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Measures of the intensity of electromagnetic radiation |
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| Nov26-12, 03:47 AM | #1 |
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Measures of the intensity of electromagnetic radiation
I've always struggled with the commonly used measures of the intensity of electromagnetic radiation and it's catching up to me lately. Suppose [itex]\bar{P}(R,\phi,\theta)[/itex] is the Poynting vector of an electromagnetic field (in spherical coordinates) with norm [itex]I(R,\phi,\theta)=\|\bar{P}(R,\phi,\theta)\|[/itex] in [itex]J.m^{-2}.s^{-1}[/itex] (strictly speaking it should be the ensemble average of the norm). Then the flux [itex]\Phi[/itex] through a surface [itex]S[/itex], for which [itex]\Psi[/itex] is the angle between Poynting vector and surface normal at [itex](R,\phi,\theta)[/itex], is given by (in [itex]J.s^{-1}[/itex])
$$\begin{align}\Phi_{S}&=\iint_{S} I(R,\phi,\theta)\cos\Psi\ \text{d}A\\ &=\iint_{S} I(R,\phi,\theta)\cos\Psi\ R^{2}\ \text{d}\Omega\\ &=\iint_{S} I(R,\phi,\theta)\cos\Psi\ R^{2}\sin\theta\ \text{d}\phi\text{d}\theta \end{align}$$ In radiometry, this is called the radiant flux. From this measure of intensity, several other measures are derived, but I'm a bit puzzled how. Take for example the radiant intensity [itex]I_{e}[/itex]: the power per solid angle (in [itex]J.sr^{-1}.s^{-1}[/itex]). From the equations above I understand that $$I_{e}=\frac{\text{d}\Phi_{S}}{\text{d}\Omega}=I(R,\phi,\theta)\cos\Ps i\ R^{2}$$ but how is this independent from surface [itex]S[/itex] (which is the point of using other measures than the radiant flux)? Secondly, while you're at it, why not using $$J=\frac{\text{d}^{2}\Phi_{S}}{\text{d}\phi \text{d}\theta}=I(R,\phi,\theta)\cos\Psi\ R^{2}\sin\theta$$ which I also don't see to be independent of [itex]S[/itex], but at least you don't need to remember the [itex]\sin\theta[/itex] when integrating it. Then there are measures like the radiance in [itex]J.m^{-2}.sr^{-1}.s^{-1}[/itex] which is defined as $$L_{e}=\frac{\text{d}^{2}\Phi_{S}}{\cos\Psi\text{d}A \text{d}\Omega}$$ I can write that down but I can't say I understand what it means. Can anyone shed some light on these issues? |
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