Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Now, I found three different definitions of Planck's law:

$$ P_1(\nu,T) = \frac{8 \pi}{c}\frac{h \nu^{3}}{c^2} \frac{1}{e^{h\nu/kT}-1} $$ $$ P_2(\nu,T) = 2\frac{h \nu^{3}}{c^2} \frac{1}{e^{h\nu/kT}-1} $$ $$ P_3(\nu,T) = \frac{h \nu^{3}}{c^2} \frac{1}{e^{h\nu/kT}-1} $$

Which of these is correct and will give me the radiated energy for a given temperature and a given frequency?

share|cite|improve this question
up vote 5 down vote accepted

Your second equation, $P(\nu,T) = \frac{2 h {\nu}^3}{c^2}$ $\frac{1}{\exp\bigl(\frac{h \nu}{kT}\bigr) - 1}$ is what is commonly referred to as Planck's law for radiation, although a more standard symbol used is $B_\nu(T)$. This is the energy radiated per time, per area, per frequency interval, per steradian. It is a formula for the 'specific intensity' of a source, which intuitively is the energy flux along a ray of radiation in a given direction, and so you must normalize by the solid angle subtended by that ray.

To get the total energy per time per area radiated by a patch of a black body, integrate over solid angle and over frequency. Be careful performing the solid angle integral, however, because you must include the geometric factor $\cos \theta$ that accounts for the projected area of the patch ($\theta = 0$ corresponds to a ray emitted in the normal direction). Rays leaving one side of a patch can only be directed into the upper hemisphere of the solid angle sphere. So the solid angle integral looks like this:

$$ F_\nu = 2 \pi \int_0^{\pi/2} B_\nu (\theta)\, \cos\theta \, \sin \theta \, d \theta$$

The $2 \pi$ out in front is for the azimuthal angle. Here, $F_\nu$ is what is commonly referred to as the specific flux ('specific' because it's still per unit frequency interval). Then, either by reading up on the Riemann $\zeta$ function, or just using a computer to tell you the answer, you can perform the frequency integral and get

$$ F = \sigma \, T^4$$

Here $F$ is what we commonly think of as the flux (energy per area per time), and $\sigma$ is the Stefan-Boltzmann constant,

$$\sigma \equiv \frac{2 \pi^5 \, k_\mathrm{B}^4}{15 \, h^3 \, c^2}$$

share|cite|improve this answer

This is easy to check. The total power frequency over the whole frequency range should match the Stephan-Boltzman law $I=\sigma T^4$ with $$\sigma=\frac{2\pi^5 k^4}{15c^2h^3}$$.

Assuming that you are asking for spectral density of radiation power in unit interval of linear frequency, we can express $I = \int P(\nu,T) d\nu$.

Since $$\int \frac{\nu^3 \, d \nu}{e^{h\nu/kT}-1}=\left (\frac{kT}{h}\right)^4 \frac{\pi^4}{15}$$ (can use any integrator, e.g., Wolfram Alpha) the correct formula is

$$P(\nu,T) =\frac{2 h \pi}{c^2} \frac{\nu^3 \, d \nu}{e^{h\nu/kT}-1}$$

Looks like none of your sources is correct. If you are looking for the radiation per steradian, see the answer by kleingodron [and it is not just a full solid angle factor $1/(4 \pi)$].

share|cite|improve this answer
The original poster asked about Planck's law, which refers to energy radiated per time per area per frequency interval per solid angle. In that case the second formula is correct. I've included more details in my answer – kleingordon Mar 10 '12 at 4:39
Ah, good catch there. (Though it wouldn't surprise me much if what the OP means by Planck's law is not exactly the standard meaning.) – David Z Mar 10 '12 at 5:00
That's a catch indeed! @kleingordon answer is definitely more useful although I must say that it is not clear from the question what normalization OP implies. – Slaviks Mar 10 '12 at 5:50

I dont think you have clearly looked at all the formulae, please do confirm your source

There are different formulae depending on the variable involved, that is which are you considering as the variable

See this for a clearer understanding

share|cite|improve this answer
I'm sorry for the downvote, but this is simply not a useful answer (a mere comment, at best). – bitmask Feb 2 '12 at 0:10
no offence taken – Jack Feb 3 '12 at 8:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.