How can vector calculus 'del' relationships be derived using other identities?

[member 428835]

hey all!

i was hoping someone could either state an article or share some knowledge with a way (if any) to derive the vector calculus "del" relationships. i.e. $$ \nabla \cdot ( \rho \vec{V}) = \rho (\nabla \cdot \vec{V}) + \vec{V} \cdot (\nabla\rho)$$

now i do understand this to be like the product rule but is there any way to derive this in a similar fashion to other vector calculus identities, perhaps using the Kronecker delta or the permutation epsilon?

as an example, i can derive almost any trig identity using $$e^{i \theta}=\cos \theta+i \sin \theta$$ by making changes to \theta, perhaps letting \theta = \alpha + \beta and then equating real and imaginary parts and doing a little bit of algebra.

is there anything like this for vector identities (perhaps not a base equation, but a method?)

thanks for your help!

 

[jedishrfu]

If you use x,y,z components for the vector then do some rearrangements you can prove it.

 

[WannabeNewton]

Pretty much every curl, divergence, and gradient identity can be proven easily and elegantly using indices. For example, consider the very useful formula $\vec{\nabla} \times (\vec{\nabla} \times \vec{\xi}) = \vec{\nabla}(\vec{\nabla}\cdot \vec{\xi}) - \vec{\nabla}^2 \vec{\xi}$.

We have $(\vec{\nabla} \times (\vec{\nabla} \times \vec{\xi}))^i \\= \epsilon^{ijk}\partial_{j}(\epsilon_{krl}\partial^{r}\xi^{l}) \\= (\delta^{i}_{r}\delta^{j}_{l} - \delta^{j}_{r}\delta^{i}_{l})\partial_{j}\partial^{r}\xi^{l} \\= \partial^{i}\partial_{l}\xi^{l} - \partial_{r}\partial^{r}\xi^{i}\\ = (\vec{\nabla}(\vec{\nabla}\cdot \vec{\xi}))^i - (\vec{\nabla}^2\vec{\xi})^i$
as desired.

 

[member 428835]

thanks wannabe Newton