2

I am not a physicist but rather an engineer / mathematician, so I've always wondered why is it that physicists use the positive sign convention in the forward Fourier transform. That is, in all of my physics classes, I've seen the forward Fourier transform written as

$$ \mathcal{F}\left\{f\left(t\right)\right\}\left(\omega\right) = \int\limits_{-\infty}^{\infty} f\left(t\right) e^{+i\omega t}dt $$

and the inverse Fourier transform written as

$$ \mathcal{F}^{-1}\left\{F\left(\omega\right)\right\}\left(t\right) = \int\limits_{-\infty}^{\infty} F\left(\omega\right) e^{-i\omega t}d\omega $$

The reason mathematicians and engineers use the opposite (and what I consider standard) convention is to connect the Fourier transform to the bilateral Laplace transform evaluated on the imaginary axis. If we used the positive sign convention, it gets a little wonky since you're technically evaluating the inverse Laplace transform, which is generally done using contour integration techniques. I believe the convergence of the forward Laplace transform integral requires a negative kernel $e^{-st}$, but I may be wrong? Please correct me if so :-)

Also, I know that physicists like to write the positive, outwardly propagating, time-harmonic wave as $e^{+i\omega t} u\left(x\right)$. In engineering (particularly electromagnetics / antennas), we use $e^{-i\omega t} u\left(x\right)$ and so the Sommerfeld condition reads

$$ \lim_{\left|x\right| \rightarrow \infty} \left|x\right|^{\frac{n-1}{2}} \left(\frac{\partial}{\partial \left|x\right|} - ik\right) u\left(x\right) = 0 $$

In the end, it doesn't really matter except maybe in connection to the Laplace transform, but I've just always been curious as to the physical explanation for using a positive sign.

Qmechanic
  • 220,844
AntennaGuy
  • 138

1 Answers1

5

This is mostly because we're usually more interested in the spatial part of a plane wave than in the temporal part, so that plane waves are most convenient when written as $$ e^{i(\mathbf k\cdot\mathbf r-\omega t)}. \tag 1 $$ The normalization follows from this choice.

In general terms, it's hard to call which factor has more weight. There are plenty of space-independent problems where you just care about the $e^{-i\omega t}$ factor, and there are plenty of stationary problems where you only care about, say matching to an asymtptotic $e^{i\mathbf k\cdot\mathbf r}$ dependence. For the general case, it's usually easier to think of $e^{i(\mathbf k\cdot\mathbf r-\omega t)}$ as a spatial plane wave with a phase that changes from moment to moment, than as a bunch of oscillators at different position with precisely synchronized phases. As I said, the spatial part tends to take precedence.

This is partly supported by the fact that the Fourier normalization in use will often depend on the nature of the variables in play. Thus, if the variable is a time, I'm likely to decompose a function$f(t)$ as $$f(t)=\int_{-\infty}^\infty \tilde f(\omega)e^{-i\omega t}\mathrm d\omega,$$ but if $g(x)$ is a function of position then I'm more likely to define $\hat g(k)=\int_{-\infty}^\infty g(x)e^{-ikx}\mathrm d x$ so that $g(x)$ will be a proper superposition of plane waves. In general, if there is a physical reason why either normalization is more convenient for a particular case (generally subject to the plane-wave convention $(1)$) then that's the way to go.

That said, the choice of normalization is essentially irrelevant. (Unlike, say, the Laplace transform, where it matters whether it's exponential decay towards the right or towards the left.) As Chris White mentioned in the comments the two square roots of $-1$, namely $i$ and $-i$, are essentially equivalent, and luckily the split in Fourier conventions mostly falls along the same lines as the split on whether the imaginary unit is denoted $i$ or $j$, so you can often simply write $j=-i$ and reconcile vast stretches of literature with one another.

Note also that the forward and inverse Fourier transforms are essentially identical, and the choice of signs is purely a matter of convention. There's nothing that can go 'wonky' by changing a convention; at worst what will happen is that the signs will change in the connection to the Laplace transform if that's what you're doing. In any case, you always need to double-check your signs to confirm that you're using the correct convention, so this shouldn't imply any additional work.

Emilio Pisanty
  • 137,480