# Effect of conductivities to the CLN parameters
**Question:**
Prove that changing the conductivities of the eddy region does not change $L_{2n+1}$ parameters. (Excitation coil is PHICOIL)

**Answer**:
We have two eddy problems with the same geometry, permeability and two different conductivities in eddy region: $\sigma_A$ and $\sigma_B$.
Conductivity of excitation coil is $\sigma_e$

The conductiviy of the whole region in case A is $\sigma = \sigma_A+\sigma_e$ and for case B is $\sigma = \sigma_B+\sigma_e$.

### $\mathbf{e}_0$ and $R_0$

Since both cases have PHICOIL excitation, they share the same $E_0$ and $R_0$.
* case A: $\mathbf{e}_{0A} =\mathbf{e}_0$ and $\dfrac{1}{R_{0A}}=\mathbf{e}_{0A}^T\sigma\mathbf{e}_{0A} = \mathbf{e}_{0}^T\sigma_e\mathbf{e}_{0} = \dfrac{1}{R_0}$
* case B: $\mathbf{e}_{0B} =\mathbf{e}_0$ and $\dfrac{1}{R_{0A}}=\mathbf{e}_{0B}^T\sigma\mathbf{e}_{0B} = \mathbf{e}_{0}^T\sigma_e\mathbf{e}_{0} = \dfrac{1}{R_0}$

Finally:
> $\mathbf{e}_{0A}=\mathbf{e}_{0B}=\mathbf{e}_0$
> $R_{0A}=R_{0B}=R_0$

### $\mathbf{a}_1$ and $L_1$:
Since $\mathbf{e}_0$ is only defined on the PHICOIL region, for case A, $\sigma\mathbf{e}_0 = (\sigma_A+\sigma_e)\mathbf{e} = \sigma_e\mathbf{e}$ and for case B, $\sigma\mathbf{e}_0 = (\sigma_B+\sigma_e)\mathbf{e} = \sigma_e\mathbf{e}$
* case A: $\mathbf{K}\tilde{\mathbf{a}}_{1A}=R_{0A}\sigma\mathbf{e}_{0A} = R_0\sigma_e\mathbf{e}_{0}$
* case B: $\mathbf{K}\tilde{\mathbf{a}}_{1B}=R_{0B}\sigma\mathbf{e}_{0B} = R_0\sigma_e\mathbf{e}_{0}$
* $\tilde{\mathbf{a}}_{1A}=\tilde{\mathbf{a}}_{1B}=\tilde{\mathbf{a}}_{1}$

then
* case A: $\mathbf{a}_{1A}=\tilde{\mathbf{a}}_{1A}=\tilde{\mathbf{a}}_1=\mathbf{a}_1$ and $L_{1A}=\mathbf{a}_{1A}^T\nu\mathbf{a}_{1A}=\mathbf{a}_{1}^T\nu\mathbf{a}_{1}=L_1$
* case B: $\mathbf{a}_{1B}=\tilde{\mathbf{a}}_{1B}=\tilde{\mathbf{a}}_1=\mathbf{a}_1$ and $L_{1B}=\mathbf{a}_{1B}^T\nu\mathbf{a}_{1B}=\mathbf{a}_{1}^T\nu\mathbf{a}_{1}=L_1$

Finally
> $\mathbf{a}_{1A}=\mathbf{a}_{1B}=\mathbf{a}_1$
> $L_{1A}=L_{1B}=L_1$

### $\mathbf{e}_2$ and $R_2$:
For second electric modes:
* case A: $\mathbf{e}_{2A}=\mathbf{e}_{0A}-\mathbf{a}_{1A}/L_{1A}=\mathbf{e}_{0}-\mathbf{a}_{1}/L_{1}=\mathbf{e}_2$
* case B: $\mathbf{e}_{2B}=\mathbf{e}_{0B}-\mathbf{a}_{1B}/L_{1B}=\mathbf{e}_{0}-\mathbf{a}_{1}/L_{1}=\mathbf{e}_2$

And for resistors:
* case A: $\dfrac{1}{R_{2A}}=\mathbf{e}_{2A}^T\sigma\mathbf{e}_{2A}=\mathbf{e}_2^T\sigma_A\mathbf{e}_2=\sigma_A\dfrac{1}{R_{2}}$ where $\dfrac{1}{R_2}=\mathbf{e}_2^T1\mathbf{e}_2$
* case B: $\dfrac{1}{R_{2B}}=\mathbf{e}_{2B}^T\sigma\mathbf{e}_{2B}=\mathbf{e}_2^T\sigma_B\mathbf{e}_2=\sigma_B\dfrac{1}{R_2}$

Finally:
> $\mathbf{e}_{2A}=\mathbf{e}_{2B}=\mathbf{e}_2$
> $\sigma_AR_{2A}=\sigma_BR_{2B}=R_2$

### $\mathbf{a}_3$ and $L_3$:
$\mathbf{e}_2$ is present on the whole region but we have to ignore the $\sigma_e$ from this stage, since there is no eddy current inside PHICOIL.
* case A: $\mathbf{K}\tilde{\mathbf{a}}_{3A}=R_{2A}\sigma\mathbf{e}_{2A} = (R_2/\sigma_A)\sigma_A\mathbf{e}_{2}=R_2\mathbf{e}_2$
* case B: $\mathbf{K}\tilde{\mathbf{a}}_{3B}=R_{2B}\sigma\mathbf{e}_{2B} = (R_2/\sigma_B)\sigma_B\mathbf{e}_{2}=R_2\mathbf{e}_2$