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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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closed as not a real question by David Z Dec 5 '12 at 3:28

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

This question(v1) appears to be caused by this post. – Qmechanic Dec 4 '12 at 8:32
Because $d(x^2)$ may be written as $2x\,dx$, as you seem to know, it would be silly to reserve the symbol $dx^2$ for that, too. Instead, $dx^2$ always means $(dx)^2$. What does the question "how is it possible" mean? What kind of an answer do you want? How would you answer a similar question "how is it possible that 2+2=4", for example? – Luboš Motl Dec 4 '12 at 10:39
@Qmechanic elected :D – Neo Dec 4 '12 at 10:43
Congrats, Qmechanic, it's good news! – Luboš Motl Dec 4 '12 at 11:00
Possible duplicate: – Qmechanic Dec 9 '12 at 19:32

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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Yes, well one could also write $\text{d}x\otimes\text{d}x$ if one wants. You add that that's not the operator precedence you were taught, but that's not really true as you already have it in $\frac{\text{d}^2 f(x)}{\text{d}x^2}$. Or maybe by "this" you mean it the other way around, but whatever. I like your dAleph, haha, is there an operator with Leibnitz rule for larger cardinals? :D – NikolajK Dec 4 '12 at 9:12
@NickKidman Alternatively $\mathrm{d}x \wedge \mathrm{d}x$ in the notation I sometimes use, since it is ultimately a wedge product of exterior derivatives. – Chris White Dec 5 '12 at 2:10
What is? Here we are concerned with the metric, right? And the line element requires no anti-symmetrization. – NikolajK Dec 5 '12 at 8:57

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