(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let [itex] F(x,y,z) [/itex] be an irrotational vector field and [itex] f(x,y,z) [/itex] a [itex] C^1 [/itex] scalar functions. Using the standard identities of vector analysis (provided in section 2 below), simplify

[itex] (\nabla f \times F) \cdot \nabla f [/itex]

2. Relevant equations

Note: The identities below require [itex] f,g,F,G [/itex] to be suitable differentiable, either order [itex] C^1 [/itex] or [itex] C^2 [/itex].

[itex] 1. \nabla (f+g) = \nabla f + \nabla g [/itex]

[itex] 2. \nabla (\lambda f) = \lambda \nabla f [/itex], where [itex] \lambda [/itex] is a constant

[itex] 3. \nabla (fg) = f \nabla g + g \nabla f [/itex]

[itex] 4. \nabla (\frac{f}{g}) = \frac{g \nabla f - f \nabla g}{g^2} [/itex]

[itex] 5. \nabla \cdot (F+G) = \nabla \cdot F + \nabla \cdot G [/itex]

[itex] 6. \nabla \times (F+G) = \nabla \times F + \nabla \times G [/itex]

[itex] 7. \nabla \cdot (fF) = f \nabla \cdot F + F \cdot \nabla f [/itex]

[itex] 8. \nabla \cdot (F \times G) = G \cdot (\nabla \times F ) - F \cdot (\nabla \times G) [/itex]

[itex] 9. \nabla \cdot (\nabla \times F) = 0 [/itex]

[itex] 10. \nabla \times (fF) = f \nabla \times F + \nabla f \times F [/itex]

[itex] 11. \nabla \times (\nabla f) = 0 [/itex]

[itex] 12. {\nabla}^2 (fg) = f{\nabla}^2 g + g{\nabla}^2 f + 2 \nabla f \cdot \nabla g [/itex]

[itex] 13. \nabla \cdot (\nabla f \times \nabla g) = 0 [/itex]

[itex] 14. \nabla (f \nabla g - g \nabla f) = f {\nabla}^2 g - g {\nabla}^2 f [/itex]

3. The attempt at a solution

I first write [itex] (\nabla f \times F) \cdot \nabla f [/itex]

= [itex] \nabla f \cdot ( \nabla f \times F ) [/itex]

However this doesn't look like any of the vector identity above. Can we write

[itex] \nabla f = f \nabla [/itex] though? As this would enable me to use the identity. Also, is the final answer 0 ?

Thanks!

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# Homework Help: Vecotr identity proof 2

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