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.
Best Answer
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.