User:Constant314/Derivation of skin depth

The full derivation of the skin depth formula is implicit in equations given in the sources. This article shows the steps of the full derivation with complex permittivity and permeability shown explicitly.

Overview
Starting with the equations and analysis on the Propagation constant page


 * $$\gamma = \alpha + j\beta = \sqrt{j \omega \mu (\sigma + j \omega \epsilon)}$$
 * $$ \delta = \frac 1 \alpha$$

with
 * $$\epsilon = \epsilon' - j\epsilon'' =$$ complex permitivity,
 * $$\mu = \mu' - j\mu'' =$$ complex permeability,

Carry out the math, gathering like terms (some cancel) and applying the formula for the square root of a complex number produces:


 * $$ \delta = \frac 1 \alpha = \sqrt{ \frac {2\rho } {\omega\mu' } } \frac  {\sqrt{\sqrt{(1+{\color{red}\xi^2+\chi^2+\xi^2\chi^2}){(\rho \omega \epsilon')}^2 +(1+2\xi\rho\omega\epsilon')(1+{\color{red}\chi^2}) } +(1 -{\color{red}\xi\chi}   )\rho \omega \epsilon'-\chi}} {1 + \rho\omega\epsilon' (\chi+\xi)}\; . $$

where
 * $$ \xi = \frac {\epsilon''}{\epsilon'} \quad $$ also known as dielectric loss tangent.
 * $$ \chi = \frac {\mu''}{\mu'} \quad$$ also known as magnetic loss tangent.

If $$ \xi < 0.1 $$ and $$ \chi < 0.1 $$, then the cross terms shown in red may be taken to be zero. That yields a simpler expression:
 * $$ \delta = \frac 1 \alpha = \sqrt{ \frac {2\rho } {\omega\mu' } } \frac  {\sqrt{\sqrt{{1+2\xi\rho\omega\epsilon'+(\rho\omega\epsilon')}^2 } +\rho\omega\epsilon'-\chi}} {1 +  (\chi+\xi)\rho\omega\epsilon'}\; . $$

If the dielectric loss is small ( $$ \xi \approx 0 $$ ), then
 * $$ \delta = \frac 1 \alpha = \sqrt{ \frac {2\rho } {\omega\mu' } } \frac  {\sqrt{\sqrt{1 +{(\rho \omega \epsilon')}^2} +\rho\omega\epsilon'-\chi}} {1 +  \chi\rho\omega\epsilon'} .$$

If the magnetic loss is also small ( $$ \chi \approx 0  $$ ) then this reduces the the more familiar form


 * $$ \delta = \frac 1 \alpha = \sqrt{ \frac {2\rho } {\omega\mu' } }   {\sqrt{\sqrt{1+{(\rho \omega \epsilon')}^2  } +\rho\omega\epsilon'}} \; . $$

Definitions
The propagation factor of a sinusoidal plane wave propagating in the x direction in a linear material may be given by two equivalent forms


 * $$ P = e^{-\gamma x} = e^{-jkx}, $$

where
 * $$\gamma = \alpha + j\beta = \sqrt{j \omega \mu (\sigma + j \omega \epsilon)}= \sqrt{(\omega \mu  + j \omega \mu ')(\sigma + \omega \epsilon  + j \omega \epsilon ')}=$$ Propagation constant ,
 * $$k = k'- jk = \sqrt{-(\omega \mu  + j \omega \mu ')(\sigma + \omega \epsilon '' + j \omega \epsilon ') }=$$  wavenumber ,
 * $$\beta = k' =$$ phase constant in the units of radians/meter,
 * $$\alpha = k'' =$$ attenuation constant in the units of nepers/meter,
 * $$\omega =$$ frequency in the units of radians/meter,
 * $$x =$$ distance traveled in the x direction,
 * $$\sigma =$$ conductivity in S/meter,
 * $$\rho =$$  resistivity in ohm-meter (Ω⋅m),
 * $$v_0 = $$ phase velocity of free space (about 3 x 108 m/s),
 * $$\lambda_0 = \frac {2\pi}{\omega} v_0 =$$ wavelength in free space,
 * $$\epsilon = \epsilon' - j\epsilon'' =$$ complex permitivity,
 * $$\mu = \mu' - j\mu'' =$$ complex permeability,
 * $$j=\sqrt{-1}$$.

The sign convention is chosen for consistency with propagation in lossy media. If the attenuation constant is positive, then the wave amplitude decreases as the wave propagates in the x direction.

