Convolution formula: $$ (f * g)(t) \mathrel{:=} \int_{-\infty}^\infty f(\tau)g(t - \tau)\mathrm d\tau. $$
I understand that convolution in the time domain is the same as multiplication in the frequency domain. But I am having problems wrapping my head around the fact that the g function gets modified to a new function by performing $g(t-\tau)$.
Yet when we perform the Fourier transform, we just ignore the $\tau$ modification of $g$ and just apply the transform to the $g$ function similar to how we apply it to the $f$ function.
Q1) Why do we ignore the tau subtraction in the $g$ function when converting to the frequency domain?
Q2) Lets assume I define an unorthodox convolution with $g(t+ \tau)$ or $g(t\tau)$, could I convert this to the frequency domain, multiply it and convert it back to the time domain? Unorthodox convolutions:
\begin{gather*} (f * g)(t) \mathrel{:=} \int_{-\infty}^\infty f(\tau)g(t + \tau)\mathrm d\tau \\ (f * g)(t) \mathrel{:=} \int_{-\infty}^\infty f(\tau)g(t\tau)\mathrm d\tau. \end{gather*}
Q3) For the unorthodox convolutions, would a new function $h(x) = g( -x)$ or $h(x) = g(\tau - \tau x)$ need to be defined such that $f*h$ produces the desired result?
Q4) Is there any current use for these unorthodox convolutions, or did I just create a new form of math that won't be used for another couple of centuries?
I tried searching for alternative forms of convolution but I didn't find anything. Maybe I don't have the right terminology for my searches.
A1) Time shift amounts to a phase factor $e^{-i \omega \tau}$ in the frequency domain. It is not ignored, it is just cancelled by the corresponding factor from $f$ when we express the latter in terms of its Fourier transform $F$ ($f(\tau) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} F(\omega) e^{i \omega \tau} d\omega$).
A2) Yes, you could. The first one is called cross-correlation. The second one can be converted into convolution by a change of variables $t = e^x$, $\tau = e^{-y}$.
A3) Are you asking about the relationship between the "unorthodox" and "orthodox" convolutions? Yes, they can be related by function redefinitions. Note that you cannot define $h(x) = g(\tau - \tau x)$, because $\tau$ is an integration variable, not a fixed parameter.
A4) Some of the uses for cross-correlation are discussed in the Wikipedia article linked above. Expressions like $\int f(\tau) g(t \tau)$ appear in the Hankel transform (essentially Fourier transform in polar/spherical coordinates) and in the Fourier transform itself ($g(\omega t) = e^{-i \omega t}$).