1
$\begingroup$

I want to express some equations in Einstein summation convention to improve readability and possibly simplify the calculations.

I have searched for 3-dimensional versions of the Kronecker Delta, without success and, in particular, the generalised version defined in wikipedia has only even numbers of indices.

Although I can just define my own methods which serve the purpose is it generally possibly to say that, where,

$$ c_i = x_i y_i \,$$ $c_i$ is a composed element of some other quantities, e.g. weights that element-wise scale other elements, I could use the formulation:

$$ c_i = \delta_i^{\mu \nu}x_\mu y_\nu\;, \quad \delta_i^{\mu \nu} = 1 \quad \text{for} \quad i=\mu=\nu. $$

If I extended this to a composition of 3 elements, can I use the formulation:

$$c_i = x_i y_i z_i = \delta_i^{\mu \nu \epsilon}x_\mu y_\nu z_\epsilon$$ or would it be more appropriate to maintain my 3D delta and compose them instead?

$$ c_i = x_i y_i z_i = \delta_i^{\epsilon \xi}\delta_\epsilon^{\mu \nu}x_\mu y_\nu z_\xi$$

$\endgroup$
1
$\begingroup$

If we pick a basis, then for any number of indices $n\geq 1$ up and $m\geq1$ down, we can define

$${\delta^{i_1\dots i_n}}_{j_1\dots j_m} = \begin{cases} 1&\text{if}\;i_1=\dots =i_n=j_1=\dots =j_m\\ 0&\text{otherwise.}\\ \end{cases}$$

The difficulty is that this definition is basis dependent. The tensor won't have the same form in any other basis (except for permutations of our current basis). The only basis-independent case is $n=m=1$, the usual Kronecker delta.

So you can use deltas with more than two indices, but you have to be careful to specify which basis you're defining them with respect to. The "all ones" vector $\delta^i$ is also sometimes useful.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.