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https://en.wikipedia.org/wiki/Wave_equation#Introduction

Does the wave equation

$$ (\partial_t{}^2 -c^2 \nabla^2) u=0 $$

imply the metric of the universe is Minkowski (-like)

$$ g= (+,-c^2, -c^2 ,-c^2 ) $$

See the d'Alembert operator https://en.wikipedia.org/wiki/D%27Alembert_operator

I came across the wave equation in my classical mechanics course in an effort to reproduce coupled oscillator behavior from an action and was flabbergasted when I just started using the d'Alembert operator without even realizing.

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    $\begingroup$ What do you mean "the metric of the universe"? There are a lot of places in the universe where the spacetime metric deviates greatly from the Minkowski metric. $\endgroup$
    – Charlie
    Commented Jan 14, 2021 at 22:07
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    $\begingroup$ Your $\partial_t$ should be $\partial_t^2$. You can write a wave equation for any metric. $\endgroup$
    – G. Smith
    Commented Jan 14, 2021 at 22:49
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    $\begingroup$ Why does your Minkowski metric have $c^2$ for the spatial components? $\endgroup$
    – G. Smith
    Commented Jan 14, 2021 at 22:51
  • $\begingroup$ @G.Smith the metric shouldn’t have units. OP is writing vectors with inconsistent units and then putting c in the metric to “fix” that and make the norm of the gradient have units of inverse time. Yes, it’s unusual... $\endgroup$
    – kaylimekay
    Commented Jan 15, 2021 at 3:54

3 Answers 3

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This wave equation merely asserts something about the scalar field $u$. A scalar field could violate this wave equation but obey the Klein-Gordon equation instead, and that would not change the structure of spacetime.

It's also possible for the universe not to be Minkowski but for this wave equation nevertheless to be the one obeyed by some field. In fact, the universe isn't Minkowskian -- it's only locally, approximately Minkowskian.

What we can say is that this wave equation is connected to the structure of spacetime in that, if spacetime is locally Minkowski, it's the only Lorentz-invariant second-order wave equation you can write for a scalar field if the only tool at your disposal is the derivatives (and not, say, a mass).

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It does not imply a Minkowski spacetime if the hypothesis is just as in the question.

However, it would imply it if making stronger the hypothesis as follows.

If we assume that

  1. $\Psi$ is a scalar
  2. in every inertial reference frame $\Psi$ satisfies the written differential equation (with the same constant $c$),

then, yes, (with some other technical hypotheses) we would conclude that the transformation laws of rest coordinates of inertial refrence frames are Poincaré ones. Hence the spacetime would be Minkowski.

However the meaning of the constant $c$ would not be the light speed in general.

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  • $\begingroup$ Actually, the maximal invariance algebra of the D'Alembert equation is the one of the conformal group, so the spacetime transformations could be more general. $\endgroup$
    – DanielC
    Commented Dec 10 at 1:57
  • $\begingroup$ Yes, I agree: in particular, if we rescale the units of space and time of the same factor the equation remains invariant. The causa structure admits a larger group of symmetry. $\endgroup$ Commented Dec 10 at 6:51
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Promoted from a comment:

You can write a wave equation for any metric.

Like the harmonic oscillator, there are lots of systems whose behavior can be modeled (especially in approximation) by "a wave equation." This similarity does not imply that all systems with wavelike behavior are somehow connected to each other, and therefore can't imply that the existence of a wave equation tells you very much about the underlying background of a system.

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