Is Permittivity Equivalent to EM Conductivity?

Thread starter Saw
Start date Mar 19, 2012
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Conductivity Em Permittivity

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The discussion explores the relationship between permittivity and electromagnetic (EM) conductivity, drawing parallels between electric current in conductors and EM wave propagation in dielectrics. It suggests that while conductivity measures a material's ability to transmit electric current, permittivity could be viewed as a form of EM conductivity, influencing how well an EM wave propagates through a dielectric. However, the analogy is nuanced; higher permittivity can slow down EM waves, contrasting with the increased current flow in more conductive materials. The conversation also touches on the differences in terms of time and frequency between conductivity and permittivity, emphasizing that while they are mathematically related, they describe different phenomena. Ultimately, the relationship between permittivity and EM wave conduction is complex and warrants careful consideration of their distinct effects on energy flow.

Mar 21, 2012

Dickfore

Saw said:

So in this case we get division by zero?

Provided that \hat{\sigma}(\omega) \neq o(\omega), \ \omega \rightarrow 0. This is true for conductors, but not for insulators (dielectrics).

EDIT:
Remember, we are working wIth the Fourier transforms of these quantities in the frequency domain. This simply tells us that there ought to be a pole of the dielectric response function at \omega = 0. Going back to time domain, we get:
 \epsilon''(t -t') = \frac{\sigma}{\epsilon_0} \, \theta(t - t') 
which gives the following relation between the polarization and the electric field:
 \mathbf{P}(t) = \sigma \, \int_{-\infty}^{t}{\mathbf{E}(t') \, dt'} 
or, the current density due to bound charges is:
 \mathbf{J}(t) = \dot{\mathbf{P}}(t) = \sigma \, \mathbf{E}(t) 
But, this is just Ohm's law in differential form, provided the "bound" charges are the free ones, as is the case for conductors.

Last edited: Mar 21, 2012