Notices
Results 1 to 20 of 20

Thread: Rayleigh vs Raman scattering

  1. #1 Rayleigh vs Raman scattering 
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    Apparently Rayleigh scattering dominates Raman scattering by several orders of magnitude. Can someone explain why this is? Is it a probabilistic thing or something to do with selection rules?


    Reply With Quote  
     

  2.  
     

  3. #2  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    Apparently Rayleigh scattering dominates Raman scattering by several orders of magnitude. Can someone explain why this is? Is it a probabilistic thing or something to do with selection rules?
    Golly, I can't say I know the proper answer to that. Speaking purely ex ano, (but trying to follow your train of thought) I am tempted to think it may be something to do with a Raman process being represented as two electric dipole processes - which is why the selection rules are different from those for simple electric dipole absorption and emission.

    But surely, as Rayleigh scattering is so strongly depenent on frequency (4th power), what you say will only be true at one end of the spectrum, won't it?

    Maybe we need a real physicist here.........


    Reply With Quote  
     

  4. #3  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    Quote Originally Posted by exchemist View Post
    I am tempted to think it may be something to do with a Raman process being represented as two electric dipole processes - which is why the selection rules are different from those for simple electric dipole absorption and emission.
    Two processes? Do you mean the frequency of the light and vibrational frequency of the molecule?

    Quote Originally Posted by exchemist View Post
    But surely, as Rayleigh scattering is so strongly depenent on frequency (4th power), what you say will only be true at one end of the spectrum, won't it?
    Interesting. But the Rayleigh scattering dependence on frequency only involves the intensity of scattered light doesn't it? Does it say it anything about probability of scattering? Actually, since you raised it I've a question about intensity: when we talk of intensity in scattering are we talking radiant or luminous intensity?

    I don't think this is true though, from what I've read so far the wavelength of Raman scattered light is independent of incoming wavelength - though I think intensity is wavelength dependent.

    Quote Originally Posted by exchemist View Post
    Maybe we need a real physicist here.........
    Haha... I appreciate all answers, even if they lead to more questions.
    Reply With Quote  
     

  5. #4  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    Quote Originally Posted by exchemist View Post
    I am tempted to think it may be something to do with a Raman process being represented as two electric dipole processes - which is why the selection rules are different from those for simple electric dipole absorption and emission.
    Two processes? Do you mean the frequency of the light and vibrational frequency of the molecule?

    Quote Originally Posted by exchemist View Post
    But surely, as Rayleigh scattering is so strongly depenent on frequency (4th power), what you say will only be true at one end of the spectrum, won't it?
    Interesting. But the Rayleigh scattering dependence on frequency only involves the intensity of scattered light doesn't it? Does it say it anything about probability of scattering? Actually, since you raised it I've a question about intensity: when we talk of intensity in scattering are we talking radiant or luminous intensity?

    I don't think this is true though, from what I've read so far the wavelength of Raman scattered light is independent of incoming wavelength - though I think intensity is wavelength dependent.

    Quote Originally Posted by exchemist View Post
    Maybe we need a real physicist here.........
    Haha... I appreciate all answers, even if they lead to more questions.
    I stress I am going on old and imperfect memory, but my recollection is that a Raman process can be modelled quantum-mechanically, i.e. is equivalent to, an absorption of a photon at one frequency and a simultaneous emission of another at a different one (Stokes if involving energy loss and anti-Stokes if energy gain), both being electric dipole processes. I can imagine the probability of a Raman process may possibly be the product of the probabilities of the two component processes in some way. But I am only speculating and a real physicist may come along and tell me this is quite wrong.

