I know that time dilation can be derived from the spacetime interval like so $$c^2d\tau^2=c^2dt^2-dx^2-dy^2-dz^2$$ Dividing both sides by $c^2dt^2$ $$\frac{d\tau^2}{dt^2}=1-\frac{1}{c^2}[(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2]=1-\frac{v^2}{c^2}$$ So $$\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2}}\implies t=\gamma\tau$$ I am struggling to do the same to prove that $s=\gamma l$ where $l^2=x^2+y^2+z^2$ and $ds^2=-c^2dt^2+dx^2+dy^2+dz^2$. Any ideas?
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The convention does not matter; it is the same argument. Just change the convention yourself. – naturallyInconsistent Dec 21 '23 at 03:57
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@naturallyInconsistent Oh right, I misread. I see it now, thank you! – Chen Weizhi Dec 21 '23 at 03:59