This so-called normal energy distribution represents something absolute, and since the reseach for absolutes has always appeared to me to be the highest form of research, I applied myself vigorously to its solution.-Max Planck

## Introduction:

In the following article, I derive Planck’s radiation formula as it may have been derived by Einstein in 1905. My reason for doing so is that Einstein’s 1905 analysis of the photoelectric effect shows that he had a much deeper insight into the discrete nature of electromagnetic radiation, unlike Planck who did not take quanta seriously.

This derivation is based on notes taken from a book which I lost, but I think the reader may check that each step of the derivation is correct.

## Statement of Planck’s law of blackbody radiation:

Generally speaking, black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, emitted by an idealised non-reflective body. For concreteness, this may be an antenna.

In this setting, Planck’s formula is fundamental as it allows us to accurately compute the energy density of radiation per unit frequency interval $$u(v)$$ for black-body radiation as follows:

$$u(\upsilon) d\upsilon = \frac{8 \pi h \upsilon^3}{c^3} \frac{1}{e^{\frac{h\upsilon}{kT}}-1} d\upsilon$$

where $$h = 6.626 \cdot 10^{34} J \cdot s$$ is Planck’s constant.

In the sections that follow, I shall explain how this formula can’t be derived using classical electromagnetic theory and how the resolution of this difficulty led to the development of an early quantum theory.

## Waves in a box:

If we consider the expression for an electromagnetic wave travelling at the speed of light in an arbitrary direction, $$\vec{r}$$, we find that if the wave has wavelength $$\lambda$$ at some instant the amplitude of the wave in the direction $$\vec{r}$$ is:

$$A(r) = A_0 \cdot \sin(\frac{2\pi r}{\lambda})$$

In terms of the wave vector $$\vec{k}$$, we have:

$$A(r) = A_0 \cdot \sin(\vec{k} \cdot \vec{r}) = A_0 \cdot \sin(kr)$$

where $$\lVert \vec{k} \rVert = \frac{2\pi}{r}$$.

Now, if the light wave travels in the $$\vec{r}$$ direction at the speed of light for a time $$t$$, its amplitude will vary so:

$$A(r,t) = A_0 \cdot \sin(\vec{k} \cdot \vec{r’}) = A_0 \cdot \sin(kr -kct)$$

since $$r = r'+ct$$.

Furthermore, if we observe the wave at a fixed value of $$r$$, we observe the amplitude to oscillate at frequency $$\upsilon = T^{-1}$$ so the time-dependence is more accurately given by:

$$\sin(\omega t)$$

and the expression of the wave simplifies to:

$$A(r,t) = A_0 \cdot \sin(kr-\omega t)$$

where $$\omega = c \cdot k$$.

## Electromagnetic modes in a box:

Let’s consider a thought experiment where a cubical box of side lengths $$L$$ has perfectly conducting walls. Therefore, the electric field of the electromagnetic waves must be zero at the walls of the box and we may only fit waves into the box that are multiples of half a wavelength.

In the $$x$$-direction, the wavelengths of the waves which may be fitted into the box are those which satisfy:

$$\forall l \in \mathbb{N^*}, \frac{l \lambda_x}{2} = L$$

and likewise, for the $$y$$ and $$z$$ directions we have:

$$\forall m \in \mathbb{N^*}, \frac{m \lambda_y}{2} = L$$

$$\forall n \in \mathbb{N^*}, \frac{n \lambda_z}{2} = L$$

and given that the wave has a point source, the wave equations in the $$x$$, $$y$$ and $$z$$ directions are given by:

$$A(x,y,z) = A_0 \cdot \sin(k_x x) \sin(k_y y) \sin(k_z z)$$

where we have:

$$k_x, k_y, k_z = \frac{\pi l}{L}, \frac{\pi m}{L}, \frac{\pi n}{L}$$

And to find the relation between $$k_x,k_y, k_z$$ and the angular frequency $$\omega$$ of the modes, we insert (10) into the electromagnetic wave equation:

