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Curl as the limit vol->0 of a surface integral

  1. Dec 20, 2015 #1
    Joos asserts on page 31 https://books.google.com/books?id=btrCAgAAQBAJ&lpg=PP1&pg=PA31#v=onepage&q&f=false that

    $$\nabla \times \mathfrak{v} = \lim_{\Delta \tau \to 0} \frac{1}{\Delta \tau }\oint d\mathfrak{S}\times \mathfrak{v}$$

    I tried to demonstrate this, and neglected to place the surface element before the vector. My result seems to show that Joos's equation should have a negative sign on one side or the other.

    My development assumes a finite cube centered on the origin with dimensions ##dx dy dz = d\tau## sufficiently small that deviations from the value of ##\mathfrak{v}(0,0,0)## are approximately linear. I sum the cross products of the field vector evaluated at the center of opposing faces with the surface element representing that face. Fiddle around with things, and end up with the curl.

    Can someone tell me if I am doing something wrong here, and if so, what that something is?

    $$\oint \mathfrak{v}\times d\mathfrak{S} \approx \left(\mathfrak{v}\left(\frac{dx}{2},0,0\right)\times \hat{\mathfrak{i}}+\mathfrak{v}\left(-\frac{dx}{2},0,0\right)\times \left(-\hat{\mathfrak{i}}\right)\right)dy dz
    +\left(\mathfrak{v}\left(0,\frac{dy}{2},0\right)\times \hat{\mathfrak{j}}+\mathfrak{v}\left(0,-\frac{dy}{2},0\right)\times \left(-\hat{\mathfrak{j}}\right)\right)dx dz
    +\left(\mathfrak{v}\left(0,0,\frac{dz}{2}\right)\times \hat{\mathfrak{k}}+\mathfrak{v}\left(0,0,-\frac{dz}{2}\right)\times \left(-\hat{\mathfrak{k}}\right)\right)dx dy$$

    $$= \left(\mathfrak{v}\left(\frac{dx}{2},0,0\right)-\mathfrak{v}\left(-\frac{dx}{2},0,0\right)\right)\times \hat{\mathfrak{i}}dy dz
    +\left(\mathfrak{v}\left(0,\frac{dy}{2},0\right)-\mathfrak{v}\left(0,-\frac{dy}{2},0\right)\right)\times \hat{\mathfrak{j}}dx dz
    +\left(\mathfrak{v}\left(0,0,\frac{dz}{2}\right)-\mathfrak{v}\left(0,0,-\frac{dz}{2}\right)\right)\times \hat{\mathfrak{k}}dx dy $$

    $$\approx \left(\left(\mathfrak{v}(0,0,0)+\frac{\partial }{\partial x}\mathfrak{v}(0,0,0)\frac{dx}{2}\right)-\left(\mathfrak{v}(0,0,0)-\frac{\partial }{\partial x}\mathfrak{v}(0,0,0)\frac{dx}{2}\right)\right)\times \hat{\mathfrak{i}}dydz
    +\left(\left(\mathfrak{v}(0,0,0)+\frac{\partial }{\partial y}\mathfrak{v}(0,0,0)\frac{dy}{2}\right)-\left(\mathfrak{v}(0,0,0)-\frac{\partial }{\partial y}\mathfrak{v}(0,0,0)\frac{dy}{2}\right)\right)\times \hat{\mathfrak{j}}dzdx
    +\left(\left(\mathfrak{v}(0,0,0)+\frac{\partial }{\partial z}\mathfrak{v}(0,0,0)\frac{dz}{2}\right)-\left(\mathfrak{v}(0,0,0)-\frac{\partial }{\partial z}\mathfrak{v}(0,0,0)\frac{dz}{2}\right)\right)\times \hat{\mathfrak{k}}dxdy$$

    $$=\left(\frac{\partial }{\partial x}\mathfrak{v}(0,0,0)dx\right)\times \hat{\mathfrak{i}}dydz
    +\left(\frac{\partial }{\partial x}\mathfrak{v}(0,0,0)dy\right)\times \hat{\mathfrak{j}}dzdx
    +\left(\frac{\partial }{\partial x}\mathfrak{v}(0,0,0)dz\right)\times \hat{\mathfrak{k}}dxdy$$