    I've just had a look at the Wiki article on Rayleigh scattering which says this - like Raman - is dependent on the polarisability of the molecule and both processes vary with 4th power of frequency, so it looks as if my remarks about that are a red herring.
    Reply With Quote  
     

  6. #5  
    KJW
    KJW is offline
    Forum Professor
    Join Date
    Jun 2013
    Posts
    1,346
    Quote Originally Posted by epimetheus View Post
    Apparently Rayleigh scattering dominates Raman scattering by several orders of magnitude. Can someone explain why this is?
    The following is somewhat of a guess, but I think it's because Rayleigh scattering is elastic whereas Raman scattering is inelastic. In addition to the scattering of photons, Raman scattering also changes the state of the molecules, resulting in a change in the energy of the scattered photons. The probability of a given photon producing a change in the state of the molecule depends on the change in the polarisability of the molecule when the molecule changes state. If the change in the polarisability of the molecule is small, then the intensity of the Raman scattering will be small. It should be noted that only particular vibration modes are Raman-active, and if these are absent due to the symmetries of the molecule, then there will be no Raman scattering at all.
    There are no paradoxes in relativity, just people's misunderstandings of it.
    Reply With Quote  
     

  7. #6  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    OK, so let me make sure I understand polarisability since you both mention it. A molecule will typically have a charge distribution - a bit negative here, more positive there - this configuration is its dipole moment which we represent as a vector? This dipole can change because the molecule vibrates - but this is not its polarisability (though important for Raman scattering). The degree to which the dipole changes due to an external field is its polarisability.

    Assuming that's correct, I can see why a molecule which does not change its polarisability (i.e. its derivative is zero) is not Raman active (so an isolated atom can never be Raman active?). But a molecule which is Raman active is also Rayleigh active. Based on what you guys are saying then elastic scattering is more common than inelastic and that has something to do with selection rules?
    Reply With Quote  
     

  8. #7  
    KJW
    KJW is offline
    Forum Professor
    Join Date
    Jun 2013
    Posts
    1,346
    Quote Originally Posted by epimetheus View Post
    Based on what you guys are saying then elastic scattering is more common than inelastic and that has something to do with selection rules?
    Of the total number of scattered photons, only a small proportion will be inelastic (Raman), the vast majority remainder will be elastic (Rayleigh). Although selection rules need to be satisfied, even if they are satisfied, the probability of inelastic scattering still depends on the magnitude of the change in polarisability.
    There are no paradoxes in relativity, just people's misunderstandings of it.
    Reply With Quote  
     

  9. #8  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    So molecules with very high polarisability will Raman scatter more photons? I thought that the polarisability of the molecule only effected the magnitude of shifted wavelength of the photon if it inelatically scattered at all, not the chance of it happening. Is there an equation to show how the probability of Raman and Rayleigh scattering is effected by polarisability?
    Reply With Quote  
     

  10. #9  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    So molecules with very high polarisability will Raman scatter more photons? I thought that the polarisability of the molecule only effected the magnitude of shifted wavelength of the photon if it inelatically scattered at all, not the chance of it happening. Is there an equation to show how the probability of Raman and Rayleigh scattering is effected by polarisability?
    I should think that logically there ought to be, seeing as if there is no change in polarisability due to the transition it will not occur at all. But I'm afraid I do not know what it would be. I see the Wiki article says the intensity is proportional to the change in polarisability:

    "For a molecule to exhibit a Raman effect, there must be a change in its electric dipole-electric dipole polarizability with respect to the vibrational coordinate corresponding to the rovibronic state. The intensity of the Raman scattering is proportional to this polarizability change. Therefore, the Raman spectrum, scattering intensity as a function of the frequency shifts, depends on the rovibronic states of the molecule."

    For Rayleigh scattering, Wiki has a formula (α is the polarisability of the molecule) :

    The expression above can also be written in terms of individual molecules by expressing the dependence on refractive index in terms of the molecular polarizability α, proportional to the dipole moment induced by the electric field of the light. In this case, the Rayleigh scattering intensity for a single particle is given in CGS-units by[10]




    Note: It would appear important to note that while Rayleigh scattering depends on the polarisability, Raman scattering depends on the difference in polarisability between the two states involved in the excitation (or de-excitation). This would be expected to be a lot less than the polarisability itself, which I think is KJW's point.

