MHB Is \(\nabla \times (f \nabla g) = \nabla f \times \nabla g\)?

[mathmari]

Hey!

I want to show that $\nabla\times (f\nabla g)=\nabla f\times \nabla g$.

We have that $f\nabla g=f\sum\frac{\partial g}{\partial x_i}\hat{x}_i$, therefore we get \begin{align*}&\nabla\times (f\nabla g)=\nabla\times \left (f\sum\frac{\partial g}{\partial x_i}\hat{x}_i \right )\\ & =\nabla\times \left (\sum f\frac{\partial g}{\partial x_i}\hat{x}_i \right ) \\ & =\nabla\times \left (f\frac{\partial g}{\partial x_1}, f\frac{\partial g}{\partial x_2} ,f\frac{\partial g}{\partial x_3} \right ) \\ & = \left [ \frac{\partial}{\partial x_2} \left (f\frac{\partial g}{\partial x_3}\right )-\frac{\partial}{\partial x_3} \left (f\frac{\partial g}{\partial x_2}\right ), \frac{\partial}{\partial x_3} \left (f\frac{\partial g}{\partial x_1}\right )-\frac{\partial}{\partial x_1} \left (f\frac{\partial g}{\partial x_3}\right ), \frac{\partial}{\partial x_1} \left (f\frac{\partial g}{\partial x_2}\right )-\frac{\partial}{\partial x_2} \left (f\frac{\partial g}{\partial x_1}\right )\right ]\\ & = \left [ \frac{\partial}{\partial x_2} \left (f\right )\frac{\partial g}{\partial x_3}+f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_3}\right )-\frac{\partial}{\partial x_3} \left (f\right )\frac{\partial g}{\partial x_2}-f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_2}\right ), \frac{\partial}{\partial x_3} \left (f\right )\frac{\partial g}{\partial x_1}+f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_1}\right )-\frac{\partial}{\partial x_1} \left (f\right )\frac{\partial g}{\partial x_3}-f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_3}\right ), \frac{\partial}{\partial x_1} \left (f\right )\frac{\partial g}{\partial x_2}+f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_2}\right )-\frac{\partial}{\partial x_2} \left (f\right )\frac{\partial g}{\partial x_1}-f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_1}\right )\right ]\\ & = \left [ \frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_3}+f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_3}\right )-\frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_2}-f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_2}\right ), \frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_1}+f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_1}\right )-\frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_3}-f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_3}\right ), \frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_2}+f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_2}\right )-\frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_1}-f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_1}\right )\right ]\\ & = \left [ \frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_3}-\frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_2}, \frac{\partial f}{\partial x_3} \frac{\partial g}{\partial x_1}-\frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_3}, \frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_2}-\frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_1}\right ]+\left [ f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_3}\right )-f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_2}\right ), f\frac{\partial}{\partial x_3} \left (\frac{\partial g}{\partial x_1}\right )-f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_3}\right ), f\frac{\partial}{\partial x_1} \left (\frac{\partial g}{\partial x_2}\right )-f\frac{\partial}{\partial x_2} \left (\frac{\partial g}{\partial x_1}\right )\right ]\end{align*}

Is everything correct so far? How could we continue? (Wondering)

Or is there an other way to show this? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

It looks correct to me, and I'm not aware of a different way to prove it.

To continue we need to realize that:
$$\pd {}{x_2}\left(\pd{g}{x_3}\right)=\pd {}{x_3}\left(\pd{g}{x_2}\right)=\frac {\partial^2g}{\partial x_2 \partial x_3}
$$
See Schwarz's Theorem in Symmetry of second derivatives. (Thinking)

 

[mathmari]

To continue we need to realize that:
$$\pd {}{x_2}\left(\pd{g}{x_3}\right)=\pd {}{x_3}\left(\pd{g}{x_2}\right)=\frac {\partial^2g}{\partial x_2 \partial x_3}
$$
See Schwarz's Theorem in Symmetry of second derivatives. (Thinking)

How could we continue using this property? I got stuck right now... (Wondering)

 

[I like Serena]

How could we continue using this property? I got stuck right now... (Wondering)

Doesn't it make the second term in your last expression become the zero vector?

And which expression are we aiming for? That is, what is $\nabla f \times \nabla g$? (Wondering)

 

[mathmari]

Doesn't it make the second term in your last expression become the zero vector?

Ah yes! (Blush)

And which expression are we aiming for? That is, what is $\nabla f \times \nabla g$? (Wondering)

The expression that is remained is $\nabla f \times \nabla g$, so we are done! (Yes)

Thank you! (Smile)