    $$=\left(\left(\frac{\partial }{\partial x}v_x\hat{\mathfrak{i}}+\frac{\partial }{\partial y}v_x\hat{\mathfrak{j}}+\frac{\partial }{\partial z}v_x\hat{\mathfrak{k}}\right)\times \hat{\mathfrak{i}}+\left(\frac{\partial }{\partial x}v_y\hat{\mathfrak{i}}+\frac{\partial }{\partial y}v_y\hat{\mathfrak{j}}+\frac{\partial }{\partial z}v_y\hat{\mathfrak{k}}\right)\times \hat{\mathfrak{j}}+\left(\frac{\partial }{\partial x}v_z\hat{\mathfrak{i}}+\frac{\partial }{\partial y}v_z\hat{\mathfrak{j}}+\frac{\partial }{\partial z}v_z\hat{\mathfrak{k}}\right)\times \hat{\mathfrak{k}}\right)d\tau$$

    $$=\left(-\frac{\partial }{\partial y}v_x\hat{\mathfrak{k}}+\frac{\partial }{\partial z}v_x\hat{\mathfrak{j}}+\frac{\partial }{\partial x}v_y\hat{\mathfrak{k}}-\frac{\partial }{\partial z}v_y\hat{\mathfrak{i}}-\frac{\partial }{\partial x}v_z\hat{\mathfrak{j}}+\frac{\partial }{\partial y}v_z\hat{\mathfrak{i}}\right)d\tau$$

    $$=\left(\left(\frac{\partial }{\partial y}v_z-\frac{\partial }{\partial z}v_y\right)\hat{\mathfrak{i}}+\left(\frac{\partial }{\partial z}v_x-\frac{\partial }{\partial x}v_z\right)\hat{\mathfrak{j}}+\left(\frac{\partial }{\partial x}v_y-\frac{\partial }{\partial y}v_x\right)\hat{\mathfrak{k}}\right)d\tau$$

    $$\lim_{d\tau \to 0} \frac{1}{d\tau }\oint \mathfrak{v}\times d\mathfrak{S}=\left(\frac{\partial }{\partial y}v_z-\frac{\partial }{\partial z}v_y\right)\hat{\mathfrak{i}}+\left(\frac{\partial }{\partial z}v_x-\frac{\partial }{\partial x}v_z\right)\hat{\mathfrak{j}}+\left(\frac{\partial }{\partial x}v_y-\frac{\partial }{\partial y}v_x\right)\hat{\mathfrak{k}}$$
     
  2. jcsd
  3. Dec 20, 2015 #2
    Here I've reversed the order of the cross product terms, and followed an abbreviation of the approach shown above. This seems to give the result Joos published. My mind has gone numb. It may all be obvious to me after a good night's sleep, but right now, I'm not seeing how these differ.

    $$dx dS_x \hat{i}\times \frac{\partial \overset{\rightharpoonup }{v}}{\partial x}+dy dS_y \hat{j}\times \frac{\partial \overset{\rightharpoonup }{v}}{\partial y}+dz dS_z \hat{k}\times \frac{\partial \overset{\rightharpoonup }{v}}{\partial z}$$

    $$d\tau \left(\hat{i}\times \frac{\partial \overset{\rightharpoonup }{v}}{\partial x}+\hat{j}\times \frac{\partial \overset{\rightharpoonup }{v}}{\partial y}+\hat{k}\times \frac{\partial \overset{\rightharpoonup }{v}}{\partial z}\right)$$

    $$d\tau \left(\frac{\partial \left(\hat{j} v_x-\hat{i} v_y\right)}{\partial z}+\frac{\partial \left(\hat{i} v_z-\hat{k} v_x\right)}{\partial y}+\frac{\partial \left(\hat{k} v_y-\hat{j} v_z\right)}{\partial x}\right)$$

    $$d\tau \left(-\hat{i} \frac{\partial v_y}{\partial z}+\hat{i} \frac{\partial v_z}{\partial y}+\hat{j} \frac{\partial v_x}{\partial z}-\hat{j} \frac{\partial v_z}{\partial x}-\hat{k} \frac{\partial v_x}{\partial y}+\hat{k} \frac{\partial v_y}{\partial x}\right)$$

    $$d\tau \left(\hat{i} \left(\frac{\partial v_z}{\partial y}-\frac{\partial v_y}{\partial z}\right)+\hat{j} \left(\frac{\partial v_x}{\partial z}-\frac{\partial v_z}{\partial x}\right)+\hat{k} \left(\frac{\partial v_y}{\partial x}-\frac{\partial v_x}{\partial y}\right)\right)$$
     
  4. Dec 20, 2015 #3
    Ok, I think I see it, but Whole Foods Market on P Street in Washington DC told me I am "camping out" and cannot not stay. I've been doing this for years, but I said something to the management about their environmental control system wasting energy and keeping the area uncomfortably warm. I'll pick this up tomorrow.
     
