Wikipedia's Mie scattering Mathematics discusses the scattering amplitudes $S_1(\theta), \ S_2(\theta)$ for each outgoing polarization of an incident EM plane wave on a uniform sphere.

It defines the scattered intensities as

$$i_1(\theta) = |S_1(\theta)|^2$$

$$i_2(\theta) = |S_2(\theta)|^2$$

and then defines the scattering intensity

$$I(\theta) = I_0 \frac{\lambda^2}{8 \pi^2 r^2} \left(i_1(\theta) + i_2(\theta)\right)$$

where r is the radius of the sphere.

The incoming intensity $I_0$ would likely have units of power per unit area, whereas I'd expect the scattered intensity or radiance $I(\theta)$ to have units of power per unit solid angle, but if that were true the units don't work.

Could $I_0$ possibly be the power incident on the geometrical cross-section $\pi r^2$?

I'm doing an approximate calculation of something similar to what's described in this question. I've found a python script that generates the scattering amplitudes $S_1(\theta), \ S_2(\theta)$, I'll have a laser pointer with $I$ watts/m^2 and a detector at some angle $\theta$ with solid angle $\Delta \Omega$ and I'd like to calculate the scattered intensity reaching that detector for a given single particle, but I'm stuck on how to correctly convert from $S$ to scattered power into $\Delta \Omega$.

The answer there says:

Wikipedia just pointed me to an English translation of the original paper by Mie, which I didn't know existed. I haven't yet had a chance to read it, so I don't know how useful it is.

but I'm not able to access that link.

  • 1
    $\begingroup$ Are the amplitudes $S$ dimensionless? $\endgroup$ – rob Jan 31 at 15:25
  • $\begingroup$ @rob I checked again in Wikipedia, they certainly look dimensionless to me. $\endgroup$ – uhoh Jan 31 at 15:31
  • $\begingroup$ Your point about expecting $I(\theta)$ to integrate to $I_0$ is a good one. Two remarks: (1) Some people learn to edit Wikipedia before they learn to properly deal with differential cross-sections. (2) The solid angle is dimensionless, so letting $I$ and $I_0$ have the same unit (as they do in your expression) isn't fatal. $\endgroup$ – rob Jan 31 at 15:36
  • $\begingroup$ @rob the only $I_0$ I can think of is power per unit area (e.g. 5mW laser, 1 mm^2 beam) since it's (essentially) a plane wave with zero solid angle. $\endgroup$ – uhoh Jan 31 at 15:37
  • $\begingroup$ @rob It's possible there's an answer here, but it's paywalled and libraries are now closed here (for a week at least!) due to Chinese New Year: osapublishing.org/ao/abstract.cfm?uri=ao-19-9-1505 $\endgroup$ – uhoh Feb 1 at 2:58

I think the intensity here in the radiometry context is called irradiance and it is power per unit area. When your detector is far enough from the scatter, power per unit area is also power per unit area per solid angle (radiance). To calculate ΔΩ, we can integrate intensity over a solid angle. There might be a cos factor, depending on whether your detector is perpendicular to the scattered light.

  • $\begingroup$ That's what I meant by "The incoming intensity $I_0$ would likely have units of power per unit area..." While the equation suggests that $I(\theta)$ has the same units as $I_0$, it doesn't make sense to me because there is no distance involved. You can't produce 1 watt/m^2 of intensity from scattering on a screen independent of distance from the scatterer to the screen. If it's possible to add some more to your answer to help me understand how the units work here that would be great. There is also the possibility that there's an omission in the original Wikipedia article. $\endgroup$ – uhoh Feb 19 at 6:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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