Physics of Continuous Media : Problems and Solutions in by Grigory Vekstein

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By Grigory Vekstein

This publication offers a superb and exemplary choice of up to date workouts and their strategies on non-stop media, masking quite a lot of themes from electro-, magnetohydro- and fluid dynamics, and from the speculation of elasticity. the writer is a global professional with decades of study and instructing event within the box. every one bankruptcy starts off with a complete precis of definitions and the Read more...

summary: This publication provides a superb and exemplary selection of up to date routines and their ideas on non-stop media, masking quite a lot of themes from electro-, magnetohydro- and fluid dynamics, and from the idea of elasticity. the writer is a world professional with decades of analysis and educating event within the box. each one bankruptcy starts off with a complete precis of definitions and the mathematical description of the actual legislation essential to comprehend and clear up the sequence of difficulties that persist with. the issues and routines are a steady outfitted up in all the subject

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31), the angle θ1 between the vector S and the optical axis is given by tan θ1 = − Ez ǫ⊥ = tan θ Ex ǫ Therefore, if ǫ⊥ = ǫ , the two angles are equal to each other only when θ = 0, π/2. 11 Derive the reflection coefficient for a linearly polarized electromagnetic wave at a plane boundary between free space and a dielectric medium with dielectric permittivity ǫ and magnetic permeability µ. 3. The geometry of reflection and refraction is completely determined by the spatial invariance of the system in the (x, y) plane and its temporal invariance (no variation with time).

One can always choose the coordinate in such a way that the z-axis is directed along the optical axis of the crystal, while the wave vector k is lying in the (x, z) plane: k = (kx , 0, kz ). 30) The respective dispersion equation, Det||Lik || = 0, takes the form −k 2 + ω2 ǫ⊥ c2 −kz2 + ω2 ǫ⊥ c2 −kx2 + ω2 ǫ c2 − kx2 kz2 = 0, which yields two possible solutions. The first one is −k 2 + (ω 2 /c2 )ǫ⊥ = 0. It corresponds to the ordinary wave, whose refractive index is equal to 1/2 n = ǫ⊥ and does not depend on the direction of the wave vector k.

Thus, only the field components By , Ex , Ez are present. 36), which in free space and in the dielectric reads −κ1 By = iω iω Ex , κ2 By = ǫ(ω)Ex c c Since the field components By and Ex are both continuous at the boundary z = 0, the solvability condition of the above relation is κ2 = −ǫ(ω)κ1 . Therefore, such a surface wave can exist only in the frequency range where the permittivity of the dielectric, ǫ(ω), is negative. 4. In the limiting case of short wavelength, when kc/ωpe ≫ 1, such a wave becomes almost electrostatic, so that the magnetic field B is much weaker than the electric one, E.

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