Kronecker 0 0 form

Question

Thanks, your explanation makes sense.

Now with all the substitution rules I have ended up with a term like

\begin{equation}{}Ricci = -6\partial_{0}{}^{0}{a} {a}^{-3}-16\partial_{0}{a} \partial^{0}{a} {a}^{-4}+16\delta_{0}{}^{0} \partial_{0}{a} \partial^{0}{a} {a}^{-4}\end{equation}

Notice the $\delta_{0}{}^{0}$.

I can't figure out how to substitute this away, it gets turned into one right away in the substitution rule and therefore the pattern can't match.

Comments

Forgot to mention that doing an eliminate_kronecker doesn't help

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You wrote as a separate question (please don't do that):

"Meant to say that simplifying to one immediately is the problem, since the /delta 0 0 term was created in a prior substitution rule somehow through some combination of things. Now I cannot get rid of it because of the immediate conversion.

Does that make sense?"

No, I don't understand. Does it, or does it not keep the \delta^{0}_{0} in your expression? If it does, that's a bug, please send me a sample notebook that shows it (try to trim it down to minimal form).

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Never mind, I see how this can happen. Will fix.

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For status updates see the bug report at https://github.com/kpeeters/cadabra2/issues/69

Answer 1

\delta{#}::KroneckerDelta;
ex:= \delta^{0}_{0};

simplifies to '1' immediately. If it doesn't do that for you, it's a bug; send me the notebook by email and I'll have a look.