I Radiance and energy density of a black body

[Airton Rampim]

How can I find the relation between the radiance and the energy density of a black body? According to Planck's law, the energy density inside a blackbody cavity for modes with frequency $\nu \in [\nu, \nu + \mathrm{d}\nu]$ is given by $$ \rho(\nu, T)\mathrm{d}\nu = \frac{8\pi\nu^2}{c^3}\frac{h\nu}{\exp\left(\frac{h\nu}{k T}\right) - 1}\mathrm{d}\nu. $$ Integrating over all the frequencies we will have

$$ \rho_T = \intop_0^{\infty}\rho(\nu, T)\mathrm{d}\nu = \frac{8\pi h}{c^3} \intop_0^\infty \frac{\nu^3}{\exp\left(\frac{h\nu}{k T}\right) - 1}\mathrm{d} \nu $$

$$ \rho_T = \frac{8\pi h}{c^3}\left(\frac{k T}{h}\right)^4 \underbrace{\intop_0^\infty \frac{u^3}{\exp\left(u\right) - 1}\mathrm{d} u}_{\zeta(4)\Gamma(4) = \pi^4/15} = \frac{8\pi^5 k^4}{15 h^3 c^3}T^4 $$

At equilibrium, these modes will have the same energy of the blackbody that is proportional to the fourth power of the temperature. This is the same result predicted by Stefan-Boltzmann law for the radiance of a black body

$$ R_T = \intop_0^\infty R(\nu, T)\mathrm{d}\nu = \sigma T^4, $$

where $R(\nu, T)\mathrm{d}\nu$ is the radiance emitted by frequencies between $\nu$ and $\nu + \mathrm{d}\nu$ and $\sigma$ is the Stefan-Boltzmann constant. My question is how can I find the following relation

$$ R(\nu, T)\mathrm{d}\nu = \frac{c}{4} \rho(\nu, T)\mathrm{d}\nu? $$

 

[Charles Link]

This is an effusion rate formula from chemistry: The number of particles per unit area per unit time that emerge from an aperture is $R=\frac{n \bar{v}}{4}$, where $n$ is the particle density, (number of particles per unit volume), and $\bar{v}$ is the average speed. $\\$ ==========================================================================$\\$ I have also derived this formula during my undergraduate days, and I still remember the derivation, which involves a double integral. Let me write out this derivation momentarily: $\\$ Let $g(v)$ be the speed distribution of the particles that have density $n$. The number of particles $N$ that emerge from an aperture of area $A$ in time $\Delta t$ is given by $N=n \int\limits_{r=0}^{+\infty} \int\limits_{\theta=0}^{\pi/2} \int\limits_{\phi=0}^{2 \pi} r^2 \, \sin{\theta} \, d \theta \, d \phi \, dr \frac{A \cos{\theta}}{4 \pi r^2} \int\limits_{\frac{r}{\Delta t}}^{+\infty} g(v) \, dv$. $\\$ The $\phi$ integral gives $2 \pi$, and the $\sin{\theta} \cos{\theta} \, d \theta$ integral gives $\frac{1}{2}$, so we have $N=\frac{nA}{4} \int\limits_{r=0}^{+\infty} dr \, \int\limits_{v=\frac{r}{\Delta t}}^{+\infty} g(v) \, dv$. $\\$ Reversing the order of integration, it becomes $N=\frac{nA}{4} \int\limits_{v=0}^{+\infty} g(v) \, dv \, \int\limits_{r=0}^{v \, \Delta t} \, dr$ , which gives the result $N=\frac{nA \, \Delta t}{4} \int\limits_{v=0}^{+\infty} g(v) v \, dv=\frac{nA \, \Delta t \, \bar{v}}{4}$. $\\$ Finally, $R=\frac{N}{A \, \Delta t}=\frac{n \bar{v}}{4}$. $\\$ Note: The reason for the $v=\frac{r}{\Delta t }$ term on the lower limit of the velocity distribution is that the particle needs to be moving fast enough to get to the aperture in time $\Delta t$. The derivation basically assumes a collisionless gas. The fraction of particles that are moving in the direction of the aperture is $\frac{A \cos{\theta}}{4 \pi r^2}$. The aperture $A$ is assumed to be small. $\\$ The coordinate system is a spherical coordinate system whose origin is at the aperture. The gas is in the upper hemisphere above the aperture, and the density is uniform, and the speed distribution function, $g(v)$ (normalized to 1), is independent of position. $\\$ ======================================================================== $\\$ The above result is very useful in doing a derivation of the Planck blackbody function, where the cavity is assumed to contain a photon gas of particles all moving at speed $c$, so that $\bar{v}=c$.

