# I Black body radiation, Planck's law, derivation

#### Max Loo Pin Mok

The following are 3 equations of Planck's law or Planck's distribution function. Are they all correct? How do they derive from each other?

Equation One:
From page 512 of http://metronu.ulb.ac.be/npauly/art_2014_2015/shockley_1961.pdf
We denote by Qs the number of quanta of frequency greater than vg incident per unit area per unit time for black-body radiation of temperature Ts. For later purposes we shall also introduce the symbol Qs(vg, Ts) in order to be able to represent situations for different values of the limiting frequency. Equation Two:
From page 1 of https://edisciplinas.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank.pdf
Energy density of radiation per unit frequency interval u(v) for black-body radiation is described by the Planck formula: Equation Three:
From https://en.wikipedia.org/wiki/Planck's_law
The spectral radiance of a body, Bv, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. The SI unit of spectral radiance in frequency is the watt per steradian per square metre per hertz (W·sr-1·m-2·Hz-1). #### Attachments

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• dextercioby and Charles Link
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Edit: Give this one an A+ for a very good question. This is one that requires a careful look at a couple of quantities. $\\$See https://www.physicsforums.com/threads/radiance-and-energy-density-of-a-black-body.956343/#post-6063569 $\\$ See also: https://www.physicsforums.com/threads/trying-to-understand-my-profs-derivation-of-plancks-law.958747/#post-6079472 $\\$ Both of these previous discussions might help answer your question. $\\$ I haven't checked every factor of $2$ and $\pi$ in the 3 equations you posted, but yes, these 3 quantities are all connected. The first "link" shows how to get from the second relation to the 3rd, with a factor of $\frac{c}{4}$. $\\$ Going from the second relation to the 3rd, it also gets divided by $\pi$. That is because a blackbody radiates as $I(\theta, \phi)=I_o \cos{\theta}$, where $I$ is the intensity in watts/steradian, and $I_o$ is the on-axis intensity. (This is because an ideal blackbody looks equally bright when viewed from an angle, but the observed area is $A \cos{\theta}$, e.g. when a circular area is observed at an angle it appears elliptical with the area reduced by the factor of $\cos{\theta}$ ). $\\$ The radiated power over a hemisphere is $P=\int\limits_{0}^{2 \pi} \int\limits_{0}^{\frac{\pi}{2}} I(\theta, \phi) \sin{\theta} \, d \theta \, d \phi=I_o \pi$. $\\$ (Notice for a blackbody radiating over a hemisphere, because of the $\cos{\theta}$ factor, the effective solid angle over which the power $P$ is radiated is $\pi$ steradians rather than $2 \pi$ steradians. This is shown by the equation $P=I_o \pi$ ). $\\$ Meanwhile the brightness $B$, which also is often indicated with the letter $L$, satisfies $P=I_o \pi=LA \pi$, so that $L=B=\frac{P}{A \pi}$. Thereby the division by $\pi$ along with the $\frac{c}{4}$ factor in going from the 2nd to the 3rd expression. $\\$ ........................................................................................................................................................................ $\\$ Another useful quantity, designated by $M$ is the power radiated per unit area. $M=L \pi =B \pi$. $\\$ $\int\limits_{0}^{+\infty} M_{\nu}(\nu, T) \, d \nu=\sigma T^4$, which is the Stefan-Boltzmann Law. $\\$ To get $M_{\nu}(\nu,T)$ from the $u_{\nu}(\nu,T)$ of the second expression, you only need to multiply by the above factor of $\frac{c}{4}$ that is derived in the first "link". $\\$ It can be shown that $\sigma= \frac{\pi^2}{60} \frac{k_b^4}{\hbar^3 c^2}$, because the $\int\limits_{0}^{+\infty} M_{\nu}(\nu,T) \, d \nu$ integral, although a difficult one, can be evaluated in closed form. $\\$ From this, you should be able to see that $\int\limits_{0}^{+\infty} B_{\nu}(\nu,T) \, d \nu=\frac{\sigma T^4}{\pi}$. $\\$ ........................................................................................................................................................................ $\\$ To get the first expression, you basically take $M_{\nu}(\nu, T)$ and divide by the photon energy $h \nu$ to get the number of photons $Q_{\nu}(\nu, T)$ that are radiated. The reason for the limits that they chose for this integral should be quite obvious. $\\$ ...................................................................................................................................................................... $\\$ Additional comment: Most textbooks are not complete enough in their treatment of the Planck function to address the subject of this post. They may give $u_{\nu}(\nu,T)$, or they may give $B_{\nu}(\nu,T)$ or $M_{\nu}(\nu, T)$, but most of them do not properly tie these all together.

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• Max Loo Pin Mok and dextercioby

#### Max Loo Pin Mok

Thanks for your reply. I'll work on this again after I've covered some other areas.

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