Obtaining vector identities

  • Thread starter bcjochim07
  • Start date
  • #1
374
0

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?
 

Answers and Replies

  • #2
311
1
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
 
  • #3
lurflurf
Homework Helper
2,440
138
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
 

Related Threads on Obtaining vector identities

  • Last Post
Replies
1
Views
862
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
851
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
690
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
1K
Top