Black body

A black body is an idealized body that absorbs all incident radiation. It is a good approximation for many bodies that only (or mostly) exhibit thermal radiation, such as stars and the human body.

A hollow spherical object with the inner surface painted black and a pinhole dug into it is an accurate real world representation of a black body. The cavity emits electromagnetic radiation from the pinhole due to thermal excitation. If the cavity is large enough, the system approximately looses the dependency on boundary conditions.

It was the physical system that spurred the start of quantum mechanics, as the classical result for the emission spectrum differs massively from reality in the higher frequencies, such as ultraviolet: this discrepancy has been called the ultraviolet catastrophe.

Classical description#

Let's assume our body is made up of many little electric dipoles (imagine atoms moving around and reorienting). These dipoles are organized in such a manner as to possess a translational symmetry, which allows us to use periodic boundary conditions. Namely, if we take a cubic box of side $L$ as our "periodicity cell" (the single volume element that gets periodically repeated), our wavenumber will look like

k_{i}=\frac{\pi n_{i}}{L}

where $i=x,y,z$ and $n\in \mathbb{N}$ . This comes from the harmonic oscillator. Mind you, this is not quantization: we have not quantized fields, matter, light or anything else. The number of possible oscillation modes can be calculated in the space of wavenumbers. Let's consider an infinitesimally thick (thickness $dk$ ) spherical shell in this space, and let's cut out only the octant where all components $k_{x},k_{y}$ and $k_{z}$ are positive.

Diagram Black body wavenumber shell.svg

Remember that the modes are discrete, as they depend on a natural number $n$ , so the density of oscillation modes in this infinitesimal "volume" (which is more of a 3D grid of points due to discreteness) is given by

\rho_{k}^{dK}=\frac{N_{k}}{V}=\frac{1}{8}4\pi k^{2}dk\ \underbrace{ \left( \frac{\pi}{L} \right)^{-3} }_{ \text{"volume"} }\ \underbrace{ 2 }_{ \text{polarizations} } \frac{1}{V}=\frac{k^{2}}{\pi ^{2}}dk

From here we can obtain the density of modes as a function of the frequency $\nu$ remembering that

k=\frac{2\pi}{\lambda}=\frac{2\pi \nu}{c}\quad\Rightarrow \quad dk=\frac{2\pi}{c}d\nu \quad\Rightarrow \quad d\nu=\frac{4\pi ^{2}\nu ^{2}}{\pi ^{2}c^{2}} \frac{2\pi}{c}d\nu=\frac{8\pi \nu ^{2}}{c^{3}}

The energy density will thus be

u(\nu) \bar{\varepsilon}=\frac{8\pi \nu ^{2}}{c^{3}}\bar{\varepsilon}\quad\text{where}\quad \bar{\varepsilon}=\frac{\int_{0}^{\infty}\varepsilon e^{-\beta \varepsilon}d\varepsilon}{\int_{0}^{\infty}e^{-\beta \varepsilon}d\varepsilon}=\frac{1}{\beta}

The relation

\boxed{\rho(\nu)=\frac{8\pi \nu ^{2}}{\beta c^{3}}}

is known as the Rayleigh-Jeans law. It is the classical description of the emission of a black body, and unfortunately it fails miserably as soon as you try to apply it: being a quadratic law in $\nu$ , it just keeps getting higher and higher as the frequencies climb up. In theory, extremely high frequencies like gamma rays would have to carry near infinite energy! This is quite obviously nonsensical, but the solutions are much less evident. In fact, they require claiming something almost just as nonsensical: that energy is quantized.

Quantum ensemble#

To fix the Rayleigh-Jeans law, we must shed classical descriptions and move on to a fully quantized world, where radiation is not just a wave, but also a Particle, namely a Photon. Since electromagnetic radiation is the exchange of photons, the whole cavity can be described as an ensemble of them. Specifically, it is a Bose gas of massless bosons. Since photons are absorbed and emitted, the total number of photons $N$ is not conserved. As such, we'll use a quantum grand canonical ensemble. The chemical potential is therefore zero¹, so the fugacity is $z=1$ .

