# Vector Calculus Problem on Gradient Cross Product

$$Problem:$$ If a vector function $$V=V(x,y,z)$$ is not irrotational, show that if there exists a scalar function $$g=g(x,y,z)$$ such that $$gV$$ is irrotational, then $$V\cdot (\nabla \times V )=0$$

Remember, $$V$$ Irrotational $$\iff \nabla\times V=0$$

My attempt:

Since $$\nabla\times V\not=0$$, $$\nabla\times V$$ is a vector that is orthogonal to $$V$$, and so $$V\cdot \nabla\times V=0$$ by definition of the dot product.

I am totally unconvinced this can be the answer as nowhere did I use the fact that there exists a scalar function $$g$$ in my solution.

Any help is greatly appreciated!

• V isn’t irrotational. gV is. – Jake Rose Oct 20 '19 at 15:12
• $V$ not irrotational means $V$ is rotational... did I miss something? @JakeRose – Kam Oct 20 '19 at 15:15
• @JakeRose So what does that tell me? – Kam Oct 20 '19 at 15:25
• Might Mathematics be better suited for this math question? – Kyle Kanos Oct 20 '19 at 15:38
• – Qmechanic Oct 20 '19 at 15:43

Since $$g\vec{V}$$ is irrotational, $$\nabla\times(g\vec{V}) = 0$$. Therefore, $$\nabla g \times \vec{V} + g\nabla\times\vec{V} = 0$$ or $$\vec{V} \times \nabla g = g\nabla\times\vec{V}$$. Now take the dot product with $$\vec{V}$$ so that $$$$\vec{V}\cdot(\vec{V} \times \nabla g) = g\vec{V} \cdot (\nabla\times\vec{V}).$$$$ The left hand side is zero. Since $$g$$ is not identically zero, it follows that $$\vec{V} \cdot (\nabla\times\vec{V}) = 0$$.
• Thank you for your answer! Two things: Why is $g$ not identically zero? And why is the left hand side zero? I appreciate any sort of help I can get :) – Kam Oct 20 '19 at 15:23
• If $g$ is zero, so will be $gV$ and such a function will not be terribly interesting to examine further. Left-hand size is zero because $\vec{V}\cdot(\vec{V} \times \nabla{g}) = (\vec{V} \times \vec{V})\cdot\nabla{g} = 0\cdot\nabla{g} = 0$. – Amey Joshi Oct 20 '19 at 15:25