# Radiance and energy density of a black body

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• Airton Rampim
In summary, the conversation discusses the relation between radiance and energy density in a black body according to Planck's law. It also introduces the Stefan-Boltzmann law and an effusion rate formula from chemistry. The conversation includes a derivation of the effusion rate formula and its application in the Planck blackbody function. The main question asked is how to find the relation between radiance and energy density, specifically the formula R(ν,T)dν=c4ρ(ν,T)dν.
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

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

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 ##.

Last edited:
Airton Rampim and dextercioby
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?

Airton Rampim said:
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.

Last edited:
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!

## 1. What is a black body?

A black body is an idealized object that absorbs all radiation that falls on it. It also emits radiation at all wavelengths, making it a perfect emitter.

## 2. How is the radiance of a black body defined?

The radiance of a black body is defined as the amount of radiation emitted per unit time, per unit solid angle, per unit area of the emitting surface, and per unit wavelength interval.

## 3. What is the relationship between the temperature of a black body and its radiance?

According to Planck's law, the radiance of a black body is directly proportional to the fourth power of its absolute temperature. This means that as the temperature increases, the radiance also increases.

## 4. What is the significance of the Stefan-Boltzmann law in relation to black body radiation?

The Stefan-Boltzmann law states that the total energy radiated by a black body is directly proportional to the fourth power of its absolute temperature. This law is important because it provides a way to calculate the total energy emitted by a black body at a given temperature.

## 5. How does the energy density of a black body vary with wavelength?

The energy density of a black body is highest at shorter wavelengths and decreases as the wavelength increases. This is known as Wien's displacement law and is a direct consequence of Planck's law.

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