As with all photons, their properties are

E=\hbar \omega,\qquad \mathbf{p}=\hbar \mathbf{k},\qquad \lvert \mathbf{k} \rvert =\frac{\omega}{c}

where $\hbar$ is the reduced Planck constant, $\omega$ is the angular frequency, $\mathbf{k}$ is the wave vector and $c$ is the Speed of light. They also all possess a polarization unit vector $\hat{\varepsilon}$ that is orthogonal to $\mathbf{k}$ .

Partition function#

Each state can be indexed like $n_{\mathbf{k},\hat{\varepsilon}}=0,1,2,\ldots$ and they represent normal modes of oscillation. The particle number is variable, so we can't rely on it like we can with massive particles. The total energy on the other had is always conserved, so we can use

E(\{ n_{\mathbf{k},\hat{\varepsilon}} \})=\sum_{\mathbf{k},\hat{\varepsilon}}\hbar \omega n_{\mathbf{k},\hat{\varepsilon}}\tag{1}

Since $z=1$ , the grand canonical partition function reduces itself to just the canonical partition function, except with a solvable sum since the constraint $\sum_{i}n_{i}=N$ is lifted:

\mathcal{Z}(z=1)=\sum_{N=0}^{\infty} Q_{N}=\sum_{N=0}^{\infty}\sum_{\substack{\{ n_{\mathbf{k},\hat{\varepsilon}} \} \\ \sum_{i}n_{i}=N}}e^{-\beta E(\{ n_{\mathbf{k},\hat{\varepsilon}} \})}=\sum_{\{ n_{\mathbf{k},\hat{\varepsilon}} \}}e^{-\beta E(\{ n_{\mathbf{k},\hat{\varepsilon}} \})}

This is solved as

\sum_{\{ n_{\mathbf{k},\hat{\varepsilon}} \}}e^{-\beta \sum_{\mathbf{k},\hat{\varepsilon}}\hbar \omega n_{\mathbf{k},\hat{\varepsilon}} }=\prod_{\mathbf{k},\hat{\varepsilon}}\sum_{n=0}^{\infty} (e^{-\beta \hbar \omega})^{n}=\prod_{\mathbf{k},\hat{\varepsilon}} \frac{1}{1-e^{-\beta \hbar \omega}}

since $e^{\sum n}=\prod_{n}e^{n}$ and using the geometric series. See Ideal gas > In the quantum grand canonical ensemble for steps on how to invert sum and product. The logarithm is then

\log Q_{N}=-\sum_{\mathbf{k},\hat{\varepsilon}}\log(1-e^{-\beta \hbar \omega})=-2\sum_{\mathbf{k}}\log(1-e^{-\beta \hbar \omega})

since there are only two possible $\hat{\varepsilon}$ states².

Occupation numbers#

Now that we know $Q_{N}$ , we can calculate everything else. The average occupation number for a given $\mathbf{k}$ is

\langle n_{\mathbf{k}} \rangle =- \frac{1}{\beta}\frac{ \partial }{ \partial (\hbar \omega) } \log Q_{N}=\frac{2}{\beta} \frac{\beta e^{-\hbar \omega \beta}}{1-e^{-\beta \hbar \omega}}=\frac{2}{e ^{\beta \hbar \omega}-1}

which is a Bose-Einstein distribution multiplied by the polarization factor, as we should expect.

Internal energy#

The internal energy is

U=-\frac{ \partial }{ \partial \beta } \log Q_{N}=\sum_{\mathbf{k}}\hbar \omega \langle n_{\mathbf{k}} \rangle

Note that this result is essentially equation $(1)$ but using the ensemble average instead³. This is expected, as we know that in a canonical ensemble, $\langle E \rangle=U$ . However, this sum can't be solved as is.

