3
$\begingroup$

Helmholtz equations for electric and magnetic fields are

$$∇^2 \mathbf{H} + k^2 \mathbf{H} = \mathbf{0}$$ $$∇^2 \mathbf{E} + k^2 \mathbf{E} = \mathbf{0}$$

Obviously, if a solution is found to satisfy the electric field equation, it must also satisfy the magnetic field equation. A wave traveling between two media has an electric field magnitude in medium one proportion the magnitude in medium two, in other words

$$ |\mathbf E_2| = T |\mathbf E_1| $$

where $T$ is the Fresnel transmission coefficient. Is this true for the magnetic field as well?

$$ |\mathbf H_2| = T |\mathbf H_1| $$

If not why? How do we explain that the Helmholtz solution for electric and magnetic field could be the same?

$\endgroup$
  • $\begingroup$ Isnt your Helmholtz equation for free space with boundary conditions? So if are two media you need to write down a different equation. $\endgroup$ – lalala Jul 23 at 5:56
1
$\begingroup$

It's not quite right. The two transmission coefficients will differ depending on the differing impedances of the two media.

This is because the relationship between the E-field and H-field magnitudes is $E = \eta H$, where $\eta$ is the impedance.

Thus for two media with impedances $\eta_1$ and $\eta_2$, the "transmission coefficient for the H-field would be $$H_2 = T\frac{\eta_1}{\eta_2} H_1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.