The time dimension is no longer represented by second (s) but by frequency (s-1); a count or measure in the first time derivative.
What is then the relationship between Fourier transform and the time derivative?
Inverse Fourier transform as defined, transforming from the time domain to the frequency domain,
\(F_{ f}^{-1}[F(f)]=f(t)=\int_{-\infty}^{\infty} { F(f).e^{ i2\pi ft } } df\)
The transform of the nth derivative in time of \(f(t)\) is,
\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ f(t) \right\} \overset { Fourier }{ \longrightarrow } (i2\pi f)^{ n }.F(f)\)
and the transform to the (n-1)th derivative in frequency of \(F(f)\) is,
\( t^{ n-1 }f(t)\overset { Fourier }{ \longrightarrow } \cfrac { 1 }{ (-i2\pi f)^{ n-1 } } \cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\} \)
Combining,
\( \cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n-1 }f(t) \right\} \overset { Fourier }{ \longrightarrow } (i2\pi f)^{ n }.\cfrac { 1 }{ (-i2\pi f)^{ n -1} } \cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\} \)
\(\cfrac { d^{ n } }{ dt^{ n } } \left\{ t^{ n -1}f(t) \right\} \overset { Fourier }{ \longrightarrow } (-1)^{(n-1)}(i2\pi f)\cfrac { d^{ n-1 } }{ df^{ n-1 } } \left\{ F(f) \right\} \)
with light speed \(c\),
\(t=\cfrac{x}{c}\)
\(dt=\cfrac{1}{c}dx\)
\(c^n\cfrac{d^n}{dx^n}\left(\left(\cfrac{x}{c}\right)^{(n-1)}f(\cfrac{x}{c})\right)\overset{Fourier}{\longrightarrow }(-1)^{(n-1)}(i2\pi f)\cfrac{d^{n-1}}{df^{n-1}}\left(F(f)\right)\)
If \(n=2\) and \(f\left(\cfrac{x}{c}\right)=\cfrac{x}{c}\)
\(c^2\cfrac{d^2}{d\,x^2}\left(\cfrac{x}{c}\right)^2\overset{Fourier}{\longrightarrow }-(i2\pi f).\cfrac{d}{df}\left(F(f)\right)\)
\(2\overset{Fourier}{\longrightarrow }-(i2\pi f)\cfrac{d}{df}\left(F(f)\right)=2\delta(f)\)
This implies,
\(\cfrac{d}{df}\left(F(f)\right)=-\cfrac{1}{i\pi f}\delta(f)\)
\(F(f)=-\cfrac{1}{i\pi}\int{\cfrac{1}{ f}\delta(f)}d\,f\)
\(F(f)=-\cfrac{1}{i\pi}\int{\delta(f)}d\,ln(f)\)
If \(y=ln(f)\) then \(e^{y}=f\)
\(dy=d\,ln(f)\)
\(F(f)=-\cfrac{1}{i\pi}\int{\delta(e^{y})}d\,y\)
We could have,
\(\left(\delta(e^{y})+A\right)^{'}=\delta^{'}(e^{y})=-\delta(\left(e^{y}\right)^{'})=-\delta(e^y)\) --- (*)
where \(A\) is a constant,
\(F(f)=\cfrac{1}{i\pi}\int{\left(\delta(e^{y})+A\right)^{'}}d\,y\)
\(F(f)=\cfrac{1}{i\pi}\delta(e^{y})+\cfrac{A}{i\pi}\)
\(F(f)=\cfrac{1}{i\pi}\delta(f)+\cfrac{A}{i\pi}\)
since, \(f\gt0\) but \(\delta(f)=0\) for \(f\ne0\)
\(F(f)=\cfrac{A}{i\pi}\) a constant
for \(f\gt0\).
Since \(A\) from (*) is arbitrary, we let \(A=i\pi\)
\(F(f)=1\)
More correctly, \(2\delta(x)\rightarrow \delta(x)\) because the delta function cannot be scaled.
\(F(f)=\cfrac{A}{i2\pi}\) a constant
for \(f\gt0\).
This shows,
\(t\overset{Fourier}{\longrightarrow }F(f)=\cfrac{A}{i2\pi}\)
\(F(f)\) in the frequency domain is measured in units of \(2\pi i\), the perimeter of a circular clock of radius \(i\). \(F(f)\) is the number of cycles around an imaginary clock.
Which is not the answer we want. Why do we take the time derivative of \(F\) obtained from the space dimension with considerations for the time dimension to obtain an equivalent expression for \(F\) in the time dimension.
\(\cfrac{d\,F_{space}}{d\,t}=F_{time}\)
still? A notion to be proven, not stated, but phrased as a question.