Pressure#

The pressure is from the Maxwell relations on Helmholtz free energy:

P=-\frac{ \partial A }{ \partial V } =\frac{1}{\beta}\frac{ \partial }{ \partial V } \log Q_{N}=\frac{1}{\beta}\frac{ \partial }{ \partial V } \left( -2\sum_{\mathbf{k}}\log(1-e^{-\beta \hbar c\ 2\pi \lvert \hat{\mathbf{n}} \rvert /V^{1/3}}) \right)=\frac{1}{3V}\sum_{\mathbf{k}}\hbar \omega \langle n_{\mathbf{k}} \rangle

where we used $\omega=c\lvert \mathbf{k} \rvert$ and $\lvert \mathbf{k} \rvert=\frac{2\pi}{L}\lvert \mathbf{n} \rvert=\frac{2\pi}{V^{1/3}}\lvert \mathbf{n} \rvert$ since quantum states of an ideal gas can be counted directly from phase space in a volume $V=L^{3}$ (see start of Thomas-Fermi approximation). Noticing the internal energy, we can state the equation of state of a photon gas

\boxed{PV=\frac{1}{3}U}

It differs from a "normal" (i.e. massive) ideal gas by a factor of $2$ . However, this relation can be found to also apply to massive particles if they move at speeds close to the speed of light. Clearly, relativity plays a role in the equation of state⁴.

Internal energy#

We can calculate $U$ more explicitly in the Thomas-Fermi approximation:

U=\sum_{\mathbf{k},\hat{\varepsilon}}\hbar \omega \langle n_{\mathbf{k}} \rangle \simeq 2\int_{0}^{\infty}\hbar \omega g(k)\langle n_{k} \rangle dk=\ldots

Again, don't miss the $2$ from the polarization. The state density is given by

G(k)=\frac{\frac{4}{3}\pi k^{3}}{\left( \frac{2\pi}{L} \right)^{3}}=\frac{k^{3}}{6\pi ^{2}}V,\qquad g(k)=\frac{ \partial G }{ \partial k } =\frac{k^{2}}{2\pi ^{2}}V

Substituting yields

\ldots= 2\hbar \int_{0}^{\infty} \frac{\omega k^{2}}{2\pi ^{2}}V \frac{1}{e^{\beta \hbar \omega}-1}dk=\frac{\hbar V}{\pi ^{2}} \int_{0}^{\infty} \frac{\omega ^{2}}{c^{2}} \frac{\omega}{e^{\beta \hbar \omega}-1} \frac{d\omega}{c}

The energy density by volume is

\frac{U}{V}=\int_{0}^{\infty}u(\omega,T)\ d\omega \quad\text{where}\quad \boxed{u(\omega,T)=\frac{\hbar}{\pi ^{2}c^{3}} \frac{\omega ^{3}}{e^{\beta \hbar \omega}-1}}

This is called the Planck radiation law.

Black_body.svg

Ideally, we want to solve the integral above. To that end, let's momentarily omit $\hbar$ for brevity and then do

\begin{align} \int_{0}^{\infty} \frac{\omega ^{3}}{e^{\beta \omega}-1}d\omega&=\int_{0}^{\infty} \frac{\omega ^{3}e^{-\beta \omega}}{1-e^{-\beta \omega}}d\omega \\ \text{(geometric series)}&=\int_{0}^{\infty}\omega ^{3}e^{-\beta \omega}\sum_{n=0}^{\infty}(e^{-\beta \omega})^{n}d\omega \\ (\text{Gamma})&=\sum_{n=1}^{\infty} \int_{0}^{\infty}\omega ^{3}e^{-\beta \omega n}d\omega \\ &=\sum_{n=1}^{\infty} \left( -\frac{ \partial ^{3} }{ \partial (\beta n) ^{3} } \right)\int_{0}^{\infty}e^{-\beta \omega n}d\omega \\ (n\beta \omega=\xi)&=\sum_{n=1}^{\infty} \left( -\frac{ \partial ^{3} }{ \partial (\beta n)^{3} } \right)\frac{1}{\beta n}\underbrace{ \int_{0}^{\infty} e^{-\xi}d\xi }_{ 1 } \\ &=- \frac{1}{\beta^{4}}\sum_{n=1}^{\infty} \frac{ \partial ^{3} }{ \partial n^{3} } \frac{1}{n} \\ &=\frac{6}{\beta^{4}}\sum_{n=1}^{\infty} \frac{1}{n^{4}} \\ &=\frac{6}{\beta^{4}}\zeta(4)=\frac{6}{\beta^{4}}\frac{\pi^{4}}{90} \end{align}

where we used the Riemann Zeta function. This integral could have also been solved by noticing that it becomes a Gamma function in step (Gamma). Using this, we can get the energy density

