Aula 1

UFABC - Física Quântica - Curso 2017.3 Prof. Germán Lugones

Aula 1

Evidências experimentais da teoria

quântica : radiação do Corpo Negro.

1

Cosmic microwave background Planck Satellite

Motivações para se estudar física quântica

2 A mecânica quântica (também conhecida como física

quântica ou teoria quântica), é a teoria física que descreve a natureza nas pequenas escalas (átomos e partículas subatômicas).

A mecânica quântica é muito diferente da física clássica, por exemplo:

• a energia, o momento e outras quantidades são freqüentemente restritas a valores discretos

(quantização),

• os objetos possuem características de partículas e ondas (ou seja, dualidade onda-partícula), e • há limites para a precisão com que as

quantidades podem ser conhecidas (princípio da incerteza).

A Física Quântica muitas vezes desafia o senso comum!

3 A física quântica é essencial em vários ramos da ciência e da tecnologia:

- química, nanotecnologia: a dinâmica dos átomos e moléculas é regida pela mecânica quântica.

- microeletrônica, componentes como transistores e diodos, que fazem parte dos dispositvos eletrônicos, funcionam com base nas leis da mecânica quântica.

- Fisica de Partículas

- Astrofísica, Cosmologia, origem do Universo - etc… etc….

A física quântica foi surgindo gradualmente a partir do início do século XX.

Os primeiros indícios do comportamento quântico da natureza foram obtidos por Max Planck em 1900 analisando o

problema da radiação do corpo negro e por Albert Einstein em 1905 analisando o efeito fotoelétrico.

A primeira pista da natureza quântica da radiação veio do estudo da radiação térmica emitida por corpos opacos.

Quando a radiação incide sobre um corpo opaco, parte dela é refletida e parte é absorvida.

Os corpos de cor clara refletem a maior parte da radiação visível incidente sobre eles, enquanto os corpos escuros absorvem a maior parte.

Corpo negro

A parte de absorção do processo pode ser descrita brevemente da seguinte forma.

• A radiação absorvida pelo corpo aumenta a energia cinética dos átomos constituintes, que oscilam sobre suas posições de equilíbrio.

• Como a energia cinética translacional média dos átomos determina a temperatura T do corpo, a energia absorvida faz com que T aumente.

• No entanto, os átomos contêm cargas (os elétrons) e são acelerados pelas oscilações.

• Cargas aceleradas devem emitir radiação eletromagnética → a emissão de radiação reduz a energia cinética das oscilações dos átomos e portanto reduz T.

• Quando a taxa de absorção é igual à taxa de emissão, T fica constante e dizemos que o corpo está em equilíbrio térmico com seus arredores.

A radiação eletromagnética emitida nessas circunstâncias é chamada de

Um corpo que absorve todas as radiações que incidem sobre isso é chamado de corpo negro ideal.

Em 1879, Josef Stefan encontrou uma relação empírica entre a potência irradiada por um corpo negro ideal e a temperatura:

R = σ T4 _{Lei de Stefan-Boltzmann}

onde:

• R é a potência irradiada por unidade de área. R também é chamada de radiância. R nos diz a taxa em que a energia é emitida pelo objeto.

• Unidades de R: W/m2

• T é a temperatura absoluta

• σ = 5.6705×10−8 _W/m2_·K4 _{é a constante de Stefan-Boltzmann}

R de um corpo negro depende apenas de T e não de qualquer outra característica do objeto, como a cor ou o material do qual é composto.

A lei de Stefan Boltzmann

Objetos que não são corpos negros ideais irradiam energia por unidade de área a uma taxa R inferior à de um corpo negro à mesma T.

Para esses objetos, a taxa R depende de propriedades além de T, como por exemplo a cor e a composição da superfície.

Os efeitos dessas dependências são combinados em um fator chamado de emissividade 𝜀, que multiplica o lado direito da Lei de Stefan-Boltzmann:

R = 𝜀 σT4

• Os valores de 𝜀 são sempre menores do que a unidade: 0 < 𝜀 < 1 • A emissividade 𝜀 pode depender da temperatura e de outros fatores.

Emissividade

Potência espectral

A radiância total R definida antes leva em consideração a radiação emitida em todos os comprimentos de onda.

Definimos agora a radiância espectral R(𝜆) (ou distribuição espectral) de maneira que dR=R(𝜆)d𝜆 seja a potência emitida por unidade de área no intervalo de comprimentos de onda entre 𝜆 e 𝜆+ d𝜆.

Pela definição anterior, dR=R(𝜆)d𝜆, temos:

e, portanto:

R( ) =

dR

d

R = Z ₁ 0 R( )d 8

• Na figura vemos a radiância espectral R(𝜆) medida a diferentes T.

• O eixo vertical está em unidades arbitrárias apenas para comparação. • O intervalo de 𝜆 está no espectro visível.

• O Sol emite radiação muito próxima da de um corpo negro a T=5800 K. 3-2 Blackbody Radiation 125

This result is known as Wien’s displacement law. It was obtained by Wien in 1893. Examples 3-1 and 3-2 illustrate its application.

EXAMPLE 3-1 How Big Is a Star? Measurement of the wavelength at which the spectral distribution R(L) from a certain star is maximum indicates that the star’s surface temperature is 3000 K. If the star is also found to radiate 100 times the power P_^ radiated by the Sun, how big is the star? (The symbol ^  Sun.) The Sun’s surface temperature is 5800 K.

