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I am in an argument with a friend from the university and we would like to clarify our problem:

We have given the following term to calculate: $$s_1 t_k \delta_{ii} \delta_{k1} \delta_{nn}$$

All indices are running from 1 to 3 (3 dimensional vectors(s,t))

My solution was the following: $$s_1 t_k \delta_{ii} \delta_{k1} \delta_{nn}$$ $$=s_1 t_k \cdot1\cdot \delta_{k1} \cdot1$$ $$=s_1 t_k \delta_{k1}$$ $$=s_1 t_1$$

However he said that the result is:

$$s_1 t_1 \cdot 3 \cdot 3$$

He did not tell me how he got the two threes but I assume it's because of the Kronecker-deltas with the same index. But as far as I see, he cannot use the sum convention by Einstein in this case.

Who got the correct solution?

Finn Eggers
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1 Answers1

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It seems like your friend is interpreting the expression as

$$\sum_{i=1}^{3}\sum_{n=1}^{3}\sum_{k=1}^{3} s_1 t_k \delta_{ii} \delta_{nn} \delta_{k1}$$

In this case it reduces to

$$s_1\sum_{i=1}^{3}\sum_{n=1}^{3}\sum_{k=1}^{3} t_k \delta_{k1} = s_1 t_1\sum_{i=1}^{3}\sum_{n=1}^{3} 1= 9 s_1t_1$$

In your friend's interpretation, all indices are running from $1$ to $3$. This is, I think, standard. The convention is to sum over every index that appears twice (on the other hand, if we had $a_{ij}b_{jk}$ that would indicate only a summation over $j$).

In your interpretation, only $k$ runs from $1$ to $3$ while $n$ and $i$ are arbitrary but fixed, which doesn't sound like it agrees with the description you gave.

Saying that he cannot use the Einstein notation is also confusing to me. This question is entirely a matter of interpreting notation, just as you interpreted the term $t_k\delta_{k1}$ as a sum from $k=1..3$. But from my understanding of index summation notations, in which $a_{ii}$ is often shorthand for the trace of a matrix, by convention you shouldn't neglect summing over the $\delta_{ii}$ and $\delta_{nn}$.

Badam Baplan
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  • So you would say he is right? Maths is usually not about interpretation. Einstein’s convention looks like this: a_c b_c = a1b1 + a2b2.... aNbN – Finn Eggers Dec 07 '17 at 09:08
  • Yes I would say he is right.

    But again this is about context.

    Knowing only that $i,k,n$ each take one of three values, it is reasonable for the expression $s_1 t_k \delta_{ii} \delta_{nn} \delta_{k1}$ to refer to any of the following:

    (1) A single one of the $27$ combinations of $i,n,k$ (2) A sum over $i$ with $n,k$ fixed; a sum over $n$ with $i,k$ fixed; a sum over $k$ with $n,i$ fixed. (3) A sum over $i,k$ with $n$ fixed. etc. etc. (4) A sum over $i,j,k$

    If the interpretation is not stated explicitly, that's where notational convention comes in. I'd have assumed (4)

     – Badam Baplan Dec 07 '17 at 09:30
  • Okay thank you for your explanation :) – Finn Eggers Dec 07 '17 at 09:32
  • Sure :) I just did a quick search and I think that this is a nicely written resource http://www.brown.edu/Departments/Engineering/Courses/En221/Notes/Index_notation/Index_notation.htm – Badam Baplan Dec 07 '17 at 09:36