Are EM waves reflected by inducing Hertzian dipoles?

AI Thread Summary
The discussion centers on the reflection of electromagnetic (EM) waves through the induction of Hertzian dipoles in materials. It explores how atomic and molecular structures, particularly dipole moments, contribute to this reflection mechanism. While some participants argue that localized dipoles are not necessary for reflection, they acknowledge that mobile electrons in metals play a crucial role in re-radiating EM waves. The conversation also touches on the reflective properties of materials like quartz compared to metals, noting that although quartz can reflect light, it is not as efficient as metals due to its lower relative permittivity. Overall, the interaction of EM waves with both bound and unbound charges in materials is key to understanding reflection.
Kara386
Messages
204
Reaction score
2
A comment a lab script for a recent experiment I did noted that the mechanism by which reflection of EM waves occurs is through the induction of Hertzian dipoles in a material.

Having read up on Hertzian dipoles, I have found nothing which discusses them in the context of reflection. I read that a Hertzian dipole is essentially two small spherical conductors connected by a wire, with charge flowing periodically between them, or something that can be approximated by this. I find it very hard to understand how, for example, a plank of wood would support or even vaguely resemble such a system.

To summarise, can someone explain how Hertzian dipoles are the mechanism by which EM waves are reflected, or point me to resources that explain this? What bit corresponds to the spheres and which bit to the wire?
 
Physics news on Phys.org
You must look at the molecular or atomic structure of the reflective material. The atom or molecule will have a dipole moment. That is, one end will be slightly more negative and the other end will be slightly more positive. The incident EM wave operates on the dipole by moving it. As it returns to its normal position it re-radiates the EM wave thus reflecting it.
 
billslugg said:
You must look at the molecular or atomic structure of the reflective material. The atom or molecule will have a dipole moment. That is, one end will be slightly more negative and the other end will be slightly more positive. The incident EM wave operates on the dipole by moving it. As it returns to its normal position it re-radiates the EM wave thus reflecting it.
I don't think the atomic structure is relevant here because, for instance, in a metal we can consider it to be filled with a dense cloud of mobile electrons. The electrons will move under the influence of the incident electric field in a uniform way over whatever area is involved, and will produce re-radiation. There are, therefore, no localised dipoles involved. However, if wishing to analyse the radiation pattern of a reflector using antenna array theory, the surface may be divided into rectangles about half x quarter wavelength, and each of these may be considered to be a dipole radiator forming part of a large array.
 
Yes, a metal has a very large number of unbound electrons which are available for collision with a photon. There are also many bound electrons. Thermal considerations determine the ratios. Any charge will be moved by an EM wave. On non dipole atomic structures, the two charges cancel out and no motion happens. A dipole atom or moleculre in an atomic lattice can collide with and re-emit photons from any of the charge centers.

If only unbound electrons could interact with photons, poor conductors such as quartz should not reflect light well. The opposite is true.
 
billslugg said:
Yes, a metal has a very large number of unbound electrons which are available for collision with a photon. There are also many bound electrons. Thermal considerations determine the ratios. Any charge will be moved by an EM wave. On non dipole atomic structures, the two charges cancel out and no motion happens. A dipole atom or moleculre in an atomic lattice can collide with and re-emit photons from any of the charge centers.

If only unbound electrons could interact with photons, poor conductors such as quartz should not reflect light well. The opposite is true.
Is the last statement really true? Is quartz a good reflector? It is, after all, used for lenses. The relative permittivity determines the reflection factor, and most dielectrics have a relative permittivity of less than ten, so the reflection cannot be as good as a metal.
Also, am I correct that a heavy charge, such as a positive ion, will not be moved easily an EM wave, but the associated electrons will respond easily?
 
Quartz is a good reflector but not as good as metals.

The incident EM wave will move positive charges bonded to the lattice. Since their effective masses are very large, dut to being locked in a lattice, their recoil velocity and distance is really small. It is essentially recoiless. Very similar to Mossbauer absorption.

The free electrons, being unbound, will respond quickly. This is why an EM wave cannot penetrate more than a few microns into any metal.
 
This is from Griffiths' Electrodynamics, 3rd edition, page 352. I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##. In matrix form, this tensor should look like this...
Thread 'Applying the Gauss (1835) formula for force between 2 parallel DC currents'
Please can anyone either:- (1) point me to a derivation of the perpendicular force (Fy) between two very long parallel wires carrying steady currents utilising the formula of Gauss for the force F along the line r between 2 charges? Or alternatively (2) point out where I have gone wrong in my method? I am having problems with calculating the direction and magnitude of the force as expected from modern (Biot-Savart-Maxwell-Lorentz) formula. Here is my method and results so far:- This...
Back
Top