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## Harmonic Oscillators

Today I was planning to post the fourth and final part of my series of blogs about the derivation of Planck’s radiation law. But, I realised on Sunday that it would not be ready, so I’m postponing it until next Thursday (17th). Parts 1, 2 and 3 are here, here and here respectively).

One of the reasons for this is that my time is being consumed by writing articles for 30-second Einstein which I talked about on Tuesday. Another reason is that I am scrambling to finish a slew of things by the 15th (next Tuesday!!), as I have been asked to go on another cruise to give astronomy lectures. More about that next week 🙂

There is a third reason, I have realised that I have not yet done a blog about harmonic oscillators, which is a necessary part of understanding Planck’s derivation. So, that is the subject of today’s blogpost.

Harmonic oscillators is another term for something which is exhibiting simple harmonic motion, and I did blog here about how a pendulum exhibits simple harmonic motion (SHM), and how this relates to circular motion. Another example of SHM is a spring oscillating back and forth. Whether the spring is vertical or horizontal, if it is displaced from its equilibrium position it will exhibit SHM. So, a spring is a harmonic oscillator.

## The frequency of a harmonic oscillator

The restoring force on a spring when it is displaced from its equilibrium position is given by Hooke’s law, which states

$\vec{F} = - k \vec{x}$

where $\vec{F} \text{ and } \vec{x}$ are the force and displacement respectively (vector quantities), and the minus sign is telling us that the force acts in the opposite direction to the displacement; that is it is a restoring force which is directed back towards the equilibrium position. The term $k$ is known as Hooke’s constant, and is basically the stiffness of the spring.

Because we can also write the force in terms of mass and acceleration (Newton’s 2nd law of motion), and acceleration is the second derivate of displacement, we can write

$m \vec{a} = m \frac{ d^{2}\vec{x} }{ dt^{2} }= - k \vec{x}$

If we divide by $m$ we get an expression for the acceleration, which is

$\boxed{ \vec{a} = - \frac{k}{m} \vec{x} }$

which, if you compare it to the equation for SHM for a pendulum, has the same form. The usual way to write equations of SMH is to write

$\vec{a} = - \omega^{2} \vec{x}$

where $\omega$ is the angular velocity, and as I mentioned in the blog I did on the pendulum, $\omega$ is related to the period of the SHM, via $T = 2 \pi / \omega$.

For our derivation of Planck’s radiation law, the parts which we need to know about are that the frequency of a harmonic oscillator, $\nu$ is given by $\nu = \omega / 2 \pi$, and so depends only on $\omega, \; \boxed{\nu \propto \omega }$. So the frequency of the spring’s oscillations depends only on k/m, we can write $\boxed{ \nu \propto k/m }$. A stiffer spring oscillates with a higher frequency, more mass (in either the spring or what is attached to it) will reduce the frequency of the oscillations.

## The energy of a harmonic oscillator

The other thing we need to know about to understand Planck’s derivation of his blackbody radiation law is the energy of the harmonic oscillator. This is always constant, but is divided between kinetic energy and potential energy. The kinetic energy is at a maximum when the spring is at its equilibrium position, at this moment it actually has zero potential energy.

The velocity of a harmonic oscillator $v$ can be found my differentiating the displacement $x$ with respect to time. The expression for displacement (see my blog here on SHM in a pendulum) is

$x(t) = A sin ( \omega t )$

where $A$ is the maximum displacement (amplitude) of the oscillstions. So

$v = \frac{dx}{dt} = A \omega cos ( \omega t)$

This will be a maximum when $cos( \omega t) = 1$ and so

$v_{max} = A \omega$

which means that the maximum kinetic energy, and hence the total energy of the harmonic oscillator is given by

$\text{Total energy} = E = \frac{1}{2}mv_{max}^{2} = \frac{1}{2} m A^{2} \omega^{2}$

As the frequency $\nu$ is just $\omega / 2 \pi$, this tells us that, for a harmonic oscillator of a given mass, the energy depends on both the square of frequency $\nu$ and the square of the size of the oscillations (larger oscillations mean more energy, double the size of the oscillations and the energy goes up by a factor of four). Mathematically we can write this as $\boxed{ E \propto A^{2} }$ and $\boxed{ E \propto \nu^{2} }$.

As we shall see, the theoretical explanation which Planck concocted to explain his blackbody curve involved assuming the walls of the cavity producing the radiation oscillated in resonance with the radiation; this is why I needed to derive these things on this blog today.

## Derivation of Planck’s radiation law – part 3

As I have outlined in parts 1 and 2 of this series (see here and here), in the 1890s, mainly through the work of the Physikalisch-Technische Reichsanstalt (PTR) in Germany, the exact shape of the blackbody spectrum began to be well determined. By mid-1900, with the last remaining observations in the infrared being completed, its shape from the UV through the visible and into the infrared was well determined for blackbodies with a wide range of temperatures.

