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## Derivation of Planck’s radiation law – part 4 (final part)

In part 3 of this blog series I explained how Max Planck found a mathematical formula to fit the observed Blackbody spectrum, but that when he presented it to the German Physics Society on the 19th of October 1900 he had no physical explanation for his formula. Remember, the formula he found was $E_{\lambda} \; d \lambda = \frac{ A }{ \lambda^{5} } \frac{ 1 }{ (e^{a/\lambda T} -1) } \; d\lambda$

if we express it in terms of wavelength intervals. If we express it in terms of frequency intervals it is $E_{\nu} \; d \nu = A^{\prime} \nu^{3} \frac{ 1 }{ (e^{ a^{\prime} \nu / T } - 1) } \; d\nu$

Planck would spend six weeks trying to find a physical explanation for this equation. He struggled with the problem, and in the process was forced to abandon many aspects of 19th Century physics in both the fields of thermodynamics and electromagnetism which he had long cherished. I will recount his derivation – it is not the only one and maybe in coming blog posts I can show how his formula can be derived from other arguments, but this is the method Planck himself used.

As we saw in the derivation of the Rayleigh-Jeans law (see part 3 here, and links in that to parts 1 and 2), blackbody radiation can be modelled as an idealised cavity which radiates through a small hole. Importantly, the system is given enough time for the radiation and the material from which the cavity is made to come into thermal equilibrium with each other. This means that the walls of the cavity are giving energy to the radiation at the same rate that the radiation is giving energy to the walls.

Using classical physics, as we did in the derivation of the Rayleigh-Jeans law, we saw that the energy density (the energy per unit volume) is $\frac{du}{d\nu} = \left( \frac{ 8 \pi kT }{ c^{3} } \right) \nu^{2}$

After trying to derive his equation based on standard thermodynamic arguments, which failed, Planck developed a model which, he found, was able to produce his equation. How did he do this?

### Harmonic Oscillators

First, he knew from classical electromagnetic theory that an oscillating electron radiates (as it is accelerating), and he reasoned that when the cavity was in thermal equilibrium with the radiation in the cavity, the electrons in the walls of the cavity would oscillate and it was they that produced the radiation.

After much trial and error, he decided upon a model where the electrons were attached to massless springs. He could model the radiation of the electrons by modelling them as a whole series of harmonic oscillators, but with different spring stiffnesses to produce the different frequencies observed in the spectrum.

As we have seen (I derived it here), in classical physics the energy of a harmonic oscillator depends on both its amplitude of oscillation squared ( $E \propto A^{2}$); and it also depends on its frequency of oscillation squared ( $E \propto \nu^{2}$). The act of heating the cavity to a particular temperature is what, in Planck’s model, set the electrons oscillating; but whether a particular frequency oscillator was set in motion or not would depend on the temperature.

If it were oscillating, it would emit radiation into the cavity and absorb it from the cavity. He knew from the shape of the blackbody curve (and, by now, his equation which fitted it), that the energy density $E d\nu$ at any particular frequency started off at zero for high frequencies (UV), then rose to a peak, and then dropped off again at low frequencies (in the infrared).

So, Planck imagined that the number of oscillators with a particular resonant frequency would determine how much energy came out in that frequency interval. He imagined that there were more oscillators with a frequency which corresponded to the maximum in the blackbody curve, and fewer oscillators at higher and lower frequencies. He then had to figure out how the total energy being radiated by the blackbody would be shared amongst all these oscillators, with different numbers oscillating at different frequencies.

He found that he could not derive his formula using the physics that he had long accepted as correct. If he assumed that the energy of each oscillator went as the square of the amplitude, as it does in classical physics, his formula was not reproduced. Instead, he could derive his formula for the blackbody radiation spectrum only if the oscillators absorbed and emitted packets of energy which were proportional to their frequency of oscillation, not to the square of the frequency as classical physics argued. In addition, he found that the energy could only come in certain sized chunks, so for an oscillator at frequency $\nu, \; E = nh\nu$, where $n$ is an integer, and $h$ is now known as Planck’s constant.

