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I took a relativity class as an undergraduate but lost contact with the theory many years ago. Recently I took some old notes to revisit some concepts. I am a layman in the subject, so I apologize in advance for the basic question but I have never seen any satisfactoric explanantion on the matter.

The metric tensor $g$ tells you how spacetime curves and, consequently, how one measures distances. In $\mathbb{R}^{3}$ the norm $\|x\|$ has the natural interpretation of the lenght of the vector $x$ or the distance between the point $x$ and the origin. In relativity, we talk about four dimensional vectors ${\bf{x}}=(ct,x,y,z)$. Whats does $g({\bf{x}}, {\bf{x}})=(ct)^2-x^2-y^2-z^2$ means? What does it measure? In other worlds, why is the metric tensor defined as: \begin{equation} g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0& -1 & 0 & 0 \\ 0& 0 & -1& 0 \\ 0& 0 & 0 & -1 \end{pmatrix} \end{equation} in flat spacetime?

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  • $\begingroup$ "Why?" is a difficult question in physics because it's likely to be a never-ending tower of why's. The most relevant answer is that we find from experiments and observations that if we define the metric $g$ to be the one above, then one can accurately predict the outcomes of said experiments. $\endgroup$
    – Prahar
    Commented Aug 25, 2022 at 6:13
  • $\begingroup$ @Prahar, I know the "why" question is difficult, but I thought there was something more about it. I remember when the professor deduced the Lorentz transformation in class. If one person, in some reference frame, send a light signal at time $t=0$, in time $t>0$ this light signal would have illuminated a whole sphere $(ct)^{2}-x^2-y^2-z^2 = 0$. This is precisely $g({\bf{x}},{\bf{x}}) = 0$. What's inside of the sphere is what you can see because nothing travels faster than $c$. I thought the metric tensor could be deduced or given a meaning out of some toy model like this one. $\endgroup$
    – JustWannaKnow
    Commented Aug 25, 2022 at 13:38

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It measures the "wristwatch time" $\tau$ between two events. This is the time read on a wristwatch that passes through both events at constant velocity. This is because $\tau$ is invariant and $dx=dy=dz=0$ in that frame so $d\tau=cdt$. (You made a little error in your metric. The space intervals should all have the same sign.) But there is a lot more to say about the metric. It's subtle. Check out Exploring Black Holes by Taylor and Wheel. The entire book is about the metric. More advanced books tend to gloss over the physical meaning of the metric. For example, that an object moving in a gravitational field takes the path of maximal aging. If there is no gravity then this path is a straight line. The metric also allows you to compare space and time intervals in different reference frames such as stationary and rotating or far away from a massive object and close to that object.

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  • $\begingroup$ Jelly, thanks for your answer. I don't think I fully understand, however. What exactly do you mean by "passes through both events"? What I mean is: the "norm" $g({\bf{x}},{\bf{x}}) = (ct)^{2}-x^{2}-y^{2}-z^{2}$ should be a property of the vector ${\bf{x}}$ (or its coordinates) alone, right? How come it tells something about two events? Do you mean that one event ocurred at the origin? $\endgroup$
    – JustWannaKnow
    Commented Aug 25, 2022 at 13:24
  • $\begingroup$ @IamWill $g(x,x)$ is the dot product in spacetime. It's square root gives the distance, in spacetime, between two events where x is the vector connecting them. $\endgroup$
    – jelly ears
    Commented Aug 25, 2022 at 19:38
  • $\begingroup$ @IamWill the spacetime "distance" between events connected by a light ray is 0. $\endgroup$
    – jelly ears
    Commented Aug 26, 2022 at 0:05

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