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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.

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    $\begingroup$ Are the amplitudes $S$ dimensionless? $\endgroup$
    – rob
    Jan 31, 2019 at 15:25
  • $\begingroup$ @rob I checked again in Wikipedia, they certainly look dimensionless to me. $\endgroup$
    – uhoh
    Jan 31, 2019 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, 2019 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, 2019 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, 2019 at 2:58

2 Answers 2

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For a mathematical answer try looking into Bohren&Huffman, "Absorption and scattering of light by small particles", §4.4.4 "Scattering Matrix". The relation between incident and scattered field amplitudes is given there as

\begin{equation}\begin{pmatrix}E_{||s} \\ E_{\perp s}\end{pmatrix}=\frac{e^{i\mathrm{k}(r-z)}}{-i\mathrm{k}r}\begin{pmatrix}S_2 & 0 \\ 0 & S_1\end{pmatrix}\begin{pmatrix}E_{||i} \\ E_{\perp i}\end{pmatrix}.\tag{4.75}\end{equation}

The $S_1$ and $S_2$ don't seem to have the same normalization as in miepython though, so you'll have to check. In particular, you need the documentation for the relevant functions:

def mie_S1_S2(m, x, mu):
    """
    Calculate the scattering amplitude functions for spheres.
    The amplitude functions have been normalized so that when integrated
    over all 4*pi solid angles, the integral will be qext*pi*x**2.
    The units are weird, sr**(-0.5)
    Args:
        m: the complex index of refraction of the sphere
        x: the size parameter of the sphere
        mu: cos(theta) or array of angles [cos(theta_i)]
    Returns:
        S1, S2: the scattering amplitudes at each angle mu [sr**(-0.5)]
    """

This is not quite clear, because the amplitudes are complex-valued, and one is not expected to integrate them directly. But this function is used in i_per and i_par functions, which are described as follows:

def i_per(m, x, mu):
    """
    Return the scattered intensity in a plane normal to the incident light.
    This is the scattered intensity in a plane that is perpendicular to the
    field of the incident plane wave. The intensity is normalized such
    that the integral of the unpolarized intensity over 4π steradians
    is equal to the single scattering albedo.
    Args:
       m: the complex index of refraction of the sphere
       x: the size parameter of the sphere
       mu: the angles, cos(theta), to calculate intensities
    Returns
       The intensity at each angle in the array mu.  Units [1/sr]
    """
    s1, _ = mie_S1_S2(m, x, mu)
    intensity = np.abs(s1)**2
    return intensity.astype('float')

This makes more sense: if you take $|S_1|^2$ or $|S_2|^2,$ as given by mie_S1_S2, you will get the intensity in $\mathrm{sr}^{-1}$, normalized to single scattering albedo. This doesn't seem to be the same normalization as you'll get by using the Bohren&Huffman expression.

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  • $\begingroup$ Aha! That's the answer! If you can add that to the answer post (presumably near the beginning rather than the end) I can accept right away. $\endgroup$
    – uhoh
    Jan 10, 2022 at 22:50
  • $\begingroup$ Looks great; thanks! $\endgroup$
    – uhoh
    Jan 11, 2022 at 0:52
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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.

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  • $\begingroup$ 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, 2019 at 6:05

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