I also described in part 2 that in 1896 Wilhelm Wien came up with a law, based on a thermodynamical argument, which almost explained the blackbody spectrum. The form of his equation (which we now know as Wien’s distribution law) is
$\boxed{ E_{ \lambda } d \lambda = \frac{ A }{ \lambda ^{5} } e^{ -a / \lambda T } d \lambda }$

Notice I said almost. Below I show two plots which I have done showing the Wien distribution law curve and the actual blackbody curve for a blackbody at a temperature of $T=4000 \text{Kelvin}$. As you can see, they are not an exact match, the Wien distribution law fails on the long-wavelength side of the peak of the blackbody curve.

Comparison of the Wien distribution law and the actual blackbody curve for a blackbody at a temperature of $T=4000 \text{Kelvin}$. Although they agree very well on the short wavelength side of the peak, the Wien law drops away too quickly on the long-wavelength side compared to the observed blackbody spectrum.

A zoomed-in view to highlight the difference between the Wien distribution law and the actual blackbody curve for a blackbody at a temperature of $T=4000 \text{Kelvin}$. Although they agree very well on the short wavelength side of the peak, the Wien law drops away too quickly on the long-wavelength side compared to the observed blackbody spectrum.

## Planck’s “act of desperation”

By October 1900 Max Planck had heard of the latest experimental results from the PTR which showed, beyond any doubt, that Wien’s distribution law did not fit the blackbody spectrum at longer wavelengths. Planck, along with Wien, was hoping that the results from earlier in the year were in error, but when new measurements by a different team at the PTR showed that Wien’s distribution law failed to match the observed curve in the infrared, Planck decided he would try and find a curve that would fit the data, irrespective of what physical explanation may lie behind the mathematics of the curve. In essence, he was prepared to try anything to get a fit.

Planck would later say of this work

Briefly summarised, what I did can be described as simply an act of desperation

What was this “act of desperation”, and why did Planck resort to it? Planck was 42 when he unwittingly started what would become the quantum revolution, and his act of desperation to fit the blackbody curve came after all other options seemed to be exhausted. Before I show the equation that he found to be a perfect fit to the data, let me say a little bit about Planck’s background.

## Who was Max Planck?

Max Karl Ernst Ludwig Planck was born in Kiel in 1858. At the time, Kiel was part of Danish Holstein. He was born into a religious family, both his paternal great-grandfather and grandfather had been distiguished theologians, and his father became professor of constitutional law at Munich University. So he came from a long line of men who venerated the laws of God and Man, and Planck himself very much followed in this tradition.

He attended the most renowned secondary school in Munich, the Maximilian Gymnasium, always finishing near the top of his class (but not quite top). He excelled through hard work and self discipline, although he may not have had quite the inherent natural ability of the few who finished above him. At 16 it was not the famous taverns of Munich which attracted him, but rather the opera houses and concert halls; he was always a serious person, even in his youth.

In 1874, aged 16, he enrolled at Munich University and decided to study physics. He spent three years studying at Munich, where he was told by one of his professors ‘it is hardly worth entering physics anymore’; at the time it was felt by many that there was nothing major left to discover in the subject.

In 1877 Planck moved from Munich to the top university in the German-speaking world – Berlin. The university enticed Germany’s best-known physicist, Herman von Helmholtz, from his position at Heidelberg to lead the creation of what would become the best physics department in the world. As part of creating this new utopia, Helmholtz demanded the building of a magnificient physics institute, and when Planck arrived in 1877 it was still being built. Gustav Kirchhoff, the first person to systematically study the nature of blackbody radiation in the 1850s, was also enticed from Heidelberg and made professor of theoretical physics.

Planck found both Helmholtz and Kirchhoff to be uninspring lecturers, and was on the verge of losing interest in physics when he came across the work of Rudolf Clausius, a professor of physics at Bonn University. Clausius’ main research was in thermodynamics, and it is he who first formulated the concept of entropy, the idea that things naturally go from order to disorder and which, possibly more than any other idea in physics, gives an arrow to the direction of time.

Planck spent only one year in Berlin, before he returned to Munich to work on his doctoral thesis, choosing to explore the concept of irreversibility, which was at the heart of Claussius’ idea of entropy. Planck found very little interest in his chosen topic from his professors in Berlin, and not even Claussius answered his letters. Planck would later say ‘The effect of my dissertation on the physicists of those days was nil.’

Undeterred, as he began his academic career, thermodynamics and, in particular, the second law (the law of entropy) became the focus of his research. In 1880 Planck became Privatdozent, an unpaid lecturer, at Munich University. He spent five years as a Privatdozent, and it looked like he was never going to get a paid academic position. But in 1885 Gottingen University announced that the subject of its prestigoius essay competition was ‘The Nature of Energy’, right up Planck’s alley. As he was working on his essay for this competition, he was offered an Extraordinary (assistant) professorship at the University of Kiel.

Gottingen took two years to come to a decision about their 1885 essay competition, even though they had only received three entries. They decided that no-one should receive first prize, but Planck was awarded second prize. It later transpired that he was denied first prize because he had supported Helmholtz in a scientific dispute with a member of the Gottingen faculty. This brought him to the attention of Helmholtz, and in November 1888 Planck was asked by Helmholtz to succeed Kirchhoff as professor of theoretical physics in Berlin (he was chosen after Ludwig Boltzmann turned the position down).

