Relative permittivity of vacuum aluminum interface?

In summary, the conversation discusses the properties of aluminum as a conductor and its reflection coefficient. The dielectric constant for a metal is usually irrelevant due to the large conductivity, which can be modeled using a simple equation. However, in optics, the dielectric function of metals can be inferred from reflectivity measurements. The conversation also mentions the use of this information in calculations for optical transition radiation, commonly used in measuring the profile of charged particle beams in vacuum.
  • #1
omete
3
0
funny but I could not reach the info from net quickly.
 
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  • #2
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)
where w = frequency (radians/sec)
e0 permittivity of free space
sigma = conductivity of metal.
See
https://www.physicsforums.com/showthread.php?t=162915
 
  • #3
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:

[tex] \epsilon = \epsilon_0+i\frac{\sigma}{\omega\epsilon_0} [/tex]
where \sigma, the conductivity, is 37.8e6 S/m.
 
  • #4
thank you very much...
 
  • #5
holy crap, you guys are awesome
 
  • #6
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 [tex] P=\int dt j(t) [/tex] or, in Fourier space, [tex]P(\omega)=-j(\omega)/{i \omega}[/tex], 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
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 :)
 

1. What is relative permittivity?

Relative permittivity, also known as dielectric constant, is a measure of the ability of a material to store electrical energy when an electric field is applied to it. It is the ratio of the permittivity of a material to the permittivity of a vacuum.

2. How is relative permittivity of a material measured?

The relative permittivity of a material is typically measured by comparing the capacitance of a capacitor filled with the material to the capacitance of the same capacitor with a vacuum between the plates. The ratio of these two capacitances gives the relative permittivity of the material.

3. What is the relative permittivity of vacuum?

The relative permittivity of vacuum is defined as 1, since vacuum has no ability to store electrical energy when an electric field is applied to it. In other words, it is the baseline value against which the relative permittivity of other materials is compared.

4. How does the relative permittivity of aluminum interface with vacuum?

The relative permittivity of aluminum interface with vacuum is typically lower than that of vacuum alone, as the presence of aluminum can reduce the ability of the electric field to penetrate the space between the plates of a capacitor. However, the exact value depends on the thickness and purity of the aluminum layer.

5. Why is the relative permittivity of vacuum aluminum interface important?

The relative permittivity of vacuum aluminum interface is important in understanding the behavior of electrical systems and devices that involve the use of aluminum as a conductor. It can also impact the performance of certain electronic components, such as capacitors, and must be taken into account in their design and manufacturing processes.

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