SOLUTION

Assuming the Sun and the star both radiate as blackbodies (astronomers nearly always make that assumption, based on, among other things, the fact that the solar spectrum is very nearly that of an ideal blackbody), their surface temperatures have been determined from Equation 3-5 to be 5800 K and 3000 K, respectively. Mea-surement also indicates that P_star  100 P_^. Thus, from Equation 3-4 we have that

Rstar  P_star (area)_star  100 P_^ 4Pr2_star  ST 4 star and R_^  P^ (area)_^  P_^ 4Pr2_^  ST 4 ^ Thus, we have r2_star  100 r2_^4 T^ T_star5 4 r_star  10 r_^4 T^ T_star5 2  1045800 30005 2 r_^ rstar  37.4 r_^

Since r_^  6.96  108 m, this star has a radius of about 2.6  1010 m, or about half the radius of the orbit of Mercury. This star is a red giant (see Chapter 13).

FIGURE 3-4 Spectral distribution function R(L) measured at different temperatures. The R(L) axis is in arbitrary units for comparison only. Notice the range of L in the visible spectrum. The Sun emits radiation very close to that of a blackbody at 5800 K. L_m is indicated for the 5000 K and 6000 K curves. 0.3 0.2 0 0.1 0 500 Visible 6000 K 4000 K 3000 K 1000 Wavelength L (nm) 1500 2000 R (L ) 5000 K L_m L_m 0.001 0 0 2000 K 1500 K 1000 K 2000 4000 L (nm) R (L ) TIPLER_03_119-152hr2.indd 125 8/22/11 11:33 AM 9

Para uma T dada, se observa que a radiância espectral R(𝜆) tem um máximo (pico).

O comprimento de onda 𝜆m do

pico da distribuição se desloca para comprimentos de onda menores à medida que T aumenta:

Wien descobriu empiricamente que m / 1 T ! mT = constante = 2.898 ⇥ 10 3_m · K

Lei de deslocamento de Wien

Este resultado é conhecido como lei de deslocamento de Wien. Foi obtido

Questão: A medição do comprimento de onda em que a radiância R(𝜆) de uma determinada estrela é máxima, indica que T = 3000K na superfície da estrela.

Também é medida a potência total emitida pela estrela, e o valor obtido é 100 P⊙, onde P⊙ é a potência irradiada pelo Sol,

Qual é o raio da estrela?

A temperatura da superfície do Sol é de 5800 K.

Resolução: Suponhamos que o Sol e a estrela irradiam como corpos negros (os astrônomos quase sempre fazem essa suposição, com base, entre outras coisas, no fato de que o espectro solar é quase o de um corpo negro ideal).

Exemplo: Determinação do tamanho de uma estrela.

Pela Lei de Stefan-Boltzmann, R = σ T4 _{, temos:} - para a estrela:

- para o Sol:

Portanto:

Como r = 6.96×108 _{m, esta estrela tem um raio de cerca de 2.6×10}10 _{m, ou}

cerca de metade do raio da órbita de Mercúrio. Esta estrela é uma gigante vermelha.

3-2 Blackbody Radiation 125

This result is known as Wien’s displacement law. It was obtained by Wien in 1893. Examples 3-1 and 3-2 illustrate its application.

EXAMPLE 3-1 How Big Is a Star? Measurement of the wavelength at which the spectral distribution R(L) from a certain star is maximum indicates that the star’s surface temperature is 3000 K. If the star is also found to radiate 100 times the power P_^ radiated by the Sun, how big is the star? (The symbol ^  Sun.) The Sun’s surface temperature is 5800 K.

SOLUTION

Assuming the Sun and the star both radiate as blackbodies (astronomers nearly always make that assumption, based on, among other things, the fact that the solar spectrum is very nearly that of an ideal blackbody), their surface temperatures have been determined from Equation 3-5 to be 5800 K and 3000 K, respectively. Mea-surement also indicates that P_star  100 P_^. Thus, from Equation 3-4 we have that

R_star  Pstar (area)_star  100 P_^ 4Pr2_star  ST 4 star and R_^  P^ (area)_^  P_^ 4Pr2_^  ST 4 ^ Thus, we have r2_star  100 r2_^4 T^ T_star5 4 r_star  10 r_^4 T^ T_star5 2  1045800 30005 2 r_^ r_star  37.4 r_^

Since r_^  6.96  108 m, this star has a radius of about 2.6  1010 m, or about half the radius of the orbit of Mercury. This star is a red giant (see Chapter 13).

FIGURE 3-4 Spectral distribution function R(L) measured at different temperatures. The R(L) axis is in arbitrary units for comparison only. Notice the range of L in the visible spectrum. The Sun emits radiation very close to that of a blackbody at 5800 K. L_m is indicated for the 5000 K and 6000 K curves. 0.3 0.2 0 0.1 0 500 Visible 6000 K 4000 K 3000 K 1000 Wavelength L (nm) 1500 2000 R (L ) 5000 K Lm Lm 0.001 0 0 2000 K 1500 K 1000 K 2000 4000 L (nm) R (L ) TIPLER_03_119-152hr2.indd 125 8/22/11 11:33 AM 3-2 Blackbody Radiation 125

This result is known as Wien’s displacement law. It was obtained by Wien in 1893. Examples 3-1 and 3-2 illustrate its application.

EXAMPLE 3-1 How Big Is a Star? Measurement of the wavelength at which the spectral distribution R(L) from a certain star is maximum indicates that the star’s surface temperature is 3000 K. If the star is also found to radiate 100 times the power P_^ radiated by the Sun, how big is the star? (The symbol ^  Sun.) The Sun’s surface temperature is 5800 K.

