# Relative permittivity of vacuum aluminum interface?

1. May 31, 2009

### omete

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

2. May 31, 2009

### Bob S

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

3. May 31, 2009

### Born2bwire

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.

4. Jun 1, 2009

### omete

thank you very much...

5. Mar 14, 2010

### outskut

holy crap, you guys are awesome

6. Mar 15, 2010

### DrDu

Born2bwire,
I don't quite understand your comment that the dielectric constant is irrelevant due to the large conductivity.
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).
Hence the equation you write down is an exact relationship and not just an approximation for any material. However, the conductivity depends generally on frequency and can be complex. In the optical region in metals, it is usually not justified to replace the conductivity by its static value.
The dielectric function of metals is usually inferred from reflectivity measurements and its values in the optical region can be found e.g. in Landolt/Boernstein.

7. Mar 20, 2010

### omete

OK. It is quite a while then, but I come across this discussion. Then I wanted to add the reason why I need this info, could be of interest to you. It is related to some calculations for OTR (optical transition radiation). You use it to measure the profile of your charged particle beam traveling in vacuum. Yes, I am an accelerator physicist :)