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in Curved space it seems $dw^2=(dw)^2$ how is it possible!? $$x^2+y^2+z^2+w^2=\kappa^{-1}R^2,$$ $$dw=w^{-1}(xdx+ydy+zdz),$$ $$\kappa^{-1}R^2-(x^2+y^2+z^2)=w^2,$$ $$dl^2 = dx^2 + dy^2 + dz^2+dw^2,$$ $$dl^2 = dx^2 + dy^2 + dz^2 +\frac{(xdx+ydy+zdz)^2}{\kappa^{-1}R^2-x^2-y^2-z^2}$$ Remark: in general as i know $dx^2=2xdx$ so $dw^2$ should be: $$dw^2=(2w)w^{-1}(xdx+ydy+zdz)=2(xdx+ydy+zdz),$$ but here $$dw^2=(dw)^2=w^{-2}(xdx+ydy+zdz)^2,$$ how is it possible? |
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That wiki article is unfortunately very poorly written. In any event, the convention in relativity is always that $dx^2$ is shorthand for $(dx)^2$. The same holds for $dw^2$, $dt^2$, $d\varsigma^2$, $d\aleph^2$ or whatever else you see. The reason is you are rarely interested in the differential of a square (getting your second line from the first is an exception to this), but squares of differentials come up all the time. If perchance $dx^2 = d(x^2)$ in some case, then it is merely coincidence. This is unfortunately not the operator precedence I was taught in high school with differentials and deltas, but that's how it is. |
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