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As I understand it the space-time interval $\sigma$ is defined as $\sigma^2=\eta_{\alpha \beta}x^\alpha x^\beta$. Why is it that some books define the metric with the signs (-,+,+,+) and some with (+,-,-,-)? In the book Spacetime and Geometry the Minkowski metric $\eta_{\alpha \beta}$ is defined as

$\eta_{\alpha \beta}=\begin{pmatrix} -1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1 \end{pmatrix}$

while Wikipedia uses reversed signs (+,-,-,-). When do you choose which definition of the metric? Also the book Spacetime and Geometry uses $\Delta \sigma^2 = -c^2t^2+(x^2+y^2+z^2) $ to calculate the space-time interval and defines the proper-time interval as $\Delta \tau^2=-\Delta \sigma^2$ while Wikipedia and the book Introduction to the Theory of Relativity define $\Delta \tau^2=-\frac {\Delta \sigma^2}{c^2}$? If I want to calculate the time a moving object is measuring, which definition of the proper time and which metric signature do I use? I am very confused about this, could somebody explain this to me?

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There is no proper metric to use, you can use either the MM (mostly minus) metric or the mostly positive metric when doing any problem in special relativity and get the right answer so long as you are consistent. The physics is contained in the fact that the space and time components have opposite sign.

Obviously, a time interval which is future directed should be a real positive number, so if you choose to use the mostly positive metric, then the proper time needs to be defined as the negative of the spacetime interval. This is why some people prefer the mostly negative metric, but aesthetically, a lot of physicists prefer to have "time be the weird one" and use the mostly positive signature.

Almost any introductory text on special relativity should have an explicit discussion about this.

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  • $\begingroup$ Haha well I read two introductory textbooks and none had a section on this. What about calculating the time a moving object is experiencing? Do I use the definition $\Delta \tau^2 = -\Delta \sigma^2$ or $\Delta \tau^2 =-\frac {\Delta \sigma^2}{c^2}?$ $\endgroup$
    – Jannik Pitt
    Commented Dec 4, 2016 at 21:38
  • $\begingroup$ @AccidentalFourierTransform Oh okay, I see. So when working in units where $c$ is not equal to $1$ the space-like interval $\Delta \sigma^2$ and the time-like interval $\Delta \tau^2$ are always connected by $\Delta \sigma^2 = -c^2\Delta \tau^2$ no matter which metric signature I'm using? $\endgroup$
    – Jannik Pitt
    Commented Dec 4, 2016 at 21:43
  • $\begingroup$ @JannikPitt as AccidentalFourierTransform said, this is another bit of convention. If you are working in so called 'natural units,' then these are both the same expression as $c=1$. Are you sure you are looking at introductory books? I believe Griffith's book on particle physics has a gentle and clear discussion about this, and also in the later chapters in his E&M book when he gets to the relativistic formulation. $\endgroup$
    – CasualScience
    Commented Dec 4, 2016 at 21:45
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    $\begingroup$ "There is no proper metric to use" Well. Of course the metric used by [my favorite book] is the right one. It's just that other authors haven't noticed yet so a full fledged physicist must be able to deal with the many incorrect variants. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Dec 4, 2016 at 21:54
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    $\begingroup$ I left it unspecified because it is a comment on the culture around metric choice. But my personal choice is trace = -2 with the speed of light suppressed. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Dec 4, 2016 at 23:20

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