How Do You Derive the Vector Identity Involving Divergence and Curl?

[bcjochim07]

Homework Statement

The vectors F and G are arbitrary functions of position. Starting w/ the relations F x (∇ x G) and G x (∇ x F), obtain the identity

∇(F . G) = (F . ∇)G + (G . ∇)F + F x (∇ x G) + G x (∇ x F)

Homework Equations

The Attempt at a Solution

I started off with the relation F x (∇ x G) and used the BAC-CAB rule:

F x (∇ x G) = ∇(F . G) - (G . ∇)F

so ∇(F . G = (G . ∇)F + F x (∇ x G) which seems to contradict the identity I am supposed to get. What am I doing wrong?

 

[praharmitra]

you can't apply the BAC-CAB rule here. That was derived for vectors by assuming that A X B = - B X A

However with the del operator del X A is a vector, but A X del is an operator...so there's a world of difference

 

[lurflurf]

The trouble is commuting an opperator adds a commutator term
In single variable calculus
D(uv)=uDv+vDu not uDv
we can use partial opperators to avoid this
let a opperant in {} be fixed
D(uv)=D({u}v)+D(u{v})={u}Dv+{v}Du=uDv+vDu

∇(F . G )=∇({F} . G )+∇(F . {G} )
∇({F} . G )=Fx(∇xG)+(F.∇)G
∇(F . {G} )=Gx(∇xF)+(G.∇)F
∇(F . G )=∇({F} . G )+∇(F . {G} )=Fx(∇xG)+(F.∇)G+Gx(∇xF)+(G.∇)F