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?

  • $\begingroup$ 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, 2019 at 5:56

1 Answer 1


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 $$H_2 = T\frac{\eta_1}{\eta_2} H_1$$


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