I am currently studying the textbook Infrared and Raman Spectroscopy, 2nd edition, by Peter Larkin. In a section entitled The Raman Scattering Process, the author says the following:

Both Rayleigh and Raman are two photon processes involving scattering of incident light ($h c \bar{\nu}_L$), from a “virtual state.” The incident photon is momentarily absorbed by a transition from the ground state into a virtual state and a new photon is created and scattered by a transition from this virtual state. Rayleigh scattering is by far the most probable event and the scattered intensity is c. $10^{-3}$ less than that of the original incident radiation. This scattered photon results from a transition from the virtual state back to the ground state and is an elastic scattering of a photon resulting in no change in energy (i.e., occurs at the laser frequency).

Raman scattering is far less probable than Rayleigh scattering with an observed intensity that is c. $10^{-6}$ that of the incident light for strong Raman scatterers. This scattered photon results from a transition from the virtual state to the first excited state of the molecular vibration. This is described as an inelastic collision between photon and molecule, since the molecule acquires different vibrational energy ($\bar{\nu}_m$) and the scattered photon now has different energy and frequency.

enter image description here

There seems to be a typesetting issue here. Is this supposed to be $c \cdot 10^{-3}$ and $c \cdot 10^{-6}$, respectively, where $c$ is the speed of light? I would appreciate it if someone would please take the time to clarify this.


1 Answer 1


c. = circa (around, about, roughly, approximately)

  • $\begingroup$ Ahh, thanks for that. I had no idea. $\endgroup$ Commented May 14, 2020 at 20:53
  • $\begingroup$ Don't worry. I didn't get that either. It's usually only use for dates, and even then it's usually ca. $\endgroup$
    – garyp
    Commented May 14, 2020 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.