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+1 vote

When calculating the Einstein equation in this tutorial

why write

t1 = action[1][0][0][2]
Thank you for any elucidation.

asked in General questions by (190 points)

2 Answers

+1 vote


First I'd like to point that action is an expression representing the variation of the action, written as an equation. Therefore, the zeroes component of the equation is the left-hand side, i.e., action[0] is $\delta S$, while the first component of the equation is the right-hand side, i.e., action[1] represents the whole integral , $\int \Big( \cdots \Big) \mathrm{d}x$.

Since the left-hand side of action is composed by other terms, you can repeat the decomposition process... and this allow you to select the piece of the expression for you to manipulate. In the notebook you mention, action[1][0][0][2] represents the term containing the variation of the metric (without the variation of the connection).

answered by (6.3k points)

Thank you a lot!

+1 vote

Hi skyfold,

as explained by doxdrum, the


command selects the part of expression we are interested to.
In particular, the command extracts the terms not represented by total derivatives.

I note that in general this approach is not very stable because it depends on the position [1][0][0][2] of the terms in the expression. If they had had a change of position or the expression had a different manipulation, the command would fail.


answered by (410 points)