Wavelength, phase velocity  , and skin depth   have simple relationships to the components of the propagation constant or wavenumber:
 * $$ \lambda = \frac {2 \pi} \beta, \qquad v_p = \frac \omega \beta , \qquad   \delta =  \frac 1 \alpha , $$
 * $$ \lambda = \frac {2 \pi} {k'}, \qquad v_p = \frac \omega {k'} , \qquad   \delta =  \frac 1 {k''}  .$$

Skin depth is the distance over which the wave attenuates by the factor $$e^{-1}$$. This is simply the reciprocal of the attenuation constant.

Algebraic rearrangement
By a straightforward, if lengthy, algebraic calculation, the expression for k can be simplified by defining some simple ratioes.

k=\sqrt{-(\omega \mu  + j \omega \mu ')(\sigma + \omega \epsilon  + j \omega \epsilon ')} = \omega\sqrt {\mu' \epsilon'} \sqrt {(1 - \kappa \chi - \xi \chi)-j(\chi+\kappa+\xi)} ,$$


 * where


 * $$ \xi = \frac {\epsilon}{\epsilon'} ,\quad  \chi = \frac {\mu}{\mu'} ,  \quad  \kappa = \frac 1 \tau = \frac {\sigma}{\omega \epsilon'}= \frac 1 {\rho \omega \epsilon'} .$$


 * Note $$ (\chi+\kappa+\xi) \ge 0  $$ in a source free region.  $$ \xi $$ is also called dielectric loss tangent. $$ \chi $$ is also called magnetic loss tangent.

general expression
Using the formula for the square root of a complex number


 * $$\beta = k'=\omega \sqrt{ \frac {\mu' \epsilon'} 2 } \sqrt{\sqrt{(1+{\color{red}\xi^2+\chi^2+\xi^2\chi^2}) +(2\xi\kappa+\kappa^2)(1+{\color{red}\chi^2}) } +1 -{\color{red}\xi\chi} -\kappa\chi } \; ,$$


 * $$ \alpha = k'' =\omega \sqrt{ \frac {\mu' \epsilon'} 2} \frac {(\kappa+\chi+\xi)} {\sqrt{\sqrt{(1+{\color{red}\xi^2+\chi^2+\xi^2\chi^2}) +(2\xi\kappa+\kappa^2)(1+{\color{red}\chi^2}) } +1 -{\color{red}\xi\chi} -\kappa\chi }} \; . $$


 * $$ \delta = \frac 1 \alpha =\frac 1 \omega \sqrt{ \frac 2 {\mu' \epsilon'} } \frac  {\sqrt{\sqrt{(1+{\color{red}\xi^2+\chi^2+\xi^2\chi^2}) +(2\xi\kappa+\kappa^2)(1+{\color{red}\chi^2}) } +(1 -{\color{red}\xi\chi} -\kappa\chi  })} {(\kappa+\chi+\xi)}\; . $$


 * $$ \delta = \frac 1 \alpha =\frac 1 \omega \sqrt{ \tau \frac 2 {\mu' \epsilon'} } \frac  {\sqrt{\tau\sqrt{(1+{\color{red}\xi^2+\chi^2+\xi^2\chi^2}) +(2\xi\kappa+\kappa^2)(1+{\color{red}\chi^2}) } +\tau(1 -{\color{red}\xi\chi} -\kappa\chi  })} {\tau (\kappa+\chi+\xi)}\; . $$


 * $$ \delta = \frac 1 \alpha = \sqrt{ \frac {2\rho } {\omega\mu' } } \frac  {\sqrt{\sqrt{(1+{\color{red}\xi^2+\chi^2+\xi^2\chi^2})\tau^2 +(2\xi\tau+1)(1+{\color{red}\chi^2}) } +(\tau -{\color{red}\xi\chi}\tau -\chi  })} {1 + \tau (\chi+\xi)}\; . $$

simplified general expression
If the dielectric loss tangent is small ($$ \xi < 0.1 $$) and magnetic loss tangent is small ($$ \chi < 0.1 $$), then the cross terms shown in red in the previous section can be replaced with zero. That yields simplified expressions as follows:


 * $$ k'=\omega \sqrt{ \frac {\mu' \epsilon'} 2 } \sqrt{\sqrt{1 +2\xi\kappa+\kappa^2 } +1 -\kappa\chi } $$


