Deromediqm

2022-07-20

How does the refractive index of air rely on the temperature? Is there a theoretical derivation of it?

Killaninl2

Beginner2022-07-21Added 20 answers

The refractive index of air is easy, because air is a dilute gas with a very small refractive index, which is given by:

$n=1-\sum {n}_{i}{\delta}_{i}(k)$

for small wavenumbers k. The ${n}_{i}$ are the number density for each species of molecule, and ${\delta}_{i}$ is the contribution to the index from this molecular species. You can just use N2 and O2 to get a good enough fit, and include CO2 and H2O for a better fit.

In the ideal gas limit, which is nearly perfect for air, $n=\frac{P}{kT}$. If you double the pressure, you double the deviation from 1. If you double the temperature, you halve the deviation from one, because all the components go with the same ideal gas law:

So the formula for the long-wavelength index of air is

$\overline{){\displaystyle n(P,T)=1+.000293\times \frac{P}{{P}_{0}}\frac{{T}_{0}}{T}}}$

Where ${P}_{0}$ is atmospheric pressure, and ${T}_{0}$ is the standard temperature of 300K. and this is essentially exact for all practical purposes, the corrections are negligible away from oxygen/nitrogen/water/CO2 resonances, and any deviation from the formula will be due to varying humidity.

The actual contributions ${\delta}_{i}$ requires the forward scattering amplitude for light on a diatomic molecule. This is just beyond what you can do with pencil and paper, but it is within the reach of simulations.

$n=1-\sum {n}_{i}{\delta}_{i}(k)$

for small wavenumbers k. The ${n}_{i}$ are the number density for each species of molecule, and ${\delta}_{i}$ is the contribution to the index from this molecular species. You can just use N2 and O2 to get a good enough fit, and include CO2 and H2O for a better fit.

In the ideal gas limit, which is nearly perfect for air, $n=\frac{P}{kT}$. If you double the pressure, you double the deviation from 1. If you double the temperature, you halve the deviation from one, because all the components go with the same ideal gas law:

So the formula for the long-wavelength index of air is

$\overline{){\displaystyle n(P,T)=1+.000293\times \frac{P}{{P}_{0}}\frac{{T}_{0}}{T}}}$

Where ${P}_{0}$ is atmospheric pressure, and ${T}_{0}$ is the standard temperature of 300K. and this is essentially exact for all practical purposes, the corrections are negligible away from oxygen/nitrogen/water/CO2 resonances, and any deviation from the formula will be due to varying humidity.

The actual contributions ${\delta}_{i}$ requires the forward scattering amplitude for light on a diatomic molecule. This is just beyond what you can do with pencil and paper, but it is within the reach of simulations.

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