1. Origins of Quantum Theory; UV Catastrophe¶

This note explains the origins of Quantum theory, blackbody radiation and includes derivations of Rayleigh-Jeans and Maxwell-Planck laws.

Origins of Quantum Theory¶

There are a few successes of classical physics, namely

Newtonian Mechanics
Maxwellian Electromagnetism
Boltzmann Statistical Mechanics

But, there are 2 catastrophic breakdowns when applied to black body radiation and atomic spectra.

Let's first discuss some historical events leading up to the origins of quantum theory

1792: Wedgward remarks that all objects heated to the same temperature grow the same color.
1800s: Improvements in spectroscopy show that solids emit continuous spectra; gases emit discrete lines
1859: Kirchoff proposes model of thermal radiation emission to be a function of wavelength and temperature

\[ R(\lambda, T) \qquad \text{emissive power per unit area} \]

Blackbodies¶

Blackbodies are objects that absorb all incident EM waves, regardless of frequencies and incident angles, and they also emit radiations with a spectrum determined by its temperature. They are called such because light will not be reflected directly. Instead, incident light has the change to reach thermal equilibrium with the blackbody.

In other words, it is a cavity where light bounces many times off the walls. They are

Perfect absorbers/Perfect emitters
Equilibrium established between walls at temperature $T$ and light field

Classical Observations of Blackbodies include

Stefan's Law (1879), which experimentally found that total radiation emitted from a glowing solid is proportional to $T^4$.

1.1 Stefans Law.jpg

\[ R(T) = \int_{0}^{\infty} R(\lambda, T)d\lambda = \sigma T^4, \qquad \text{where }\sigma = 5.67\times 10^{-8} \frac{W}{m^2K^4} \]

Wien's Law: From spectroscopic data, found $\lambda_{max}T = 2.898\times10^{-3}mK$

1.2 Wiens Law.jpeg

Rayleigh-Jeans Law¶

This law attempts to fit a functional form to the end part of the spectrum graph of $R$ vs. $\lambda$.

Derivation of Classical and Quantum Blackbody Radiation Formulae¶

Cavity of Standing Waves¶

Let's start with a simple object: a cube with sidelength $L$. The wave equation states

\[ \nabla^2\Psi(\vec{r}, t) = \frac{1}{c^2}\frac{\partial^2}{\partial^2t}\Psi(\vec{r}, t) \]

where $\Psi$ is a function that describes wave amplitude. Since we have a PDE, we can specify a boundary condition, where $\Psi = 0$ at the cavity walls, i.e.

\[ \Psi(x=0, y, z, t) = \Psi(x = L, y, z, t) = \cdots = 0 \]

The solution is known:

\[ \Psi(\vec{r}, t) = A(t) \sin(k_xx) \sin(k_yy)\sin(k_zz) \]

and the boundary conditions require

\[ k_i = \frac{n_i\pi}{L}, \qquad i = x, y, z; \text{and n is an integer} \]

Notice the form of the solution. It is in a standing wave form, with a time varying amplitude

\[ A(t)B(x, y, z) \]

this is what we will substitute into the wave equation. Upon simiplification, I get

\[ -(n_1^2 + n_2^2 + n_3^3) \frac{\pi^2}{L^3} A(t) B(x, y, z) = \frac{1}{c^2}B(x, y, z)\frac{\partial^2}{\partial^2t}A(t) \]

We will guess the form for $A(t)$ to be $A_0\cos(\omega t) + \phi$, where $\omega^2 = \frac{c^2\pi^2}{L}(n_1^2 + n_2^2 + n_3^2)$. This form makes sense because the function $A(t)$ is related to the second derivative.

Now, we need to count how many different ways (called modes) a given $\omega$ can be distributed over $n_i$.

Let $g(\omega) = \frac{dN(\omega)}{d\omega}$, where $g$ is the number of modes per unit frequency. Thus

\[ N(\omega) = \int_{0}^{\omega}g(\omega)d\omega \]

For a given $\omega$, the possible values of $n_i$ are bound and must obey

\[ n_1^2 + n_2^2 + n_3^2 \leq \frac{\omega^2L^2}{c^2\pi^2} \]

Isn't this just a sphere of radius $\frac{\omega L}{c\pi}$?

Therefore, let's just consider one quadrant of the whole sphere and count that much of it

\[ N(\omega) = \frac{1}{8}\left(\frac{4}{3}\pi\frac{\omega^3L^3}{c^3\pi^3}\right) = \frac{\omega^3V}{6c^3\pi^2} \]

where $V$ is the volume. Now let's convert to linear frequency using $\omega = 2\pi f$.

\[ N(f) = \frac{8\pi^3f^3V}{6c^3 \pi^2} \]

\[ g(f) = \frac{dN(f)}{df} = \frac{4\pi f^2 V}{c^3} \]

And since each mode has 2 polarization directions, we double it

$$ g(f) = \frac{8\pi f^2}{c3} V $$ This is true for both classical and quantum, since we just used a geometrical argument.

Classical Calculation of The Result¶

We are looking for energy density per volume. Where it will be

\[ \rho(f) = \frac{g(f)}{V} \times Energy \]

In Classical physics, each mode has energy $k_BT$ by the equipartition theorem. Thus, the energy density between $f$ and $f + df$ is

\[ g(f)k_BT df = \frac{8\pi f^2 V}{c^3} k_BTdf \]

And we get energy density by dividing by $V$.

\[ \rho(f) = \frac{8\pi f^2}{c^3} k_BTdf \]

Now, let's convert to $\lambda$. $f = c/\lambda$, and $df = -cd\lambda / \lambda$, so we get

\[ \rho(\lambda) = \frac{8\pi k_BT}{\lambda^4} \]

which is Rayleigh-Jeans.

Let's analyze this result. The formula works for $k_BT >> \omega$. But as $f\rightarrow \infty$, $\rho\rightarrow\infty$, which makes no sense. This is the UV Catastrophe.

Quantum Calculation of The Result¶

Let's fix the UV Catastrophe using Quantum. Planck postulated that each oscillator cannot take on all modes of energy. He said that energy is gained/lost in discrete units of $hf$, or simply

\[ E_n = nhf \]

Now let's find the expectation value of the energy

\[ \overline{E} = \frac{\sum_{n=0}^{\infty} nhf e^{-\frac{nhf}{k_BT}}}{\sum_{n=0}^{\infty}e^{-\frac{-nhf}{k_BT}}} \]

We know this sum by looking it up on a table, and find

\[ \overline{E} = \frac{hfe^{-\frac{hf}{k_BT}}}{1 - e^{-\frac{hf}{k_BT}}} = \frac{hf}{e^{\frac{hf}{k_BT}} - 1} \]

This is the energy that we use, according to Planck, instead of the Equipartition Theorems (the Classical way). Now we can say the energy density per volume is

\[ \rho(f, T) = \frac{g(f)}{V} \overline{E} = \frac{8\pi f^2}{c^3}\frac{hf}{e^{\frac{hf}{k_BT}} - 1} \]

And finally, converting to $\lambda$, we get

\[ \rho(\lambda, T) = \frac{8\pi hc}{\lambda^5} \frac{hf}{e^{\frac{hf}{k_BT}} - 1} \]

which actually matches the $R(\lambda, T)$ perfectly! The Classical result matches only the end part of it, while the Quantum result matches the whole curve.