# Raman: Why does only some light scatter inelastically?

1. Jun 12, 2012

### perels

Hi,

I have been involved in a Raman Scattering project and have grasped the fundamentals of Raman Scattering. I'm no physics student, but rather a business student.

1. Can somebody explain to me why only a few protons scatter inelastically and not all of them?
2. Is there a method for defining how many protons will scatter inelastically, i.e. can it be predicted?
3. Any suggested reading material on this subject?

Hoping for some help on at least some of the questions :-)

Cheers
Per

2. Jun 13, 2012

### mathman

3. Jun 13, 2012

### perels

Hi,

Thanks for the link - I will look into this - usually I try to avoid wikipedia as the information on this is not verified.

Yes, I mean photons, not protons - thanks for the correction :)

4. Jun 14, 2012

### Zarqon

Actually the information on wikipedia is very accurate. Studies show (think one was published in nature even) that wikipedia has about as few errors as Encyclopædia Britannica.

Also, on topic, the reason why not all photons scatter is easily answered, it's because this process is probabilistic, as most processes at that quantum level are. One can compute the probability for a scattering event to occur however, but depending on the material they scatter from, this may be a very involved calculation.

5. Aug 11, 2012

### aoner

We can describe the distribution of these energy states with the Boltzmann Distribution with:

$\frac{{{N}_{i}}}{N}=\frac{{{exp}({\frac{{{\epsilon }_{i}}}{{{k}_{B}}T}}})}{Z(T)}\$

Where the partition function Z(T) is defined as:

$Z(T)=\sum{{{e}^{-\frac{{{\varepsilon }_{i}}}{{{k}_{B}}T}}}}$

This means that there are more electrons in the ground state, thus more Stokes-shifted photons. It also means that a smaller fraction of the molecules are not in the ground state, but in vibrational excited states. Scattered photons from these molecules have higher energy compared to the incident photons. This is called anti-Stokes scattering. This blue shifted scattering is always weaker, due to the Boltzmann distribution, than the Stokes-shifted scattering.