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## 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.

### 15 Responses

1. Thank you for the thorough explanation! It helped me out a lot for a homework assignment. I had to calculate the solution in one and two dimensions. As far as I could tell, the solutions to the wave equation from part II don’t change, but the term N of course has a different expression for its volume (area or linear) as well as different coefficients derived from polarization modes and isolating the positive values.

• I’m glad you found it useful 🙂

2. Great job! After my working years I’ve decided to study Quantum Mechanics. This is a wonderful first step!

• Thank you, and good decision !!

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

4. Thanks. Just the type of derivations I need to get into something!

5. Do we consider UV catastrophe obsurd simply because we have not created a perfect black body for experiment or because the law is wrong?

• The “full” spectrum from UV to infrared was first measured for blackbodies in the 1890s, and so we knew the shape of the blackbody curve. The UV catastrophe is so-called because classical physics predicts that the energy at shorter wavelengths (higher frequencies) – i.e. the UV, goes off to infinity, which is absurd as it would mean a blackbody would radiate an infinite amount of energy. The so-called Rayleigh-Jeans law (this blog is the third part of the derivation of that law) theoretically predicted infinite energy in the UV, which is clearly absurd.

The experimentally determined curve did not show the UV catastrophe, and in 1900, after much trial and error, Max Planck found a formula which fitted the observed curve; then he had to find a physical explanation for the formula that he had found.

6. To understand the classical derivation of the Rayleigh Jeans law, it seems to me that the key is grasping the whole equipartition of energy concept. According to classical physics, energy must be conserved. So, classical physics should not (in my mind) predict obtaining an infinite amount of energy from a finite amount of energy. That would be a perpetual motion machine, no? So, we have our “Rayleigh-Jeans” cube. In it, we have some dust particles of an ideal blackbody. The blackbody dust has a certain temperature, which is to say that the blackbody has a certain finite energy. The blackbody is radiating. Ordinarily, by the conservation of energy, this radiation would cause the temperature (and internal energy) of the blackbody to decrease. However, we are in a “Rayleigh-Jeans” cube, so the radiation is bouncing all over the place and is not absorbed by the walls of the cube. After a transient period of time, the blackbody and the radiation are in thermal equilibrium. Again, no energy has left the cube, and there is a finite amount of energy in the cube. So, it seems to me that the equipartition of energy would suggest that the energy radiated from the blackbody is partitioned equally over all possible modes (field configurations) supported by the cavity. There are an infinite number of discrete modes supported by the cavity, because there is always that next harmonic. Thus, the energy in each mode would be Etot/infinity, no? This looks like zero, but we would be dealing with “levels of zero” and “levels of infinity.” So that the infinite number of modes times the energy in each mode equals the total finite energy in the cube. In other words: (infinity # of modes)*(Etot/infinity) = Etot. i.e. Infinities cancel, and the conservation of energy (predicted by classical physics) holds.

But in the Rayleigh-Jeans derivation, it predicts that the energy in each mode is proportional to temperature (since we are in thermal equilibrium). Emode = kT. What I don’t quite understand is why this k is specifically Boltzmann’s constant. Also, since k is finite and positive and T is finite and positive, then there is a nonzero (non-epsilon) energy in each mode. But I just said that the total energy must be partitioned by an infinite number of modes. So Emode = Etot/(infinite # of modes).

If we say that the number of modes (which is understood to be infinite) is N, then we would say that Emode = Etot/N. Or equivalently, Etot = N*Emode = NkT. But again, I’m having a conservation of energy crisis here. N is infinite. And kT is finite. So, NkT is infinite. Classical physics should not predict a violation of the conservation of energy. How is this issue resolved?

Let’s say that we don’t know Boltzmann’s constant. Then, maybe we can say that Etot = NkT such that N is very large, and kT is very small. Or equivalently, Emode = kT.

Okay, so N is an infinite quantity, but there are levels of infinity.

As has been derived (which I think I follow)
N = 8pi*L^3*f^3/(3c^3).

As f increases to infinity, N increases to infinity. That is, if we consider the total number of frequencies up to a certain maximum frequency, the number of modes supported by the cavity (over which the finite energy in the cavity would have to be partitioned) is given by the expression above. As this maximum frequency increases to infinity, the number of modes increases to infinity.

From here, by doing some straightforward calculus (as you did), we arrive at the Rayleigh-Jeans law, which is giving us the volumetric energy density per unit frequency, or we can say the specific energy per unit frequency. This energy density per frequency increases as the square of frequency. But this is an energy density per differential frequency. du/df. The energy density per frequency explodes to infinity, but when we multiply infinity by a differential (df), we get a finite value, no? What I mean to say (I think) is that the catastrophe of the ultraviolet catastrophe is not a violation of the conservation of energy. Classical physics should not and would not predict a violation of the conservation of energy, right?

However, I think I’m still running into a problem with a finite positive value for Boltzmann’s constant. If Boltzmann’s constant could be arbitrarily small, then I think that I could affirm with more assurance that the ultraviolet catastrophe is not a violation of the conservation of energy law. However, since Boltzmann’s constant is a definite positive value, it does seem like the Rayleigh-Jeans law predicts that if you have a blackbody in a Rayleigh-Jeans cube, you end up with infinite energy – the most absurd of all perpetual motion machines (which are all absurd anyway).

7. […] 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 […]

8. […] 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 […]

9. This is extremely useful and informative! Thank you

10. Thank you so much! This was very clear and very helpful.

11. thank you so much, sir. It is a very helpful derivation

12. great explanation, better than others I have found on line