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Apr 27, 2019 at 21:37 comment added Cryo Hmm. You could do the same thing I did in four-dimensions. However, if you want to work in four dimensions shouldn't you work with four-current density and the 4d fourier transform? Also, it would help to define the quantities are working with. For example, based on you initial post I can define $F\left(\mathbf{q}\right)=\int d^3 r \rho\left(\mathbf{r}\right)\exp\left(i\mathbf{q.r}\right)$. How do you define $F$ as a function of the magnitude of $Q^2$?
Apr 27, 2019 at 8:17 comment added ersbygre1 I mean the four momentum $Q^\mu$, such that $Q^2 = (Q^0)^2-\textbf{Q}^2$.
Apr 26, 2019 at 8:50 history answered Cryo CC BY-SA 4.0