    .......if I've understood all this after a 40 year gap........



    Further note after more digging:
    I came across an expression for the Raman transition probability which is just a standard type of QM integral or matrix element: <ψn l α l ψm> . Given that different states have orthogonal wavefunctions, this implies that if α is a constant then the integral vanishes.
    Last edited by exchemist; February 6th, 2018 at 05:30 AM.
    Reply With Quote  
     

  11. #10  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    Bear with me: I'm slow on the uptake so appreciate your patience.

    OK. I can see by that equation that the greater the polarisibility the greater the intensity (which if I'm reading it right means the least scattering occurs at right angles, didn't know that). Does the exact same equation hold for Raman scattering? Can it be if there is a change in wavelength in a scattered photon then that would change the intensity compared to a elactically scattered photon wouldn't it...


    I came across this equation which is said to be a derived from classical physics in a text book relating the induced dipole, P, the electric field E_0cos(...) and polarisibility:



    The first term relates to Rayleigh scattering, while the two later terms refer to Stokes and anti-Stokes Raman scattering. I can see here how the dipole depends on nu_1 which is said to be the vibrational frequency of the molecule, and nu_0, the frequency of the light. I can also see how there must be a change in polarisability is necessary for Raman to be active as if then the two Raman terms disappear. But it doesn't seem to suggest anything about how often Rayleigh scattering occurs compared to Raman. I suspect a quantum model may be needed for this?


    Quote Originally Posted by exchemist View Post
    Further note after more digging:
    I came across an expression for the Raman transition probability which is just a standard type of QM integral or matrix element: <ψn l α l ψm> . Given that different states have orthogonal wavefunctions, this implies that if α is a constant then the integral vanishes.
    That is interesting, though I don't understand it yet. Can you give a link/ref to where you came across this?
    Reply With Quote  
     

  12. #11  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    Whoops, double post.
    Reply With Quote  
     

  13. #12  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    Bear with me: I'm slow on the uptake so appreciate your patience.

    OK. I can see by that equation that the greater the polarisibility the greater the intensity (which if I'm reading it right means the least scattering occurs at right angles, didn't know that). Does the exact same equation hold for Raman scattering? Can it be if there is a change in wavelength in a scattered photon then that would change the intensity compared to a elactically scattered photon wouldn't it...


    I came across this equation which is said to be a derived from classical physics in a text book relating the induced dipole, P, the electric field E_0cos(...) and polarisibility:



    The first term relates to Rayleigh scattering, while the two later terms refer to Stokes and anti-Stokes Raman scattering. I can see here how the dipole depends on nu_1 which is said to be the vibrational frequency of the molecule, and nu_0, the frequency of the light. I can also see how there must be a change in polarisability is necessary for Raman to be active as if then the two Raman terms disappear. But it doesn't seem to suggest anything about how often Rayleigh scattering occurs compared to Raman. I suspect a quantum model may be needed for this?


    Quote Originally Posted by exchemist View Post
    Further note after more digging:
    I came across an expression for the Raman transition probability which is just a standard type of QM integral or matrix element: <ψn l α l ψm> . Given that different states have orthogonal wavefunctions, this implies that if α is a constant then the integral vanishes.
    That is interesting, though I don't understand it yet. Can you give a link/ref to where you came across this?
    Haha yes we're on the same page - almost literally: I have virtually the same equation in my old Atkins QM text from university! (PhDemon will like this: Atkins was his college tutor.)

    I think KJW's point, at least if I understand it properly, is a qualitative one. If one thinks about what polarisability of a molecule is, physically, i.e. the deformability of its electron clouds by a passing electric field, then it seems intuitively unlikely that this will change greatly as a result of merely promoting a molecule to the 1st vibrational excited state. So that would mean the α' factor outside the bracket would be expected to be small, and so the contribution from the terms inside the bracket would not be that great, compared to the Rayleigh term.

