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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

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This question(v1) appears to be caused by this post. –  Qmechanic Dec 4 '12 at 8:32
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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: physics.stackexchange.com/q/31594/2451 –  Qmechanic Dec 9 '12 at 19:32

1 Answer 1

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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