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Do they work the same way? Do metal waveguides in MW also work through total internal reflection like in optical waveguides?

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There is reflection at the boundary in both cases, but the reason is fundamentally different.

The reflection at the boundary of a perfect conductor arises from the boundary conditions $$\mathbf{E} = \mathbf{0}$$ $$\frac{\partial\mathbf{B}}{\partial t} = \mathbf{0}$$ inside the conductor, and this in turn arises from the infinite conductivity of the material, $\sigma \to \infty$.

The reflection occurring during total internal reflection at a dielectric interface arises from the four dielectric boundary conditions for $\mathbf{E}$ and $\mathbf{B}$, which can be written (assuming linear dielectrics and no surface charge or current sheets) as:

$$(\varepsilon_1\mathbf{E}_1 - \varepsilon_2\mathbf{E}_2)\cdot\hat{\mathbf{n}}_{12} = 0$$ $$(\mathbf{E}_1 - \mathbf{E}_2)\times\hat{\mathbf{n}}_{12} = \mathbf{0}$$ $$(\mathbf{B}_1 - \mathbf{B}_2)\cdot\hat{\mathbf{n}}_{12} = 0$$ $$\left(\frac{\mathbf{B}_1}{\mu_1} - \frac{\mathbf{B}_2}{\mu_2}\right)\times\hat{\mathbf{n}}_{12} = \mathbf{0}$$

In both the interior of metal waveguide and interior of each dielectric, we assume the incident, reflected and transmitted waves have a plane wave form (in a uniform lossless medium, the solutions to Maxwell's equations can be decomposed into plane waves). So we can write each plane wave as

$$\mathbf{E}(\mathbf{x}, t) = \mathbf{E}_0e^{\mathbf{k}\cdot\mathbf{x} - j\omega t + \varphi_0}$$ $$\mathbf{H}(\mathbf{x}, t) = \mathbf{H}_0e^{\mathbf{k}'\cdot\mathbf{x} - j\omega' t + \varphi_0'}$$

Using this form for the fields, and applying the metal or dielectric boundary conditions, it is possible to obtain relationships between the amplitudes, phases, frequencies and wavevectors of incident, reflected and transmitted waves. For metals, we find that the reflected wave always has the same amplitude and $\pi$ radian phase shift. For dielectrics, the general result is given by the Fresnel equations. You will obtain total internal reflection only for large values of the incidence angle, when the argument of some $\arcsin$ function reaches $1$ (angle of transmitted wave according to Snell's law).

As you can see, the reason for total internal reflection is more complicated that the reflection in a conductor. In particular there is no condition on the incidence angle to have reflection in a perfect metal waveguide, the reflection is always happening for any nonzero frequency.


Note: for real conductors where $\sigma$ is finite, the boundary conditions at the metal interface are different. You can have nonzero electric fields inside the metal, but they decay exponentially (see the Skin Effect). In that case the reflection is not perfect, because some power is lost in the metal conductor. The boundary conditions are also more similar to the dielectric boundary conditions in that case, because you have to assume nonzero fields on both sides of the interface.

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