# Reflection of electromagnetic waves from dielectric

I have several questions regarding EMW reflection. If it will be helpful, I am thinking about the reflection of powerful EMW and problems like heating associated with it.

Does EMW reflect from dielectric materials?

If so, does it heat up or stay cool due to the absence of current? And if there is no current, where does the energy of non-reflected waves go?

• Light reflects off of glass, so, yes. Commented Feb 29 at 19:58

So, most dielectric materials are modeled with a variable dielectric constant (or tensor) that accounts for polarization of the underlying medium (induced dipoles, etc.) Because energy is required to disturb molecules in a dielectric medium from equilibrium, the dielectric will tend to absorb a certain amount of energy from the field, and this energy will include both elastic (reversible) forms, akin to the energy stored in a harmonic oscillator, as well as inelastic (irreversible/dissipative) forms, akin to the power absorbed by a conductor, except caused by (i) radiation from perturbed electronic orbitals and (at higher intensities) (ii) charge cascades on a molecular or microscopic scale. Explicitly, assuming an isotropic and homogeneous dielectric medium, the reflection and transmission coefficients $$A_r$$ and $$A_t$$ are given (in terms of the incident coefficient $$A_{inc}$$) by \begin{align*} A_{r} = \frac{1 - \alpha}{1+\alpha} A_{inc}\\ A_t = \frac{2}{1 + \alpha} A_{inc} \end{align*} where $$\alpha$$ is the factor by which the wave vector is multiplied under transmission (i.e. the index of refraction: $$\alpha = \frac{c}{c'}$$, where $$c'$$ is the reduced speed of light in the dielectric). Letting $$\alpha = 1+\eta$$, the total reflected and transmitted power is then proportional to \begin{align*} A_r^2+A_t^2 = \frac{4 + \eta^2}{4+2\eta + \eta^2}A_{inc}^2 \end{align*} which approaches one when $$\eta \rightarrow 0$$ (i.e. perfect transmission) and when $$\eta \rightarrow \infty$$ (i.e. perfect reflection), with a minimum at $$\eta = 2$$, which is roughly consistent with what you might expect from the microscopic picture (i.e. that most energy would be lost/absorbed/thermally-reemitted/radiated when the dielectric is responsive, but not so responsive that it screens itself completely, and that energy loss from dipole radiation would be greater at higher frequencies.)

EDIT: After briefly reviewing Griffiths, I (re)learned that it is a happy mathematical accident that the internal energy of the perturbed molecular dipoles in the dielectric medium can be accounted for simply by using the material permittivity and permeability instead of their vacuum counterparts in the normal expression for the energy of an electromagnetic field. Accounting for this energy, the reflected and transmitted power combine to match the incident power. The slightly lower power in the bare field (i.e. not accounting for the internal energy of the dielectric) can be attributed to the process of the wave penetrating the dielectric medium (this can be verified by checking that the reflected and transmitted power also matches the incident power in steady state conditions when the dielectric has finite thickness.)

References:

Introduction to Electrodynamics (3rd. Edition), David J. Griffiths, section 9.3.2

Does EMW reflect from dielectric materials?

Obviously yes. Any time you have a mismatch in material properties between two places where waves can propagate, you will have reflections. Here, the index of refraction $$n=\frac cv$$ is different, so there will be reflections coming just from this.

If so, does it heat up or stay cool due to the absence of current? And if there is no current, where does the energy of non-reflected waves go?

You seem to have misconceptions here. A wave can be transmitted and reflected. If these are the only two possibilities, then the material will not heat up. The energy in the waves will be transmitted and reflected respectively, carrying away all of the initial energy. The transmission will carry the non-reflected waves to wherever it goes.

However, in reality, the medium will absorb a bit of the EM waves on it. It is this absorption that makes the medium heat up, since it is absorbing the energy from the EM waves. This absorption can be, say, due to the resistance of the charged particles inside the material not fully following the EM wave; that then appears as a phase shift from the ideal case, and the index of refraction gains a complex part. Things like these. Phenomena at this level of detail are covered in condensed matter physics, which might be available for you to learn at the end of a university degree. No part of it is easy to learn.