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Focus on the integral $$I_{ij}(k) = \int k_i k_j\ \mathrm{d}\Omega_k.$$ This is a rank 2 symmetric tensor which can only depend on $\vec{k}$ through its magnitude $k^2$, since the direction has been integrated over. So the only possibility is that $I_{ij}$ is proportional to the unit tensor (Kronecker delta): $$I_{ij}(k) = f(k^2) \delta_{ij},$$ where ...