SOLUTION

Assuming the Sun and the star both radiate as blackbodies (astronomers nearly always make that assumption, based on, among other things, the fact that the solar spectrum is very nearly that of an ideal blackbody), their surface temperatures have been determined from Equation 3-5 to be 5800 K and 3000 K, respectively. Mea-surement also indicates that P_star  100 P_^. Thus, from Equation 3-4 we have that

R_star  Pstar (area)_star  100 P_^ 4Pr2_star  ST 4 star and R_^  P^ (area)_^  P_^ 4Pr2_^  ST 4 ^ Thus, we have r2_star  100 r2_^4 T^ T_star5 4 r_star  10 r_^4 T^ T_star5 2  1045800 30005 2 r_^ r_star  37.4 r_^

Since r_^  6.96  108 m, this star has a radius of about 2.6  1010 m, or about half the radius of the orbit of Mercury. This star is a red giant (see Chapter 13).

FIGURE 3-4 Spectral distribution function R(L) measured at different temperatures. The R(L) axis is in arbitrary units for comparison only. Notice the range of L in the visible spectrum. The Sun emits radiation very close to that of a blackbody at 5800 K. L_m is indicated for the 5000 K and 6000 K curves. 0.3 0.2 0 0.1 0 500 Visible 6000 K 4000 K 3000 K 1000 Wavelength L (nm) 1500 2000 R (L ) 5000 K L_m L_m 0.001 0 0 2000 K 1500 K 1000 K 2000 4000 L (nm) R (L ) TIPLER_03_119-152hr2.indd 125 8/22/11 11:33 AM 3-2 Blackbody Radiation 125

This result is known as Wien’s displacement law. It was obtained by Wien in 1893. Examples 3-1 and 3-2 illustrate its application.

EXAMPLE 3-1 How Big Is a Star? Measurement of the wavelength at which

the spectral distribution R(L) from a certain star is maximum indicates that the star’s surface temperature is 3000 K. If the star is also found to radiate 100 times the power P_^ radiated by the Sun, how big is the star? (The symbol ^  Sun.) The Sun’s surface temperature is 5800 K.

SOLUTION

Assuming the Sun and the star both radiate as blackbodies (astronomers nearly always make that assumption, based on, among other things, the fact that the solar spectrum is very nearly that of an ideal blackbody), their surface temperatures have been determined from Equation 3-5 to be 5800 K and 3000 K, respectively. Mea-surement also indicates that P_star  100 P_^. Thus, from Equation 3-4 we have that

R_star  Pstar (area)_star  100 P_^ 4Pr2_star  ST 4 star and R_^  P^ (area)_^  P_^ 4Pr2_^  ST 4 ^ Thus, we have r2_star  100 r2_^4 T^ T_star5 4 r_star  10 r_^4 T^ T_star5 2  1045800 30005 2 r_^ r_star  37.4 r_^

Since r_^  6.96  108 m, this star has a radius of about 2.6  1010 m, or about half the radius of the orbit of Mercury. This star is a red giant (see Chapter 13).

FIGURE 3-4 Spectral distribution function R(L) measured at different temperatures. The R(L) axis is in arbitrary units for comparison only. Notice the range of L in the visible spectrum. The Sun emits radiation very close to that of a blackbody at 5800 K. L_m is indicated for the 5000 K and 6000 K curves. 0.3 0.2 0 0.1 0 500 Visible 6000 K 4000 K 3000 K 1000 Wavelength L (nm) 1500 2000 R (L ) 5000 K L_m L_m 0.001 0 0 2000 K 1500 K 1000 K 2000 4000 L (nm) R (L ) TIPLER_03_119-152hr2.indd 125 8/22/11 11:33 AM 12

O corpo negro

Radia¸c˜ao de corpo negro

Aula 3 4 / 21

O corpo negro ´e um objeto ideal que absorve toda a radia¸c˜ao incidente sobre ele e irradia com emissividade ε = 1 a chamada radia¸c˜ao do corpo negro.

radia¸c˜ao Modelo de um corpo negro ideal

■ A radia¸c˜ao eletromagn´etica de qualquer compri-mento de onda entra por um pequeno orif´ıcio de uma cavidade. Ao bater nas suas paredes, uma parte dela ´e absorvida e uma outra ´e refletida. Eventualmente, toda a radia¸c˜ao ser´a absorvida pela cavidade;

■ Os ´atomos das paredes da cavidade absorvem a radia¸c˜ao. Na situa¸c˜ao de equil´ıbrio t´ermico, as paredes possuem uma temperatura T e emitem radia¸c˜ao t´ermica. Eventualmente, parte dessa radia¸c˜ao pode escapar da cavidade

atrav´es do orif´ıcio. Alguns materiais pretos como o lampblack se aproximam de

um corpo negro ideal, mas a melhor realização prática de um corpo negro ideal é uma cavidade com um pequeno orifício. A radiação que incide no buraco entra na cavidade e é refletida inúmeras vezes até ser completamente absorvida.

A catástrofe ultravioleta

Os átomos das paredes da

cavidade absorvem a radiação. Na situação de equilíbrio térmico, as paredes possuem uma

temperatura T e emitem radiação térmica.

Parte dessa radiação pode

escapar da cavidade através do orifício e ser observada.

A potência irradiada para fora do furo é proporcional à densidade de

energia total U (a energia por unidade de volume da radiação na cavidade). É possível mostrar que a constante de proporcionalidade é c/4, onde c é a velocidade da luz.

Da mesma forma, a distribuição espectral da potência, R(𝜆), emitida a partir do furo é proporcional à distribuição espectral da densidade de energia na cavidade, u(𝜆).

