퍼텐셜 에너지(Potential Energy)

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<중력과 전기력 비교>

3.3.3. Covariant Differentiation and Christoffel Symbol (2) - 3.3.3.3 Generalization of Covariant Differentiation & 3.3.3.4 Christoffel Transformation

3.3.3.3 Generalization of Covariant Differentiation (Moore의 17장 및 Problem 참조)

<Reference>

• https://physicspages.com/Moore%20Relativity.html Moore Problem 17.2 COVARIANT DERIVATIVE OF A GENERAL TENSOR”

■  앞장(chapter 3.3.3.2)의 공변 미분은 Tensor Type (1,0) 혹은 Tensor Type(0,1) 경우에 해당한다. 다음은 Mixed Tensor Type(1,2)의 공변 미분을 유도한다. 동 방법은 Moore 참조

$D_{bc}^aA_a^{}B_{}^bC_{}^c = scalar$ 이므로 ${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = {\partial _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right)$가 성립

${\partial _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}C_{}^c} \right)$ ${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = \left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\nabla _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\nabla _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\nabla _d}C_{}^c} \right)$ ${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = \left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\nabla _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\nabla _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\nabla _d}C_{}^c} \right)$ $ = \left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{} - \;\Gamma _{ad}^m{A_m}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}{B^b} + \Gamma _{md}^b\;{B^m}} \right)C_{}^c$ $ + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}{C^c} + \Gamma _{md}^c\;{C^m}} \right)$

${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = {\partial _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right)$ 이므로 $\left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{} - \;\Gamma _{ad}^m{A_m}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}{B^b} + \Gamma _{md}^b\;{B^m}} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}{C^c} + \Gamma _{md}^c\;{C^m}} \right)$ $ = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}C_{}^c} \right)$

$\left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\Gamma _{ad}^m{A_m}B_{}^bC_{}^c - D_{bc}^a\Gamma _{md}^bA_a^{}{B^m}C_{}^c - D_{bc}^a\Gamma _{md}^cA_a^{}B_{}^b\;{C^m}$
$D_{bc}^a\Gamma _{ad}^m{A_m}B_{}^bC_{}^c\;$ 항은 $m \to a,\;a \to m$ 이면, $D_{bc}^m\Gamma _{md}^a{A_a}B_{}^bC_{}^c$
$D_{bc}^a\Gamma _{md}^bA_a^{}{B^m}C_{}^c$ 항은 $m \to b,\;b \to m$ 이면, $D_{mc}^a\Gamma _{bd}^mA_a^{}{B^b}C_{}^c$.
$D_{bc}^a\Gamma _{md}^cA_a^{}B_{}^b\;{C^m}$ 항은 $m \to c,\;c \to m$ 이면, $D_{bm}^a\Gamma _{cd}^mA_a^{}B_{}^b\;{C^c}$

$\left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^m\Gamma _{md}^a{A_a}B_{}^bC_{}^c - D_{mc}^a\Gamma _{bd}^mA_a^{}{B^b}C_{}^c - D_{bm}^a\Gamma _{cd}^mA_a^{}B_{}^b\;{C^c}$ ${\nabla _d}D_{bc}^a = {\partial _d}D_{bc}^a + \Gamma _{md}^aD_{bc}^m - D_{mc}^a\Gamma _{bd}^m - D_{bm}^a\Gamma _{cd}^m$

■  공변 미분의 일반화

3.3.3.4 Christoffel Transformation

<Reference>

• Ryder, “Introduction to General Relativity 3.11. Some relations involving connection coefficients””

■  ${\nabla _{\rm{c}}}{\rm{A}}{{\rm{'}}_{\rm{a}}} = {\rm{\;A'}}_{{\rm{a\;\;}};{\rm{c}}}^{} = {\rm{\;\;A'}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} - {\rm{\;\Gamma }}_{{\rm{ac}}}^{\bf{b}}{\rm{A}}{{\rm{'}}_{\bf{b}}}$

