itensor -> ic_convert



Try

 eqn:ishow(td([],[i,j]) = canform(rename(expand(liediff(v,g([],[i,j]))))))$

instead of what you were using. The variable on the LHS of the assigment
must have the same free indices as the expression on the RHS.

(As a practical suggestion, unrelated to your problem, I often find it
useful to initialize the variable on the LHS as a matrix before running the
code produced by ic_convert. That is, do a td:zeromatrix(dim,dim) before
evaluating the code returned by ic_convert; this way, the result will be
stored as a matrix as opposed to a two-dimensional list.)


Viktor


-----Original Message-----
From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu]
On Behalf Of Leo Butler
Sent: Sunday, September 27, 2009 3:49 PM
To: 'Maxima List'
Subject: itensor -> ic_convert

In maxima I enter the following:

if get('itensor,'version)=false then load(itensor);
if get('ctensor,'version)=false then load(ctensor);
dim:3;
remcomps(g);
imetric(g);
eqn:ishow(td = canform(rename(expand(liediff(v,g([],[i,j]))))));

Which outputs
                            i %1  j      %1  i j    %1 j  i
(%t32)             td = - g     v    + v   g    - g     v
                                  ,%1        ,%1          ,%1
(%o32) td = - g([], [i, %1]) v([], [j], %1) + v([], [%1]) g([], [i, j],%1)
                                                 - g([], [%1, j]) v([],
                                                   [i], %1)
Then I enter:


ic_convert(eqn);

Improper indices in - g([], [i, %1]) v([], [j], %1)
                 + v([], [%1]) g([], [i, j], %1) - g([], [%1, j]) v([],
[i], %1)


Aside from the definition of eqn in terms of a Lie derivative, this is
the example from itensor8.dem, which works fine.

What am I doing wrong?

Leo

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