Vector Product Rule Deviation

[John Creighto]

Wikipedia gives an extensive amount of vector identities:

http://en.wikipedia.org/wiki/Vector_calculus_identities
http://en.wikipedia.org/wiki/List_of_vector_identities

Does anyone know of a link where most of these are proved. I'm particularly interested in the product rules but would like to know how all these identities are derived.

 

[John Creighto]

I found a link that might be helpful:

Vector algebra is a powerful and needful tool for Physics but unfortunately, due to lack of mathematical skills, it becomes misleading for first undergraduate courses of science and engineering studies. Standard vector identities are usually proved using Cartesian components or geometrical arguments, accordingly. Instead, this work presents a new teaching strategy in order to derive symbolically vector identities without analytical expansions in components, either explicitly or using indicial notation. This strategy is mainly based on the correspondence between three-dimensional vectors and skew-symmetric second-rank tensors. Hence, the derivations are performed from skew tensors and dyadic products, rather than cross products. Some examples of skew-symmetric tensors in Physics are illustrated.

http://www.citeulike.org/user/pak/article/4524046
http://arxiv.org/PS_cache/arxiv/pdf/0904/0904.1814v1.pdf

 

[daudaudaudau]

I don't think there is anything wrong with proving these identities in cartesian coordinates. If you have an identity such as $$\nabla (fg)=f\nabla g+g\nabla f$$, this just tells you that one vector is equal to another vector. If this is true in one coordinate system, it is true in all coordinate systems.

 

[John Creighto]

I don't think there is anything wrong with proving these identities in cartesian coordinates. If you have an identity such as $$\nabla (fg)=f\nabla g+g\nabla f$$, this just tells you that one vector is equal to another vector. If this is true in one coordinate system, it is true in all coordinate systems.

If is a lot less efficient and less intuitive to prove vector relationships by resorting to expressing the vectors as individual components. People are much more likely to be able to see simplifications in vector algebra if they have some higher level algebra tools available to them.

 

[lurflurf]

The identities of vector calculus are easily derived and proven by algebraic mean given a few lemmas. It would be bad to prove them in cartesian coordinates because it would be messy, lengthly, inelegant, and would exagerate the importance of coordinates.
Anyone who disagrees should post a cartesian coordinates proof of
$$\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}$$
from another thread

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 an 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