$$\nabla^2 A = \frac{1}{c^2} \frac{\partial^2 A}{\partial t^2} \equiv \frac{\partial^2 A}{\partial x^2} + \frac{\partial^2 A}{\partial y^2} + \frac{\partial^2 A}{\partial z^2} = \frac{\partial^2 A}{\partial t^2}$$

and given (5), we may find the following dispersion relation of the wave:

$$\lVert \vec{k} \rVert^2 = k_x^2 + k_y^2 + k_z^2 = \frac{\omega^2}{c^2}$$

where $$\lVert \vec{k} \rVert^2$$ must satisfy:

$$k_x^2 + k_y^2 + k_z^2 = \frac{\pi^2}{L^2}(l^2 + m^2 + n^2) = \frac{\pi^2 p^2}{L^2}$$

and we obtain $$p^2 = l^2 + m^2 + n^2$$, an important pythagorean relation.

## Counting the modes:

If we consider the number of modes of oscillation in the frequency interval $$\upsilon$$ to $$\upsilon + d\upsilon$$, we find that we only need to count the number of lattice points in the interval of $$k$$-space $$k$$ to $$k+dk$$ that correspond to $$\upsilon + d\upsilon$$. Now, the number density of lattice points is one per unit volume of $$(l,m,n)$$ space and since we are only interested in positive values of $$l$$, $$m$$, and $$n$$ we only need to consider an eighth of the sphere of radius $$p$$.

The volume of a spherical shell of radius $$p$$ and thickness $$dp$$ is $$4\pi p^2 dp$$ so the number of modes in the octant is given by:

$$dN(p) = N(p)dp = \frac{1}{8} 4 \pi p^2 dp$$

Since $$k = \frac{\pi p}{L}$$ and $$dk = \frac{\pi dp}{L}$$, we find:

$$dN(p) = \frac{L^3}{2\pi^2} k^2 dk$$

Given that the volume of the box is $$L^3 = V$$ and $$k=\frac{2\pi \upsilon}{c}$$, we have:

$$dN(p) = \frac{V}{2 \pi^2} k^2 dk = \frac{V}{2 \pi^2} \frac{8 \pi^3 \upsilon^2}{c^3} d\upsilon = \frac{4\pi \upsilon^2 V}{c^3} d\upsilon$$

Now, since for electromagnetic waves always have two independent modes or polarisations per state, we must double this result:

$$dN = \frac{8\pi \upsilon^2 V}{c^3} d\upsilon$$

or per unit volume,

$$dN = \frac{8\pi \upsilon^2}{c^3} d\upsilon$$

## The average energy per mode and the Ultraviolet Catastrophe:

If we wait long enough, each mode of oscillation should attain the same average energy $$\bar{E}$$ when the system is in thermodynamic equilibrium. Therefore, the energy density of radiation per unit frequency interval per unit volume is:

$$u(\upsilon) = \frac{8\pi \upsilon^2}{c^3} \bar{E}$$

and given that the average energy of a harmonic oscillator in thermal equilibrium is $$\bar{E} = kT$$, the spectrum of black-body radiation is expected to be:

$$u(\upsilon) = \frac{8\pi \upsilon^2}{c^3} \bar{E} = \frac{8 \pi \upsilon^2 kT}{c^3}$$

and although this result, known as the Rayleigh-Jeans Law, holds for the measured spectrum at low frequencies and high temperatures, this energy density diverges at high frequencies:

$$\int_{0}^\infty u(\upsilon) d\upsilon = \int_{0}^\infty \frac{8 \pi \upsilon^2 kT}{c^3} d\upsilon = \infty$$

as discovered by Einstein during his investigations of the photo-electric effect.

