The workings of ev() are mysterious. Basically, ev does the applying of
kdelta before it evaluates lc2kdt, which is not what you want. This might
work instead:
ishow(apply(ev,[contract(expand(lc2kdt(expr))),kdelta]))$
By using apply(), you ensure that ev is applied only after its arguments
have been evaluated.
In many cases, though, it's much easier to just do the evaluation in two
steps, as in
contract(expand(lc2kdt(expr)))$
ishow(ev(%,kdelta))$
Viktor
-----Original Message-----
From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu]
On Behalf Of Nicola Cabibbo
Sent: Thursday, June 12, 2008 7:18 AM
To: maxima at math.utexas.edu
Cc: Nicola Cabibbo
Subject: Re: [Maxima] How to simplify kdelta?
After more investigation, if after the commands
if get('itensor, 'version)=false then load (itensor) ;
idim(3);
dim:3;
defcon(q,q,qq);
expr:'levi_civita([i,k,m],[])*'levi_civita([],[i1,k,m])*q([],
[i])*q([i1],[])$
expr1:contract(expand(lc2kdt(expr)));
I enter
ishow(ev(expr1,kdelta));
the result is the expected " 2qq ", while if I enter
ishow(ev(contract(expand(lc2kdt(expr))),kdelta));
the result is " 5*qq-kdelta*qq ". Why the difference?
Thanks
On Jun 12, 2008, at 1:12 PM, Nicola Cabibbo wrote:
> Just trying to learn Maxima, with the commands:
>
> if get('itensor, 'version)=false then load (itensor) ;
> idim(3);
> dim:3;
> defcon(q,q,qq);
> expr:'levi_civita([i,k,m],[])*'levi_civita([],[i1,k,m])*q([],
> [i])*q([i1],[])$
> ishow(contract(expand(lc2kdt(expr))));
>
> I obtain " 5*qq-kdelta*qq ". This is correct with kdelta=3, but
> how can I directly obtain the simplified result "2 qq"?
>
> Thanks for the attention
>
> PS: my setup is (on a Mac)
>
> wxMaxima 0.7.5 http://wxmaxima.sourceforge.net
> Maxima 5.15.0 http://maxima.sourceforge.net
> Using Lisp CMU Common Lisp Snapshot 2008-03 (19E)
>
>
>
>
>
>
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