Se u(𝜆)d𝜆 é a fração da energia por unidade de volume na cavidade no intervalo d𝜆, então u(𝜆) e R(𝜆) estão relacionados por

126 Chapter 3 Quantization of Charge, Light, and Energy

Rayleigh-Jeans Equation

The calculation of the distribution function R(L) involves the calculation of the energy density of electromagnetic waves in a cavity. Materials such as black velvet or lampblack come close to being ideal blackbodies, but the best practical realiza-tion of a ideal blackbody is a small hole leading into a cavity (such as a keyhole in a closet door; see Figure 3-5). Radiation incident on the hole has little chance of being reflected back out of the hole before it is absorbed by the walls of the cavity. The power radiated out of the hole is proportional to the total energy density U (the energy per unit volume of the radiation in the cavity). The proportionality constant can be shown to be c4, where c is the speed of light.8

R  1

4 cU 3-6

Similarly, the spectral distribution of the power emitted from the hole is proportional to the spectral distribution of the energy density in the cavity. If u(L) d L is the frac-tion of the energy per unit volume in the cavity in the range d L, then u(L) and R(L) are related by

R(L)  1

4 cu(L) 3-7

The energy density distribution function u(L) can be calculated from classical physics in a straightforward way. The method involves finding the number of modes of oscillation of the electromagnetic field in the cavity with wavelengths in the inter-val d L and multiplying by the average energy per mode. The result is that the number of modes of oscillation per unit volume, n(L), is independent of the shape of the cav-ity and is given by

n(L)  8PL 4 3-8

According to classical kinetic theory, the average energy per mode of oscillation is kT, the same as for a one-dimensional harmonic oscillator, where k is the Boltzmann constant. Classical theory thus predicts for the energy density distribution function

uL  kT nL  8PkT L 4 3-9

This prediction, initially derived by Lord Rayleigh,9 is called the Rayleigh-Jeans equation. It is illustrated in Figure 3-6.

At very long wavelengths the Rayleigh-Jeans equation agrees with the experi-mentally determined spectral distribution, but at short wavelengths this equation pre-dicts that u(L) becomes large, approaching infinity as L 4 0, whereas experiment shows (see Figure 3-4) that the distribution actually approaches zero as L 4 0. This enormous disagreement between the experimental measurement of u(L) and the pre-diction of the fundamental laws of classical physics at short wavelengths was called the ultraviolet catastrophe. The word catastrophe was not used lightly; Equation 3-9 implies that

)

R 0

uL d L 4 @ 3-10

That is, every object would have an infinite energy density, which observation assures us is not true.

CCR FIGURE 3-5 A small hole in the wall of a cavity

approximating an ideal blackbody. Radiation entering the hole has little chance of leaving before it is completely absorbed within the cavity.

16

TIPLER_03_119-152hr4.indd 126 9/8/11 11:18 AM

14

126 Chapter 3 Quantization of Charge, Light, and Energy

Rayleigh-Jeans Equation

The calculation of the distribution function R(L) involves the calculation of the energy density of electromagnetic waves in a cavity. Materials such as black velvet or lampblack come close to being ideal blackbodies, but the best practical realiza-tion of a ideal blackbody is a small hole leading into a cavity (such as a keyhole in a closet door; see Figure 3-5). Radiation incident on the hole has little chance of being reflected back out of the hole before it is absorbed by the walls of the cavity. The power radiated out of the hole is proportional to the total energy density U (the energy per unit volume of the radiation in the cavity). The proportionality constant can be shown to be c4, where c is the speed of light.8

R  1

4 cU 3-6

Similarly, the spectral distribution of the power emitted from the hole is proportional to the spectral distribution of the energy density in the cavity. If u(L) d L is the frac-tion of the energy per unit volume in the cavity in the range d L, then u(L) and R(L) are related by

R(L)  1

4 cu(L) 3-7

The energy density distribution function u(L) can be calculated from classical physics in a straightforward way. The method involves finding the number of modes of oscillation of the electromagnetic field in the cavity with wavelengths in the inter-val d L and multiplying by the average energy per mode. The result is that the number of modes of oscillation per unit volume, n(L), is independent of the shape of the cav-ity and is given by

n(L)  8PL 4 3-8

According to classical kinetic theory, the average energy per mode of oscillation is

kT, the same as for a one-dimensional harmonic oscillator, where k is the Boltzmann

constant. Classical theory thus predicts for the energy density distribution function

uL  kT nL  8PkT L 4 3-9

This prediction, initially derived by Lord Rayleigh,9 is called the Rayleigh-Jeans

equation. It is illustrated in Figure 3-6.

At very long wavelengths the Rayleigh-Jeans equation agrees with the experi-mentally determined spectral distribution, but at short wavelengths this equation pre-dicts that u(L) becomes large, approaching infinity as L 4 0, whereas experiment shows (see Figure 3-4) that the distribution actually approaches zero as L 4 0. This enormous disagreement between the experimental measurement of u(L) and the pre-diction of the fundamental laws of classical physics at short wavelengths was called the ultraviolet catastrophe. The word catastrophe was not used lightly; Equation 3-9 implies that

)

R 0

uL d L 4 @ 3-10

That is, every object would have an infinite energy density, which observation assures us is not true.

CCR

FIGURE 3-5 A small hole in the wall of a cavity

approximating an ideal blackbody. Radiation

entering the hole has little chance of leaving before it is completely absorbed within the cavity.

