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In all physics courses we are taught that the divergence of a curl is always zero: $\nabla \cdot(\nabla \times\vec{V}) = 0$

So to prove this to myself I simply solve it to get $0$, but I am not coming up to zero. Can someone please point out where my mistake is?

\begin{align*} \nabla \times \vec{V}&= \begin{vmatrix}\hat{x} &\hat{y} &\hat{z} \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix}\\ &=\left( \frac{\partial}{\partial y}V_z - \frac{\partial}{\partial z}V_y\right)\hat{x} + \left( \frac{\partial}{\partial x}V_z - \frac{\partial}{\partial z}V_x\right)\hat{y} + \left( \frac{\partial}{\partial x}V_y - \frac{\partial}{\partial y}V_x\right)\hat{z}\\ &=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}V_z - \frac{\partial}{\partial z}V_y\right)\hat{x} + \frac{\partial}{\partial y}\left( \frac{\partial}{\partial x}V_z - \frac{\partial}{\partial z}V_x\right)\hat{y} + \frac{\partial}{\partial z}\left( \frac{\partial}{\partial x}V_y - \frac{\partial}{\partial y}V_x\right)\hat{z}\\ &= \frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_y}{\partial x \partial z} + \frac{\partial^2 V_z}{\partial y \partial x} - \frac{\partial^2 V_x}{\partial y \partial z} + \frac{\partial^2 V_y}{\partial x \partial z} - \frac{\partial^2 V_x}{\partial z \partial y} \end{align*}

Only one term cancels out the $\frac{\partial^2 V_y}{\partial x \partial z}$? Leaving me with:

$$ \nabla\cdot(\nabla \times\vec{V}) = 2\left (\frac{\partial^2 V_z}{\partial x \partial y} - \frac{\partial^2 V_x}{\partial z \partial y}\right ). $$

I know I have screwed up somewhere, but I have been staring at my work for a while now and I can't seem to figure out where I went wrong. All terms should cancel out...

Thank you.

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  • $\begingroup$ In your expression for the curl, should the second term be negated? $\endgroup$ – littleO Apr 19 '17 at 14:39
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    $\begingroup$ It's much easier to do this sort of thing in index notation with summation convention: $(\nabla \times V)_i = \epsilon_{ijk} \partial_j V_k$, then $\nabla \cdot (\nabla \times V) = \epsilon_{ijk} \partial_i \partial_j V_k = 0$ because the $\epsilon$ is antisymmetric when swapping $i$ and $j$ and the $\partial_i\partial_j$ is symmetric since partial derivatives commute. Moreover, it tells you the signs of the middle term correctly: $(\nabla \times V)_2 = \epsilon_{2jk} \partial_j V_k = \epsilon_{231}\partial_3V_1-\epsilon_{213}\partial_1V_3 = \partial_3 V_1 - \partial_1 V_3$. $\endgroup$ – Chappers Apr 19 '17 at 14:46
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The correct form for curl is $$ \nabla \times \vec{V}=\left( \frac{\partial}{\partial y}V_z - \frac{\partial}{\partial z}V_y\right)\hat{x} + \underbrace{\left(\frac{\partial}{\partial z}V_x - \frac{\partial}{\partial x}V_z\right)}_{\text{note the switch}}\hat{y} + \left( \frac{\partial}{\partial x}V_y - \frac{\partial}{\partial y}V_x\right)\hat{z} $$ Another version of curl (which is convenient for this proof) is $$ \nabla \times \vec{V} = \hat x \times \frac{\partial V}{\partial x} + \hat y \times \frac{\partial V}{\partial y} + \hat z \times \frac{\partial V}{\partial z} $$ Similarly, divergence is given as $$ \nabla \cdot \vec{V} = \hat x \cdot \frac{\partial V}{\partial x} + \hat y \cdot \frac{\partial V}{\partial y} + \hat z \cdot \frac{\partial V}{\partial z} $$

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  • $\begingroup$ Ohhhhhh my goodness. Thank you I have been switching that middle one.. Thank you that clears it all up. I appreciate it! $\endgroup$ – macas Apr 19 '17 at 14:45
  • $\begingroup$ @macas glad to help. $\endgroup$ – Omnomnomnom Apr 19 '17 at 14:50
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You have a sign error in your middle term in $\nabla \times \vec{V}$ (the coefficient of $\hat{y}$). If you're calculating the determinant by expansion by minors, remember that the signs of the terms need to alternate; see eq. (2) in the above link.

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