1
$\begingroup$

I have a question regarding refractive index dependency on the density of a dielectric, specifically air.

Background

Let us start from Newton's second law form of driven harmonic oscillators $$-kx-bv+F_e=ma.$$

Knowing that $F_e=eE$ the equation may be rewritten as $$m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+b\frac{\mathrm{d}{x}}{\mathrm{d}t}+kx=eE.$$

The relevant solution is given by $$x(t)=\frac{eE}{\sqrt{(-m\omega^2+k)^2+(2b\omega)^2}}\cos({\omega}t+\phi)$$

Assume a dielectric material. Let incoming light be monochromatic. In this case, I have shown that refractive index $n$ is dependant on the frequency $\omega$ via

$$n=\sqrt{\mu+\frac{e^2}{m}\frac{n_c}{\epsilon_0}\frac{\mu}{\sqrt{(\omega_0^2-\omega^2)^2+(2\beta\omega)^2}}\cos({\omega}t+\phi)}$$

where $\omega_0=\sqrt{\frac{k}{m}}$, $\beta\equiv\frac{b}{2m}$, and $n_c$ is the number of charge dipoles per unit volume.

Question

The next step is to show that refractive index varies as density changes. Intuitively, the dipole concentration $n_c$ is related to the density of a material. Any further than that, I am clueless.

  1. How is refractive index related to the density of the medium, e.g., density of air? Please give a short, simple derivation if possible.

  2. When we have the equation from question (1), am I allowed to make ideal gas substitution in order to directly derive how the refractive index depends on air temperature? In other words, I would like to substitute density $\rho$ for $\frac{MP}{RT}$.

Extra

I have glanced over the following links:

Nevertheless, these seem not to be exactly what I am looking for.

$\endgroup$
  • $\begingroup$ en.wikipedia.org/wiki/Refractive_index#Density $\endgroup$ – Fabrice NEYRET Feb 6 '16 at 22:12
  • $\begingroup$ Are you after a robust equation, or a derivation? Because your first link has a good equation for $n_{tp}$ that shows there's a very small deviation from a straight line relationship. $\endgroup$ – Floris Feb 6 '16 at 22:28
  • $\begingroup$ @FabriceNEYRET I have now added a reply which pretty much outlines the answer I am looking for (plus a new question). In other words, an equation and an intuitive grasp would be ideal. I love equations which I can understand :=). $\endgroup$ – Linear Christmas Feb 6 '16 at 22:58
  • $\begingroup$ @Floris The comment comment above is also directed at you. Furthermore, thanks for pointing out the linearity of $n_{tp}$. I had missed that. $\endgroup$ – Linear Christmas Feb 6 '16 at 22:58
  • $\begingroup$ The cosine should not appear in the refractive index $\endgroup$ – Jannick Feb 8 '16 at 16:23
1
$\begingroup$

You have to account for the local fields and how they change as the density changes. Look up the Clausius–Mossotti Equation and its derivation.

$\endgroup$
  • $\begingroup$ Thank you for the reply! I have given it a go, and posted the result. Could you verify? $\endgroup$ – Linear Christmas Feb 6 '16 at 23:09
  • $\begingroup$ Chose this as the answer because the reference was enough to set me on the right track. For those viewing in the future, check my (edited) reply as well. $\endgroup$ – Linear Christmas Feb 7 '16 at 23:48
0
$\begingroup$

So, per JQK's suggestion, I began from the equation $\text(a)$

$$\frac{\frac{n^2}{\mu}-1}{\frac{n^2}{\mu}+2}=\frac{1}{3\epsilon_0}\frac{N}{V}\alpha$$

where $N$ is the number of gas particles, and $\alpha$ is molecular polarizability. From the ideal gas law $$PV=zRT$$ it follows that $\frac{N}{V}=\frac{N_A\rho}{M}$. Thus $$\frac{\frac{n^2}{\mu}-1}{\frac{n^2}{\mu}+2}=\frac{1}{3\epsilon_0}\frac{N_A\rho}{M}\alpha.$$

Algebraic manipulation yields $\text{(b)}$ $$n=\sqrt{\frac{\mu(1+2G)}{1-G}}$$ where $$G=\frac{1}{3\epsilon_0}\frac{N_A\rho}{M}\alpha.$$

Is this $\text{(b)}$ correct? Also, if someone could further explain the derivation of $\text{(a)}$, it would be helpful. Thanks!

Edit

This research (http://arxiv.org/pdf/0907.0782v1.pdf) suggests an alternative formula. Namely $$n\approx1+\frac{{P\alpha'}}{2k_BT}.$$ Notice that I redefined $\gamma_{mol}$ as $\alpha'$. Now since $\frac{P}{k_BT}=\frac{N}{V}=\frac{N_A\rho}{M}$, we can rewrite $$n\approx1+\frac{N_A\rho}{2M}\alpha'.\text{ eqn. (c)}$$

If we evaluate our own $\text{(b)}$ by a order one Maclaurin series expansion (take $\mu\approx1$), we reach $$n\approx1+\frac{3}{2}G.$$ Plugging in $G$ gives $\text{(d)}$ $$n\approx1+\frac{N_A\rho}{2M\epsilon_0}\alpha.$$ Discrepancy between $\text{(c)}$ and $\text{(d)}$

This is a result of different definitions of the dipole moment $p$. Our formula assumes $p=\alpha E$ whereas the research defines $p=\epsilon_0\alpha'E$. It follows that $$\alpha=\epsilon_0\alpha'.$$ Substition into $\text{(d)}$ is sufficient to show that $\text{(c)}=\text{(d)}$; the equations are equivalent. Furthermore, according to the study, the formula is compatible with experiments.

$\endgroup$
  • $\begingroup$ I think your algebraic manipulation is wrong - it should be $n=\sqrt{\frac{1+2G}{1-G}}$ $\endgroup$ – Floris Feb 6 '16 at 23:14
  • $\begingroup$ @Floris, you are correct. Thank you. I will edit in a moment. However, other than that? $\endgroup$ – Linear Christmas Feb 6 '16 at 23:18
  • $\begingroup$ I think what you've done is correct. To fully appreciate the derivation of (a) you need to consider the microscopic view of molecules and their influence on each other and how it relates to the index of refraction. $\endgroup$ – JQK Feb 6 '16 at 23:37
  • $\begingroup$ @JQK, I did some more research and edited the post. Of course, in case you're interested. $\endgroup$ – Linear Christmas Feb 7 '16 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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