What does this mean? Well, in classical physics, an oscillator can have any energy, which for a particular oscillator vibrating at a particular frequency can be altered by changing the amplitude. Suppose we have an oscillator vibrating with an amplitude of 1 (in abitrary units), then because the energy goes as the square of the amplitude its energy is $E=1^{2} =1$. If we increase the amplitude to 2, the energy will now be $E=2^{2} = 4$. But, if we wanted an energy of 2, we would need an amplitude of $\sqrt{2} = 1.414$, and if we wanted an energy of $3$ we would need an amplitude of $\sqrt{3} = 1.73$.

In classical physics, there is nothing to stop us having an amplitude of 1.74, which would give us an energy of 3.0276 (not 3), or an amplitude of 1.72 whichg would give us an energy of 2.9584 (not 3). But, what Planck found is that this was not allowed for his oscillators, they did not seem to obey the classical laws of physics. The energy could only be integers of $h\nu$, so $E=0h\nu, 1h\nu, 2h\nu, 3h\nu, 4h\nu$ etc.

Then, as I said above, he further assumed that the total energy at a particular frequency was given by the energy of each oscillator at that frequency multiplied by the number of oscillators at that frequency. The frequency of a particular oscillator was, he imagined, determined by its stiffness (Hooke’s constant). The energy of a particular oscillator at a particular frequency could be varied by the amplitude of its oscillations.

Let us assume, just to illustrate the idea, that the value of h is 2. If the total energy in the blackbody at a particular frequency of, say, 10 (in arbitrary units) were 800 (also in arbitrary units), this would mean that the energy of each chunk ( $E=h \nu$) was $E = 2 \times 10 = 20$. So, the number of chunks at that frequency would then be $800/20 = 40$. 40 oscillators, each with an energy of 20, would be oscillating to give us our total energy of 800 at that frequency.

Because of this quantised energy, we can write that $E_{n} = nh \nu$, where $n=0,1,2,3, \cdots$.

### The number of oscillators at each frequency

The next thing Planck needed to do was derive an expression for the number of oscillators at each frequency. Again, after much trial and error he found that he had to borrow an idea first proposed by Austrian physicist Ludwig Boltzmann to describe the most likely distribution of energies of atoms or molecules in a gas in thermal equilibrium. Boltzmann found that the number of atoms or molecules with a particular energy $E$ was given by $N_{E} \propto e^{-E/kT}$

where $E$ is the energy of that state, $T$ is the temperature of the gas and $k$ is now known as Boltzmann’s constant. The equation is known as the Boltzmann distribution, and Planck used it to give the number of oscillators at each frequency. So, for example, if $N_{0}$ is the number of oscillators with zero energy (in the so-called ground-state), then the numbers in the 1st, 2nd, 3rd etc. levels ( $N_{1}, N_{2}, N_{3},\cdots$) are given by $N_{1} = N_{0} e^{ -E_{1}/kT }, \; N_{2} = N_{0} e^{ -E_{2}/kT }, \; N_{3} = N_{0} e^{ -E_{3}/kT }, \cdots$

But, as $E_{n} = nh \nu$, we can write $N_{1} = N_{0} e^{ -h \nu /kT }, \; N_{2} = N_{0} e^{ -2h \nu /kT }, \; N_{3} = N_{0} e^{ -3h \nu /kT }, \cdots$ Planck modelled blackbody radiation as a series of harmonic oscillators with equally spaced energy levels

To make it easier to write, we are going to substitute $x = e^{ -h \nu / kT }$, so we have $N_{1} = N_{0}x, \; N_{2} = N_{0} x^{2}, \; N_{3} = N_{0} x^{3}, \cdots$

The total number of oscillators $N_{tot}$ is given by $N_{tot} = N_{0} + N_{1} + N_{2} + N_{3} + \cdots = N_{0} ( 1 + x + x^{2} + x^{3} + \cdots)$

