Intensity of stokes and anti-stokes lines?

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Discussion Overview

The discussion revolves around the intensity of Stokes and anti-Stokes lines in the context of the Raman effect, exploring the relationship between wavelength and intensity, as well as the underlying mechanisms influencing these phenomena.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the intensity of Raman lines is directly proportional to the fourth power of the wavelength, questioning why Stokes lines, with higher wavelengths, are more intense than anti-Stokes lines.
  • Another participant suggests that the initial question implies the answer, asserting that higher wavelengths correspond to greater intensity.
  • A different participant notes that anti-Stokes lines are near absorption bands, leading to lower gain and thus lower intensity compared to Stokes lines.
  • One participant corrects the earlier claim, stating that intensity is inversely proportional to the fourth power of wavelength, introducing potential confusion.
  • A participant questions whether the lambda^-4 dependence is characteristic of Rayleigh scattering, indicating a possible misunderstanding of the principles involved.
  • Another participant explains that Raman anti-Stokes transitions require a transition from an excited vibrational level to the ground level, while Stokes transitions involve the opposite, emphasizing the role of the Boltzmann distribution in determining the populations of these levels.
  • A subsequent contribution adds that vibrational eigenmodes are bosons and their distribution is governed by the Bose occupation number, particularly at low temperatures, which affects the populations of different levels.
  • A later reply confirms that the explanation provided answers the initial question.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between intensity and wavelength, with some asserting direct proportionality and others suggesting inverse relationships. The discussion includes multiple competing explanations regarding the mechanisms behind Stokes and anti-Stokes line intensities, indicating that consensus has not been reached.

Contextual Notes

There are unresolved aspects regarding the dependence of intensity on wavelength and the specific conditions under which Stokes and anti-Stokes transitions occur. The discussion also reflects varying interpretations of the Raman effect and related scattering phenomena.

itari1985
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According to Raman effect, the intensity is directly proportional to the 4th power of the wavelength. Then how come stokes lines, which have higher wavelengths than anti-stokes lines, are more intense than the anti-stokes lines?
 
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uh? I think you answered your own question. If I is propotional to lambda to power 4 and the stokes line has higher wavelengths of course they're more intense then.
 
If I remember well, the anti-stokes line is near absorption bands. Its gain is too low there, so it is less intense than stokes line.
 
Sorry intensity is inversely proportional to the 4th power of wavelength.:biggrin:
 
Isn't the lambda^-4 dependence characteristic of Rayleigh scattering?

Claude.
 
To have a Raman anti-stoke diffusion, we need to have a transition of an atom initially in a excited vibrationnal level to the ground level (if we forget the intermediate virtual state). The stoke diffusion, instead, relies on a transition from the ground level to an excited vibrationnal level.

At thermal equilibrium, the populations of the different levels follows the Boltzmann distribution. So the population of the ground level is higher that the excited level. So the stokes transition is much more probable that the anti-stokes transition.

Does it answer to the question ?

Barth
 
To add a touch to Claude's and Barth's posts : vibrational eigenmodes (or phonons in solids) are bosons. Their distribution among different levels is given by the Bose occupation number. At low temperatures, the different levels have significantly different populations.
 
Last edited:
yes it does answer my question. Thanx
 

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