Scattering Probability and Particle Size in the Atmosphere

[nhmllr]

I heard from a lecture by Walter Lewin that the sky is blue because the chance of a photon of λ wavelength scattering is 1 / λ⁴ and blue's higher wavelength gives it a higher chance of scattering than red, making there more blue in the sky than red. This explanation might be incomplete, but this is not my question.

What partcles exactly can a photon scatter on? Would it scatter on any atoms in the atmosphere, (eg nitrogen, oxygen) or on bigger dust particles or what? How does the size of the particle affect the probability? Obviously, a macroscopic object has an almost 100% chance of scattering light, which is why we see it.

 

[Meir Achuz]

A photon will scatter off of anything.
In addition to the $$1/\lambda^4$$, the scattering cross section is proportional to the 'polarizability' of the scatterer squared, and the polarizability usually is proportional to the volume of the scatterer.

 

[nhmllr]

So it's
Volume / Wavelength^4 ?
Larger atoms have a higher probability of scattering then, right?

 

[Meir Achuz]

Yes, but its volume^2. The light scattering is mainly off of groups of molecules.

 

[nhmllr]

Yes, but its volume^2. The light scattering is mainly off of groups of molecules.

Wait- I'm doing a problem where I need the probability of light scattering on an object with a given volume. So I need a probability a/b. What units for volume and wavelength do I use? What happens with high values of volume where a > b, as probabilities can't be over 100%? I don't think the probability is ever 100%, right?

 

[Meir Achuz]

The cross section is given by $$\sigma=(8\pi/3)\eta^2 k^4$$, where $$\eta$$ is the polarizability, which is proportional to the volume, and k is the wave number.
The cross section has units of area.
The number of photons scattered equals the number per unit area in the incident beam times the cross section.

 

[nhmllr]

The cross section is given by $$\sigma=(8\pi/3)\eta^2 k^4$$, where $$\eta$$ is the polarizability, which is proportional to the volume, and k is the wave number.
The cross section has units of area.
The number of photons scattered equals the number per unit area in the incident beam times the cross section.

Sorry to be replying so late to these

What I don't understand is that that there are no specified units. It seems to me that a photon has a probability of scattering off of something. The probability can be expressed a/b, where if you throw b photons at the object, and average of a will scatter. So is the volume measured in meters³, or cm² or what? By what unit do you measure the wavelength? But if that were the case, then it seems that the probability would exceed 100%, a > b, and more photons would scatter than hit the object! Is there something fundamental that I'm not understanding here? Thanks.

 

[Meir Achuz]

How can a volume be in cm^2? It has to be something^3. If the volume is measured in cm^3 and the wave number in cm^-1, the cross section will be in cm^2, the units of area. The probability of a photon scattering will be this area divided by the area of the incident beam. The probability will always be <<1 because the approximations that led to the formula I gave break down if the beam area is not much larger than the cross section area.

 

[nhmllr]

How can a volume be in cm^2? It has to be something^3. If the volume is measured in cm^3 and the wave number in cm^-1, the cross section will be in cm^2, the units of area. The probability of a photon scattering will be this area divided by the area of the incident beam. The probability will always be <<1 because the approximations that led to the formula I gave break down if the beam area is not much larger than the cross section area.

Sorry -the cm^2 was just a mistake because I was typing quickly.

So the probability = area of cross section/(area of incident beam x [wavelength cm]^4) ? Where do you take the cross section of the 3d object? Do you take it of a sphere of equal volume to the object?

Sorry I'm asking all these questions- I'm not used to this terminology.

 

[Meir Achuz]

Sorry -the cm^2 was just a mistake because I was typing quickly.

So the probability = area of cross section/(area of incident beam x [wavelength cm]^4) ? Where do you take the cross section of the 3d object? Do you take it of a sphere of equal volume to the object?

Where did you get "area of cross section/(area of incident beam x [wavelength cm]^4)"?
The probability is just area/area.
The word 'cross section' is just used because sigma has the units of area.
It has nothing to do any particular shape, but could be interpreted as the cross-sectional area you would see if you looked at the scatterer.
The cross section is calculated by the formula I gave you, but only if the size of the scatterer
is much smaller than the wavelength and the width of the beam.

For the specific case of a dielectric sphere, the polarizability is given by
$$\eta=[(\epsilon-1)/(\epsilon+2)][3V/4\pi].$$.

Actually, the probability is not what is most important. The intensity of the scattered radiation is usually of most interest.

 

[nhmllr]

I'll back up for a second and explain what I'm trying to do here. Above I explained the explanation of the sky being blue (which was extended to why sunsets are red) by Lewin. I wanted to demonstrate that mathematically. I thought of it as the light passing through a lot of atoms, and at each atom has a probability of scattering. I thought of this as probability problem. After about 30 minutes, I came up with a little equation that found the probability of a photons scattering off of b atoms when there are c photons in total. To calculate it, I need the probability of one photon with a certain wavelength scattering off an atom.

I found a way to calculate the number of atoms in one "slice" of the atmosphere, using the total mass of the atmosphere (found on wikipedia) and the mass of the "average" atom. (A weighted average of the average masses of oxygen and nitrogen). Because I wanted to find the probabilities at different times of the day, I needed to know how much length of the atmosphere the sunlight had to penetrate, because when it's lower in the sky it goes through more. That was just a clever trig problem.

The last piece of information I need to know if the probability that a photon with a wavelength will scatter off my average atom with a volume (for which it's easy to find a weighted average of volume). So when you say

"The probability is just area/area."

I'm confused for two reasons. One is that photons don't really have area, and the other is that the wavelength is not included in that explanation.

But now that you mention intensity, not only can I not understand the probability, but I'm worried that I'm doing the problem with a completely silly approach!

Thanks for being patient, I'm in way over my head.

 

[Meir Achuz]

What may help you is the probability of scattering of a photon passing through a density
$$\rho$$ of scatterers, that is $$\rho$$=the number of scatterers per cm^3.
This is $$dP=\rho\sigma dx$$, whee dP/dx is the probability of scattering per cm.
This is usually <<1. If not small, then dP must be integrated to give
$$P=1-exp[-\rho\sigma x]$$.