\boxed{\frac{U}{V}=\frac{\hbar}{\pi ^{2}c^{3}} \frac{1}{(\beta \hbar)^{4}} \frac{6\pi^{4}}{90}=\frac{\pi ^{2}}{15} \frac{(k_{B}T)^{4}}{(\hbar c)^{3}}}

Heat capacity#

We can use this to get the heat capacity:

C_{V}=\frac{1}{V}\frac{ \partial U }{ \partial T } =\frac{4\pi ^{2}}{15} \frac{k_{B}^{4}T^{3}}{(\hbar c)^{3}}

Irradiance#

The irradiance by temperature can be expressed in terms of the irradiance by frequency for all frequencies:

I(T)=\int_{0}^{\infty}I(\omega,T)d\omega

Then the irradiance by frequency is the energy emitted over a solid angle $\Omega$ :

I(\omega,T)=c\int \frac{1}{4\pi}u(\omega,T)\cos \theta \ d\Omega

where $u$ is the energy density by frequency. The appearance of light speed can be see from dimensional considerations: the units of $u$ are

[u]=\frac{\text{Energy}}{\text{Volume}\times\text{Frequency}}

and those of the irradiance are

[I]=\frac{\text{Energy}}{\text{Area}}=\frac{\text{Energy}}{\text{Volume}\times\text{Frequency}}\times \frac{\text{Length}}{\text{Time}}

We can see that the irradiance is an energy density multiplied by a velocity. Since photons move at light speed, the velocity must be the speed of light.

The solid angle we integrate over is not a full sphere, but rather just one hemisphere. This is because the emission is from a pinhole on the surface of the black body. The surface is massive compared to the pinhole, so we can consider it locally flat. Thus, half of space is covered by the black body itself and radiation is not propagated in that direction.

center

$u$ is not dependent on any angle, so it comes out. We are left with

\int \frac{\cos\theta}{4\pi}d\Omega=\int_{0}^{2\pi}d\phi\int_{0}^{\pi/2} \frac{\sin\theta}{4\pi}\cos \theta\ d\theta=\frac{1}{4}

As such, we get

I(T)=\frac{c}{4}\int_{0}^{\infty}u(\omega,T)d\omega

We solved this integral before, so our final result is

\boxed{I(T)=\frac{\pi ^{2}k_{B}^{4}}{60\hbar ^{3}c^{2}}T^{4}=\sigma T^{4}}

This is known as the Stefan-Boltzmann law for irradiation, where $\sigma$ is the Stefan-Boltzmann constant.

This might seem weird as photons are being added and removed all the time. The real reason is that photons have no mass and only carry kinetic energy, so when they are absorbed, they give away their energy and vanish. As such, the energy increases, but the particles do not (since the photon vanished). Photons really just move energy around and once they deliver their energy, the disappear. In fact, all ensembles of massless particles have zero chemical potential. ↩
If the photon had more than two polarizations, the factor would be different. However, all experimental tests point to there being only two. This is one way of proving theoretically that a photon can only have two polarizations. (Consider that at the time this derivation was first made, the number of photon polarizations was still a matter of debate.) ↩
It's like taking the average of both sides. However, do be careful of the factor 2 from the sum over polarization states. ↩
This is due to the energy dispersion relation: massive non-relativistic particles possess kinetic energy $E=p^{2}/2m$ , so $\propto p^{2}$ , but relativistic particles behave like $E=cp$ , so $\propto p$ . This drops the $2/3$ to a $1/3$ . More generally, if $E\propto p^{s}$ , then $PV=(s/3)U$ . ↩

Classical description#

Quantum ensemble#

Partition function#

Occupation numbers#

Internal energy#

Pressure#

Internal energy#

Heat capacity#

Irradiance#

Footnotes#