In flat Minkowski space with the metric signature (-,+,+,+) the spacetime interval $ds^2$ is defined as

$ds^2 = -c^2 dt^2 + x^2 + y^2 + z^2$

This interval (or better the square of the distance) can become negative (for timelike intervals) so the spacetime distance $ds$ can be complex.

In quantum mechanics a complex quantity is something which cannot be measured. Only real quantities can be measured. Does that mean that the spacetime distance becomes unmeasurable if it becomes complex? Or how is the interpretation in relativity for complex spacetime distances?

What confuses me is that one can also choose a different metric $(+,-,-,-)$. In that case the spacetime interval is

$ds^2 = c^2 dt^2 - x^2 - y^2 - z^2$

Then it is not timelike distances that become complex, but instead spacelike distances will be complex. So it seems to depend ob the convention of the metric signature when the space-time distance will be complex. So do complex spacetime distances (or negative spacetime intervals) have any physical meaning at all?

  • $\begingroup$ In QM, complex quantities are not automatically unmeasurable. But a complex observable quantity can always be decomposed into real observables (e.g. the real and imaginary parts of a complex observable are real observables) and complex observables make the math annoyingly complicated so we choose by convention to not deal with complex observables. Cf. physics.stackexchange.com/a/82616/107092 $\endgroup$
    – HTNW
    Sep 16, 2021 at 19:55

1 Answer 1


The scalar $ds^2:=dx^\mu dx_\mu$ is real, but its sign changes when you switch metric convention. Whether $\sqrt{ds^2}$ is real or imaginary is therefore not purely a physical question.

On the other hand, if $ds^2\ne0$ then $dx^\mu$ satisfies either $dt=0$ or $dx^i=0$ in a suitable reference frame, and which is applicable is very much physically relevant. If we can impose $dx^i=0$, the path is causally traversable; if we can impose $dt=0$, the path's endpoints can be regarded as simultaneous events, making them outside each other's light cone. (But you probably know all that.)

If you're interested in the real-complex distinction, it can help to work in proper time $d\tau^2=dt^2-d\vec{x}^2/c^2$, which settles the convention question. Traversable paths have $d\tau\in\Bbb R$; between events we can take to be simultaneous, $d\tau\in i\Bbb R$. But in the latter case, we don't normally consider it convenient to work with proper time.

  • $\begingroup$ I'm wondering, does a complex proper time then maybe have the same physical meaning as a complex speed of light (I mean like a complex phase velocity, which means the amplitude of a wave will decay)? $\endgroup$
    – asmaier
    Sep 18, 2021 at 20:50
  • $\begingroup$ @asmaier No, it means a proper distance of $\sqrt{d\vec{x}^2-c^2dt^2}$. $\endgroup$
    – J.G.
    Sep 18, 2021 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.