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Divergence as the limit of a surface integral a volume->0

  1. Dec 16, 2015 #1
    The following is my interpretation of the development of the divergence of a vector field given by Joos:

    $$dy dz dv_x=dy dz\left(v_x(dx)-v_x(0)\right)=dy dz\left(v_x(0)+dx\frac{\partial v_x}{\partial x}(0)- v_x(0)\right)$$
    $$=dy dz dx\frac{\partial v_x}{\partial x}(0)=d\tau \frac{\partial v_x}{\partial x}(0)$$
    $$\oint \mathfrak{v}\cdot d\mathfrak{S}=\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)d\tau$$
    $$\oint \mathfrak{v}\cdot d\mathfrak{S}=d\tau \overset{\rightharpoonup }{\nabla}\cdot \mathfrak{v}$$

    I know I am picking nits here, but I want to understand what a rigorous development would be. I contend that ##dx, dy, dz## are independent real number variables of arbitrary magnitude, and ##dv_x\equiv dx \frac{\partial v_x}{\partial x}##. Also ##\Delta v_x \equiv v_x(dx)-v_x(0)##.

    Introducing ##\varepsilon_x \equiv \frac{\Delta v_x}{\Delta x}-\frac{\partial v_x}{\partial x}## and ##\Delta x = dx##, we can write the first equation as:

    $$dy dz dv_x=dy dz\left(v_x(dx)-v_x(0)-\varepsilon_x dx\right)=dy dz\left(v_x(0)+dx\frac{\partial v_x}{\partial x}(0)- v_x(0)\right)$$

    Using ##\Delta \mathfrak{v} \equiv \mathfrak{v}(dx \hat{ i } + dy \hat{ j } + dz \hat{ k })-\mathfrak{v}(0)##, the surface integral should be:

    $$\oint \mathfrak{v}\cdot d\mathfrak{S}=d\tau \Delta \mathfrak{v} \cdot ( \hat{ i } + \hat{ j } + \hat{ k })$$

    Using the same approach as above, we could write this as:

    $$\oint \mathfrak{v}\cdot d\mathfrak{S}=\left(\frac{\partial v_x}{\partial x}+\varepsilon_x+\frac{\partial v_y}{\partial y}+\varepsilon_y+\frac{\partial v_z}{\partial z}+\varepsilon_z\right)d\tau$$

    In order to make the final original equation rigorous, we would need to express it as a limit.
    $$\lim_{d\tau \rightarrow 0}\frac{1}{d\tau}\oint \mathfrak{v}\cdot d\mathfrak{S}= \overset{\rightharpoonup }{\nabla}\cdot \mathfrak{v}$$

    This would be much easier to follow if I could provide drawings, etc. Is my reasoning correct in what I have presented here?
  2. jcsd
  3. Dec 21, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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