- #1
Safinaz
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- TL;DR Summary
- A question about a Kroneker Delta identity
Is there a Kroneker Delta identity:
$$ \delta^3 (k) \delta^3 (k) = \frac{2\pi^2}{k^3} ~ \delta (k_1- k_2) $$?
Where k is a wave number. In this Paper: Equations 26 and 27, I think this identity is used to make ##k_1=k_2## and there is an extra negative sign.
$$ \delta^3 (k) \delta^3 (k) = \frac{2\pi^2}{k^3} ~ \delta (k_1- k_2) $$?
Where k is a wave number. In this Paper: Equations 26 and 27, I think this identity is used to make ##k_1=k_2## and there is an extra negative sign.