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Disappearing terms in electrodynamics boundary conditions

  1. Apr 6, 2015 #1
    In the derivation of the boundary conditions we apply the integral form of maxwell's equations, but once we take a very small volume we find that some terms disappear like the displacement current aswell as the time derivative of the magnetic field. Why do these terms disappear? For reference the terms that disappear: first derivative of D wrt time and first derivative of B wrt time, but the conduction current and surface charge dont disappear.

    I looked at many books (Hayt, Pozar, and some online books), and the only argument they put forward is that these terms are finite and the volume/surface/contour is infinitesimal, but this same argument can be used for the terms that dont disappear as well! Please provide an argument that can nit be used against those terms that dont disappear.

    Thank you.
     
  2. jcsd
  3. Apr 6, 2015 #2
    http://imgur.com/ubkgClz image.jpg
     
  4. Apr 6, 2015 #3

    vanhees71

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    The fundamental laws of electromagnetism are the Maxwell equations in differential form.

    Take as an example Gauss's Law for the electric field, here written in terms of macroscopic electrodynamics in Heaviside-Lorentz units
    $$\vec{\nabla} \cdot \vec{D}=\rho$$
    and consider the case of a boundary of some some surface. The singular piece in Gauss's Law is then a surface charge ##\sigma## (charge per unit area) which has a ##\delta##-distribution like singularity with respect to the direction perpendicular to the surface, while a usual "bulk" charge distribution ##\rho'## (charge per unit volume) is not more singular than perhaps a jump along the surface (e.g., a charged body in air or in a vacuum). Then you integrate along an infinitesimal box with two of its surfaces parallel to the surface, one on each side of the surface. Then using Gauss's integral theorem for such a box gives
    $$\int_{\partial \delta V} \mathrm{d} \vec{F} \cdot \vec{D}=\vec{n} \delta F \cdot (\vec{D}_{>}-\vec{D}_{<})=\delta F \sigma + \delta F \delta x_{\perp} \rho'.$$
    Now you divide by ##\delta F## and let ##\delta x \rightarrow 0##, which gives the boundary condition
    $$\vec{n} \cdot (\vec{D}_{>}-\vec{D}_{<})=\sigma.$$
    Above the symbols have the following meaning: ##\delta F## is the piece of the boundary of the infinitesimal box parallel to the surface under investigation, ##\vec{n}## is the unit vector perpendicular to the surface on one side (in Gauss's integral theorem it's always pointing out of the volume over which you integrate), then ##-\vec{n}## is the corresponding surface normal unit vector on the other side of the surface, and ##\vec{D}_{>}## and ##\vec{D}_{<}## are the values of ##\vec{D}## on the surface taken as limits from the one and the other side, respectively. Finally ##\delta x_{\perp}## is the infinitesimal height of the infinitesimal box perpendicular to the surface. As you see, the normal component of ##\vec{D}## jumps along the surface, and the size of this jump is given by the surface charge (charge per unit area), while the "bulk charge distribution" doesn't contribute.

    The same technique shows that ##\vec{D}## itself has at most jumps. The only exception is the presence of a point charge (making a singularity in ##\rho## of the type ##\delta^{(3)}(\vec{x}-\vec{x}_0##) or a line-charge distribution (charge along a line with unit charge per length, where you have a ##\delta^{(2)}##-distribution-like singularity in ##\rho##), in which case you have singularities at the position of the point charge and along the charged line.
     
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