# Does the quadrupole moment tensor contracted with itself yield Kronecker delta?

I have trouble understanding the Kronecker delta and how it comes up in tensor equations. I know the metric contracted with itself gives the Kronecker delta which either is 0 or 1 depending on if the indices are equal or not. However, I'm not sure if any two given tensors that contract each other's indices gives the Kronecker delta. Like two quadrupole moment tensors.

No, in general a contraction that leaves a two-index tensor will not produce the Kronecker delta.

For example, contracting the appropriate two indices of the four-index Riemann tensor gives the Ricci tensor, not the Kronecker delta. (Contracting other pairs gives the zero tensor.)

Contracting the product of two quadrupole tensors will not give the Kronecker delta either.

The metric tensor is unusual in this regard. In general, the fully contra-variant form of a two-index tensor is not the inverse of the fully-covariant form.

Simple way to get a Kroenecker delta is to consider coordinate transformation where you tranform into the same coordinate system you started with

So usually if you have a (e.g.) vector $$V^\alpha$$ in coordinates $$S$$ and you change to $$\tilde{S}$$, the vector components will change as

$$\tilde{V}^\alpha=\Lambda^\alpha_{\:\beta} V^\beta$$, where $$\Lambda^\alpha_{\:\beta}=\frac{\partial \tilde{x}^\alpha}{\partial x^\beta}$$. If $$\tilde{S}=S$$ then $$\Lambda^\alpha_{\:\beta}=\delta^\alpha_{\:\beta}=\frac{\partial x^\alpha}{\partial x^\beta}$$

Using the definition $$\delta^\alpha_{\:\beta}=\frac{\partial x^\alpha}{\partial x^\beta}$$ it is also easy to check that delta is indeed a tensor:

$$\tilde{\delta}^\alpha_{\:\beta}=\frac{\partial\tilde{x}^\alpha}{\partial{\tilde{x}^\beta}}=\frac{\partial x^\gamma}{\partial\tilde{x}^\beta}\frac{\partial\tilde{x}^\alpha}{\partial x^\gamma}=\frac{\partial x^\gamma}{\partial\tilde{x}^\beta}\frac{\partial\tilde{x}^\alpha}{\partial x^\eta}\delta^\eta_{\:\gamma}$$