# Is there a $\pi$ in Planck's Radiation Law?

In Sear and Zemansky's University Physics with Modern Physics 13th edition and other sources like this post, it says the formula for Planck's radiation is given by:

$$I(\lambda)=\frac{2\pi hc^{2}}{\lambda^{5}\left(e^{hc/\lambda kt}-1\right)}$$

But other sources, most notably wikipedia say otherwise:

$$B_{\lambda}\left(\lambda, T\right)=\frac{2hc^{2}}{\lambda^{5}}\frac{1}{e^{hc/(\lambda k_{B}T)}-1}$$

Even moreso, other forums are saying there should be a $$-1$$ factor and other things. Why is there all these discrepancies? The same missing $$\pi$$ factor is also the case with Rayleigh-Jeans Law

• Wikipedia correctly cites that formula as the one appropriate for specific intensity or spectral radiance (same thing). The answer you quote incorrectly identifies the formula as spectral radiance. I don't know what Sears and Zemansky define $I_{\lambda}$ as (in words) - could you add this? – Rob Jeffries Aug 11 '19 at 7:48
• It calls it spectral emittance see G smith's answer below or this link – sab hoque Aug 11 '19 at 8:05
• In which case, both are correct. – Rob Jeffries Aug 11 '19 at 12:28
• yep they are both correct. It was I that did an incorrect assumption that spectral radiance and spectral exitance/emittance is the same thing – sab hoque Aug 11 '19 at 12:53

$$B_\lambda$$ and $$I_\lambda$$ are different but related physical quantities.

$$B_\lambda$$ is spectral radiance. It is the power per unit projected area per unit wavelength per unit solid angle radiated by the blackbody. “Projected” means that the area is to be measured perpendicularly to the direction of observation.

$$I_\lambda$$ is spectral exitance. It is the power per unit area per unit wavelength radiated by the blackbody.

The relationship between them is thus that the exitance is the radiance, times the cosine of the angle between the radiating surface and the direction of emission (to account for the projected area), integrated over all possible solid angles into which the radiation can be emitted. This is the hemisphere outside the radiating surface. So

$$I_\lambda=\int_h B_\lambda \cos\theta\,d\Omega,$$

where $$h$$ is a hemisphere.

As you can see, $$B_\lambda$$ is independent of the angle of emission, so this becomes

$$I_\lambda=B_\lambda\int_h\cos\theta\,d\Omega$$

If you work out this elementary integral, you will find that it is equal to $$\pi$$, explaining the “discrepancy” that you noticed. So

$$I_\lambda=\pi B_\lambda.$$

To summarize: $$I_\lambda$$ integrates over solid angles, $$B_\lambda$$ does not. It is a subtle but important difference.

You can also express these quantities in terms of other properties of the radiation, such as its frequency $$\nu$$ rather than its wavelength $$\lambda$$. The usual notation is $$B_\nu$$ and $$I_\nu$$. The relationship is

$$B_\nu\,d\nu=B_\lambda\,d\lambda$$

and similarly for $$I$$.

There is no good reason to put in a factor of -1. I suppose some people do this because if these quantities are expressed in terms of frequency rather than wavelength, the spectrum “goes the other way”.

• That makes a lot more sense, thank you – sab hoque Aug 11 '19 at 5:46
• Note: There is a lot of inconsistency in the symbols used for various radiant quantities. You won’t always see $B$ for radiance and $I$ for exitance. Even various Wikipedia articles do not use consistent notation. – G. Smith Aug 11 '19 at 6:00