I'm currently studying SR and GR and I noticed that all the books introduce spacetime interval without any motivation as $$(\Delta s)^2 = -c^2 (\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$$ and than motivate that expression defining the Minkowski metric $\eta_{\mu\nu}$.

Is there a physical reason why the metric of spacetime is the Minkowski metric?

  • $\begingroup$ I very much doubt that "all the books" do not motivate this quantity at all (e.g. by showing it is the interval preserved by Lorentz transformations). $\endgroup$
    – ACuriousMind
    Mar 29, 2020 at 10:57
  • $\begingroup$ You can easily show that this interval is invariant under lorentz transformations. $\endgroup$
    – bemjanim
    Mar 29, 2020 at 11:59

1 Answer 1


Simple explanation: Its the only type of metric that conserves the speed of light.

Mathematical explanation: Let us assume an inertial observer, $O$, measures the speed of light as $dx/dt = c$. Let us assume another inertial observer, $O'$, measuring the speed of light in his coordinates, $dx'/dt' = c$.

Naturally at this point one can ask, What is the connection between $(t,x)$ and $(t',x')$ ?

The crucial point is to keep $c$ a constant while doing this transformation. Calculations show that Galilean transformation does not satisfy this condition since $c$ varies.

Let us take $dx/dt = c$ and write in the form of $dx^2 - c^2dt^2=0$.

(PS: We need to square because light can move in either poisitive or in negative direction. Such that the equations should both satisfy for $dx/dt = c$ and $dx/dt = -c$. For this reason we should square both sides)

Similarly by using $dx'/dt' = c$, we can write $dx'^2 - c^2dt'^2=0$

This implies that $$dx^2 - c^2dt^2=dx'^2 - c^2dt'^2$$

This transformation is the one that conserves the speed of light and that's what we're looking for.

In general we write this as $$ds^2 = -c^2dt^2 + dx^2$$

And later on, you can also show that

$ds^2 = -c^2dt^2 + dx^2 = -c^2dt'^2 + dx'^2$

  • 1
    $\begingroup$ I think you need to say why you are squaring, otherwise your argument applies without squaring anything and gives the wrong answer. $\endgroup$
    – m4r35n357
    Mar 29, 2020 at 11:55
  • 1
    $\begingroup$ @m4r35n357 I edited my post $\endgroup$
    – seVenVo1d
    Mar 29, 2020 at 12:25
  • $\begingroup$ that deals with my comment, thanks! $\endgroup$
    – m4r35n357
    Mar 29, 2020 at 19:15

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