# Relative permittivity of vacuum aluminum interface?

## Main Question or Discussion Point

funny but I could not reach the info from net quickly.

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Aluminum is a conductor, so the EM wave is attenuated and absorbed. A good conductor reflects the incident wave. the reflection coefficient R is approximately
R = 1 -2 sqrt(2 w e0/sigma)
e0 permittivity of free space
sigma = conductivity of metal.
See

Born2bwire
Gold Member
The dielectric constant for a metal is usually hard to find (I don't know if it really is greater than unity) but for most purposes it is irrelevant due to the large conductivity of the metal that will dominate the reflection and transmission properties. Wikipedia gives the conductivity to be 37.8e6 mhos/m (interestingly enough this does not match the given resistivity, oh Wikipedia!). So you can model the permittivity of alumin(i)um simply as:

$$\epsilon = \epsilon_0+i\frac{\sigma}{\omega\epsilon_0}$$
where \sigma, the conductivity, is 37.8e6 S/m.

thank you very much...

holy crap, you guys are awesome

DrDu
In optics one usually choses the polarization $$P=\int dt j(t)$$ or, in fourier space, $$P(\omega)=-j(\omega)/{i \omega}$$, where j is the induced current in the material (see, e.g. Landau Lifshetz, electrodynamics of continuous media).