○ ${\rm{A'}}_{{\rm{a\;\;}};{\rm{c}}}^{}$와 ${\rm{A}}{{\rm{'}}_{\bf{b}}}$는 Tensor인 반면, ${\rm{A'}}_{{\rm{\;a\;}},{\rm{\;c}}}^{}$와 ${\rm{\Gamma }}_{{\rm{ac}}}^{\bf{b}}$는 Tensor가 아님

○ ${\rm{A'}}_{{\rm{a\;\;}};{\rm{c}}}^{}$와 ${\rm{A}}{{\rm{'}}_{\bf{b}}}$는 Tensor 이므로 각각 다음을 만족한다

♦ A a ;c' =∂ xd∂ x'a∂ xe∂ x'cAd ;e

♦ A'b=∂ xf∂ x'bAf

○ ${\rm{A'}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} = \frac{\partial }{{\partial {x^{{\rm{'c}}}}}}\left( {{\rm{A'}}_{{\rm{\;a}}}^{}} \right) = \frac{\partial }{{\partial {x^{{\rm{'c}}}}}}\left( {\frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}} \right) = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{{\rm{'c}}}}}} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{\bf{d}}}}} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}{\rm{\;}}$ $ = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}$

♦ Partial Derivative 과정과 동일하나 다음과 같이 변환함을 주의

♦ ${\rm{A}}{{\rm{'}}_{\bf{b}}} = \frac{{\partial {\rm{\;}}{x^{\rm{f}}}}}{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{\rm{A}}_{\bf{f}}^{}$에서 ${\rm{A}}_{\bf{f}}^{}$와 동일하게 하기 위하여 ${\rm{A}}_{{\rm{\;a}}}^{} = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}$로 공변 변환

♦ ${\rm{A}}_{{\rm{\;a\;\;}};{\rm{c}}}^{{\rm{'\;}}} = \frac{{\partial {\rm{\;}}{x^{\rm{d}}}}}{{\partial {\rm{\;}}{x^{{\rm{'a}}}}}}\frac{{\partial {\rm{\;}}{x^{\rm{e}}}}}{{\partial {\rm{\;}}{x^{{\rm{'c}}}}}}{\rm{A}}_{{\rm{d\;\;}};{\rm{e}}}^{}$의 d가 나타나도록 $\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{{\rm{'c}}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{\bf{d}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}$게 변환

○ ${\rm{A}}_{{\rm{a\;\;}};{\rm{c}}}^{} = {\rm{\;\;A}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} - {\rm{\;\Gamma }}_{{\rm{ac}}}^{\bf{b}}{{\rm{A}}_{\bf{b}}}$에 위에서 산출한 식들을 대입하면,

♦ $\frac{{\partial {\rm{\;}}{x^{\rm{d}}}}}{{\partial {\rm{\;}}{x^{{\rm{'a}}}}}}\frac{{\partial {\rm{\;}}{x^{\rm{e}}}}}{{\partial {\rm{\;}}{x^{{\rm{'c}}}}}}{\rm{A}}_{{\rm{d\;\;}};{\rm{e}}}^{} = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} - {\rm{\Gamma }}_{{\rm{ac}}}^{\bf{b}}\frac{{\partial {\rm{\;}}{x^{\rm{f}}}}}{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{\rm{A}}_{\bf{f}}^{}$