16

u(𝜆) pode ser calculada a partir da física clássica de maneira direta:

• O método envolve encontrar o número de modos de oscilação do campo eletromagnético na cavidade com comprimentos de onda no intervalo d𝜆 e multiplicar pela energia média por modo (ver livro do Eisberg)

• O resultado é que o número de modos de oscilação por unidade de volume, n(𝜆), é independente da forma da cavidade e é dado por: n(𝜆) = 8π𝜆-4

• De acordo com a teoria cinética clássica, a energia média por modo de oscilação é Ē = kT onde k é a constante de Boltzmann (o mesmo que para um oscilador harmônico unidimensional). Assim, a teoria clássica prevê:

Esta previsão, inicialmente derivada por Lord Rayleigh, é chamada de equação

de Rayleigh-Jeans. ₁₅

• Para 𝜆 grande, a eq. de RJ concorda com a distribuição espectral determinada experimentalmente.

• Para 𝜆 pequeno (região do ultravioleta), a eq. de RJ prevê que u(𝜆) se torne grande.

• Mais ainda, de acordo com a eq. de RJ, u(𝜆) diverge quando 𝜆 → 0,

• No entanto, os dados experimentais mostram que a distribuição realmente se aproxima de zero quando 𝜆 → 0.

A equação de Rayleigh-Jeans (RJ) e a distribuição espectral de energia determinada experimentalmente.

Esse enorme desacordo entre a medida experimental de u(𝜆) e a predição das leis fundamentais da física clássica em comprimentos de onda curtos foi chamado de

A equação de Rayleigh-Jeans apresenta outro severo problema. Se calculamos a radiância total obtemos:

Ou seja, o corpo negro teria uma radiância infinita, em contradição com os experimentos.

R =

Z

₁ 0

R( )d =

Z

₁ 0

1

4

cu( )d = 2⇡kT c

Z

₁ 0

d

4

! 1

17

Em 1900, o físico alemão Max Planck anunciou que, ao fazer pressupostos um

tanto estranhos, poderia derivar uma função u(𝜆) que concordava com os dados

experimentais.

Sabemos que, para qualquer cavidade, quanto menor o comprimento de onda, mais ondas estacionárias (modos) serão possíveis.

Planck raciocinou que, para que u(𝜆) se aproxime de zero quando 𝜆 é pequeno, a energia média por modo devia depender de 𝜆_{, em vez de ser igual ao valor kT previsto} pela teoria clássica.

A Lei de Planck

Max Planck 1858-1947

Planck considerou que os átomos das paredes da cavidade emitiam radiação com apenas alguns valores de energia, especificamente com valores 0, 𝟄, 2𝟄, ... , n𝟄,…. , com n inteiro.

Assim, a energia emitida por um átomo pode ser:

onde h é uma constante universal e f é a frequência da radiação eletromagnética.

A constante h é conhecida hoje como a constante de Planck e possui o valor h=6.626×10−34 _{J·s = 4.136×10}−15 _eV·s.

E

_n

= n✏ = nhf

n = 0, 1, 2,

_{· · ·}

Para obter a energia média, usamos a função de distribuição de Boltzmann. Com a hipótese de quantização da energia emitida pelo corpo negro, a função de distribuição de Boltzmann se torna discretizada:

onde k=1,38×10−23 _{J/K é a constante de Boltzmann.}

Como f(E) é uma função de distribuição de probabilidade, ela deve ser normalizada: Portanto:

f

_n

= Ae

En/kT

_{= Ae}

n✏/kT 1

X

n=0

f

_n

= A

1

X

n=0

e

n✏/kT

= 1

A =

P

₁

1

n=0

e

n✏/kT 20

A energia média dos modos de radiação é:

Realizando as somas acima obtemos (ver Apêndice):

Para obter a função de distribuição de densidade de energia u(𝜆) da radiação na cavidade, devemos fazer:

isto é, multiplicamos a energia média pelo número de modos de oscilação por unidade de volume.

¯

E =

1

X

n=0

E

n

f

n

=

1

X

n=0

E

n

Ae

n✏/kT

=

P

₁ n=0

n✏ e

n✏/kT

P

₁ n=0

e

n✏/kT

¯

E =

✏

e

✏/kT

₁

=

hf

e

hf /kT

₁

=

hc/

e

hc/ kT

₁

21

u( ) = ¯

E n( )

22 No slide 15, mostramos que n(𝜆) = 8π𝜆-4_{; logo obtemos:}

Portanto:

Esta função é chamada Lei de Planck. É claro a partir da figura que o resultado se adequa muito bem aos dados experimentais.

u( ) = ¯

E n( ) =

✓

hc/

e

hc/ kT

₁

◆ ✓

8⇡

4

◆

u( ) =

8⇡hc

₅

1

e

hc/ kT

1

23

3-2 Blackbody Radiation

127

Planck’s Law

In 1900 the German physicist Max Planck

10

announced that by making somewhat

strange assumptions, he could derive a function u(L) that agreed with the

experimen-tal data. He first found an empirical function that fit the data and then searched for a

way to modify the usual calculation so as to predict his empirical formula. We can see

the type of modification needed if we note that, for any cavity, the shorter the

wave-length, the more standing waves (modes) there will be possible. Therefore, as L

4 0

the number of modes of oscillation approaches infinity, as evidenced in Equation 3-8.

In order for the energy density distribution function u(L) to approach zero, we expect

the average energy per mode to depend on the wavelength L and approach zero as L

approaches zero, rather than be equal to the value kT predicted by classical theory.

Parenthetically, we should note that those working on the ultraviolet catastrophe

at the time—and there were many besides Planck—had no a priori way of knowing

whether the number of modes n(L) or the average energy per mode kT (or both) was

the source of the problem. Both were correct classically. Many attempts were made to

re-derive each so as to solve the problem. As it turned out, it was the average energy

per mode (that is, kinetic theory) that was at fault.

Classically, the electromagnetic waves in the cavity are produced by accelerated

electric charges in the walls of the cavity vibrating as simple harmonic oscillators.

