If the radiant intensity $I(\phi,\theta)$ of electromagnetic radiation is given in spherical coordinates ($\phi$ the azimuth, $\theta$ the polar angle, indenpendent of $R$) with units $Js^{-1}sr^{-1}$ and I integrate over the azimuth $I(\theta)=\int I(\phi,\theta)\ \text{d}\phi$, in what units is $I(\theta)$ expressed? An example is the "intensity" used in the Rietveld formula for powder diffraction.
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It's ${\rm J}\cdot {\rm s}^{-1}$, you might add "per unit $\cos\theta$" as well, but the latter has no standardized symbol because it is unphysical because the physical dimension of angles and their cosines is "one": they are dimensionless.
After all, this is true even for radians and steradians. You may have added ${\rm sr}^{-1}$ to improve the readability and purpose of the quantity $I(\phi,\theta)$ but from purely dimensional considerations, $1\,{\rm sr}\equiv 1$ so you may omit the powers of ${\rm sr}$ as you like, too.
When we talk about the angular frequency $\omega$, we don't say that the unit is ${\rm rad}\cdot {\rm s}^{-1}$, radians per second, although this is also "more comprehensible" a way to write the unit. We just call the unit ${\rm s}^{-1}$. The same holds for any powers of steradians. There's a natural unit of angle and solid angle, namely radian and steradian, respectively, and they are dimensionally as well as numerically (as long as one uses the natural units for geometry) equal to one.
It's only the dimensionful units – those for which there is no "preferred, culturally independent unit" – for which we must be careful to add the units we use. Angles and solid angles are dimensionless and don't need any units with new names.
answered Nov 21, 2012 at 15:59
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$\begingroup$ I understand that you can just omit $\text{rad}$ and $sr$ but if I would choose to write them anyway, would $I(\theta)$ be expressed in $Js^{-1}\text{rad}^{-1}$? $\endgroup$– WoxNov 21, 2012 at 16:32
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$\begingroup$ No, it's misleading because $d\Omega / d\phi$ isn't $d\theta$ so that it would give you a "unit angle". Instead, $d\Omega / d\phi = \sin\theta\cdot d\theta = d(\cos\theta)$, so as I wrote, the quantity after integrating per $\phi$ is the radiant intensity per unit $\cos\theta$, not per unit angle $\theta$. So as long as you consider $1\,{\rm rad}=1$, you don't spoil anything by adding powers of radians anywhere. But if you want to "improve" the readability of the formula much like by the steradian at the beginning, your addition of the inverse radian is wrong. $\endgroup$ Nov 21, 2012 at 16:37