♦ 우변에서 A Vector가 f로 일치되어 있음을 알 수 있음

♦ 공변 미분(${\rm{A}}_{{\rm{a\;\;}};{\rm{c}}}^{} = {\rm{\;\;A}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} - {\rm{\;\Gamma }}_{{\rm{ac}}}^{\bf{b}}{{\rm{A}}_{\bf{b}}}{\rm{\;}}$)을 이용하여 $\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} = {\rm{A}}_{{\rm{\;f\;}},{\rm{\;d}}}^{}$를 표현하면, ${\rm{A}}_{{\rm{f\;\;}};{\rm{d}}}^{} = {\rm{\;\;A}}_{{\rm{\;f\;}},{\rm{\;d}}}^{} - {\rm{\;\Gamma }}_{{\rm{fd}}}^{\rm{e}}{{\rm{A}}_{\bf{e}}}$ 이므로 ${\rm{A}}_{{\rm{\;f\;}},{\rm{\;d}}}^{} = {\rm{A}}_{{\rm{f\;\;}};{\rm{d}}}^{} + {\rm{\;\Gamma }}_{{\rm{fd}}}^{\bf{e}}{{\rm{A}}_{\bf{e}}}$를 위 식에 대입
∂ xd∂ x'a∂ xe∂ x'cAd ;e=∂xf∂x'a ∂xd∂x'cAf ;d+ ΓfdeAe+∂2xf∂x'c∂x' aA f-Γacb∂ xf∂ x'bAf
=∂xf∂x'a ∂xd∂x'cAf ;d+∂xf∂x'a ∂xd∂x'cΓfdeAe+∂2xf∂x'c∂x' aA f-Γacb∂ xf∂ x'bAf
Γacb∂ xf∂ x'bAf=∂xf∂x'a ∂xd∂x'cΓfdeAe+∂2xf∂x'c∂x' aA f ∵∂ xd∂ x'a∂ xe∂ x'cAd ;e=∂xf∂x'a ∂xd∂x'cAf ;d , (d→f, e→d)

By relabelling (${\rm{A}}_{\bf{f}}^{}$ Vector가 임의의 Vector이므로 label 혹은 Index를 변경 가능)
Γacb∂ xf∂ x'bAf=∂xe∂x'a ∂xd∂x'cΓedfAf+∂2xf∂x'c∂x' aA f=∂xe∂x'a ∂xd∂x'cΓedfAf+∂2xf∂x'c∂x' aA f
Γacb∂ xf∂ x'b=∂xe∂x'a ∂xd∂x'cΓedf+∂2xf∂x'c∂x' a
양변에 $\frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}$를 곱하면,
$\therefore {\rm{\Gamma }}_{{\rm{ac}}}^{\bf{b}} = \frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}\frac{{\partial {x^{\bf{e}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\rm{d}}}}}{{\partial {x^{{\rm{'c}}}}}}{\rm{\Gamma }}_{{\rm{ed}}}^{\rm{f}} + \frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}\frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}$

■  ∇cAa= A' ;ca= ∂cA'a+Γbca A'b

○ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}}$와 ${{\rm{A}}^{\rm{b}}}$는 Tensor인 반면, ${\rm{A}}_{{\rm{\;\;}},{\rm{\;c}}}^{\rm{a}}$와 ${\rm{\Gamma }}_{{\rm{bc}}}^{\rm{a}}$는 Tensor가 아님

○ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}}$와 ${{\rm{A}}^{\rm{b}}}$는 Tensor 이므로 각각 다음을 만족한다

♦ A ;c' a=∂ x'a∂ xd∂ xe∂ x'cA ;ed

♦ A' b=∂ x'b∂ xfAf

○ A' , ca=∂∂x'cA' a=∂∂x'c∂x'a∂xfA f=∂2x'a∂x'c∂xfA f+∂x'a∂xf ∂A f∂x'c=∂2x'a∂x'c∂xfA f+∂x'a∂xf∂xd∂x'c∂A f∂xd
=∂x'a∂xf∂xd∂x'c∂dA f+∂2x'a∂x'c∂xfA f

♦ Partial Derivative 과정과 동일하나 다음과 같이 변환함을 주의

♦ ${A^{{\rm{'\;}}{\bf{b}}}} = \frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}{\rm{A}}_{}^{\rm{f}}$에서 ${\rm{A}}_{}^{\bf{f}}$와 동일하게 하기 위하여 ${\rm{A'}}_{}^{\rm{a}} = \frac{{\partial {x^{{\rm{'a}}}}}}{{\partial {x^{\rm{f}}}}}{\rm{A}}_{\rm{\;}}^{\bf{f}}$로 반변 변환