And so Planck returned to Berlin in the spring of 1889, eleven years after he had spent a year there, but this time not as a graduate student but as an Extraordinary Professor. In 1892 Planck was promoted to Ordinary (full) Professor. In 1894 both Helmholtz and August Kundt, the head of the department, died within months of each other; leaving Planck at just 36 as the most senior physicist in Germany’s foremost physics department.

Max Planck who, in 1900 at the age of 42, found a mathematical equation which fitted the entire blackbody spectrum correctly.

As part of his new position as the most senior physicist in the Berlin department, he took over the duties of being adviser for the foremost physics journal of the day – Annalen der Physik (the journal in which Einstein would publish in 1905). It was in this role of adviser that he became aware of the work being done at PTR on determining the true spectrum of a blackbody.

Planck regarded the search for a theoretical explanation of the blackbody spectrum as nothing less than the search for the absolute, and as he later stated

Since I had always regarded the search for the absolute as the loftiest goal of all scientific activity, I eagerly set to work

When Wien published his distribution law in 1896, Planck tried to put the law on a solid theoretical foundation by deriving it from first principles. By 1899 he thought he had succeeded, basing his argument on the second law of thermodynamics.

## Planck finds a curve which fits

But, all of this fell apart when it was shown conclusively on the 2nd of February 1900, by Lummer and Pringsheim of the PTR, that Wien’s distribribution law was wrong. Wien’s law failed at high temperatures and long wavelengths (the infrared); a replacement which would fit the experimental curve needed to be found. So, on Sunday the 7th of October, Planck set about trying to find a formula which would reproduce the observed blackbody curve.

He was not quite shooting in the dark, he had three pieces of information to help him. Firstly, Wien’s law worked for the intensity of radiation at short wavelengths. Secondly, it was in the infrared that Wien’s law broke down, at these longer wavelengths it was found that the intensity was directly propotional to the temperature. Thirdly, Wien’s displacement law, which gave the relationship between the wavelength of the peak of the curve and the blackbody’s temperature worked for all observed blackbodies.

After working all night of the 7th of October 1900, Planck found an equation which fitted the observed data. He presented this work to the German Physical Society a few weeks later on Friday the 19th of October, and this was the first time others saw the equation which has now become known as Planck’s law.

The equation he found for the energy in the wavelength interval $d \lambda$ had the form
$\boxed{ E_{\lambda} \; d \lambda = \frac{ A }{ \lambda^{5} } \frac{ 1 }{ (e^{a/\lambda T} - 1) } \; d\lambda }$

(compare this to the Wien distribution law above).

After presenting his equation he sat down; he had no explanation for why this equation worked, no physical understanding of what was going on. That understanding would dawn on him over the next few weeks, as he worked tirelessly to explain the equation on a physical basis. It took him six weeks, and in the process he had to abandon some of the ideas in physics which he held most dear. He found that he had to abandon accepted ideas in both thermodynamics and electromagnetism, two of the cornerstones of 19th Century physics. Next week, in the fourth and final part of this blog-series, I will explain what physical theory Planck used to explain his equation; the theory which would usher in the quantum age.

## Derivation of Planck’s radiation law – part 1

One of my most popular blogposts is the series I did on the derivation of the Rayleigh-Jeans law, which I posted in three parts (part 1 here, part 2 here and part 3 here). I have had many thousands of hits on this series, but several people have asked me if I can do a similar derivation of the Planck radiation law, which after all is the correct formula/law for blackbody radiation. And so, never one to turn down a reasonable request, here is my go at doing that. I am going to split this up into 2 or 3 parts (we shall see how it goes!), but today in part 1 I am going to give a little bit of historical background to the whole question of deriving a formula/law to explain the shape of the blackbody radiation curve.

## ‘Blackbody’ does not mean black!

When I first came across the term blackbody I assumed that it meant the object had to be black. In fact, nothing could be further from the truth. As Kirchhoff’s radiation laws state

A hot opaque solid, liquid, or gas will produce a continuum spectrum

(which is the spectrum of a blackbody). The key word in this sentence is opaque. The opaqueness of an object is due to the interaction of the photons (particles of light) with the matter in the object, and it is only if they are interacting a great deal (actually in thermal equilibrium) that you will get blackbody radiation. So, examples of objects which radiate like blackbodies are stars, the Cosmic Microwave Background, (which is two reasons why astronomers are so interested in blackbody radiation), a heated canon ball, or even a canon ball at room temperature. Or you and me.

Kirchhoff’s 3 radiation laws, which he derived in the mid-1800s

Stars are hot, and so radiate in the visible part of the spectrum, as would a heated canon ball if it gets up to a few thousand degrees. But, a canon ball at room temperature or you and me (at body temperature) do not emit visible light. But, we are radiating like blackbodies, but in the infrared part of the spectrum. If you’ve ever seen what people look like through a thermal imaging camera you will know that we are aglow with infrared radiation, and it is this which is used by Police for example to find criminals in the dark as the run across fields thinking that they cannot be seen.

The thermal radiation (near infrared) from a person. The differences in temperature are due to the surface of the body having different temperatures in different parts (e.g. the nose is usually the coldest part).

Kirchhoff came up with his radiation laws in the mid-1800s, he began his investigations of continuum radiation in 1859, long before we fully knew the shape (spectrum) of a blackbody.

