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for example, i have a vague notion that the actual answer is that the permittivity and permisivity are different in each different material, so all waves refract at every boundary, but we only call it that if it makes it out with any real magnitude left which depends on the skin depth or something like that, but is there a simple off the cuff way to estimate based on the ratio of the wavelength and that of the object?

for example, will a radio wave refract through a tree?

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Do you want to know just whether the EMW will be refracted by the object (spoiler: yes) or estimate the refraction index? Is direct measurement allowed? –  Maksim Zholudev Jan 8 '12 at 16:19
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From one engineer to (I suspect) another, I'll give you my practical perspective in this largely theoretical forum.

If you're talking about radio refraction by objects in the real world, the "back of the envelope" or intuitive way to think about it is in terms of conductivity or lossiness of the medium. The concept of skin depth you mention is also useful. In classical EM theory, an EM wave will refract through anything that's not a perfect conductor. However, in real everyday materials, there is the concept of skin depth, which is the spatial rate of exponential falloff of the fields that penetrate the object. It's like the fields go as $e^{-x/\delta}$ as you go $x$ units into a material with $\delta$ skin depth. This is because the wave is robbed of energy through ohmic losses, whereby the wave's energy is used to move around electron currents in the material, which jostle around creating heat energy. Since energy is conserved, the heat generated takes away energy from the EM wave.

The skin depth is related to the conductivity or lossiness of the material you're talking about. You can calculate it for materials if you know the material permittivity, permeability, conductivity, and the frequency of the EM wave. If you're used to decibels, a rule of thumb is that each skin-depth you penetrate into the material takes about 8.7 dB out of the wave (the exact number is $20 \log_{10}(e)$=8.685889638065...).

So, if you're concerned with effects, say, within 90 dB of dynamic range of the incident power levels, then about 10 skin depths is about the largest thing through which you have to worry about refraction. It boils down to picking a cutoff for "what's weak enough" to neglect, then figuring how many skin depths corresponds to that level of attenuation in your particular material and frequency. If your tree is larger than that, you can neglect the refractive propagation through the tree.

Incidentally, very lossy materials also reflect waves at their interface in addition to attenuating what does penetrate them. To be more rigorous, you'd have to include that effect as well. This reflection coefficient is again related to permittivity, permeability, and conductivity. And in fact, higher conductivity means less penetration (more reflection) to begin with, so the bottom line is that if the material is even slightly conductive and the frequency is VHF or higher, you don't usually consider refraction through the material as a propagation mechanism of concern. This doesn't mean that it doesn't physically occur, just that people working in radio don't usually consider this effect when thinking of radio propagation.

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Like rajb245 explained, the refractive properties if a material result from several properties, including conductivity. Other factors include atomic energy levels, dielectric properties (how much the material tends to make dipoles), density, and maybe a few more I can't think about.

If you're not looking for a physical explanation, rather a quick way to know what will happen, you need only look up the complex index if refraction of that material which accounts for all of those factors. Look at this site: http://refractiveindex.info. A material's index of refraction is different for varying wavelengths, so you have to take that into consideration. The real part, $n$, represents how much light is slowed (or sped up, if $n<1$) in the medium, which determines how much refraction occurs as well. The imaginary part, listed there as $k$, relates how much the radiation is absorbed in the medium. The skin depth is how far a wave penetrates before the magnitude of the wave falls off by a factor of $e$. A material with a high $k$ for a given wavelength doesn't just absorb the light, however, it's usually reflected. Silver, for example, has a $k$ of around 3 to 4 for visible light, which it's why used in mirrors. Absorption occurs more due to resonance than conductive properties, but I'm near the limit of my understanding there.

Now to correct a few things about your question: the comparison between the wavelength of the radiation and the physical size of the object it will be passing through doesn't have anything to do with refraction. That is, the refractive properties of a material are independent of how thick or how big the material is. Now, you're thinking about this because you're asking about radio waves. Radio waves have wavelengths on the order of meters to kilometers. While they can pass through materials, and also be reflected and/or refracted by them, they're more likely to bend around the objects than anything else. This, however, is called diffraction, not refraction (see also this wikipedia article about radio propagation).

So, in summary, yes, there are ways to think about materials that can clue us in on how they behave optically, and measured values that contain this information. However, refraction isn't necessarily responsible for the phenomena you're thinking of.

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The photon field is the vector carrier field that bears the electromagnetic force, so electrical charge must and will always interact with any photons it may be coupled to. Likewise for anything with a magnetic moment, for example, a neutron.

So all everyday matter must alter the electromagnetic field and thus can be thought of as having a refractive index, although a refractive index is not really a helpful way of describing this interaction when the wavelength is comparable to the widths of the atoms and molecules that make a substance up. Refractive index as it is wontedly thought of is only meaningful when the photons concerned are coupled to many atoms and molecules in the substance they are interacting with at a time. So if the wavelength is optical, a lone photon is coupled to all the atoms in a region at least of the order of a micron wide, therefore replacing the material by a continuum is a good approximation. But hard X-ray or $\gamma$ photons are coupled to very few matter particles at a time and the material is not a continuum with a refractive index but rather a lattice whose links interact one at a time with the photon concerned.

Some matter absorbs light. It can be thought of as having a refractive index with a big imaginary part. Physically, what is going on is most often that photons are being absorbed by electrons coupled to many other quantum oscillators within the matter. So the photon is less likely to be re-emitted but rather the energy is taken up into the matter as phonons, molecular vibrational energy and the like. The photon is thus lost from the light field.

A tree interacting with a radio wave is an interesting case. Here the wavelength is huge, and, from the farfield, the interaction can be modelled accurately by assuming that every atom in the tree is at the same point. You won't have to account for inhomogeneities and other complex variation of refractive index within the tree, it can be modelled as a point, isotropic scatterer with an amplitude to couple with the field modes; the coupling co-efficient with simply be the sum of all the coupling co-efficients of the atoms involved. Locally, however, things are quite different. The tree's complex spatial matter distribution, which can in principle be accurately represented by a variation of complex refractive index with position, will imprint a detailed "shape" into the scattered field. But, since the length scales over which the tree's matter varies are much, much smaller than the wavelength, these detailed spatial field variations beget evanescent waves, which cannot propagate far from the tree itself. So their effect is confined to the near field. From afar, the tree looks like an dielectric "speck".

The detailed science of calculating refractive indices is not really wonted to me, but it can be done for simple substances by enumerating all the vibrational and other energy states in the matter and working out the scattering matrix elements for the incoming / outgoing photons from these. Some raised states will not re-emit photons, and thus contribute to the refractive index's imaginary part. Others scatter photons elastically, thus contributing to the real part. For most substances, however, one needs to resort to tables such as those at the http://refractiveindex.info site or from glass manufacturer's data, specified as co-efficients for the Sellmeier Formulas, for example: it is too complex to calculate refractive indices for most materials. Refractive index does not only depend on composition, it also depends on the heating and annealing cycle a glass undergoes in its making. When you buy a block of, say Schott N-LASF31A glass, you are not only getting something made to a composition formula, it has to have undergone a detailed heatup cooldown annealing cycle to achieve the refractive index specifications the manufacturer claims.

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