2 ⇒ n = Lω πc Note. This n has the same form as the 1 dimensional case. The difference is that now n is defined to be a square root of a sum of squares. 3.3.4 Density of modes Note. There are 2 indepenant polarizations for any EM radiation. Any other polarization can be written as a linear combination of these two orthogonal polarizations. Consider n spcae: a coordinate system with nx plotted on the x̂ axis, etc. Since ni ∈ N, the space is an octant of a sphere of radius n. Any state of the EM radiation system can be represented as a point in n space. Definition. Consider an octant of a shell of width ω to ω+dω. The number of modes in this shell is the density of modes, D(ω). Note. D(w) is defined per unit frequency range. It must be integrated over a range of frequencies to give the actual number of modes in that frequency range. 6 Integrate to find an expression for D(ω), D(ω)︸ ︷︷ ︸ density of modes dω = 2︸︷︷︸ polarization × 1 8︸︷︷︸ octant × 4πn2︸︷︷︸ surface area of sphere dn = πn2dn D(ω) is given by, D(ω) = π n2︸︷︷︸ L2ω2 π2c2 dn dω︸︷︷︸ L πc Write L3 = V , the volume of the cavity. D(w) = V ω2 π2c3 This is the density of modes per unit frequency range in a volume V . 7