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In classical electrodynamics, we introduce the electric quadrupole moment $$D^{ij}\equiv\int y^i y^j \rho \mathrm{d}^3y$$ and its reduced (trace-less) version $$\mathcal{D}^{ij}\equiv D^{ij} - \frac{1}{3}\delta^{ij}D^{kk}, \;\;\;\; \mathcal{D^{ii}}=0.$$ How could I prove that $$\dddot{D}^{ij}\dddot{D}^{ij}-\frac{1}{3}\dddot{D}^{ii} \dddot{D}^{jj} = \dddot{\mathcal{D}}^{ij}\dddot{\mathcal{D}}^{ij}?$$


I tried to compute $$\dot{D}^{ij}=\int y^i y^j \dot{\rho}\mathrm{d}^3y=\int y^i y^j\partial_k j^k\mathrm{d}^3y=\int\partial_k\left(y^i y^j\right)j^k \mathrm{d}^3y=\int\left(y^j j^i + y^i j^j\right)\mathrm{d}^3y,$$ using the continuity equation, integrating by parts and since we can assume that $j^i$ has compact spatial support. But I'm not sure this is a smart way to proceed.

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I'm really sorry if I've missed something, and hence oversimplified the problem.

Setting $X=\dddot{D}$ and $\mathcal{X}=\dddot{\mathcal{D}}$ just to avoid having to write all the dots, both $X$ and $\mathcal{X}$ are symmetric tensors, and in matrix form we can write (just by differentiating your starting equation) $$\mathcal{X}=X-\frac{1}{3} \text{Tr}(X) \mathbb{1}$$ where $\mathbb{1}$ is the unit matrix. So $$ \mathcal{X}^2 =X^2 - \frac{2}{3} \text{Tr}(X)\, X + \frac{1}{9}[\text{Tr}(X)]^2 \mathbb{1} $$ and hence taking the trace $$ \text{Tr} (\mathcal{X}^2) =\text{Tr}(X^2) - \frac{1}{3} [\text{Tr}(X)]^2 $$ which is your target equation.

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