♦ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{{\rm{'\;a}}} = \frac{{\partial {\rm{\;}}{x^{{\rm{'a}}}}}}{{\partial {\rm{\;}}{x^{\rm{d}}}}}\frac{{\partial {\rm{\;}}{x^{\rm{e}}}}}{{\partial {\rm{\;}}{x^{{\rm{'c}}}}}}{\rm{A}}_{{\rm{\;\;}};{\rm{e}}}^{\rm{d}}$의 d가 나타나도록 $\frac{{\partial {\rm{A}}_{\rm{\;}}^{\bf{f}}}}{{\partial {x^{{\rm{'c}}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\frac{{\partial {\rm{A}}_{\rm{\;}}^{\bf{f}}}}{{\partial {x^{\bf{d}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{\rm{\;}}^{\bf{f}}$게 변환

○ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}} = {\rm{\;\;A}}_{{\rm{\;\;}},{\rm{\;c}}}^{\rm{a}} - {\rm{\;\Gamma }}_{{\bf{b}}{\rm{c}}}^{\rm{a}}{\rm{A}}_{}^{\bf{b}}$에 위에서 산출한 식들을 대입하면,

♦ ∂ x'a∂ xd∂ xe∂ x'cA ;ed=∂x'a∂xf∂xd∂x'c∂dA f+∂2x'a∂x'c∂xfA f+Γbca∂ x'b∂ xfAf

♦ 우변에서 A Vector가 f로 일치되어 있음을 알 수 있음

♦ 공변 미분(${\rm{A'}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}} = {\rm{\;\;A'}}_{{\rm{\;\;}},{\rm{\;c}}}^{\rm{a}} - {\rm{\;\Gamma }}_{{\bf{b}}{\rm{c}}}^{\rm{a}}{\rm{A'}}_{}^{\bf{b}}{\rm{\;}}$)을 이용하여 $\partial _{\rm{d}}^{}{\rm{A}}_{\rm{\;}}^{\bf{f}} = {\rm{A}}_{{\rm{\;\;}},{\rm{\;d}}}^{\rm{f}}$를 표현하면, ${\rm{A}}_{{\rm{\;\;}};{\rm{d}}}^{\rm{f}} = {\rm{\;\;A}}_{{\rm{\;\;}},{\rm{\;d}}}^{\rm{f}} + {\rm{\;\Gamma }}_{{\bf{e}}{\rm{d}}}^{\rm{f}}{\rm{A}}_{}^{\bf{e}}$ 이므로 $\partial _{\rm{d}}^{}{\rm{A}}_{\rm{\;}}^{\bf{f}} = {\rm{A}}_{{\rm{\;\;}},{\rm{\;d}}}^{\rm{f}} = {\rm{A}}_{{\rm{\;\;}};{\rm{d}}}^{\rm{f}} - {\rm{\;\Gamma }}_{{\bf{e}}{\rm{d}}}^{\rm{f}}{\rm{A}}_{}^{\bf{e}}$를 위 식에 대입
∂ x'a∂ xd∂ xe∂ x'cA ;ed=∂x'a∂xf∂xd∂x'cA ;df- ΓedfAe+∂2x'a∂x'c∂xfA f+Γbca∂ x'b∂ xfAf
=∂x'a∂xf∂xd∂x'cA ;df-∂x'a∂xf∂xd∂x'cΓedfAe+∂2x'a∂x'c∂xfA f+Γbca∂ x'b∂ xfAf
Γbca∂ x'b∂ xfAf=∂x'a∂xf∂xd∂x'cΓedfAe-∂2x'a∂x'c∂xfA f ∵∂ x'a∂ xd∂ xe∂ x'cA ;ed=∂x'a∂xf∂xd∂x'cA ;df , (d→f, e→d)