Recall that the radiation emitted by such an oscillator has the same frequency as the

oscillation itself. The average energy for a one-dimensional simple harmonic

oscil-lator is calculated classically from the energy distribution function, which in turn is

found from the Maxwell-Boltzmann distribution function. That energy distribution

function has the form (see Chapter 8)

f (E)

 Ae

 EkT

3-11

where A is a constant and f (E) is the fraction of the oscillators with energy equal to E.

The average energy E is then found, as is any weighted average, from

E



)

R 0

E f (E) dE



)

R 0

EAe

 EkT

dE

3-12

with the result E

 kT, as was used by Rayleigh and others.

u(

)

0

2000

Planck’s

law

Rayleigh-Jeans

law

4000

6000


nm

FIGURE 3-6

Comparison of Planck’s law and the Rayleigh-Jeans equation with experimental

data at T  1600 K obtained by W. W. Coblenz in about 1915. The u(L) axis is linear. [

Adapted

from F. K. Richmyer, E. H. Kennard, and J. N. Cooper, Introduction to Modern Physics, 6th ed.,

McGraw-Hill, New York (1969), by permission.

]

Casos limite da lei de Planck

Para comprimentos de onda grandes, podemos expandir em série de potências a exponencial que aparece na lei de Planck:

Logo,

Substituindo na Lei de Planck temos:

Portanto, para comprimentos de onda grandes, a Lei de Plank apresenta o mesmo comportamento que a Lei de Rayleigh-Jeans.

e

x

_{⇡ 1 + x + · · ·}

x

_{⌧ 1}

e

hc/ kT

_{⇡ 1 + hc/ kT + · · ·}

hc/ kT

_{⌧ 1}

128

Chapter 3 Quantization of Charge, Light, and Energy

Planck found that he could derive his empirical formula by calculating the

aver-age energy E assuming that the energy of the oscillating charges, and hence the

radia-tion that they emitted, was a discrete variable; that is, that it could take on only the

values 0, E, 2E, . . . nE, where n is an integer, and further, that E was proportional to the

frequency of the oscillators and, hence, to that of the radiation. Planck therefore wrote

the energy as

E

_n

 nE  nhf n  0, 1, 2,

c

3-13

where the proportionality constant h is now called Planck’s constant. The

Maxwell-Boltzmann distribution (Equation 3-11) then becomes

f

_n

 Ae

 EnkT

 Ae

 nEkT

3-14

where A is determined by the normalization condition that the sum of all fractions f

_n

must, of course, equal one, that is,

=

@ n 0

f

_n

 A =

@ n 0

e

 nEkT

 1

3-15

The average energy of an oscillator is then given by the discrete sum equivalent of

Equation 3-12:

E

 =

@ n 0

E

_n

f

_n

 =

@ n 0

E

_n

Ae

 EnkT

3-16

Calculating the sums in Equations 3-15 and 3-16 (see Problem 3-60) yields the

result:

E



E

e

EkT

 1



hf

e

hfkT

 1



hc

L

e

hcLkT

 1

3-17

Multiplying this result by the number of oscillators per unit volume in the interval d L

given by Equation 3-8, we obtain for the energy density distribution function of the

radiation in the cavity:

u

L 

8PhcL

 5

e

hcLkT

 1

3-18

This function, called Planck’s law, is sketched in Figure 3-6. It is clear from the

fig-ure that the result fits the data quite well.

For very large L, the exponential in Equation 3-18 can be expanded using e

x



1

 x  … for x

7 1, where x  hcLkT. Then

e

hcLkT

 1 

hc

LkT

and

u

L 4 8PL

 4

kT

which is the Rayleigh-Jeans formula. For short wavelengths, we can neglect the 1 in

the denominator of Equation 3-18, and we have

u

L 4 8PhcL

 5

e

 hcLkT

4 0

as L

4 0. The value of the constant in Wien’s displacement law also follows from

Planck’s law, as you will show in Problem 3-23.

TIPLER_03_119-152hr2.indd 128 8/22/11 11:33 AM

u( ) =

8⇡hc

₅

1

e

hc/ kT

₁

⇡

8⇡hc

5

kT

hc

= 8⇡kT

4 24

Para comprimentos de onda pequenos, podemos negligenciar o 1 no denominador da Lei de Plank:

Portanto, a Lei de Planck resolve o problema da catástrofe ultravioleta.

u( ) = 8⇡hc₅ 1 ehc/ kT ₁ ⇡ 8⇡hc 5 e hc/ kT ! !0 0 25

Lei de Stefan-Boltzmann a partir da lei de Planck

Usando a lei de Planck, é possível mostrar que a densidade de energia em uma cavidade de corpo negro é proporcional a T4 _{de acordo com a lei de}

Stefan-Boltzmann (Lista 1, questão 6).

A densidade de energia total U é obtida a partir da função de distribuição u(𝜆). Lembremos que u(𝜆)d𝜆 representa a energia por unidade de volume considerando apenas a radiação com comprimento de onda entre 𝜆 e 𝜆+d𝜆. Portanto, integrando sobre 𝜆, obtemos a energia por unidade de volume U que leva em consideração as ondas com todos os comprimentos de onda possíveis:

130 Chapter 3 Quantization of Charge, Light, and Energy

EXAMPLE 3-4 Stefan-Boltzmann from Planck Show that the total energy density in a blackbody cavity is proportional to T 4 in accordance with the Stefan-Boltzmann law.

