Derivation of Divergence in Cartesian Coordinates

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23
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In section 1-5 of the third edition of Foundations of Electromagnetic Theory by Reitz, Milford and Christy, the authors give a coordinate-system-independent definition of the divergence of a vector field:

$$\nabla\cdot\mathbf{F} = \lim_{V\rightarrow 0}\frac{1}{V}\int_S\mathbf{F\cdot n}da$$

with ##S## the surface enclosing the volume ##V##.

In deriving the Cartesian form of the divergence they use ##\Delta x\Delta y\Delta z## as a volume element over which to make what appear to be first order Taylor expansions:

$$F_x(x_0 + \Delta x, y, z) = F_x(x_0, y, z) + \Delta x\frac{\partial F_x}{\partial x}\Bigg\arrowvert_{x_0, y, z}$$
$$F_y(x, y_0 + \Delta y, z) = F_y(x, y_0, z) + \Delta y\frac{\partial F_y}{\partial y}\Bigg\arrowvert_{x, y_0, z}$$
$$F_z(x, y, z_0 + \Delta z) = F_z(x, y, z_0) + \Delta z\frac{\partial F_z}{\partial z}\Bigg\arrowvert_{x, y, z_0}$$

I say "appear to be" because the zero order term and all derivatives in a Taylor expansion are evaluated at the initial point ##(x_0, y_0, z_0)##. In the next step they plug the above and ##\mathbf{F}(x_0, y_0, z_0)## into the divergence equation, with the six sides of ##\Delta x\Delta y\Delta z## as the surface:

$$\nabla\cdot\mathbf{F} = \lim_{V\rightarrow 0}\frac{1}{\Delta x\Delta y\Delta z}\Bigg\{\int F_x(x_0, y, z)dydz + \Delta x\Delta y\Delta z\frac{\partial F_x}{\partial x}\\ + \int F_y(x, y_0, z)dxdz + \Delta x\Delta y\Delta z\frac{\partial F_y}{\partial y}\\ + \int F_z(x, y, z_0)dxdy + \Delta x\Delta y\Delta z\frac{\partial F_z}{\partial z}\\ - \int F_x(x_0, y, z)dydz - \int F_y(x, y_0, z)dxdz - \int F_z(x, y, z_0)dxdy\Bigg\}$$

Is there anything to leaving terms in the form ##F_x(x_0, y, z)## instead of ##F_x(x_0, y_0, z_0)## and ##\int F_x(x_0, y, z)dydz## instead of ##F_x(x_0, y_0, z_0)\Delta y\Delta z##, especially since the partials are treated like constants which get multiplied by the areas of the sides outright?
 

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  • #2
Orodruin
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If you replaced then you would need to worry about your integration introducing terms that do not vanish in the limit, but the integrals cancel so why worry about doing an approximation that you will need to justify when you already have the result?
 
  • #3
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If, in composing the Taylor expansion, you evaluate all the zero-order terms and partials at the point ##(x_0, y_0, z_0)## then would not the integrals have only constants over which to integrate, leaving only those constants multiplied by the area? But the expansions I copied above show the terms as functions varying over their respective surfaces. I understand that the integrals get negated but I balked at seeing the terms presented as functions varying over a surface instead of evaluated at a point and therefore constant.

Note that they show the partials in the first set of equations also as functions of two of the three space dimensions (eg ##\frac{\partial F_x}{\partial x}\Big\arrowvert_{x_0, y, z}## instead of ##\frac{\partial F_x}{\partial x}\Big\arrowvert_{x_0, y_0, z_0}##), but they get treated as constants in the second set.

Have I read too much into the symbols here? Two other similar discussions of the divergence, in Arfken and Boas, don't have this problem.
 
  • #4
Orodruin
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then would not the integrals have only constants over which to integrate
No. The first order corrections in y and z would depend on y and z for the constant x sides and so on. The point renains that there is zero reason to introduce those approximations because those terms cancel exactly with each other anyway.
 
  • #5
23
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OK. The ##F_x(x_0, y, z)## term must subsume ##(y-y_0)\frac{\partial F_x}{\partial y}\Big\arrowvert_{x_0, y_0, z_0}## and ##(z-z_0)\frac{\partial F_x}{\partial z}\Big\arrowvert_{x_0, y_0, z_0}##. Thanks.
 

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