> Thanks for your interest in itensor.
Thanks for working with it & supporting my newbie questions.
> >exp : u([],[i],i)$
> >show(exp)$
> >would print (div u) or (del dot u)?
> I still don't understand what a sence to present in such a form. As to me
> notation U^i_{,i} is equally well as div U.
For that simple case. But, for some of the more complex cases I have been
working with, it is more difficult to tell.
> Lc does not understand the symbolic indices. lch does it, however it is
still
> unsatisfactory. Actually it is possible to do without these functions. I
mean
> you migth use the kdelta instead them.
> As I said you can kdelta here or use the rule which will be replace all
the
> single lch with relevant kdelta.
> v cross v in covariant notation will be
> kdelta([i,j],[i1,j1])*v([i1],[])*v([j1],[]). You can check that this is
> exactly zero.
> (C7) 'kdelta([i,j],[i1,j1])*v([i1])*v([j1])$
> (C8) show(%)$
> I1 J1
> (D8) KDELTA V V
> I/J I1 J1
> (C9) canform(ev(%,kdelta));
> (D9) 0
How is this expression equivalent to (v cross v)? It looks to me like it
has two free indices, where the result should only have one.
If instead, I enter.
ex : kdelta([i,j],[i1,j1])*v([i1])*v([j1])$
indices(ex);
returns
[[I, J], [I1, J1]]
This is indeed the case.
This also brings up the question; what exactly do the quotes ' ' do? They
seem to supress expansion of the kdelta function.
> This way is very usefull if you want to apply conditions like a
curl-free
> to expression. Say, lets introduce the two-form like
> (C18) tf: 'kdelta([i,j],[i1,j1])*'diff(v([i1]),j1)$
> (C19) show(tf)$
> I1 J1 d
> (D19) KDELTA ( --- (V ))
> I/J dJ1 I1
>
> Then we can use the property D D Forma =0,
>
> (C20) matchdeclare(I1,true);
> (D20) DONE
> (C21) defrule(grad, V([I1],[]),diff(FV([],[]),I1));
> (D21) GRAD : V([I1], []) -> DIFF(FV([], []), I1)
> (C22) APPLYB1(tf,grad);
> d
> (D22) (--- (FV([], [], I1))) KDELTA([I, J], [I1, J1])
> dJ1
> (C23) contract(canform(ev(ev(%,diff),kdelta)));
> (D23) 0
This part is going to take a while longer for me to digest..
Thanks for all the help,
Rob
>