SOLUTION

The total energy density is obtained from the distribution function (Equation 3-18) by integrating over all wavelengths:

U  ) R 0 uL d L  ) R 0 8PhcL 5 ehcLkT  1 d L

Define the dimensionless variable x  hcLkT. Then dx  (hcL2kT ) d L or d L  L2(kThc) dx. Then U   ) R 0 8PhcL 3 ex  1 4 kT hc5 dx  8Phc4 kT hc5 4 ) R 0 x3 ex  1 dx

Since the integral is now dimensionless, this shows that U is proportional to T 4. The value of the integral is P415. Then U  (8P5k 415h3c3)T 4. This result can be com-bined with Equations 3-4 and 3-6 to express Stefan’s constant in terms of P, k, h, and c (see Problem 3-13).

A dramatic example of an application of Planck’s law on the current frontier of physics is in tests of the Big Bang theory of the formation and present expansion of the universe. Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion of space, one consequence of which was to fill the infant universe with radiation whose spectral distribution must surely have been that of an ideal blackbody. Since that time, the universe has expanded to its present size and cooled to its present temperature T_now. However, it should still be filled with radiation whose spectral distribution should be that characteristic of a blackbody at T_now.

FIGURE 3-7 The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck’s law with T  2.725 K. These measurements (the black dots) were made by the COBE satellite. 1500 1000 500 0 Energy density u (f ) 0 100 200 300 Frequency (s 109 _Hz) 400 500 600 700 TIPLER_03_119-152hr2.indd 130 8/22/11 11:33 AM 26

Para realizar a integração, introduzimos a variável adimensional x:

Substituindo na integral, temos:

Como a integral é adimensional, isso mostra que U é proporcional a T4_.

O valor da integral é 𝜋4_{/15. Portanto:}

130

Chapter 3 Quantization of Charge, Light, and Energy

EXAMPLE 3-4 Stefan-Boltzmann from Planck

Show that the total energy

density in a blackbody cavity is proportional to T

4

in accordance with the

Stefan-Boltzmann law.

SOLUTION

The total energy density is obtained from the distribution function (Equation 3-18)

by integrating over all wavelengths:

U



)

R 0

u

L d L 

)

R 0

8PhcL

 5

e

hcLkT

 1

d L

Define the dimensionless variable x  hc

LkT. Then dx  (hcL

2

kT ) d L or

d L  L

2

(kT

hc) dx. Then

U





)

R 0

8PhcL

 3

e

x

 1

4

kT

hc

5 dx  8Phc4

kT

hc

5

4

)

R 0

x

3

e

x

 1

dx

Since the integral is now dimensionless, this shows that U is proportional to T

4

. The

value of the integral is P

4

15. Then U  (8P

5

k

4

15h

3

c

3

)T

4

. This result can be

com-bined with Equations 3-4 and 3-6 to express Stefan’s constant in terms of P, k, h,

and c (see Problem 3-13).

A dramatic example of an application of Planck’s law on the current frontier of

physics is in tests of the Big Bang theory of the formation and present expansion of

the universe. Current cosmological theory suggests that the universe originated in an

extremely high-temperature explosion of space, one consequence of which was to fill

the infant universe with radiation whose spectral distribution must surely have been that

of an ideal blackbody. Since that time, the universe has expanded to its present size and

cooled to its present temperature T

_now

. However, it should still be filled with radiation

whose spectral distribution should be that characteristic of a blackbody at T

_now

.

FIGURE 3-7

The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck’s law with T  2.725 K. These measurements (the black dots) were made by the COBE satellite. 1500 1000 500 0 Energy density u (f ) 0 100 200 300 Frequency (s 109 Hz) 400 500 600 700 TIPLER_03_119-152hr2.indd 130 8/22/11 11:33 AM

130 Chapter 3 Quantization of Charge, Light, and Energy

EXAMPLE 3-4 Stefan-Boltzmann from Planck Show that the total energy density in a blackbody cavity is proportional to T 4 in accordance with the Stefan-Boltzmann law.

SOLUTION

The total energy density is obtained from the distribution function (Equation 3-18) by integrating over all wavelengths:

U  ) R 0 uL d L  ) R 0 8PhcL 5 ehcLkT  1 d L

Define the dimensionless variable x  hcLkT. Then dx  (hcL2kT ) d L or d L  L2(kThc) dx. Then U   ) R 0 8PhcL 3 ex  1 4 kT hc5 dx  8Phc4 kT hc5 4 ) R 0 x3 ex  1 dx

Since the integral is now dimensionless, this shows that U is proportional to T 4. The value of the integral is P415. Then U  (8P5k 415h3c3)T 4. This result can be com-bined with Equations 3-4 and 3-6 to express Stefan’s constant in terms of P, k, h, and c (see Problem 3-13).

A dramatic example of an application of Planck’s law on the current frontier of physics is in tests of the Big Bang theory of the formation and present expansion of the universe. Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion of space, one consequence of which was to fill the infant universe with radiation whose spectral distribution must surely have been that of an ideal blackbody. Since that time, the universe has expanded to its present size and cooled to its present temperature T_now. However, it should still be filled with radiation whose spectral distribution should be that characteristic of a blackbody at T_now.

FIGURE 3-7 The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck’s law with T  2.725 K. These measurements (the black dots) were made by the COBE satellite. 1500 1000 500 0 Energy density u (f ) 0 100 200 300 Frequency (s 109 _Hz) 400 500 600 700 TIPLER_03_119-152hr2.indd 130 8/22/11 11:33 AM

130 Chapter 3 Quantization of Charge, Light, and Energy

EXAMPLE 3-4 Stefan-Boltzmann from Planck Show that the total energy

density in a blackbody cavity is proportional to T 4 in accordance with the Stefan-Boltzmann law.