Remember, this is the number of oscillators at each frequency, so the energy at each frequency is given by the number at each frequency multiplied by the energy of each oscillator at that frequency. So $E_{1}=N_{1} h \nu , \; E_{2} = N_{2} 2h \nu , \; E_{3} = N_{3} 3h \nu, \cdots$

which we can now write as $E_{1} = h \nu N_{0}x, \; E_{2} = 2h \nu N_{0}x^{2}, \; E_{3} = 3h \nu N_{0}x^{3}, \cdots$

The total energy $E_{tot}$ is given by $E_{tot} = E_{0} + E_{1} + E_{2} + E_{3} + \cdots = N_{0} h \nu (0 + x + 2x^{2} + 3x^{3} + \cdots)$

The average energy $\langle E \rangle$ is given by $\langle E \rangle = \frac{ E_{tot} }{ N_{tot} } = \frac{ N_{0} h \nu (0 + x + 2x^{2} + 3x^{3} + \cdots) }{ N_{0} ( 1 + x + x^{2} + x^{3} + \cdots ) }$

The two series inside the brackets can be summed. The sum of the series in the numerator, which we will call $S_{1}$ is given by $S_{1} = \frac{ x - (n+1)x^{n+1} + nx^{n+2} }{ (1-x)^{2} }$

(for the proof of this, see for example here)

The series in the denominator, which we will call $S_{2}$, is just a geometric progression. The sum  of such a series is simply $S_{2} = \frac{ 1 - x^{n} }{ (1-x) }$

Both series  are in $x$, but remember $x = e^{-h \nu / kT}$. Also, both series are from a frequency of $\nu = 0 \text{ to } \infty$, and $e^{-h \nu /kT} < 1$, which means the sums converge and can be simplified. $S_{1} \rightarrow \frac{x}{ (1-x)^{2} } \text{ and } S_{2} \rightarrow \frac{ 1 }{(1-x)}$

which means that $\langle E \rangle = (h \nu S_{1})/S_{2}$ is given by $\langle E \rangle = \frac{ h \nu x }{ (1-x)^{2} } \times \frac{ (1-x) }{1} = \frac{h \nu x}{ (1-x) }$

and so we can write that the average energy is $\boxed{ \langle E \rangle = \frac{h \nu}{( 1/x - 1) } = \frac{h \nu}{ (e^{h \nu/kT} - 1) } }$

## The radiance per frequency interval

In our derivation of the Rayleigh-Jeans law (in this blog here), we showed that, using classical physics, the energy density $du$ per frequency interval was given by $du = \frac{ 8 \pi }{ c^{3} } kT \nu^{2} \, d \nu$

where $kT$ was the energy of each mode of the electromagnetic radiation. We need to replace the $kT$ in this equation with the average energy for the harmonic oscillators that we have just derived above. So, we re-write the energy density as $du = \frac{ 8 \pi }{ c^{3} } \frac{ h \nu }{ (e^{h\nu/kT} - 1) } \nu^{2} \; d\nu = \frac{ 8 \pi h \nu^{3} }{ c^{3} } \frac{ 1 }{ (e^{h\nu/kT} - 1) } \; d\nu$ $du$ is the energy density per frequency interval (usually measured in Joules per metre cubed per Hertz), and by replacing $kT$ with the average energy that we derived above the radiation curve does not go as $\nu^{2}$ as in the Rayleigh-Jeans law, but rather reaches a maximum and turns over, avoiding the ultraviolet catastrophe.

It is more common to express the Planck radiation law in terms of the radiance per unit frequency, or the radiance per unit wavelength, which are written $B_{\nu}$ and $B_{\lambda}$ respectively. Radiance is the power per unit solid angle per unit area. So, as a first step to go from energy density to radiance we will divide by $4 \pi$, the total solid angle. This gives $\frac{ 2 h \nu^{3} }{ c^{3} } \frac{ 1 }{ (e^{h\nu/kT} - 1) } \; d\nu$

We want the power per unit area, not the energy per unit volume. To do this we first note that power is energy per unit time, and second that to go from unit volume to unit area we need to multiply by length. But, for EM radiation, length is just $ct$. So, we need to divide by $t$ and multiply by $ct$, giving us that the radiance per frequency interval is $\boxed{ B_{\nu} = \frac{ 2h \nu^{3} }{ c^{2} } \frac{ 1 }{ (e^{h\nu/kT} - 1) } \; d\nu }$

which is the way the Planck radiation law per frequency interval is usually written.