## Germans derive the complete blackbody spectrum

We actually did not know the complete shape of a blackbody spectrum until the 1890s. And the motivation for experimentally determining it is quite surprising. In the 1880s German industry decided they wanted to develop more efficient lighting than their British and American rivals. And so they set about deriving the complete spectrum of heated objects. In 1887 the German government established a research centre, the Physikalisch-Technische Reichsandstalt (PTR) – the Imperial Institute of Physics and Technology, one of whose aims was to fully determine the spectrum of a blackbody.

PTR was set up on the outskirts of Berlin, on land donated by Werner von Siemens, and it took over a decade to build the entire facility. Its research into the spectrum of blackbodies began in the 1890s, and in 1893 Wilhelm Wien found a simple relationship between the wavelength of the peak of a blackbody and its temperature – a relationship which we now call Wien’s displacement law.

Wien’s displacement law states that the wavelength of the peak, which we will call $\lambda_{peak}$ is simply given by

$\lambda_{peak} = \frac{ 0.0029 }{ T }$

if the temperature $T$ is expressed in Kelvin. This will give the wavelength in metres of the peak of the curve. That is why, in the diagram below, the peak of the blackbody shifts to shorter wavelengths as we go to higher temperatures. Wien’s displacement law explains why, for example, an iron poker changes colour as it gets hotter. When it first starts glowing it is a dull red, but as the temperature increases it becomes more yellow, then white. If we could make it hot enough it would look blue.

The blackbody spectra for three different temperatures, and the Rayleigh-Jeans law, which was behind the term “the UV catastrophe”

By 1898, after a decade of experimental development, the PTR had developed a blackbody which reached temperatures of 1500 Celsius, and two experimentalists working there Enrst Pringsheim and Otto Lummer (an appropriate name for someone working on luminosity!!) were able to show that the blackbody curve reached a peak and then dropped back down again in intensity, as shown in the curves above. However, this pair and others working at the PTR were pushing the limits of technology of the time, particularly in trying to measure the intensity of the radiation in the infrared part of the spectrum. By 1900 Lummer and Pringsheim had shown beyond reasonable doubt that Wien’s ad-hoc law for blackbody radiation did not work in the infrared. Heinrich Rubens and Ferdinand Kurlbaum built a blackbody that could range in temperature from 200 to 1500 Celsius, and were able to accurately measure for the first time the intensity of the radiation into the infrared. This showed that the spectrum was as shown above, so now Max Planck knew what shape curve he had to find a formula (and hopefully a theory) to fit.

In part 2 next week, I will explain how he went about doing that.

## Measuring the sizes of stars

Last week I blogged about how we measure the brightnesses of stars, and how at optical (and near infrared) wavelengths we use something called the magnitude system. Remember, the magnitude system has its zero point defined by Vega, so there is a conversion between the magnitude of a star and its luminosity. Astronomers tend to observe and measure the brightness of stars through a set of standard filters, for example the Johnson UBV system (U – ultraviolet, B – blue and V – visual). The diagram below shows the transmission curves for these three filters.

One of the standard set of filters used in astronomy is the Johnson UBV system.

The conversion between the apparent magnitude of a star when measured in the V-band $m_{V}$ and its actual flux density (energy flowing per second per unit area per unit frequency interval) is

$m_{V} = 0 \rightarrow L = 3640 \times 10^{-26} \text{ W/m}^{2} \text{/Hz}$

Referring back to blackbody curves, which I discussed in my blog about the OBAFGKM spectral classification system, we see that to get the total power per unit area from a blackbody we have to first of all sum over all frequencies (or wavelengths). This is just the area under the blackbody curve. The formula for this is the Steffan-Boltzmann law which is

$\text{ Power per unit area } = \frac{ P }{ A } = \sigma T^{4}$

where $\sigma$ is the so-called Steffan-Boltzmann constant. So, we can see that the power emitted per unit area by a blackbody, such as a star, is simply related to its temperature. Remember also that the wavelength of the peak of the blackbody curve depends on temperature, as given by Wien’s displacement law

$\lambda_{peak} = \frac{ 0.029 }{ T }$

where we need to measure the wavelength $\lambda$ in metres and the temperature $T$ in then given in Kelvin.

The figure below shows three blackbody curves for blackbodies of different temperatures, $T=7000, 5270 \text{ and } 4000 \text{ Kelvin}$. These temperatures correspond to a blue star ($T=7000 \text{K}$), a yellow star like the Sun ($T=5270 \text{K}$) and a red star ($T=4000 \text{K}$.

This figure shows the curves for three different blackbodies, at temperatures of T=7000, 5270 and 4000 Kelvin. Notice that not only does the position of the peak change, but also the height of the peak and the total area under the curve.

## Red Giants

Notice that the area under the red star’s curve is much smaller than the area under the other two. Yet, when we calcualted the intrinsic luminosities of stars in my blog about measuring the brightnesses of stars, we saw that Betelgeuse, a red star, is 10,000 times brighter than the Sun. How can it be both cooler and brighter?

The answer is that the area under the curve is the power per unit area, and the area we are talking about is the surface area of the star, given by $A=4 \pi R^{2}$ where $R$ is the radius of the star. Therefore, the only way a cool star can be brighter than a hotter star is if its surface area is bigger, so we know that Betelgeuse must be physically a much larger star than the Sun.

