독학 아재의 물리 이야기: (The Limits of Classical Physics-2) Derivation of Planck’s Formula (Gasiorowicz chapter 1)

< Quotation >  Robert Eisberg, Robert Resnick, “Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles”, 1.3. LASSICAL THEORY OF CAVITY RADIATION

■ Planck realized that, in the circumstances that prevail for the case of blackbody radiation, the average energy of the standing waves is a function of frequency having the properties indicated by

\bar{ℇ} \underset{v \to 0}{\to} k T a n d \bar{ℇ} \underset{v \to \infty}{\to} 0

○ This is in contrast to the law of equipartition of energy which assigns to the average energy $\bar{ℇ}$ a value independent of frequency.

■ Let us look at the origin of the equipartition law. It arises, basically, from a more comprehensive result of classical statistical mechanics called the Boltzmann distribution. Here we shall use a special form of the Boltzmann distribution

P (ℇ) = \frac{e^{- ℇ / k T}}{k T}

Such an abrupt change in potential is In which P(ℇ)dℇ is the probability of finding a given entity of a system with energy in the interval between ℇ and ℇ+dℇ, when the number of energy states for the entity in that interval is independent of ℇ.

\bar{ℇ} = \frac{\int_{0}^{\infty} ℇ P (ℇ) d ℇ}{\int_{0}^{\infty} P (ℇ) d ℇ}

Such an abrupt change in potential is The integrand in the numerator is the energy, ℇ, weighted by the probability that the entity will be found with this energy. The integral in the numerator can be evaluated, and the result is just the law of equipartition of energy

\bar{ℇ} = k T

■ Planck's great contribution came when he realized that he could obtain the required cutoff, indicated in $\bar{ℇ} \underset{v \to \infty}{\to} 0$ , if he modified the calculation leading from Pℇ to by treating the energy ℇ as if it were a discrete variable instead of as the continuous variable that it definitely is from the point of view of classical physics. Quantitatively, this can be done by rewriting $\bar{ℇ} = \int_{0}^{\infty} ℇ P (ℇ) d ℇ / \int_{0}^{\infty} P (ℇ) d ℇ$ in terms of a sum instead of an integral.

○ Planck assumed that the energy ℇ could take on only certain discrete values, rather than any value, and that the discrete values of the energy were uniformly distributed; that is, he took

ℇ = 0, ∆ ℇ, 2 ∆ ℇ, 3 ∆ ℇ, 4 ∆ ℇ \dots

Such an abrupt change in potential is as the set of allowed values of the energy. Here ∆ℇ is the uniform interval between successive allowed values of the energy

○ Recapitulating, Planck discovered that he could obtain $\bar{ℇ} ≃ k T$ when the difference in adjacent energies ∆ℇ is small, and $\bar{ℇ} ≃ 0$ when∆ℇ is large. Since he needed to obtain the first result for small values of the frequency $v$,and the second result for large values of $v$, he clearly needed to make ∆ℇ an increasing function of $v$. Numerical work showed him that he could take the simplest possible relation between ∆ℇ and $v$ having this property. That is, he assumed these quantities to be proportional

∆ ℇ \propto v

Such an abrupt change in potential is Written as an equation instead of a proportionality, this is

∆ ℇ = h v

where h is the proportionality constant

○ The formula Planck obtained for ℇ ̅ by evaluating the summation analogous to the integral in $\bar{ℇ} = \int_{0}^{\infty} ℇ P (ℇ) d ℇ / \int_{0}^{\infty} P (ℇ) d ℇ$

\bar{ℇ} (v) = \frac{\sum_{n = 0}^{\infty} ℇ P (ℇ)}{\sum_{n = 0}^{\infty} P (ℇ)} = \frac{h v}{e^{h v / k T} - 1}

Sums must be used because with Planck's postulate the energy ℇ becomes a discrete variable that takes on only the values ℇ=0,hv,2hv,3hv,⋯. That is, ℇ=nhv where n=0,1,2,3,⋯

\bar{ℇ} = \frac{\sum_{n = 0}^{\infty} ℇ P (ℇ)}{\sum_{n = 0}^{\infty} P (ℇ)} = \frac{\sum_{n = 0}^{\infty} \frac{ℇ}{k T} e^{- ℇ / k T}}{\sum_{n = 0}^{\infty} \frac{1}{k T} e^{- ℇ / k T}} = \frac{\sum_{n = 0}^{\infty} \frac{n h v}{k T} e^{- n h v / k T}}{\sum_{n = 0}^{\infty} \frac{1}{k T} e^{- n h v / k T}} (∵ P (ℇ) = \frac{e^{- ℇ / k T}}{k T} = \frac{e^{- n h v / k T}}{k T})

= k T \frac{\sum_{n = 0}^{\infty} n α e^{- n α}}{\sum_{n = 0}^{\infty} e^{- n α}} w h e r e α = \frac{h v}{k T}

∵ - α \frac{d}{d α} l n \sum_{n = 0}^{\infty} e^{- n α} = \frac{- α \frac{d}{d α} \sum_{n = 0}^{\infty} e^{- n α}}{\sum_{n = 0}^{\infty} e^{- n α}} = \frac{- \sum_{n = 0}^{\infty} α \frac{d}{d α} e^{- n α}}{\sum_{n = 0}^{\infty} e^{- n α}} = \frac{\sum_{n = 0}^{\infty} n α e^{- n α}}{\sum_{n = 0}^{\infty} e^{- n α}}

\bar{ℇ} = k T (- α \frac{d}{d α} l n \sum_{n = 0}^{\infty} e^{- n α}) = - h v \frac{d}{d α} l n \sum_{n = 0}^{\infty} e^{- n α} (∵ α = \frac{h v}{k T})

\sum_{n = 0}^{\infty} e^{- n α} = 1 + e^{- α} + e^{- 2 α} + e^{- 3 α} + \dots = 1 + X + X^{2} + X^{3} + \dots w h e r e X = e^{- α}

{(1 - X)}^{- 1} = 1 + X + X^{2} + X^{3} + \dots

\bar{ℇ} = - h v \frac{d}{d α} l n \sum_{n = 0}^{\infty} e^{- n α} = - h v \frac{d}{d α} l n {(1 - e^{- α})}^{- 1}

= \frac{- h v}{{(1 - e^{- α})}^{- 1}} (- 1) {(1 - e^{- α})}^{- 2} e^{- α} = \frac{h v e^{- α}}{1 - e^{- α}} = \frac{h v}{e^{α} - 1} = \frac{h v}{e^{\frac{h v}{k T}} - 1}

∴ 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 = \frac{h v}{e^{\frac{h v}{k T}} - 1} ∙ 8 π \frac{υ^{2}}{c^{3}} d υ

u (λ, T) d λ = \frac{h}{e^{\frac{h c}{λ k T}} - 1} ∙ \frac{c}{λ} ∙ 8 π \frac{{(\frac{c}{λ})}^{2}}{c^{3}} (- \frac{c}{λ^{2}}) d λ = \frac{8 π h c}{λ^{5}} \frac{d λ}{e^{\frac{h c}{λ k T}} - 1} (∵ v = \frac{c}{λ})

독학 아재의 물리 이야기

2020년 9월 10일 목요일

(The Limits of Classical Physics-2) Derivation of Planck’s Formula (Gasiorowicz chapter 1)

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