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< Reference >  I referred to Professor Joon-Gon Choi's lecture on quantum mechanics 1 at Korea University. https://www.youtube.com/watch?v=kIA9mleY0AI&list=PLIoJAwnfA7S8dYxnS_78q4VbOLZuBeZHK  Robert Eisberg, Robert Resnick, “Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles”, 1.3. LASSICAL THEORY OF CAVITY RADIATION

■ Theoretical research in the field of thermal radiation began in 1859 with the work of Kirchhoff, who showed that for a given λ, the ratio of the emissive power E to the absorptivity A, defined as the fraction of incident radiation of wavelength λ that is absorbed by the body, is the same for all bodies. Kirchhoff considered two emitting and absorbing parallel plates and showed from the equilibrium condition that the energy emitted was equal to the energy absorbed (for each λ), that the ratio E/A must be the same for the two plates. Soon thereafter, he observed that for a black body, defined as a surface that totally absorbs all radiation that falls on it, so that A=1, the function \(E\left( {\lambda ,T} \right)\) is a universal function.

○ The energy from the outside of the blackbody is measured in proportion to the amount of energy from the inside of the blackbody through the hole dA of the blackbody.

○ The total energy (\( \in \left( {{\rm{\lambda }},T} \right)\)) generated by the energy density inside the black body comes out through the hole dA

○ \(E\left( {{\rm{\lambda }},T} \right)\) : energy(\( \in \left( {{\rm{\lambda }},T} \right)\)) emitted per unit area(\(dA\)) per unit time (\({\rm{\Delta }}t\))

○ \(u\left( {{\rm{\lambda }},T} \right)\) : Energy density in blackbody

E (λ, T) = \frac{\in (λ, T)}{d A ∙ Δ t}

\in (λ, T) = \int_{0}^{c Δ t} \int_{0}^{π / 2} \int_{0}^{2 π} u (λ, T) ∙ r^{2} \sin θ d r d θ d ϕ ∙ d A ∙ \frac{\cos θ}{4 π r^{2}}

= \int_{0}^{c Δ t} r^{2} d r \int_{0}^{π / 2} \sin θ \cos θ d θ \int_{0}^{2 π} d ϕ u (λ, T) ∙ d A ∙ \frac{1}{4 π r^{2}} = \int_{0}^{c Δ t} d r \int_{0}^{π / 2} \sin θ \cos θ d θ \int_{0}^{2 π} d ϕ u (λ, T) ∙ d A ∙ \frac{1}{4 π}

= c Δ t ∙ \frac{1}{2} ∙ 2 π ∙ u (λ, T) ∙ d A ∙ \frac{1}{4 π} = \frac{c Δ t}{4} ∙ d A ∙ u (λ, T)

E (λ, T) = \frac{c}{4} ∙ u (λ, T)

■ Energy Density

u (λ, T) = u (ν, T) |\frac{d λ}{d ν}| = u (ν, T) \frac{c}{ν^{2}}

\frac{d λ}{d ν} = - \frac{c}{ν^{2}} (∵ λ ν = c, λ = \frac{c}{ν})

■ Classical Derivation of Rayleigh-Jeans Law

○ Boundary Condition

♦ Periodic boundary condition

f (x, y, 0) = f (x, y, L)

f (x, 0, z) = f (x, L, z)

f (0, y, z) = f (L, y, z)

○ The function that satisfies the periodic boundary condition is typically an oscillator.

♦ One dimension

\frac{1}{\sqrt{L}} e^{i k x}, e^{i k L} = 1, k = \frac{2 π}{L} n, n = i n t e g e r

♦ Three dimension

\frac{1}{L^{3 / 2}} e^{i (k_{1} x + k_{2} y + k_{3} z)}, k_{1} = \frac{2 π}{L} n_{1}, k_{2} = \frac{2 π}{L} n_{2}, k_{3} = \frac{2 π}{L} n_{3},

k_{i} L = 2 π n_{i} (i = 1,2, 3)

○ Equipartition Theorem(Robert Eisberg, Robert Resnick)

♦ The prediction comes from classical kinetic theory, and it is called the law of equipartition of energy. This law states that for a system of gas molecules in thermal equilibrium at temperature T, the average kinetic energy of a molecule per degree of freedom

\frac{1}{2} k T, k = B o l t z m a n n c o n s t a n t

♦ However, each oscillating standing wave has a total energy which is twice its average kinetic energy. This is a common property of physical systems which have a single degree of freedom that execute simple harmonic oscillations in time; familiar cases are a pendulum or a coil spring.

♦ Rayleigh assumed the classical law of equipartition energy. He said, “one dimensional waves always have 2 degrees of freedom, one for potential energy (x) and the other for kinetic energy (v). In case of electromagnetic wave, these two degree of freedom are derived from electric field and magnetic field.” ( http://www.pa.uky.edu/~kwng/phy361/class/class10.pdf )

\bar{ℇ} = a v e r a g e t o t a l E n e r g y p e r u n i t d e g r e e o f f r e e d o m = 2 ∙ \frac{1}{2} k T = k T

○ degree of freedom

♦ Since k is a function of n and n is an integer, it is easy to count n. Therefore, we calculate the degrees of freedom using n.

4 π n^{2} d n = 4 π {(\frac{k L}{2 π})}^{2} \frac{L}{2 π} d k = 4 π {(\frac{L}{2 π})}^{3} k^{2} d k ∵ k L = 2 π n, n = \frac{k L}{2 π}

= 4 π {(\frac{L}{2 π})}^{3} {(\frac{2 π}{c} υ)}^{2} \frac{2 π}{c} d υ ∵ ω = c k = 2 π υ, k = \frac{2 π}{c} υ = 4 π L^{3} \frac{υ^{2}}{c^{3}} d υ

♦ This completes the calculation except that we must multiply these results by a factor of 2. because, for each of the allowed frequencies we have enumerated, there are actually two independent waves corresponding to the two possible states of polarization of electromagnetic radiation.

∴ u (υ, T) d υ = \bar{ℇ} ∙ 2 ∙ \frac{d e g r e e o f f e e d o m}{V} d υ = k T ∙ 2 ∙ \frac{d e g r e e o f f e e d o m}{V} d υ = k T ∙ 8 π \frac{υ^{2}}{c^{3}} d υ ∵ V = L^{3}

독학 아재의 물리 이야기

2020년 9월 10일 목요일

(The Limits of Classical Physics-1) Classical Derivation of Rayleigh-Jeans Formula (Gasiorowicz chapter 1)

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