The luminosity of a star, let us call it $L$, is going to be dependent on both its temperature and its surface area (from the Steffan-Boltzmann law), so, we can write

$L \propto R^{2} T^{4}$

To compare the luminosity of e.g. Betelgeuse $L_{Bet}$ to that of the Sun $L_{\odot} \text{ (} \odot$ is the standard symbol used in astronomy for the Sun) we can write

$\frac{ L_{Bet} }{ L_{\odot} } = \frac{ R_{Bet}^{2} T_{Bet}^{4} }{ R_{\odot}^{2} T_{\odot}^{4} }$

Re-arranging, this gives that the ratio of the radius of Betelgeuse compared to the radius of the Sun $\frac{ R_{Bet} }{ R_{\odot} }$ as

$\frac{ R_{Bet} }{ R_{\odot} } = \frac{ T_{\odot}^{2} }{ T_{Bet}^{2} } \sqrt{ \frac{ L_{Bet} }{ L_{\odot} } }$

Betelgeuse has a surface temperature of about $T=3000 \text{K}$ as measured from its blackbody spectrum, and as we saw in this blog, the ratio of its luminosity to the luminosity of the Sun is 10,000 (Betelgeuse is intrinsically 10,000 times brighter than the Sun). The surface temperature of the Sun is $T_{\odot} = 5800 \text{K}$, so plugging these numbers in gives us

$\boxed{ \frac{ R_{Bet} }{ R_{\odot} } = \left( \frac{ 5800 }{ 3000} \right)^{2} \sqrt{ \frac{ 1 \times 10^{4} }{ 1 } } = 373.77 \approx 375!!! }$

As we know the distance from the Earth to the Sun, and can measure the angular size of the Sun as seen from the Earth (it is about half a degree), a simple big of trigonometry gives the radius of the Sun as about 700 million metres, so the radius of Betelgeuse is about $2.63 \times 10^{11} \text{m}$. An AU, the average distance from the Earth to the Sun, is about 150 million km, or $150 \times 10^{9} \text{m}$, so the radius of Betelgeuse is approximately $\boxed{ 1.75 \text{AUs!!} }$.

This means that, if we were to replace the Sun with Betelgeuse, even Mars would lie within the outer envelope of the star. This is why we call such intrinsically bright, red stars red giants, they truly are enormous!

## Derivation of the Rayleigh-Jeans law – part 3 (final part)

In part 1 of this blog, I showed that the 3-dimensional wave equation for an electromagnetic (EM) wave can be written (ignoring the magnetic component $\vec{B}$ as it is much smaller than the electric component $\vec{E}$ ) as

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

In part 2 I showed that, for EM waves in a cubic cavity with sides of length $L,$ the only modes which can exist have to satisfy the equation

$n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = \frac{ 4 L^{2} }{ \lambda^{2} }$

where $n_{x}$ is the number of modes allowed in the cavity in the x-direction, etc. This is the so-called “standing wave solution to the wave equation for a cubical cavity with sides of length L”.

In this third part and final part of the derivation of the Rayleigh-Jeans law, I will calculate the total number $N$ of allowed modes in the cavity given this standing wave solution, secondly I will calculate the number of modes per unit wavelength in the cavity, and finally I will calculate the energy density per unit wavelength and per unit frequency of the EM waves. This final part is the famous Rayleigh-Jeans law.

## The total number of modes in our cubic cavity

In order to calculate the total number of allowed modes $N$ in our cubic cavity we need to sum over all possible values of $n_{x}, n_{y} \text{ and } n_{z}$. To do this we use a mathematical trick of working in “n-space”, that is to say we determine the volume of a sphere where the x-axis is given by $n_{x}$, the y-axis by $n_{y}$ and the z-axis by $n_{z}$. The value of $n = \sqrt{ n_{x}^{2} + n_{y}^{2} + n_{z}^{2} }$. We can determine the value of $N$ by considering the volume of a sphere with radius $n$, which is of course just

$N = \frac{ 4 \pi }{ 3 } n^{3}$

But, as we can see in the diagram below, if we sum over $n$ for an entire sphere we will be including negative values of $n_{x}, n_{y} \text{ and } n_{z}$, whereas we only have positive values of each.

To correct for this, to only consider the positive values of $n_{x}, n_{y} \text{ and } n_{z}$, we just need to divide the volume above by 8, as the part of a sphere in the positive part of the diagram is one eighth of the total volume. But, we also need to make another correction. Light can exist independently in two different polarisations at right angles to each other, so we need to double the number of solutions to our standing wave equation to account for this .We therefore can write

$N = ( \frac{ 4 \pi }{ 3 } n^{3} ) ( \frac{ 1 }{ 8 } ) 2 = \frac{ \pi }{ 3 } n^{3} = \frac{ \pi }{ 3 } ( n_{x}^{2} + n_{y}^{2} + n_{z}^{2} )^{3/2} = \frac{ \pi }{ 3 } ( \frac{ 4 L^{2} }{ \lambda^{2} } )^{3/2}$

which gives the number of modes in the cavity as

$\boxed{ N = \frac{ 8 \pi L^{3} }{ 3 \lambda^{3} } }$

## The number of modes per unit wavelength

The expression above is the total number of modes in the cavity summed over all wavelengths. The number of modes per unit wavelength can be found by differentiating this expression with respect to $\lambda$, i.e. we find $\frac{ dN }{ d\lambda }$.