 * $$ k'' =\omega \sqrt{ \frac {\mu' \epsilon'} 2} \frac {(\kappa+\chi+\xi)} {\sqrt{\sqrt{1 +2\xi\kappa+\kappa^2 } +1 -\kappa\chi  }}   $$


 * $$ \delta =\frac 1 \omega \sqrt{ \frac 2 {\mu' \epsilon'} }

\frac {\sqrt{\sqrt{1+2\xi\kappa+\kappa^2} +1 -\kappa\chi  }} {(\kappa+\chi+\xi)}  $$

alternate simplified general expression
Multiplying the previous expressions by $$ (\rho\omega\epsilon') / (\rho\omega\epsilon') $$ yields these alternate expressions.


 * $$ k'=\sqrt{ \frac {\omega \mu' } {2\rho} } \; \sqrt{ \sqrt{{(\rho\omega\epsilon')}^2 +2\xi\rho\omega\epsilon'+1 } +\rho\omega\epsilon' -\chi  } $$


 * $$ k''=

\sqrt{ \frac{\omega\mu'}{2\rho}}\;\frac{1+\rho\omega\epsilon'(\chi+\xi)}{\sqrt{\sqrt{{(\rho\omega\epsilon')}^2 +2\xi\rho\omega\epsilon'+1} +\rho\omega\epsilon'-\chi}}$$


 * $$ \delta = \sqrt{ \frac {2\rho} {\omega\mu' } } \;

\frac {\sqrt{\sqrt{{(\rho\omega\epsilon')}^2 +(2\xi\rho\omega\epsilon'+1) } +\rho\omega\epsilon' -\chi }} {1+\rho\omega\epsilon'(\chi+\xi)}  $$

Insulator
Using the simplified general expressions for low dielectric and magnetic losses, the skin depth is given by


 * $$ \delta= \frac 1 {k''} = \frac  { \sqrt{ 2 }\sqrt{\sqrt{1 +2\xi\kappa+\kappa^2 } +1  -\kappa\chi  }} {\omega \sqrt{\mu' \epsilon'} \; (\kappa+\chi+\xi)}  \quad  $$ low loss tangents insulator form suitable for $$ \kappa < 1$$

good insulator
For a good insulators at typical frequencies of interest, $$ \kappa = \frac {\sigma}{\omega \epsilon'} $$ is very small. For example, at 1 mHz for polyethylene $$ \; \kappa \approx 0.0005 \; $$ and gets smaller at higher frequencies.

The expression for skin depth can be simplified by setting the cross terms $$ \; \chi\kappa, \; \xi\kappa, \; \kappa^2 \; $$ to zero.


 * $$ \delta=  \frac  2 {\omega\sqrt{\mu' \epsilon'} \; (\chi+\kappa+\xi)}  \quad  $$ expression for skin depth with low material losses.  If there are no losses (such as vacuum), then skin depth is infinite.

Conductor
If the dielectric and magnetic losses are small ($$ \xi < 0.1 $$ and $$ \chi < 0.1 $$) then the product of those terms in the alternate general expression can be taken to be zero.


 * $$ \delta= \frac 1 {k''}= \sqrt{ \frac {2\rho} {\omega\mu'} } \frac  {

\sqrt{\sqrt{{(\rho\omega\epsilon')}^2 +2\xi\rho\omega\epsilon'+1 } +\rho\omega\epsilon' -\chi } } { 1+(\chi+\xi)\rho\omega\epsilon'} \quad  $$ low loss tangents conductor form

If the magnetic loss and dielectric loss are sufficiently small, the formula simplifies to the formula from skin effect article.
 * $$ \delta=\sqrt{ \frac {2\rho} {\omega\mu' } }

{\sqrt{\sqrt{1+{(\rho\omega\epsilon')}^2} +\rho\omega\epsilon' }} \quad $$

good conductor
$$ \rho\omega\epsilon' \;$$ is typically small for good conductors. For example, copper at 1 THz $$ \; \rho\omega\epsilon' \approx 10^{-6}$$.

Since $$ \rho\omega\epsilon' $$ is small, $${(\rho\omega\epsilon')}^2$$ and $$\;\xi\rho\omega\epsilon'$$ can be taken to be zero.


 * $$ \delta= \sqrt{ \frac {2 \rho} {\omega \mu'}  } {\sqrt{1+\rho\omega\epsilon' -\chi }}  \approx  \sqrt{ \frac {2 \rho} {\omega \mu'}  }$$