Vector multiply that is NOT dot or cross?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 3K views
ognik
Messages
626
Reaction score
2
Hi - just working through my text (studying by correspondence) on Del operator - so Curl, div etc. Came across some identities parts of which which have me confused. what does it mean when a vector is shown as multiplying something - but without dot or cross? For example F(∇.G) or ∇(F.G) or (G.∇)F ...

I get that something like (G.∇) expands to each component of G times each component of ∇ - which is a scalar; also ∇.G is a normal dot product. So I understand f.(∇.G) and ∇.(F.G) and (G.∇).F and but am confused when the 'dot' outside the bracket is missing - how do we multiply those?

Thanks
Alan
 
Physics news on Phys.org
These are all vectors, an example identity is
∇x(FXG) = F(∇.G) - G(∇.F) + (G.∇)F - (F.∇)G
 
ognik said:
Hi - just working through my text (studying by correspondence) on Del operator - so Curl, div etc. Came across some identities parts of which which have me confused. what does it mean when a vector is shown as multiplying something - but without dot or cross? For example F(∇.G)
∇.G is a scalar function so F(∇.G) is "scalar multiplication"- each component of F multiplied by ∇.G

or ∇(F.G)
F.G is a scalar function so ∇(F.G) is the gradient of F.G

or (G.∇)F ...0
This is the same as G(∇.F)


I get that something like (G.∇) expands to each component of G times each component of ∇ - which is a scalar; also ∇.G is a normal dot product. So I understand f.(∇.G) and ∇.(F.G) and (G.∇).F and but am confused when the 'dot' outside the bracket is missing - how do we multiply those?

Thanks
Alan
 
  • Like
Likes   Reactions: ognik
Nice explanation thanks hallsofivy, I could see dotting them was the only way to get anything done, but its nice to understand why.
 
I'm not sure this is what you're looking for, but you might want to have a look at the definition of "geometric product" in Clifford algebras, plus the concept of "geometric derivative" proposed by D. Hestenes, which generalizes div,grad,curl operators.