$\frac{ dN }{ d\lambda } = \frac{ d }{ d\lambda } ( \frac{ 8 \pi L^{3} }{ 3 \lambda^{3} } ) = - 3 \frac{ 8 \pi L^{3} }{ 3 \lambda^{4} } = - \frac{ 8 \pi L^{3} }{ \lambda^{4} }$

The minus sign is telling is that the number of modes decreases with increasing wavelength.

We can also derive the number of modes per unit wavelength in the cavity volume by dividing by the volume of the cavity

$\frac{ \text{Number of modes per unit wavelength} }{ \text{cavity volume} } = - \frac{ 1 }{ L^{3} } \frac{ dN }{ d\lambda } = - \frac{ 1 }{ L^{3} } \frac{ 8 \pi L^{3} }{ \lambda^{4} } = -\frac{ 8 \pi }{ \lambda^{4} }$

## The energy per unit volume per unit wavelength and per unit frequency

Because the matter and radiation are in thermal equilibrium with each other, we can say that the energy of each mode of the EM radiation is $E = kT$ where $k$ is Boltzmann’s constant and $T$ is the temperature in Kelvin of the radiation. This comes from the principle of the Equipartition of Energy.We write the energy per unit volume (also called the energy density) with the symbol $u$ so we have that the energy per unit volume per unit wavelength is given by

$\frac{ du }{ d\lambda } = \frac{ 1 }{ L^{3} } \frac{ dE }{ d\lambda} = \frac{ 1 }{ L^{3} } ( \frac{ dN }{d \lambda } ) kT = \frac{ 8 \pi }{ \lambda^{4} } kT$

To write this in terms of frequency we remember that

$\lambda = \frac{ c }{ \nu } \text{ so } \frac{ d \lambda }{ d \nu } = - \frac{ c }{ \nu^{2} } \rightarrow \frac{ 1 }{ \lambda^{4} } = \frac{ \nu^{4} }{ c^{4} }$

and, from the chain rule we can write that

$\frac{ du }{ d \nu } = \frac{ du }{ d\lambda } \frac{ d\lambda }{ d\nu } \text{ and } \frac{d \lambda }{ d \nu } = -\frac{ c }{ \nu^{2} }$

(the minus sign is just telling is that as $\lambda \text{ increases } \nu \text{ decreases )}$.

This gives us that

$\frac{ du }{ d \nu} = (8 \pi kT) \frac{ \nu^{4} }{ c^{4} } ( \frac{ c }{ \nu^{2} } )$

So, finally we have the Rayleigh-Jeans law, that the energy density of the radiation is given by

$\boxed{ \frac{ du }{ d \nu } = \left( \frac{ 8 \pi kT }{ c^{3} } \right) \nu^{2} }$

So, using Classical Physics, we find that the energy density is proportional to the frequency squared ($\frac{ du }{ d\nu } \propto \nu^{2}$), which means the energy density plotted as a function of frequency should look like the purple curve below.

The purple curve shows the so-called “ultraviolet” catastrophe”, because the energy density $\frac{ du }{ d \nu } \propto \nu^{2}$.

Of course, physicists already knew that the energy density of blackbodies followed the blackcurve, not the purple curve. If it were to follow the purple curve (the Rayleigh-Jeans law) the blackbody would get brighter and brighter at shorter and shorter wavelengths, the so-called ultraviolet catastrophe. In a future blog I will outline how Max Planck resolved this problem, and in so doing heralded in the dawn of Quantum Mechanics, an entirely new way of thinking about the sub-atomic world.

Part 1 of this blog is here.

Part 2 of this blog is here.

## Derivation of the Rayleigh-Jeans law – part 2

In part 1 of this blog I showed that the 1-dimensional (let’s say the x-direction) wave equation for the electric field $\vec{E}(x,t) = E \sin(kx - \omega t)$ is

$\frac{ d^{2}E }{dx^{2} } = \frac{ 1 }{ c^{2} } \frac{ d^{2} E }{dt^{2} }$

and that this can be generalised to the 3-dimensional wave equation

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

By analogy to the 1-D electric field, the 3-D electric field (that is, travelling in any direction) can be written as

$\vec{E}(r,t) = E \sin( k\vec{r} - \omega t )$

The direction $\vec{r}$ of the electric field can be split into its x,y and z components so that $\vec{r} = \vec{x} + \vec{y} + \vec{z}$, and the wavenumber $k$ can be split into its x,y and z-components such that $k = \sqrt{k_{x}^2 + k_{y}^2 + k_{z}^2}$ ($k_{x},k_{y} \text{ and } k_{z}$ are not generally the same value, unless the wave is travelling at 45 degrees to each of the axes).

