Fourier Transform - Electrodynamics

Question

I'm studying electrodynamics and am looking at Fourier transforms, especially for Greens functions. Normally, Fourier transforms are given by the following:$$G(x)=\int dk\ \tilde{G}(k)e^{ikx}$$ However, in some of the functions I'm looking at, the fourier transform is given as: $$G(t)=\int dw\ \tilde{G}(k)e^{-iwt}$$ My question is, why does the sign of the power change for the time fourier transform?

Answer 1

It's partially just arbitrary convention, but it becomes a somewhat more significant thing in the context of special relativity. There, the proper "dot product" between a frequency and vector has the sign-flip on the time component. That is to say, when it's just space, you have $$k = (k_x,k_y,k_z)$$ $$x = (x_x,x_y,x_Z)$$ and work with $$x\cdot k = k_x x_x + k_y x_y + k_z x_z$$ in your Fourier transform. When you use special relativity, you want the proper time with the time counting negative, or

$$k = (k_t,k_x,k_y,k_z)$$ $$x = (x_t,x_x,x_y,x_Z)$$ and then use $$x\cdot k = k_x x_x + k_y x_y + k_z x_z - k_t x_t $$

So when you do a Fourier transform only in time and ignore the spatial components, this negative time convention can emerge. Of course it doesn't matter in a relativity-ignoring situation, but the convention can bleed over!