By relabelling (${\rm{A}}_{\bf{f}}^{}$ Vector가 임의의 Vector이므로 label 혹은 Index를 변경 가능)
Γbca∂ x'b∂ xfAf=∂x'a∂xe∂xd∂x'cΓfdeAf-∂2x'a∂x'c∂xfA f=∂x'a∂xe∂xd∂x'cΓfde-∂2x'a∂x'c∂xfA f (e→f, f→e)
Γbca∂ x'b∂ xf=∂x'a∂xe∂xd∂x'cΓfde-∂2x'a∂x'c∂xf

양변에 $\frac{{\partial {\rm{\;}}{x^{\rm{f}}}}}{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}$. 를 곱하면,
∴Γbca=∂xd∂x'cΓfde-∂ xf∂ x'b∂2x'a∂x'c∂xf=∂x'a∂xe∂ xf∂ x'b∂xd∂x'cΓfde-∂ xf∂x'b∂ xe∂x'c∂2x'a∂ xe∂xf =∂x'a∂xe∂ xf∂ x'b∂xd∂x'cΓfde-∂ xf∂x'b∂ xd∂x'c∂2x'a∂ xd∂xf

■  Ryder 책 연습문제에서는 쉽게 풀린다고 하는데…. 난 왜 이렇게 힘들게? 다른 책은?

기초적인 수학 개념(2) - 등식의 성질

 

기초적인 수학 개념(1) - 나누기와 분수의 나눗셈

(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

ℇ-→v→0kT and ℇ-→v→∞0

○ This is in contrast to the law of equipartition of energy which assigns to the average energy 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

Pℇ=e-ℇ/kTkT

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.

ℇ-=∫0∞ℇPℇdℇ∫0∞Pℇdℇ

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

ℇ-=kT

■  Planck's great contribution came when he realized that he could obtain the required cutoff, indicated in ℇ-→v→∞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 ℇ-=∫0∞ℇPℇdℇ/∫0∞Pℇdℇ 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 ⋯

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 ℇ-≃kT when the difference in adjacent energies ∆ℇ is small, and 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∝ v

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

∆ℇ=hv

where h is the proportionality constant

○ The formula Planck obtained for ℇ ̅ by evaluating the summation analogous to the integral in ℇ-=∫0∞ℇPℇdℇ/∫0∞Pℇdℇ

ℇ-v=∑n=0∞ℇPℇ∑n=0∞Pℇ=hvehv/kT-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,⋯

ℇ-=∑n=0∞ℇPℇ∑n=0∞Pℇ=∑n=0∞ℇkTe-ℇ/kT∑n=0∞1kTe-ℇ/kT=∑n=0∞nhvkTe-nhv/kT∑n=0∞1kTe-nhv/kT ∵Pℇ=e-ℇ/kTkT=e-nhv/kTkT =kT∑n=0∞nαe-nα∑n=0∞e-nα where α=hvkT ∵-αddαln∑n=0∞e-nα=-αddα∑n=0∞e-nα∑n=0∞e-nα=-∑n=0∞αddαe-nα∑n=0∞e-nα=∑n=0∞nαe-nα∑n=0∞e-nα ℇ-=kT-αddαln∑n=0∞e-nα=-hvddαln∑n=0∞e-nα ∵α=hvkT ∑n=0∞e-nα=1+e-α+e-2α+e-3α+⋯=1+X+X2+X3+⋯ where X=e-α 1-X-1=1+X+X2+X3+⋯ ℇ-=-hvddαln∑n=0∞e-nα=-hvddαln1-e-α-1 =-hv1-e-α-1-11-e-α-2e-α=hve-α1-e-α=hveα-1=hvehvkT-1 ∴uυ,Tdυ=ℇ-∙2∙degree of feedomVdυ=hvehvkT-1∙8πυ2c3dυ uλ,Tdλ=hehcλkT-1∙cλ∙8πcλ2c3-cλ2dλ=8πhcλ5dλehcλkT-1 ∵v=cλ