We can then write

$\vec{E}(r,t) = \vec{E}(x,y,z,t) = E \sin (k_{x}x + k_{y}y + k_{z}z - \omega t)$

## Radiation in a cubical cavity

Remember, for blackbody radiation we need to enclose the radiation in a cavity with perfectly reflective walls, so that the radiation and the matter (thin gas in our example) can come into thermal equilibrium with each other with no loss of energy. For simplicity, the cavity will be a cube where each side will be of length $L$.

We are going to define the vertical direction as the z-direction, the direction to the right as the y-direction, and the direction towards us as the x-direction, as shown in the figure above. We need the electric field to be zero at the walls, because if it were non-zero the electric field would impart energy to the cavity walls and lose energy itself. If the electric field needs to be zero at the walls, this means its x,y and z-components $E_{x}, E_{y} \text{ and } E_{z}$ also need to be zero at the walls.

Thus, only electric fields with suitable wavelengths can exist in the cavity. This is analogous to standing waves on a string, if we have two fixed ends (such as a guitar string), only certain wavelengths can exist. In the diagram below I show the first four possible “modes” for the electric field between the two walls a distance $L$ apart [first four in the sense that the one with the lowest frequency (longest wavelength) is the yellow curve, then the next lowest frequency is the blue curve, then the green, and finally the black].

In the graph above, the x-axis is in units of $L$, so $1$ corresponds to $L=1$. The yellow curve is the lowest frequency (longest wavelength) wave that can exist between the two walls. Again, using the analogy of standing waves on a string, this would be the fundamental wave, the lowest note the string could produce. The equation which describes this yellow curve is $y=\sin( \frac{ \pi x }{ L })$, but remember we can also write, for any wave, $y=\sin(k x)$ where $k$ is the wavenumber which we introduced in part 1 of this blog. $k$ is the number of waves per unit wavelength, and is related to the wavelength via the equation $k = \frac{ 2\pi }{ \lambda } \text{ where } \lambda$ is the wavelength.

So, if $k$ for the green curve is equal to $\frac{ \pi }{ L }$, we can write $\lambda = \frac{ 2\pi }{ k } = \frac{ 2\pi L }{ \pi } = 2L$, so half of the wave fits between the two walls.

For the blue curve, we have $y = \sin ( \frac{ 2 \pi x }{ L } )$, and so $k = \frac{ 2 \pi }{ L }$ which leads to $\lambda = \frac{ 2 \pi }{ k } = \frac{ 2 \pi L}{ 2 \pi } = L$, and as we can see the blue curve has one complete wavelength between the two walls.

For the green curve, we have $y = \sin ( \frac{ 3 \pi x }{ L } )$, and so $k = \frac{ 3 \pi }{ L }$ which leads to $\lambda = \frac{ 2 \pi }{ k } = \frac{ 2\pi L }{ 3 \pi } = \frac{ 2L }{ 3 }$; the wavelength of the green curve is 2/3rds of the distance L, or to put it another way there are one and a half waves of the green curve between the walls.

Finally, for the black curve, we have $y = \sin( \frac{ 4\pi x }{ L } )$ and so $k = \frac{ 4 \pi }{ L }$ which leads to $\lambda = \frac{ 2\pi L }{ 4\pi } = \frac{ 2 L }{ 4 } = \frac{ L }{ 2 }$, so the wavelength is half the distance between the two walls, meaning there will be two complete waves between the walls.

I have only shown the first four possible waves, but of course in reality we can have $y=\sin( \frac{ 5\pi x }{ L } ), y=\sin( \frac{ 6\pi x }{ L } )$ etc., so in general we can write $y=\sin( \frac{ n \pi x }{ L } )$ where $n=1,2,3,4,5,6$ etc. In theory, $n \text{ can take any value from } 1 \text{ to } \infty$.

So, in the x-direction we can write that

$E_{x}=E \sin ( \frac{ n \pi x }{ L })$.

Although the wavelength of the x,y and z-components of the EM are all the same, there may not be the same number of wavelengths in the x,y and z-directions between the walls. For example, if the wave is travelling wholly horizontally then there may be no electric field component in the vertical (z) direction, and if it is travelling horizontally at e.g. $30^{\circ}$ to the left hand wall then the distance the wave is travelling in the y-direction will be $\sqrt{ 3 }$ times the distance it travels in the x-direction. Because of this, although we can write similar equations for $E_{y} \text{ and } E_{z}$, the values of $n$ will be different.

Therefore, we are going to write

$E_{x} = E \sin ( \frac{ n_{x} \pi x }{ L }), E_{y} = E \sin ( \frac{ n_{y} \pi y }{ L }) \text{ and } E_{z} = E \sin ( \frac{ n_{z} \pi z }{ L })$

where $n_{x}, n_{y} \text{ and } n_{z}$ are related to the number of waves between the walls in each of the x,y and z-directions via $k_{x}=\frac{ n_{x} \pi }{ L }, k_{y}=\frac{ n_{y} \pi }{ L } \text{ and } k_{z}\frac{ n_{z} \pi }{ L }$ in each of the spatial directions.