    The last equation came from Moore's Physical Chemistry, again sitting (mostly) on my shelf the last 40 years! It's my bad representation of Dirac's bracket notation for the standard QM integral that would determine the transition probability. It means ⎰ψ*n.α.ψm dτ , i.e. the size of the interaction between the two states m and n arising due to a polarisability difference between them, or something like that. I can't recite the exact details for you as it's all too long ago, but this ought be enough to put you on the right track I should think, if you know or have access to a bit of QM. I am far too rusty to work through it myself, I regret to say.
    Reply With Quote  
     

  14. #13  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    Quote Originally Posted by exchemist View Post

    Haha yes we're on the same page - almost literally: I have virtually the same equation in my old Atkins QM text from university! (PhDemon will like this: Atkins was his college tutor.)

    I think KJW's point, at least if I understand it properly, is a qualitative one. If one thinks about what polarisability of a molecule is, physically, i.e. the deformability of its electron clouds by a passing electric field, then it seems intuitively unlikely that this will change greatly as a result of merely promoting a molecule to the 1st vibrational excited state. So that would mean the α' factor outside the bracket would be expected to be small, and so the contribution from the terms inside the bracket would not be that great, compared to the Rayleigh term.

    The last equation came from Moore's Physical Chemistry, again sitting (mostly) on my shelf the last 40 years! It's my bad representation of Dirac's bracket notation for the standard QM integral that would determine the transition probability. It means ⎰ψ*n.α.ψm dτ , i.e. the size of the interaction between the two states m and n arising due to a polarisability difference between them, or something like that. I can't recite the exact details for you as it's all too long ago, but this ought be enough to put you on the right track I should think, if you know or have access to a bit of QM. I am far too rusty to work through it myself, I regret to say.
    This might be what I'm looking for. What page was that - I'll try to find a copy. It'll take me a while to work through and I'll probably come back later with more questions than answers.
    Reply With Quote  
     

  15. #14  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    Quote Originally Posted by exchemist View Post

    Haha yes we're on the same page - almost literally: I have virtually the same equation in my old Atkins QM text from university! (PhDemon will like this: Atkins was his college tutor.)

    I think KJW's point, at least if I understand it properly, is a qualitative one. If one thinks about what polarisability of a molecule is, physically, i.e. the deformability of its electron clouds by a passing electric field, then it seems intuitively unlikely that this will change greatly as a result of merely promoting a molecule to the 1st vibrational excited state. So that would mean the α' factor outside the bracket would be expected to be small, and so the contribution from the terms inside the bracket would not be that great, compared to the Rayleigh term.

    The last equation came from Moore's Physical Chemistry, again sitting (mostly) on my shelf the last 40 years! It's my bad representation of Dirac's bracket notation for the standard QM integral that would determine the transition probability. It means ⎰ψ*n.α.ψm dτ , i.e. the size of the interaction between the two states m and n arising due to a polarisability difference between them, or something like that. I can't recite the exact details for you as it's all too long ago, but this ought be enough to put you on the right track I should think, if you know or have access to a bit of QM. I am far too rusty to work through it myself, I regret to say.
    This might be what I'm looking for. What page was that - I'll try to find a copy. It'll take me a while to work through and I'll probably come back later with more questions than answers.
    p783 of 5th ed. But this is an undergraduate Phys Chem textbook from the 1970s, and not a specialist QM book at that, so it does not go into a lot of detail.

    But I would think a textbook on quantum chemistry or molecular spectroscopy might well give you the sort of detail you seem to be after.
    Reply With Quote  
     

  16. #15  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    OK, I think I'm going to have to learn second order perturbation theory to get to the bottom of this: might take a while. I also came across Fermi's Golden rule which may be relevant. I'll be back.

