[Math] Is line element mathematically rigorous

calculusdifferentialdifferential-formsinfinitesimalsmultivariable-calculus

I know differentials (in a way of standard analysis) are not very rigorous in mathematics, there are a lot of amazing answers here on the topic.

But what about line element?

$$ds^2 = dx^2 + dy^2 +dz^2 $$

The way I think about this line element is being geometrically constructed from Pythagoras theorem as:
$$\Delta s^2 =\Delta x^2 + \Delta y^2 +\Delta z^2$$
and then we assume that we can 'get' these quantities ($\Delta x$) to be infinitesimally small (as small as we like) and represent as $dx$ instead, right?

Now then lets take line element on a sphere:
$$ds_2 ^2=r^2sin^2(\theta)d\phi^2 + r^2d\theta ^2$$

It is geometrically constructed again using Pythagoras theorem and assuming that sides of a 'triangle' are small:

$$\Delta s_2 ^2 \approx (rsin(\theta)\Delta \phi)^2 + (r\Delta\theta)^2$$

But this approximation never really becomes equality, the smaller the angles the better it works, but still never equality! People just replace $\Delta->d$ and say $ds$ and say it's differential.

I guess my question is this:

when we write something like
$$ds_2 ^2=r^2sin^2(\theta)d\phi^2 + r^2d\theta ^2$$
we actually have in mind that this quantity contains higher order terms, but they will vanish after we parametrise?
I think about parametrisation in a way:
$$\frac{ds_2^2}{dt^2}=r^2sin^2(\theta)\frac{d\phi^2}{dt^2} + r^2\frac{d\theta^2}{dt^2}$$

Best Answer

differentials (in a way of standard analysis) are not very rigorous in mathematics

Standard or not has nothing to do with it.

Operations on differential notation are fully rigorous in standard mathematics, as a well-defined representation of properties of differentiable functions. If you want the differentials themselves to be mathematical objects that satisfy the rules of the differential notation, that can be done (and is done, in several different useful ways) in standard mathematics. Robinson NSA is only one option.

this approximation never really becomes equality, the smaller the angles the better it works, but still never equality!

That is true in all forms of infinitesimal reasoning except those with nilpotents (which can be defined by asserting that the order $k$ Taylor approximations are an equality, for a selected value of $k$). Calculations are always "up to higher order terms".

we actually have in mind that this quantity contains higher order terms, but they will vanish after we parametrise?

We have in mind that only a finite leading-order term is being calculated at the end, and the rest is an error term that is qualitatively smaller in the sense that it vanishes in the limit, or is infinitesimal or nilpotent or non-non-zero (which are some of the meanings of "qualitatively smaller" that are assigned in different formalizations of infinitesimal reasoning). For example, the product rule $d(FG) = (dF)G + F(dG)$ is true only up to second-order terms, which means ignoring $(dF)(dG)$ as something that will not contribute to the end result when calculating some quantity such as $d(FG)/dt$. If you do not want the limiting or infinitesimal answer, but a finite change in $FG$ over a finite interval of $t$, then the product rule is false and the higher order terms need to be accounted for. The calculus rules are only for handling the principal terms in asymptotic calculations that become "exact in the limit".

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