Average total energy up to frequency $\omega$ in a blackbody

Question

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

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

I have a couple of questions about this:

1. How did the author get from $\int^\omega d^3 \vec{n} \dfrac{\hbar \omega_n}{e^{\hbar \omega_n \beta} - 1}$ to $\int_{-1}^{1} d \cos(\theta) \int^{2\pi}_0 d \phi \int_0^\omega d |\vec{n}| \dfrac{|\vec{n}|^2 \hbar \omega_n}{e^{\hbar \omega_n \beta} - 1}$?

2. How did the author get from $\int_{-1}^{1} d \cos(\theta) \int^{2\pi}_0 d \phi \int_0^\omega d |\vec{n}| \dfrac{|\vec{n}|^2 \hbar \omega_n}{e^{\hbar \omega_n \beta} - 1}$ to $4 \pi \hbar \dfrac{L^3}{8\pi^3} \int_0^\omega d\omega' \dfrac{\omega'^3}{e^{\hbar \omega' \beta} - 1}$?

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

Answer 1

1.My understanding is that one assumption you do is isotropy, that is you don't care in what orientation the radiation goes ($\omega(r,\theta,\varphi)=\omega(r)$), so it is more more natural to use a sphere as your blackbody instead of a cube or other surface and accordingly use spherical coordinates instead of other coordinates, making use of its radial symmetry as has been pointed in the comments ($\omega(\sqrt{x^2+y^2+z^2})=\omega(r)$).
Usually integrating over $d^3 x$ doesn't make much sense but you integrate three times when going over the infinitesimal of volume $dV=dx\cdot dy \cdot dz$ so $d^3 \vec{n}$ should be understand I believe also as a shorthand notation for volume in this case for spherical coordinates so $$d^3 \vec{n}=dV=r^2 \sin(\theta)d\theta d \phi d r\ (=d(-r\cos(\theta)) \cdot d(r\phi) \cdot d r), $$ so integrating one could see what the notation $\int^\omega d^3 \vec{n}$ means $$\int^\omega d^3 \vec{n}=\int_V dV=\int\limits_{\theta=0}^{\pi}\int\limits_{\varphi=0}^{2 \pi}\ \int\limits_{w'=0}^{w} r^2 \sin(\theta)d\theta d \phi d r=\int\limits_{\theta=0}^{\pi}d(-\cos(\theta))\int\limits_{\varphi=0}^{2 \pi}d\phi\ \int\limits_{w'=0}^{w} r^2 d r. $$ (From what I can see from amazon the book does a mistakes when writing $\int_{-1}^{1} d \cos(\theta)$instead of $\int_{0}^{\pi} d (-\cos(\theta))$).

2.As you wrote $\hbar \omega_n = \dfrac{2\pi}{L} \hbar |\vec{n}|$ or simply $|\vec{n}|=\dfrac{L}{2\pi}\omega_n$ and from this you have two things, that $d|\vec{n}|= \dfrac{L}{2\pi}d\omega_n$ and also that $|\vec{n}|^2=\dfrac{L^2}{4\pi^2}\omega_n^2$ and multiplying this last two you get $d |\vec{n}| |\vec{n}|^2=\dfrac{L^3}{8\pi^3}\omega_n^2 d\omega_n$, and so you can verify the result, $$E(\omega)= \int^\omega d^3 \vec{n} \dfrac{\hbar \omega_n}{e^{\hbar \omega_n \beta} - 1} = \int_{0}^{\pi} d(-\cos(\theta)) \int^{2\pi}_0 d \phi \int_0^\omega d |\vec{n}| \dfrac{|\vec{n}|^2 \hbar \omega_n}{e^{\hbar \omega_n \beta} - 1}=4\pi\hbar \int_0^\omega d |\vec{n}| \dfrac{|\vec{n}|^2 \omega_n}{e^{\hbar \omega_n \beta} - 1}= 4 \pi \hbar \dfrac{L^3}{8\pi^3} \int_0^\omega d\omega_n \dfrac{\omega_n^3}{e^{\hbar \omega_n \beta} - 1}= 4 \pi \hbar \dfrac{L^3}{8\pi^3} \int_0^\omega d\omega' \dfrac{\omega'^3}{e^{\hbar \omega' \beta} - 1}$$ (From what I can see from amazon you did a mistake when writing $\int_0^\omega d\omega' \dfrac{\omega'}{e^{\hbar \omega' \beta} - 1}$instead of $\int_0^\omega d\omega' \dfrac{\omega'^3}{e^{\hbar \omega' \beta} - 1}$).