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I define a metric:

$$ (d s)^2=(d x)^2+(d y)^2 $$

As per the pythagorean theorem, all terms of the infinitesimal metric $(ds)^2$, $(dx)^2$ and $(dy)^2$ are (infinitesimal) areas. I am trying to connect the terms to their corresponding area, but I am having trouble. For instance, say I define an area as

$$ A[x] $$

Then taking its derivative produces $dA[x]=\frac{dA[x]}{dx}dx$. For example with $A[x]=x^2$, I get $dA[x]=2xdx$. In this case $dx$ is not squared. Is there any $dA[x]$ for which $dx$ is squared?

What I am asking is what is the anti-derivative of the infinitesimal area $(dx)^2$? What expression of area $A[x]$ has the differential form of $(dx)^2$?

Anon21
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  • One way to think about a 2D integral of a surface with an orientation like $\iint\mathrm dx\mathrm dy$ is with differential forms as $\int \mathrm dx\wedge\mathrm dy$. If you try to interpret $\left(\mathrm dx\right)^2$ in this sort of way, you end up with $\mathrm dx\wedge\mathrm dx=0$ so it doesn't really work. The closest thing you can do is something like $\int\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^{2}\mathrm dt$, but there it's crucial that we don't end with $(\mathrm dt)^2$. – Mark S. Mar 30 '19 at 21:19

1 Answers1

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In fact $dA=dxdy$, so $\frac{dA}{dx}=dy$. This looks like a strange equation unless you accept that infinitesimals in calculus can have different orders. Similarly, in polar coordinates $dA=rdrd\theta$, so $\frac{dA}{dr}=rd\theta,\,\frac{dA}{d\theta}=rdr$.

But quantities such as $dx^2$ aren't meant to be integrated as such. With respect to what would you integrate them? I can write $\int 2x dx=x^2$, but $\int dx^2 dx$ would be... What? Well, it's not $dx^3$ because of the $\int$, and we'd learn nothing new from saying it's $\int dx^3$ (which a reader might interpret to mean something very different, such as integration over a variable called $x^3$ where the superscript isn't an exponent, or as a triple integral $\int d^3 x$).

The way to understand $ds^2=dx^2+dy^2$ is an infinitesimal variant of squares erected on sides of a right-angled triangle, viz. Pythagoras. Note this doesn't require you to integrate the terms in this equation. A more careful way to think about it is $$\int ds=\int\sqrt{\left(\frac{dx}{dp}\right)^2+\left(\frac{dy}{dp}\right)^2}dp$$for a parameter $p$ (provided we can and do see all contributions to the arc length as positive).

J.G.
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