Radiance and energy density of a black body

  • #1
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

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

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  • #2
Charles Link
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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 ##.
 
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  • #3
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?
 
  • #4
Charles Link
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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.
 
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  • #5
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!
 
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