Remember the electric field in 3-D can be written as

$\vec{E} = E\sin(k_{x}x + k_{y}y +k_{z}z - \omega t) \text{. But } k_{x}x = \frac{ n_{x} \pi x }{ L } \text{ etc. and } \omega = 2\pi \nu = \frac{ 2 \pi c }{ \lambda }$

so we can write the electric field in 3-dimensions as

$\vec{E} = E \sin( \frac{ n_{x} \pi x }{ L } + \frac{ n_{y} \pi y }{ L } + \frac{ n_{z} \pi z }{ L } - \frac{ 2 \pi c t }{ \lambda } )$

You may remember from trigonometry that $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$ (the so-called compound angle formula. Although it is pretty tedious to prove it (see for example this link), we can write

$\sin(X+Y+Z) = \sin(X)\cos(Y)\cos(Z) + \cos(X)\sin(Y)\cos(Z) + \cos(X)\cos(Y)\sin(Z) - \sin(X)\sin(Y)\sin(Z)$

(where we have written $X,Y \text{ and } Z$ to represent $\frac{ n_{x} \pi x }{ L }$ etc.). To find $\sin(X+Y+Z-T)$ (where $T= \frac{ 2 \pi c t }{ \lambda }$) is even more tedious, and also involves working out $\cos(X+Y+Z)$, but if you go through it you will find (eventually!) that

$\sin(X+Y+Z-T) = \sin((X+Y+Z)-T) = \sin(X+Y+Z)\cos(T) -\cos(X+Y+Z)\sin(T)$

where $\cos(X+Y+Z)$ can be written

$\cos(X+Y+Z) = \cos(X)\cos(Y)\cos(Z) - \sin(X)\sin(Y)\cos(Z) - \sin(X)\cos(Y)\sin(Z) - \cos(X)\sin(Y)\sin(Z)$

and so

$\sin(X+Y+Z-T) = \sin(X)\cos(Y)\cos(Z)\cos(T) + \cos(X)\sin(Y)\cos(Z)\cos(T) + \cos(X)\cos(Y)\sin(Z)\cos(T) - \sin(X)\sin(Y)\sin(Z)\cos(T) - \cos(X)\cos(Y)\cos(Z)\sin(T) + \sin(X)\sin(Y)\cos(Z)\sin(T) + \sin(X)\cos(Y)\sin(Z)\sin(T) + cos(X)\sin(Y)\sin(Z)\sin(T)$

Remember, however, that the electric field has to be zero at the walls, and in particular at the origin where $X=Y=Z=0$. This means that all the terms with $\cos(X) \text{ or } \cos(Y) \text{ or } \cos(Z)$ will disappear, and so the above expression will simplify to

$\sin(X+Y+Z-T) = - \sin(X)\sin(Y)\sin(Z)\cos(T)$

But, unlike the values of $X=0,Y=0 \text{ and } Z=0$ which are determined by the origin of the cavity, the value of $T=0$ is completely arbitrary, so we can just shift it by $\omega = \pi/2$ (one quarter of a period) to give us $\cos(T+\pi/2) = -\sin(T)$ and then we have

$\sin(X+Y+Z-T) = \sin(X)\sin(Y)\sin(Z)\sin(T)$

Replacing our $X,Y,Z \text{ and } T$ we have

$\vec{E} = E\sin(\frac{ n_{x} \pi x }{ L }) \sin(\frac{ n_{y} \pi y }{ L }) \sin(\frac{ n_{z} \pi z }{ L }) \sin(\frac{ 2 \pi ct }{ \lambda })$
Then,using the 3-dimensional wave equation we had earlier

$\frac{ \partial^{2} E }{ \partial x^2 } + \frac{ \partial^{2} E }{ \partial y^{2} } + \frac{ \partial^{2} E }{ \partial z^{2} } = \frac{1}{c^2} \frac{ \partial^{2} E }{ \partial t^{2} }$
we will need to work out $\frac{ \partial^{2} E }{ \partial x^{2} }$ etc.

$\frac{ \partial E }{ \partial x } = \frac{ \partial }{ \partial x } E\sin(\frac{ n_{x} \pi x }{ L }) = E \frac{ n_{x} \pi }{ L } \cos(\frac{ n_{x} \pi x }{ L })$
and
$\frac{ \partial^{2} E }{ \partial x^{2} } = -E (\frac{ n_{x} \pi }{ L })^{2} \sin( \frac{ n_{x} \pi x }{ L }) = -E_{x} (\frac{ n_{x} \pi }{ L })^{2}$

If we do the same thing for $\frac{ \partial^{2} E }{ \partial y^{2} }, \frac{ \partial^{2} E }{ \partial z^{2} } \text{ and } \frac{ 1 }{ c^{2} } \frac{ \partial^{2} E }{ \partial t^{2} }$ we get

$(\frac{ n_{x} \pi }{ L })^{2} + (\frac{ n_{y} \pi }{ L })^{2} + (\frac{ n_{z} \pi }{ L })^{2} = \frac{ 1 }{ c^{2} } (\frac{ 2 \pi c }{ \lambda })^{2} = (\frac{2 \pi }{ \lambda })^{2}$

which simplifies to

$n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = \frac{ 4 L^{2} }{ \lambda^{2} }$.

This is the so-called “standing wave solution to the wave equation for a cubical cavity with sides of length L”. In part 3 of this blog I will explain what this solution means in terms of the number of allowed EM-wave modes that can exist in the cavity.

Part 1 of this blog can be found here, and Part 3 is here.