## Derivation of Planck’s law:

As discovered by Einstein, the quantisation of electromagnetic radiation means that the energy of a particular mode of frequency $$\upsilon$$ may not have an arbitrary value but only those energies which are multiples of $$h\upsilon$$. It follows that if there are $$n$$ photons associated with a particular mode, then the energy of that mode is given by:

$$E(\upsilon) = n h \upsilon$$

Now, let’s consider all the modes(and photons) to be in thermal equilibrium at temperature $$T$$. In order for equilibrium to be reached, there must be ways of exchanging energy between modes(and photons) and this may occur through interactions with any particles or oscillations within the volume or walls of the enclosure.

We may now use the Boltzmann distribution to determine the expected occupancy of the modes in thermal equilibrium. In fact, the probability that a single mode has energy $$E_n = n h \upsilon$$ is given by the Boltzmann factor:

$$p(n) = \frac{\exp(-E_n/kT)}{\sum_{n=0}^\infty \exp(-E_n/kT)}$$

which is the probability that the state contains $$n$$ photons of frequency $$\upsilon$$, and so the average energy of the mode of frequency $$\upsilon$$ is given by:

$$\bar{E} = \sum_{n=0}^\infty E_n \cdot p_n = \sum_{n=0}^\infty \frac{E_n \exp(-E_n/kT)}{\sum_{n=0}^\infty \exp(-E_n/kT)}= \sum_{n=0}^\infty \frac{n h \upsilon \cdot \exp(-E_n/kT)}{\sum_{n=0}^\infty \exp(-E_n/kT)}$$

To simplify the formula for $$\bar{E}$$, we may use the substitution $$x= \exp(-h\upsilon/kT)$$:

$$\bar{E} = \frac{h\upsilon \sum_{0}^\infty nx^n}{\sum_{0}^\infty x^n} = \frac{h\upsilon (x+2x^2+3x^3+…)}{1+x+x^2+…} = \frac{h\upsilon x (1+2x+3x^2+…)}{1+x+x^2+…}$$

and if we use the following series expansions:

$$\frac{1}{1-x} = 1+x+x^2+x^3+…$$

$$\frac{1}{(1-x)^2} = 1+2x+3x^2+…$$

then the mean energy of the mode is given by:

$$\bar{E} = \frac{h\upsilon x}{1-x} = \frac{h\upsilon}{x^{-1}-1} = \frac{h\upsilon}{\exp(h\upsilon/kT)-1}$$

Given $$\bar{E}$$, we may now complete the derivation of Planck’s law:

$$u(\upsilon) d\upsilon = \frac{8\pi \upsilon^2}{c^3} \bar{E} = \frac{8 \pi h \upsilon^3}{c^3} \frac{1}{\exp(h\upsilon/kT)-1}$$

## Rayleigh-Jeans and another corollary:

Given (30), we find that if we use the Maclaurin series expansion of the exponential:

$$\exp(h\upsilon/kT)-1 = 1 + \frac{h\upsilon}{kT} + \frac{1}{2!}(\frac{h\upsilon}{kT})^2 + … -1$$

so to a first-order approximation,

$$\exp(h\upsilon/kT)-1 \approx \frac{h\upsilon}{kT}$$

so for low frequencies the following approximation holds:

$$\bar{E} = \frac{h\upsilon}{\exp(h\upsilon/kT)-1} \approx kT$$

which gives us the Rayleigh-Jeans law:

$$u(\upsilon) = \frac{8\pi \upsilon^2}{c^3} \bar{E} = \frac{8 \pi \upsilon^2 kT}{c^3}$$

It is also worth noting that given $$E_n = nh \upsilon$$, we find that the average number of photons in a single mode of frequency $$\upsilon$$ in thermal equilibrium is given by:

$$\bar{n} = \frac{1}{\exp(h\upsilon/kT)-1}$$

which is also known as the photon occupation number in thermal equilibrium.

## References:

1. Albert Einstein. On a heuristic point of view about the creation and conversion of light. 1905.
2. Max Planck. On the Theory of the Energy Distribution Law of the Normal Spectrum. 1900.