Planck Blackbody Radiation Formula
The spectral energy density of blackbody radiation as a function of frequency is given by Planck's radiation law:
\[ u(\nu)d\nu = \frac{8\pi h\nu^3}{c^3} \frac{1}{e^{h\nu/kT}-1} \, d\nu \]where
- \(h = 6.626 \times 10^{-34} \, \text{J·s}\) is Planck's constant
- \(u(\nu)\) is the energy density of radiation per unit frequency interval
- \(\nu\) is the frequency
- \(T\) is the temperature
- \(k\) is the Boltzmann constant
Waves in a Box
Consider an electromagnetic wave travelling with the speed of light in some arbitrary direction represented by the position coordinate \(x\). If the wavelength of the wave is \(\lambda\), the amplitude of the wave along the \(x\)-direction can be written as
\[ A(x) = A_0 \sin \left(\frac{2\pi x}{\lambda}\right) \]This expression can also be written in terms of the wave number \(k\):
\[ A(x) = A_0 \sin(kx) \]where the wave number is defined as
\[ k = \frac{2\pi}{\lambda} \]Time Dependence of the Wave
After a time \(t\), the wave propagates in space and the position becomes
\[ x' = x - ct \]Therefore the wave function becomes
\[ A(x) = A_0 \sin[k(x-ct)] \] or equivalently \[ A(x) = A_0 \sin(kx - kct) \]Relation Between Frequency and Wave Number
\[ kc = \frac{2\pi c}{\lambda} = 2\pi\nu = \omega \]Hence the wave can be written in the standard form
\[ A(x) = A_0 \sin(kx - \omega t) \]and therefore the speed of the wave is
\[ c = \frac{\omega}{k} \]Electromagnetic Modes in a Box
Consider a cubical box of side \(L\) and imagine electromagnetic waves bouncing back and forth inside it. The walls of the box are assumed to be perfectly conducting and rigid.
Because the walls are conducting, the electric field of the electromagnetic wave must vanish at the boundaries of the box. Therefore, only standing waves that satisfy the boundary conditions can exist inside the box.
This means that the waves that fit inside the box must be integer multiples of half a wavelength.
Allowed Wavelengths
Thus the allowed wavelengths along each direction are quantized.
Standing Wave Solution
The spatial dependence of the wave along the \(x\)-direction can be written as
\[ A(x) = A_0 \sin(k_x x) \]The wave number is related to the wavelength by
\[ k_x = \frac{2\pi}{\lambda_x} \]Using the boundary condition \(L = \ell \lambda_x /2\), we obtain
\[ \lambda_x = \frac{2L}{\ell} \]Substituting into the expression for the wave number gives
\[ k_x = \frac{\pi \ell}{L} \]where
\[ \ell = 1,2,3,\dots \]Wave Numbers in Three Dimensions
Similarly, the allowed wave numbers in the \(y\) and \(z\) directions are
\[ k_y = \frac{\pi m}{L} \] \[ k_z = \frac{\pi n}{L} \]where \(m,n = 1,2,3,\dots\)
Electromagnetic Modes in Three Dimensions
The standing wave solution inside the cubical box can be written as
\[ A(x,y,z) = A_0 \sin(k_x x)\sin(k_y y)\sin(k_z z) \]The electromagnetic wave satisfies the wave equation
\[ \frac{\partial^2 A}{\partial x^2} + \frac{\partial^2 A}{\partial y^2} + \frac{\partial^2 A}{\partial z^2} = \frac{1}{c^2} \frac{\partial^2 A}{\partial t^2} \]For the standing wave solution we obtain
\[ |k|^2 = k_x^2 + k_y^2 + k_z^2 \]Using the relation between frequency and wave number
\[ |k|^2 = \frac{\omega^2}{c^2} \]Substituting the quantized wave numbers
\[ k_x = \frac{\pi \ell}{L}, \qquad k_y = \frac{\pi m}{L}, \qquad k_z = \frac{\pi n}{L} \] we obtain \[ \frac{\pi^2}{L^2}(\ell^2+m^2+n^2) = \frac{\omega^2}{c^2} \] Let \[ p^2 = \ell^2 + m^2 + n^2 \] Thus \[ \frac{\omega^2}{c^2} = \frac{\pi^2 p^2}{L^2} \]Counting the Modes
To count the number of electromagnetic modes, we consider the points \((\ell,m,n)\) in the positive octant of a sphere in mode space.
