Fourier transform in Minkowski space

In summary, it is possible to represent a function in Minkowski space by a superposition of some convenient functions that propagate at speeds LESS than c. This may be useful for representing an arbitrary charge distribution as a superposition of charge distributions moving inertially.
  • #1

In Fourier analysis, we can decompose a function into sine waves with different wavenumbers that travel at different speeds (i.e., for a given wavenumber k they can have different frequencies ω and therefore different speeds v = ω/k). There is no upper bound on the speed of propagation v for an ordinary Fourier transform. I am wondering whether it is possible to similarly represent a function in Minkowski space by a superposition of some convenient functions (e.g., we could still use sine waves if we only have 1 spatial and 1 temporal dimension) that propagate at speeds LESS than c.

The specific application I have in mind is to represent an arbitrary charge distribution as a superposition of charge distributions moving inertially (ideally, at speeds less than c).

Thanks for your help.
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  • #2
Hi zlasner,

A true Fourier Transform cannot be accomplished in Minkowski space. The FT is essentially a mapping of a displacement along one or more spatial coordinates with respect to time into frequency and phase components. Each spatial dimension must be orthogonal to any other and all must be orthogonal to (independent of) the time dimension. However, applying the Lorentz Transformation destroys the orthogonal relationships between the dimensions (you no longer work within Euclidean geometry)

There are some proposals for remedial measures that would allow some types of pseudo-Fourier Transforms I believe. But it's unclear what there usefulness is and how the results should be interpreted. However, within each Galilean reference frame that you get after after applying the LT you can apply the FT. Maybe you want to find a transformation that is equivalent to the LT for values in the frequency domain?
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  • #3
While different frequencies / wavenumbers can have different speeds (in a dispersive medium, for example), there isn't anything that says they HAVE to have different speeds.

So if you're doing a Fourier transform in Minkowski space, you'd want all the waves to travel at 'c', because that's how fast waves travel in Minkowski space.
  • #4
What??! A Fourier transform in Minkowski space is done all the time in field theory, where configuration space is hardly ever used in favor of momentum space.

ψ(x) = (2π)-2 ∫ φ(k) d4k
φ(k) = (2π)-2 ∫ ψ(x) d4x

When a dispersion formula is known, ω2 = f(k2) it is handled just by inserting a delta function in φ(k).
  • #5
As Bill says, it is a standard thing to do Lorentz-invariant Fourier transforms in Minkowski space. He forgot the kernel in his integrals, though:

[tex]\varphi(x) = \int \frac{d^4 k}{(2 \pi)^4} \, \hat \varphi(k) \, e^{-i k_\mu x^\mu} = \int \frac{dt}{2 \pi} \int \frac{d^3 k}{(2 \pi)^3} \, \hat \varphi(t, \vec k) \, e^{i \omega t - i \vec k \cdot \vec x}[/tex]

(various signs depend on your metric conventions)
  • #6
I'm wondering how that can work. Is the FT effectively applied after the LT has been applied so that you're in a new Galilean reference frame? The right hand side of the kernel has an independent time dimension, doesn't it?

Apparently the inner product is re-defined so that the transform is actually a pseudo-Fourier Transform:

Because of this, two vectors satisfying <x | y> = 0 are called orthogonal. An important variant of the standard dot product is used in Minkowski space: R4 endowed with the Lorentz product

[itex]<x | y> = x_1 y_1 + x_2 y_2 + x_3 y_3 - x_4 y_4[/itex].[49]

In contrast to the standard dot product, it is not positive definite: <x | x> also takes negative values, for example for x = (0, 0, 0, 1). Singling out the fourth coordinate—corresponding to time, as opposed to three space-dimensions—makes it useful for the mathematical treatment of special relativity.
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  • #7
Phil, if you look carefully at the formulas I wrote, you will see that the argument of the exponential gives [itex]\omega[/itex] and [itex]\vec k[/itex] opposite signs. This is why it is Lorentz-invariant.

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