The four dimensional Fourier transform, at least in a physics context is usually (e.g. here) defined as follows. Given a function $G(\mathbf{x},t)$, then its Fourier transform is defined as
$$ g(\mathbf{k},\omega) = \int e^{-i(\mathbf{k}\cdot\mathbf{x}-\omega t)}G(\mathbf{x},t) \mathrm{d}^3x \,\mathrm{d}t $$
and the inverse Fourier transform as
$$ G(\mathbf{x},t) = \int e^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}g(\mathbf{k},\omega) \mathrm{d}^3k \,\mathrm{d}\omega $$
I was wondering, is there a definite mathematical reason for the sign difference between $\mathbf{k}\cdot\mathbf{x}$ and $\omega t$ in the exponent? Could we not define it with both terms having the sign as well (I mean $\mathbf{k}\cdot\mathbf{x}+\omega t$)? It seems to me that the only reason why there is a sign difference is because one would like to expand in "plane waves" for physical reasons ($\mathbf{k}$ may be taken to be the actual wave vector of the plane wave and $\omega$ to be the actual frequency). Also another reason seems to be that the expression in the exponent
$$ \mathbf{k}\cdot \mathbf{x} -\omega t = x_\mu k^\mu$$
is invariant under Lorentz transformations. Could someone confirm/refute my ideas and tell me whether it would be admissible to change the signs in the exponent in the definition of the Fourier transform?
Best Answer
I know it has been some time since you asked the question, but I was thinking about that the other day, too, which is why I would like to share my answer:
No, there is not. Suppose you want to solve a differential equation using Fourier Transforms. You get a mathematically valid solution for both cases, using the same relative sign or a different one. However, in a physics context, this is different, because it introduces a meaning to "time", which mathematics does not care about.
The mathematical solution for the $i(\mathbf k\cdot \mathbf x+\omega t)$ convention corresponds to a universe where time goes backwards compared to our physical universe:
If you look, for example, at an object moving with velocity $c$ in one dimension, its position $r$ is given by something like: \begin{equation}r = r_0+ct \end{equation} The physical picture at time $t=t_0$ is equivalent to the situation at $\tilde t_0=0$ for an object shifted in the opposite direction.
\begin{equation} r = r_0+ct_0 \iff r - ct_0 = r_0 \iff \tilde r = r_0 +c\tilde t_0 \end{equation} where $\tilde r := r-ct_0$.
In a way, you could say that in our universe time has the opposite direction to space. This is accounted for in the Minkowski metric, which is why the explanation with four-vectors works, too. However, as I tried to show with the example above, people knew this before special relativity and Lorentz transformations were a thing.
I hope this was helpful.