 

[Airton Rampim]

This is an effusion rate formula from chemistry: The number of particles per unit area per unit time that emerge from an aperture is $R=\frac{n \bar{v}}{4}$, where $n$ is the particle density, (number of particles per unit volume), and $\bar{v}$ is the average speed. $\\$ ==========================================================================$\\$ I have also derived this formula during my undergraduate days, and I still remember the derivation, which involves a double integral. Let me write out this derivation momentarily: $\\$ Let $g(v)$ be the speed distribution of the particles that have density $n$. The number of particles $N$ that emerge from an aperture of area $A$ in time $\Delta t$ is given by $N=n \int\limits_{r=0}^{+\infty} \int\limits_{\theta=0}^{\pi/2} \int\limits_{\phi=0}^{2 \pi} r^2 \, \sin{\theta} \, d \theta \, d \phi \, dr \frac{A \cos{\theta}}{4 \pi r^2} \int\limits_{\frac{r}{\Delta t}}^{+\infty} g(v) \, dv$. $\\$ The $\phi$ integral gives $2 \pi$, and the $\sin{\theta} \cos{\theta} \, d \theta$ integral gives $\frac{1}{2}$, so we have $N=\frac{nA}{4} \int\limits_{r=0}^{+\infty} dr \, \int\limits_{v=\frac{r}{\Delta t}}^{+\infty} g(v) \, dv$. $\\$ Reversing the order of integration, it becomes $N=\frac{nA}{4} \int\limits_{v=0}^{+\infty} g(v) \, dv \, \int\limits_{r=0}^{v \, \Delta t} \, dr$ , which gives the result $N=\frac{nA \, \Delta t}{4} \int\limits_{v=0}^{+\infty} g(v) v \, dv=\frac{nA \, \Delta t \, \bar{v}}{4}$. $\\$ Finally, $R=\frac{N}{A \, \Delta t}=\frac{n \bar{v}}{4}$. $\\$ Note: The reason for the $v=\frac{r}{\Delta t }$ term on the lower limit of the velocity distribution is that the particle needs to be moving fast enough to get to the aperture in time $\Delta t$. The derivation basically assumes a collisionless gas. The fraction of particles that are moving in the direction of the aperture is $\frac{A \cos{\theta}}{4 \pi r^2}$. The aperture $A$ is assumed to be small. $\\$ The coordinate system is a spherical coordinate system whose origin is at the aperture. The gas is in the upper hemisphere above the aperture, and the density is uniform, and the speed distribution function, $g(v)$ (normalized to 1), is independent of position. $\\$ ======================================================================== $\\$ The above result is very useful in doing a derivation of the Planck blackbody function, where the cavity is assumed to contain a photon gas of particles all moving at speed $c$, so that $\bar{v}=c$.

Thank you very much! It's a really nice demonstration. The only thing I still didn't get it is the fraction of particles being $\frac{A\cos\theta}{4\pi r^2}$. The $cos(\theta)$ term came from the flux of particles emerging from A? Also, why $4\pi r^2$ and not $2\pi r^2$ if the gas is in the upper hemisphere?

 

[Charles Link]

Thank you very much! It's a really nice demonstration. The only thing I still didn't get it is the fraction of particles being $\frac{A\cos\theta}{4\pi r^2}$. The $cos(\theta)$ term came from the flux of particles emerging from A? Also, why $4\pi r^2$ and not $2\pi r^2$ if the gas is in the upper hemisphere?

The calculation is simply computing what goes out the aperture. As viewed from a location at $(r, \theta, \phi)$, the aperture looks like it has area $A \cos{\theta}$. (The circular aperture will look like an oval or ellipse when viewed from angle $\theta$). Meanwhile from that point at $(r,\theta, \phi)$, the particles will emanate uniformly in all directions, and the computation basically computes what fraction passes through the aperture of area $A$, (projected area of $A \cos{\theta}$), that cross the surface at radius $r$ with surface area $4 \pi r^2$. That fraction is $\frac{A \cos{\theta}}{4 \pi r^2}$. $\\$ The contributions of all of the various locations in the entire hemisphere are then summed in an integral.

 

[Airton Rampim]

The calculation is simply computing what goes out the aperture. As viewed from a location at $(r, \theta, \phi)$, the aperture looks like it has area $A \cos{\theta}$. Meanwhile from that point at $(r,\theta, \phi)$, the particles will emanate uniformly in all directions, and the computation basically computes what fraction passes through the aperture of area $A$, (projected area of $A \cos{\theta}$), that cross the surface at radius $r$ with surface area $4 \pi r^2$. That fraction is $\frac{A \cos{\theta}}{4 \pi r^2}$. $\\$ The contributions of all of the various locations in the entire hemisphere are then summed in an integral.

Ok, now I got it. Thank you again!