The volume of a spherical shell of radius \(p\) and thickness \(dp\) is
\[ 4\pi p^2 dp \]Since only the positive octant is allowed,
\[ dN(p) = \frac{1}{8} 4\pi p^2 dp \]Relation Between \(p\) and \(k\)
\[ k = \frac{\pi p}{L} \] \[ dk = \frac{\pi}{L} dp \]The wave number is related to frequency by
\[ k = \frac{\omega}{c} = \frac{2\pi \nu}{c} \]The volume of the box is
\[ V = L^3 \]Number of Modes
Substituting these relations we obtain \[ dN(p) = \frac{L^3}{2\pi^2} k^2 dk \] or \[ dN = \frac{V}{2\pi^2} k^2 dk \] Using \[ k = \frac{2\pi \nu}{c} \] we obtain \[ dN = \frac{V}{2\pi^2} \left(\frac{2\pi \nu}{c}\right)^2 \left(\frac{2\pi}{c} d\nu\right) \] Finally, \[ dN = \frac{4\pi \nu^2 V}{c^3} d\nu \]Mode Density per Unit Volume
From the previous result, the number of modes in the frequency interval \( \nu \) to \( \nu + d\nu \) inside a volume \(V\) is
\[ dN = \frac{4\pi \nu^2 V}{c^3}\, d\nu \]Since each electromagnetic mode has two possible polarization states, we multiply by 2. Hence
\[ dN = \frac{8\pi \nu^2 V}{c^3}\, d\nu \]For unit volume, \(V=1\), so the number of modes per unit volume is
\[ dN = \frac{8\pi \nu^2}{c^3}\, d\nu \]Average Energy per Mode and Ultraviolet Catastrophe
The energy density of radiation per unit frequency interval per unit volume is
\[ du = u(\nu)\, d\nu = \frac{8\pi \nu^2}{c^3}\,\bar{E}\, d\nu \]where \( \bar{E} \) is the average energy of each mode. In classical physics, by the equipartition theorem,
\[ \bar{E} = kT \]Substituting this into the above expression, we obtain
\[ u(\nu)\, d\nu = \frac{8\pi \nu^2}{c^3} kT\, d\nu \]Therefore,
\[ u(\nu) = \frac{8\pi \nu^2}{c^3} kT \]This is the Rayleigh–Jeans law.
Ultraviolet Catastrophe
The total energy density is obtained by integrating over all frequencies:
\[ \int_0^\infty u(\nu)\, d\nu = \int_0^\infty \frac{8\pi \nu^2}{c^3} kT \, d\nu \to \infty \]Thus, according to classical theory, the energy density of blackbody radiation diverges at high frequencies. This failure of classical physics is known as the ultraviolet catastrophe.
Planck's Law
Planck proposed that the energy of each electromagnetic mode is quantized. The allowed energies are
\[ E_n = n h \nu, \qquad n = 0,1,2,3,\dots \]where
- \(h\) is Planck's constant
- \(\nu\) is the frequency
Boltzmann Distribution
According to the Boltzmann distribution, the probability of a state with energy \(E_i\) is proportional to
\[ P_i \propto e^{-E_i/kT} \]Including the normalization factor, the probability becomes
\[ P_i = \frac{1}{Z} e^{-E_i/kT} \] where \(Z\) is the partition function. Thus, \[ P(n) = \frac{e^{-E_n/kT}} {\sum_{n=0}^{\infty} e^{-E_n/kT}} \]Average Energy of a Mode
The average energy of a mode is given by
\[ \bar{E}_\nu = \sum_{n=0}^{\infty} E_n P(n) \] Substituting the Boltzmann probability, \[ \bar{E}_\nu = \frac{\sum_{n=0}^{\infty} E_n e^{-E_n/kT}} {\sum_{n=0}^{\infty} e^{-E_n/kT}} \] Since \(E_n = nh\nu\), \[ \bar{E}_\nu = \frac{\sum_{n=0}^{\infty} nh\nu \, e^{-nh\nu/kT}} {\sum_{n=0}^{\infty} e^{-nh\nu/kT}} \]Simplifying the Expression
Let \[ x = e^{-h\nu/kT} \] Then \[ \bar{E}_\nu = h\nu \frac{\sum_{n=0}^{\infty} n x^n} {\sum_{n=0}^{\infty} x^n} \] Expanding the series, \[ \bar{E}_\nu = h\nu \frac{x + 2x^2 + 3x^3 + \dots} {1 + x + x^2 + \dots} \]Evaluation of the Average Energy
Using the series expansions
\[ 1 + x + x^2 + x^3 + \dots = \frac{1}{1-x} \] and \[ 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(1-x)^2} \] we can simplify the expression for the average energy.From the previous result
\[ \bar{E}_\nu = h\nu \frac{\sum_{n=0}^{\infty} n x^n} {\sum_{n=0}^{\infty} x^n} \] Substituting the series identities, \[ \bar{E}_\nu = h\nu \frac{\frac{x}{(1-x)^2}} {\frac{1}{1-x}} \] \[ \bar{E}_\nu = \frac{h\nu x}{1-x} \]Since \(x = e^{-h\nu/kT}\),
\[ \bar{E}_\nu = \frac{h\nu}{e^{h\nu/kT}-1} \]Classical Limit
In the classical limit where
\[ \frac{h\nu}{kT} \ll 1 \] the exponential can be approximated as \[ e^{h\nu/kT} \approx 1 + \frac{h\nu}{kT} \] Thus, \[ \bar{E}_\nu = \frac{h\nu}{\frac{h\nu}{kT}} = kT \]This reproduces the classical equipartition result.
Planck's Radiation Law
Substituting the average energy into the expression for the energy density,
\[ u(\nu)d\nu = \frac{8\pi\nu^2}{c^3}\,\bar{E}_\nu\, d\nu \] we obtain \[ u(\nu)d\nu = \frac{8\pi\nu^2}{c^3} \frac{h\nu}{e^{h\nu/kT}-1} \, d\nu \]Therefore the final result is
\[ u(\nu)d\nu = \frac{8\pi h\nu^3}{c^3} \frac{1}{e^{h\nu/kT}-1} \, d\nu \]This is Planck's Blackbody Radiation Law that we wanted to derive.
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