## Radiance per unit wavelength interval

If you would prefer the radiance per wavelength interval, we note that $\nu = c/\lambda$ and so $d\nu = -c/\lambda^{2} \; d\lambda$. Ignoring the minus sign (which is just telling us that as the frequency increases the wavelength decreases), and substituting for $\nu$ and $d\nu$ in terms of $\lambda$ and $d\lambda$, we can write $B_{\lambda} = \frac{ 2h }{ c^{2} } \frac{ c^{3} }{ \lambda^{3} } \frac{ 1 }{ ( e^{hc/\lambda kT} - 1 ) } \frac{ c }{ \lambda^{2} } \; d\lambda$

Tidying up, this gives $\boxed{ B_{\lambda} = \frac{ 2hc^{2} }{ \lambda^{5} } \frac{ 1 }{ ( e^{hc/\lambda kT} - 1 ) } \; d\lambda }$

which is the way the Planck radiation law per wavelength interval is usually written.

## Summary

To summarise, in order to reproduce the formula which he had empirically derived and presented in October 1900, Planck found that he he could only do so if he assumed that the radiation was produced by oscillating electrons, which he modelled as oscillating on a massless spring (so-called “harmonic oscillators”). The total energy at any given frequency would be given by the energy of a single oscillator at that frequency multiplied by the number of oscillators oscillating at that frequency.

However, he had to assume that

1. The energy of each oscillator was not related to either the square of the amplitude of oscillation or the square of the frequency of oscillation (as it would be in classical physics), but rather to the square of the amplitude and the frequency, $E \propto \nu$.
2. The energy of each oscillator could only be a multiple of some fundamental “chunk” of radiation, $h \nu$, so $E_{n} = nh\nu$ where $n=0,1,2,3,4$ etc.
3. The number of oscillators with each energy $E_{n}$ was given by the Boltzmann distribution, so $N_{n} = N_{0} e^{-nh\nu/kT}$ where $N_{0}$ is the number of oscillators in the lowest energy state.

In a way, we can imagine that the oscillators at higher frequencies (to the high-frequency side of the peak of the blackbody) are “frozen out”. The quantum of energy for a particular oscillator, given by $E_{n}=nh\nu$, is just too large to exist at the higher frequencies. This avoids the ultraviolet catastrophe which had stumped physicists up until this point.

By combining these assumptions, Planck was able in November 1900 to reproduce the exact equation which he had derived empirically in October 1900. In doing so he provided, for the first time, a physical explanation for the observed blackbody curve.

• Part 1 of this blogseries is here.
• Part 2 is here.
• Part 3 is here.

### 6 Responses

1. on 18/01/2016 at 16:15 | Reply shab

Thank you so much, Dr. Evans! I thought you probably forgot it, but you did it finally. Thanks once again

2. […] have blogged several times about blackbody radiation. Here I derived Planck’s radiation law using his original arguments of 1900, here I blogged about the fact that the Cosmic Microwave […]

• on 11/05/2017 at 12:38 | Reply pd

A wonderfully clearly explained and well written series on the explanation/derivation of the uv catastrophe and planck.
I’ve really enjoyed it and also learned a lot.
Thank you so much.

• on 11/05/2017 at 12:40 RhEvans

You’re welcome, I’m glad that you found them useful.

3. on 13/11/2017 at 22:20 | Reply Classical to Quantum Mechanics | Vinaire's Blog

[…] Thus, Planck had to postulate that “electromagnetic radiation could only be emitted or absorbed in discrete packets.” These discrete packets, called quanta, consisted of energy that was proportional to the frequency of radiation: E (quanta) = hf. (See Derivation of Planck’s radiation law) […]

4. on 20/08/2018 at 16:35 | Reply Sutia

This is an excellent blog series. It is much recommended to those who wants to know the classical physics and quantum physics collide. It includes statistical distributions through thorough analysis of the Planck’s Radiation Law.