SOLUTION

The total energy density is obtained from the distribution function (Equation 3-18) by integrating over all wavelengths:

U  ) R 0 uL d L  ) R 0 8PhcL 5 ehcLkT  1 d L

Define the dimensionless variable x  hcLkT. Then dx  (hcL2kT ) d L or d L  L2(kThc) dx. Then U   ) R 0 8PhcL 3 ex  1 4 kT hc 5 dx  8Phc4 kT hc 5 4 ) R 0 x3 ex  1 dx

Since the integral is now dimensionless, this shows that U is proportional to T 4. The value of the integral is P415. Then U  (8P5k 415h3c3)T 4. This result can be com-bined with Equations 3-4 and 3-6 to express Stefan’s constant in terms of P, k, h, and c (see Problem 3-13).

A dramatic example of an application of Planck’s law on the current frontier of physics is in tests of the Big Bang theory of the formation and present expansion of the universe. Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion of space, one consequence of which was to fill the infant universe with radiation whose spectral distribution must surely have been that of an ideal blackbody. Since that time, the universe has expanded to its present size and cooled to its present temperature T_now. However, it should still be filled with radiation whose spectral distribution should be that characteristic of a blackbody at T_now.

FIGURE 3-7 The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck’s law with T  2.725 K. These measurements (the black dots) were made by the COBE satellite. 1500 1000 500 0 Energy density u (f ) 0 100 200 300 Frequency (s 109 _Hz) 400 500 600 700 TIPLER_03_119-152hr2.indd 130 8/22/11 11:33 AM

130 Chapter 3 Quantization of Charge, Light, and Energy

EXAMPLE 3-4 Stefan-Boltzmann from Planck Show that the total energy density in a blackbody cavity is proportional to T 4 in accordance with the Stefan-Boltzmann law.

SOLUTION

The total energy density is obtained from the distribution function (Equation 3-18) by integrating over all wavelengths:

U  ) R 0 uL d L  ) R 0 8PhcL 5 ehcLkT  1 d L

Define the dimensionless variable x  hcLkT. Then dx  (hcL2kT ) d L or d L  L2(kThc) dx. Then U   ) R 0 8PhcL 3 ex  1 4 kT hc 5 dx  8Phc4 kT hc 5 4 ) R 0 x3 ex  1 dx

Since the integral is now dimensionless, this shows that U is proportional to T 4. The value of the integral is P415. Then U  (8P5k 415h3c3)T 4. This result can be com-bined with Equations 3-4 and 3-6 to express Stefan’s constant in terms of P, k, h, and c (see Problem 3-13).

A dramatic example of an application of Planck’s law on the current frontier of physics is in tests of the Big Bang theory of the formation and present expansion of the universe. Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion of space, one consequence of which was to fill the infant universe with radiation whose spectral distribution must surely have been that of an ideal blackbody. Since that time, the universe has expanded to its present size and cooled to its present temperature T_now. However, it should still be filled with radiation whose spectral distribution should be that characteristic of a blackbody at T_now.

FIGURE 3-7 The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck’s law with T  2.725 K. These measurements (the black dots) were made by the COBE satellite. 1500 1000 500 0 Energy density u (f ) 0 100 200 300 Frequency (s 109 _Hz) 400 500 600 700 TIPLER_03_119-152hr2.indd 130 8/22/11 11:33 AM 0 ∞ U = 8⇡ 5_k4 15h3_c3 T 4 27

Lembrando que a radiância total é R = 1/4 c U (slide 14) temos:

Comparando com a Lei de Stefan R = 𝜎 T4 _{descobrimos que a constante de}

Stefan-Boltzmann é: R = c 4U = 2⇡5k4 15h3_c2 T 4 = 2⇡ 5_k4 15h3_c2 28

O fundo de radiação cósmica

130 Chapter 3 Quantization of Charge, Light, and Energy

EXAMPLE 3-4 Stefan-Boltzmann from Planck Show that the total energy density in a blackbody cavity is proportional to T 4 in accordance with the Stefan-Boltzmann law.

SOLUTION

The total energy density is obtained from the distribution function (Equation 3-18) by integrating over all wavelengths:

U  ) R 0 uL d L  ) R 0 8PhcL 5 ehcLkT  1 d L

Define the dimensionless variable x  hcLkT. Then dx  (hcL2kT ) d L or d L  L2(kThc) dx. Then U   ) R 0 8PhcL 3 ex  1 4 kT hc5 dx  8Phc4 kT hc5 4 ) R 0 x3 ex  1 dx

Since the integral is now dimensionless, this shows that U is proportional to T 4. The value of the integral is P415. Then U  (8P5k 415h3c3)T 4. This result can be com-bined with Equations 3-4 and 3-6 to express Stefan’s constant in terms of P, k, h, and c (see Problem 3-13).

A dramatic example of an application of Planck’s law on the current frontier of physics is in tests of the Big Bang theory of the formation and present expansion of the universe. Current cosmological theory suggests that the universe originated in an extremely high-temperature explosion of space, one consequence of which was to fill the infant universe with radiation whose spectral distribution must surely have been that of an ideal blackbody. Since that time, the universe has expanded to its present size and cooled to its present temperature T_now. However, it should still be filled with radiation whose spectral distribution should be that characteristic of a blackbody at T_now.

FIGURE 3-7 The energy density spectral distribution of the cosmic microwave background radiation. The solid line is Planck’s law with T  2.725 K. These measurements (the black dots) were made by the COBE satellite. 1500 1000 500 0 Energy density u (f ) 0 100 200 300 Frequency (s 109 _Hz) 400 500 600 700 TIPLER_03_119-152hr2.indd 130 8/22/11 11:33 AM 29

Estrelas de nêutrons emitem como corpos negros