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Vector calc identities

  1. Oct 11, 2013 #1


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    Gold Member

    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 [itex]\theta[/itex], perhaps letting [itex]\theta = \alpha + \beta[/itex] 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!
  2. jcsd
  3. Oct 11, 2013 #2


    Staff: Mentor

    If you use x,y,z components for the vector then do some rearrangements you can prove it.
  4. Oct 11, 2013 #3


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    Science Advisor

    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.
  5. Oct 12, 2013 #4


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    thanks wannabe newton
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