I'm learning vector calculus on my own and sometimes strange things happen that I don't know how I should explain them. We have this famous equality:


Now, if we set $j=q$ we get $$\epsilon_{ijk}\epsilon_{pjk}=\delta_{ip}\delta_{jj}-\delta_{ij}\delta_{jp}=\delta_{ip}-\delta_{ip}=0$$

But apparently the correct equality is $$\epsilon_{ijk}\epsilon_{pjk}=2\delta_{ip}$$

Why is it so? Where's my mistake? :|


In the second equation, you sum over $j$, so $\delta_{ip}\delta_{jj}$ should be evaluated as $3\delta_{ip}$.

  • $\begingroup$ Ah... Thanks.. I was going crazy. You saved me :P $\endgroup$ – math.n00b Jul 20 '14 at 8:37
  • $\begingroup$ No problem. As you see, the world still works and holds together :) $\endgroup$ – Peter Franek Jul 20 '14 at 8:40
  • $\begingroup$ So, whenever we have $\delta_{jj}$ it's equal to $3$ instead of $1$? because according to my book, whenever that we have repeated indices we sum over it. So, by keeping that in mind, $\delta_{jj}=3$. Right? $\endgroup$ – math.n00b Jul 20 '14 at 8:57
  • $\begingroup$ It depends on whether $j$ is fixed or you sum over it. In your computations, you first substituted the fixed $q$ by a fixed $j$ and then summed over $j$. Clearly, $\delta_{11}+\delta_{22}+\delta_{33}=3$. (If you are not sure what to do, it might help to always write the $\sum$ symbols) $\endgroup$ – Peter Franek Jul 20 '14 at 9:06
  • $\begingroup$ Yes, but according to my book, which is about elasticity theory, it says that whenever that we had something like $A_{jj}$ we'd sum over $j$. So, by this convention, we'll always get $\delta_{jj}=3$. Right? $\endgroup$ – math.n00b Jul 20 '14 at 9:09

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