  5. Dec 21, 2015 #4
    OK, I think this is correct. My original error makes me wonder at what level of thinking I transposed ideas. I'm leaving it as a puzzle for the curious.

    Some preliminary identities:
    $$\hat{\mathfrak{i}}\times \mathfrak{v}=\hat{\mathfrak{k}} v_y-\hat{\mathfrak{j}} v_z \text{; }
    \hat{\mathfrak{j}}\times \mathfrak{v}=\hat{\mathfrak{i}} v_z-\hat{\mathfrak{k}} v_x \text{; }
    \hat{k}\times \mathfrak{v}=\hat{\mathfrak{j}} v_x-\hat{\mathfrak{i}} v_y $$

    I want to resolve this expression onto rectangular Cartesian coordinates:

    $$\nabla \times \mathfrak{v}\equiv \lim_{\Delta \tau \to 0} \frac{1}{\Delta \tau }\oint d\mathfrak{S}\times \mathfrak{v}$$

    Sum the contributions ##d\mathfrak{S}_i\times \mathfrak{v}## of each face of a small, finite rectangular solid ##dx dy dz = d \tau##, centered at the origin.

    $$\oint d\mathfrak{S}\times \mathfrak{v}\approx dS_x \left(\hat{\mathfrak{i}}\times \mathfrak{v}\left(\frac{dx}{2},0,0\right)-\hat{\mathfrak{i}}\times \mathfrak{v}\left(-\frac{dx}{2},0,0\right)\right)+dS_y \left(\hat{\mathfrak{j}}\times \mathfrak{v}\left(0,\frac{dy}{2},0\right)-\hat{\mathfrak{j}}\times \mathfrak{v}\left(0,-\frac{dy}{2},0\right)\right)+dS_z \left(\hat{\mathfrak{k}}\times \mathfrak{v}\left(0,0,\frac{dz}{2}\right)-\hat{\mathfrak{k}}\times \mathfrak{v}\left(0,0,-\frac{dz}{2}\right)\right)$$

    (* Replace the values on the faces with differential approximations.*)

    $$\approx \hat{\mathfrak{i}}\times \left(\left(\mathfrak{v}+\frac{\partial }{\partial x}\mathfrak{v}\frac{dx}{2}\right)-\left(\mathfrak{v}-\frac{\partial }{\partial x}\mathfrak{v}\frac{dx}{2}\right)\right)dydz+\hat{\mathfrak{j}}\times \left(\left(\mathfrak{v}+\frac{\partial }{\partial y}\mathfrak{v}\frac{dy}{2}\right)-\left(\mathfrak{v}-\frac{\partial }{\partial y}\mathfrak{v}\frac{dy}{2}\right)\right)dzdx+\hat{\mathfrak{k}}\times \left(\left(\mathfrak{v}+\frac{\partial }{\partial z}\mathfrak{v}\frac{dz}{2}\right)-\left(\mathfrak{v}-\frac{\partial }{\partial z}\mathfrak{v}\frac{dz}{2}\right)\right)dxdy|_{\mathfrak{v}=\mathfrak{v}(0,0,0)}$$

    (*Simplify terms.*)

    $$=\hat{\mathfrak{i}}\times \frac{\partial }{\partial x}\mathfrak{v}dxdydz+\hat{\mathfrak{j}}\times \frac{\partial }{\partial y}\mathfrak{v}dydzdx+\hat{\mathfrak{k}}\times \frac{\partial }{\partial z}\mathfrak{v}dzdxdy$$

    (*Since the basis vectors are constant, the cross products can be preformed prior to partial differentiation.*)

    $$=d\tau \left(\frac{\partial \left(\hat{\mathfrak{j}} v_x-\hat{\mathfrak{i}} v_y\right)}{\partial z}+\frac{\partial \left(\hat{\mathfrak{i}} v_z-\hat{\mathfrak{k}} v_x\right)}{\partial y}+\frac{\partial \left(\hat{\mathfrak{k}} v_y-\hat{\mathfrak{j}} v_z\right)}{\partial x}\right)$$
    Rearranging terms and and taking the limit gives the desired result.

    $$\lim_{d\tau \to 0} \frac{1}{d\tau }\oint d\mathfrak{S}\times \mathfrak{v}=\left(\frac{\partial }{\partial y}v_z-\frac{\partial }{\partial z}v_y\right)\hat{\mathfrak{i}}+\left(\frac{\partial }{\partial z}v_x-\frac{\partial }{\partial x}v_z\right)\hat{\mathfrak{j}}+\left(\frac{\partial }{\partial x}v_y-\frac{\partial }{\partial y}v_x\right)\hat{\mathfrak{k}}$$

    Any critiques are welcome.
     
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