There are a lot of papers on optical rotation which cite Rosenfeld's (German) 1928 paper "Quantum mechanical theory of natural optical rotation..." [Quantenmechanische Theorie der naturlichen optischen Aktivitat von Flussigkeiten und Gasen, Zeitschrift für Physik, Volume 52, Issue 3-4, pp. 161-174] as background. Here is a paid link--

Apologies that I cannot link to the paper. I have a paper copy and can't find a paywall-free version.

The symbols below are: $\lambda$--wavelength; $c$ is the speed of light; $\nu$ is frequency; $n$ is the index of refraction; $\rho$ is a quantity that does not change between equations (42) and (43).

The last two equations in that paper (42), (43), are a conclusion about an angle of rotation $\theta:$

$$\Delta n = \frac {2\pi ~\rho}{\lambda~ n} \tag{42}$$


$$\theta = \frac {2\pi \nu~\cdot \Delta n}{c~\cdot ~ 2} = \frac{2 \pi^2\rho}{\lambda^2} \tag{43}$$

My question is pretty dumb. If we insert the expression for $\Delta n$ into the equality in (43) we get:

$$\frac{2\pi\nu}{c}\cdot \frac{2\pi \rho}{2\lambda~ n} = \frac{2\pi^2\rho}{\lambda^2}\to \frac{\nu}{c~n}=\frac{1}{\lambda}\to\lambda =\frac{cn}{\nu} $$

We know already--and Rosenfeld mentions this a few lines up and at eq. (21)--that

$$\lambda = \frac{c}{\nu~n}$$

which forces $n$ to be $1$. In general $n \neq 1$.

Can someone tell me what, if anything, has gone awry here?

Edit: After looking at several versions of Rosenfeld's equation I see $\theta$ is proportional to the trace of the rotation matrix...the constant of prop. can take various forms and is not the focus of this approach.

  • $\begingroup$ In the medium, both the speed of light in vacuo and the wavelength are smaller by a factor of 1/n, but the frequency, nu, is independent of n. Does this help in your substitution? $\endgroup$ – Ed V Feb 8 at 23:10
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    $\begingroup$ Of course so many landmark papers had typos and those great scientists readily accepted their mistakes. I recently wrote an article on the history of the term "shot noise" and highlighted a mistake by Schottky (1920). When people used his value the charge on the electron was completely wrong but that paper was a landmark paper. Just curious why are you reading a 1928 paper? I usually go from modern to the oldest. $\endgroup$ – M. Farooq Feb 8 at 23:22
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    $\begingroup$ So in eqn 42, lambda is the wavelength in the medium and on the RHS of eqn 43, that lambda is the wavelength in vacuo. It equals n times lambda in the medium. $\endgroup$ – Ed V Feb 8 at 23:52
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    $\begingroup$ And here is what I have: chemistry.meta.stackexchange.com/a/4552/79678 $\endgroup$ – Ed V Feb 9 at 17:54
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    $\begingroup$ Assuming you do not get a better answer (and you are the judge of that) and assuming it is not unethical or in violation of any rules here, then that is fine with me! So please post the answer and I will award you the bounty! Thanks, daniel! $\endgroup$ – Ed V Feb 16 at 23:10

This answer was suggested by Ed V. It resolves the apparent problem nicely and readers with some fluency in the subject would probably not find the transition from $\lambda$ (medium) to $\lambda$ (vacuum) too troublesome.

We begin with $c = \nu \lambda_o$ in a vacuum and $\frac{c}{n} =\nu\lambda $ in a medium, in which $n$ is th refractive index.

In equation (42) of Rosenfeld's paper we have (using $\lambda n =\lambda_o$):

$$\Delta n = \frac{2\pi\rho}{\lambda n}= \frac{2\pi\rho}{\lambda_o}.\hspace 74 mm ~~~(42)$$

Then in equation (43) above we get $\lambda_o^2$ in the denominator:

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{\Delta n}{2}=\frac{2\pi\nu}{c}\cdot\frac{\pi\rho}{\lambda_o}=\frac{2\pi^2\nu\rho}{c\lambda_o}=\frac{2\pi^2\rho}{\lambda_o^2}\hspace 30 mm (43) $$

and finally substitution of (42) into (43) gives

$$\theta = \frac{2\pi\nu}{c}\cdot \frac{2\pi\rho}{2\lambda n}= \frac{2\pi^2\rho}{\lambda_o\lambda n}=\frac{2\pi^2\rho}{\lambda_o^2}. $$

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