# Why is a spacetime interval defined as $(\Delta s)^2$ = $\eta_{\mu\nu}\Delta x^{\mu} \Delta x^{\nu}$?

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?

• 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). Mar 29, 2020 at 10:57
• You can easily show that this interval is invariant under lorentz transformations. Mar 29, 2020 at 11:59

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

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

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