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If the four-vector $x^\mu$ is defined as $x^\mu\equiv(ict,{\bf x})$, instead of $x^\mu\equiv (ct,{\bf x})$, the Lorentz group will be the compact(?) ${\rm SO}(4, \mathbb{C})$ group. But the Lorentz group is regarded as the noncompact group ${\rm SO}(3,1)$. But I could never figure out what is the real problem of using $ict$ instead of $ct$? In short, what will go wrong if I choose to work with $ict$? Does it pose a problem from the point of view of representation theory? I am only interested in flat spacetime.

SRS
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2 Answers2

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${\rm SO}(4, \mathbb{C})$ allows all four coordinates of spacetime to be complex. It doesn’t just allow time to be imaginary.

G. Smith
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Misner, Thorne, & Wheeler (MTW) offer arguments in
“Farewell to ict" on Gravitation, p.51.

Reasons for using ict:

  1. It makes spacetime geometry look like Euclidean geometry.
  2. It make a Lorentz transformation look like a rotation.
  3. It allows one to avoid distinguishing components of a vector from its metric-dual one-form.

Reasons NOT to use ict:

  1. A vector is a very different geometric object from a one-form.
  2. The Euclidean angle is periodic, whereas the Minkowski-angle, better known as the rapidity ("velocity parameter"), increases monotonically without bound.
  3. Hiding the Lorentzian signature (- + + +) hides the light-cones that encode the causal structure.
  4. No one has discovered a way to use this in general relativity for a general curved spacetime manifold.
    and thus conclude:
    "If '$x^4 = ict$' cannot be used there, it will not be used here" [in this book Gravitation].

See page 19 of this 20-page excerpt at http://laplace.physics.ubc.ca/000-People-matt/200/gravitation.pdf.

In my opinion, disadvantage #3 concerning the causal structure is the most important reason not to use it.

robphy
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