    Also found this excellent book which goes into quite some detail on the Raman effect - just doesn't explicitly derive what I want.
    Reply With Quote  
     

  17. #16  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    OK, I think I'm going to have to learn second order perturbation theory to get to the bottom of this: might take a while. I also came across Fermi's Golden rule which may be relevant. I'll be back.

    Also found this excellent book which goes into quite some detail on the Raman effect - just doesn't explicitly derive what I want.
    Yes, you need to get into transition probabilities, so indeed perturbation theory, matrix elements, Fermi's Golden Rule and so forth seem to be the right sorts of things to look at. And that's about as far as I can take it, these days.
    Reply With Quote  
     

  18. #17  
    KJW
    KJW is offline
    Forum Professor
    Join Date
    Jun 2013
    Posts
    1,346
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by epimetheus View Post
    OK, I think I'm going to have to learn second order perturbation theory to get to the bottom of this: might take a while. I also came across Fermi's Golden rule which may be relevant. I'll be back.

    Also found this excellent book which goes into quite some detail on the Raman effect - just doesn't explicitly derive what I want.
    Yes, you need to get into transition probabilities, so indeed perturbation theory, matrix elements, Fermi's Golden Rule and so forth seem to be the right sorts of things to look at. And that's about as far as I can take it, these days.
    You also should look at books about group theory written specifically for chemists.
    There are no paradoxes in relativity, just people's misunderstandings of it.
    Reply With Quote  
     

  19. #18  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by KJW View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by epimetheus View Post
    OK, I think I'm going to have to learn second order perturbation theory to get to the bottom of this: might take a while. I also came across Fermi's Golden rule which may be relevant. I'll be back.

    Also found this excellent book which goes into quite some detail on the Raman effect - just doesn't explicitly derive what I want.
    Yes, you need to get into transition probabilities, so indeed perturbation theory, matrix elements, Fermi's Golden Rule and so forth seem to be the right sorts of things to look at. And that's about as far as I can take it, these days.
    You also should look at books about group theory written specifically for chemists.
    There was a very good one from my era written by F Albert Cotton, who also wrote the Cotton & Wilkinson Inorganic Chemistry textbook. But I've no idea if this is still used.
    Reply With Quote  
     

  20. #19  
    Forum Freshman
    Join Date
    Jan 2018
    Posts
    14
    Quote Originally Posted by KJW View Post

    You also should look at books about group theory written specifically for chemists.
    Oh no, group theory... this may take a very long time.
    Reply With Quote  
     

  21. #20  
    exchemist
    Join Date
    May 2013
    Location
    London
    Posts
    2,972
    Quote Originally Posted by epimetheus View Post
    Quote Originally Posted by KJW View Post

    You also should look at books about group theory written specifically for chemists.
    Oh no, group theory... this may take a very long time.
    Well, the thing about symmetry groups in theoretical chemistry is that symmetry considerations alone can save you a lot of number crunching, by telling you what interactions can occur and what cannot. They can't give you numerical answers but can simplify things a great deal, when you are dealing with a molecule with several atoms and several tens of electrons.

    But as we don't know exactly what you want to explore, these are just suggestions for where you may want to look.
    Reply With Quote  
     

Similar Threads

  1. Compton scattering
    By whizkid in forum Physics
    Replies: 11
    Last Post: November 27th, 2013, 09:56 AM
  2. Can Rayleigh Scattering Explain the Sky’s Blue Colour?
    By galexander in forum Pseudoscience
    Replies: 171
    Last Post: May 8th, 2011, 05:56 AM
  3. Raman Spectroscopy
    By emetzner in forum Chemistry
    Replies: 7
    Last Post: May 6th, 2009, 11:53 AM
  4. Compton Scattering
    By VMStudent in forum Physics
    Replies: 2
    Last Post: July 1st, 2008, 10:40 PM
  5. Selection Rule For Raman Scaterring
    By Yash in forum Chemistry
    Replies: 3
    Last Post: March 5th, 2008, 04:03 PM
Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •