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Problem with solution Simplify 6 Expressions with Kronecker Delta

Level 3 (with higher mathematics)
Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.

Simplify the following expressions using the rules for calculating with Kronecker delta:

  1. \(\delta_{31}\,\delta_{33}\)

  2. \(\delta_{ji}\,T_{ink}\)

  3. \(\delta_{j1}\,\delta_{ji}\,\delta_{2i}\)

  4. \(\delta_{ik}\,\delta_{i3}\,\delta_{3k}\)

  5. \(\delta_{jj}\) with \(j ~\in~ \{ 1,2,3,4 \} \)

  6. \(\delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} \) with \(s ~\in~ \{ 1,2 \} \)

Solution tips

Use the properties of Kronecker delta that you learned in the lesson. In all the following expressions you should keep in mind that \(\delta_{ik}\,\delta_{ij}\) can be summed up to \(\delta_{kj}\) and that you have to sum over equal indices:$$ \delta_{ii} ~=~ 1~+~1~+~ ... ~+~ 1 ~=~ n $$

Solution for (a)

Here we simplify the following expression:\begin{align} \delta_{31}\,\delta_{33} ~&=~ 0 \cdot 1 ~=~ 0 \end{align}

We exploited that \( \delta_{31} = 0 \) is because the indices have two different values and \( \delta_{33} = 1 \) has two equal indices.

Solution for (b)

We want to simplify the following expression:\begin{align} \delta_{ji} \, T_{ink} ~&=~ T_{jnk} \end{align}

We exploited the rule that the Kronecker delta \( \delta_{ji} \) eliminates everything except \( T_{jnk} \), where \( i = j \).

Solution for (c)

Here we simplify the following expression:\begin{align} \delta_{j1} \, \delta_{ji} \, \delta_{2i} \end{align}

For example, first combine the index \(j\) in \(\delta_{j1} \, \delta_{ji}\) and then combine the index \(i\):\begin{align} \delta_{j1} \, \delta_{ji} \, \delta_{2i} &~=~ \delta_{i1} \, \delta_{2i} \\\\ &~=~ \delta_{12} \\\\ &~=~ 0 \end{align}

Of course, you could just as well first combine the index \(i\) in \(\delta_{ji} \, \delta_{2i}\):\begin{align} \delta_{j1} \, \delta_{ji} \, \delta_{2i} &~=~ \delta_{j1} \, \delta_{2j} \\\\ &~=~ \delta_{21} \\\\ &~=~ 0 \end{align}

You get the same result.

Solution for (d)

Proceed analogously to (c), except that at the end you get not 0 but 1 according to the Kronecker delta definition:\begin{align} \delta_{ik} \, \delta_{i3} \, \delta_{3k} &~=~ \delta_{k3} \, \delta_{3k} \\\\ &~=~ \delta_{33} \\\\ &~=~ 1 \end{align}

Again, the order of simplification does not matter.

Solution for (e)

Here we simplify the following expression, summing over the index \( j \) up to 4:\begin{align} \delta_{jj} \end{align}We get a sum with four summands:\begin{align} \delta_{jj} &~=~ \delta_{11} ~+~ \delta_{22} ~+~ \delta_{33} ~+~ \delta_{44} \\\\ &~=~ 1~+~1~+~1~+~1 \\\\ &~=~ 4 \end{align}

Solution for (f)

Here we simplify the following expression, summing over the index \( s \) up to 2:\begin{align} \delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} \end{align}

Let us first combine \( \delta_{k\mu} \, \varepsilon_{kmn} \) and then write out the sum \(\delta_{ss}\):\begin{align} \delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} &~=~ \varepsilon_{\mu mn} \, \delta_{ss} \\\\ &~=~ \varepsilon_{\mu mn} \, \left( \delta_{11} ~+~ \delta_{22} \right) \\\\ &~=~ \varepsilon_{\mu mn} \, \left( 1~+~ 1 \right) \\\\ &~=~ 2\, \varepsilon_{\mu mn} \end{align}