Page 4 of the textbook Quantum Field Theory and the Standard Model by Schwartz says the following:

The incompatibility of observations with the classical prediction led Planck to postulate that the energy of each electromagnetic mode in the cavity is quantised in units of frequency:

$$E_n = \hbar \omega_n = \dfrac{2\pi}{L} \hbar |\vec{n}| = |\vec{p}_n|,$$

where $h$ is the Planck constant and $\hbar \equiv \dfrac{h}{2\pi}$.

On page 5, the author then goes on to say the following:

Now let us take the continuum limit, $L \to \infty$. In this limit, the sums turn into integrals and the average total energy up to frequency $\omega$ in the blackbody is

$$\begin{align} E(\omega) &= \int^\omega d^3 \vec{n} \dfrac{\hbar \omega_n}{e^{\hbar \omega_n \beta} - 1} \\ &= \int_{-1}^{1} d \cos(\theta) \int^{2\pi}_0 d \phi \int_0^\omega d |\vec{n}| \dfrac{|\vec{n}|^2 \hbar \omega_n}{e^{\hbar \omega_n \beta} - 1} \\ &= 4 \pi \hbar \dfrac{L^3}{8\pi^3} \int_0^\omega d\omega' \dfrac{\omega'^3}{e^{\hbar \omega' \beta} - 1} \end{align}$$

Thus, the intensity of light as a function of frequency is (adding a factor of 2 for the two polarizations of light)

$$I(\omega) = \dfrac{1}{V} \dfrac{dE(\omega)}{d \omega} = \dfrac{\hbar}{\pi^2} \dfrac{\omega^3}{e^{\hbar \omega \beta} - 1}$$

When I attempt to follow how the author got $I(\omega) = \dfrac{1}{V} \dfrac{dE(\omega)}{d \omega} = \dfrac{\hbar}{\pi^2} \dfrac{\omega^3}{e^{\hbar \omega \beta} - 1}$, I get

$$I(\omega) = \dfrac{1}{V} \dfrac{\hbar L^3}{\pi^2} \dfrac{\omega^3}{e^{\hbar \omega \beta} - 1}$$

What happens here with the $V$ and the $L^3$? We know that $L$ is the size of the "box" (as the author describes it), and I'm guessing that $V$ is volume (of phase space, as the author implies). So since we take the continuum limit, $L \to \infty$, the size of the "box" is growing to infinity; and if $V$ is the volume of phase space, I'm assuming that this volume also grows to infinity? And so, based on my best attempt at inferring what's going on here, it seems that $V$ and $L$ cancel out due the "box" becoming infinitely large? But, from a mathematical point of view, would this not result in the indeterminate form $\dfrac{\infty}{\infty}$, since both $L$ and $V$ blow-up to infinity?

I would greatly appreciate it if people could please take the time to clarify this.


1 Answer 1


$V$ is the volume of the box, $V=L^3$. It cancels out, as you say.

  • $\begingroup$ Are you sure about this? The textbook says the following: By the (classical) equipartition theorem, blackbodies should emit light equally in all modes with the intensity growing as the differential volume of phase space: $$I(\omega) \equiv \dfrac{1}{V} \dfrac{d}{d\omega} E(\omega) = \text{const} \times c^{-3}\omega^2 k_B T \ \text{(classical)}$$ $\endgroup$ Jun 30, 2019 at 10:45
  • 1
    $\begingroup$ The differential volumes of the phase space is in the derivative of E. $\endgroup$
    – nasu
    Jun 30, 2019 at 12:55
  • $\begingroup$ @nasu Ok, thanks for the clarification. $\endgroup$ Jun 30, 2019 at 12:56

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