The rigor of using differentials in line elements

[axmls]

Greetings all,

In Hartle's Gravitation, it seems like a given the way he manipulates the "differentials" when he gives a line element. The thing is that it works, and I know it's not mathematically rigorous, but first of all, I'm wondering when treating the differentials in the line elements in this way breaks down, and I'm also wondering if there's some kind of rigorous mathematical framework behind framing line elements in this way.

Example: the standard Euclidean line element $ds^2 = dx^2 + dy^2$. When I was in calculus II, we covered arc lengths, but there was never a good explanation as to where the formula comes from, but it seems so intuitive to say $$s = \int_a ^b ds = \int_a ^b \sqrt{dx^2 + dy^2} = \int_a ^b \sqrt{dx^2\left(1+\frac{dy^2}{dx^2}\right)} = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} \ dx$$

I guess what I'm asking is: is there anything that justifies these operations with line elements without going into nonstandard analysis in a way where it's not so less-than-rigorous?

 

[SteamKing]

Example: the standard Euclidean line element $ds^2 = dx^2 + dy^2$. When I was in calculus II, we covered arc lengths, but there was never a good explanation as to where the formula comes from, but it seems so intuitive to say $$s = \int_a ^b ds = \int_a ^b \sqrt{dx^2 + dy^2} = \int_a ^b \sqrt{dx^2\left(1+\frac{dy^2}{dx^2}\right)} = \int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2} \ dx$$

The arc length formula is just an application of Euclidean geometry to a curve which has been discretized using an arbitrary number of points. The lengths of the line segments created between pairs of points on the curve can be calculated using the Pythagorean relation, ds² = dx² + dy², which holds even as the number of points increases and dx, dy, and ds get smaller and smaller.

http://www.intmath.com/applications-integration/11-arc-length-curve.php

 

[lavinia]

One can get this formula without thinking about infinitesimals.

If $c(u) = (x(u),y(u))$ is a curve with derivative $dc/cu = (dx/du,dy/du)$ then the arc length is

$s = ∫_{a}^{b} \sqrt{dc/du⋅dc/du}du$ where $dc/du⋅dc/du$ is the dot product of $dc/du$ with itself.

- Note that if one thinks of $u$ as time, then $dc/du$ is the velocity vector of $c$ and $\sqrt{dc/du⋅dc/du}$ is its speed. So its integral is the distance traveled along $c$ i.e. it is the length of the curve.

$dc/du$ may be rewritten as $(dx/du)^2 + (dy/du)^2$. In your case, $u = x$ so $dx/du = 1$ and $du = dx$. Substituting into the equation gives your formula i.e.

$s = ∫_{a}^{b} \sqrt{1 + (dy/dx)^2}dx$

The derivatives, $dc/du$, $dx/du$, and $dy/du$ are just derivatives - computed as limits of Newton quotients. One does not need to think of them as ratios of infintesimals. The equation could have just as easily been written as

$s = ∫_a^b \sqrt {x'(u)^2 + y'(u)^2}du$ Then $x'(u) = 1$ and $y'(u)$ = $y'(x)$. Same thing.

What is ds?

One can write arclength as a function $s(u) = ∫_{a}^{u} \sqrt{(dx/du)^2 + (dy/du)^2}du$

$ds$ is its differential. Differentiating gives $ds = \sqrt{(dx/du)^2 + (dy/du)^2}du$ or equivalently
$ds = \sqrt{(x'(u))^2 + (y'(u)^2}du$

Here $u$ is also a function and $du$ is its differential.

 

[leright]

As SteamKing pointed out, it is just an application of the pythagorean theorem.

 

[Ssnow]

$d\,x,d\,y,...$ can be seen as a basis of the cotangent space of your ''natural interested space'' ..., it is rigorous ...

 

[zinq]

The rigorous idea that includes differentials is called "1-dimensional differential forms" in general. At any point of the space you are considering, a differential form is a linear map from tangent vectors to the real numbers. If tangent vectors have each been assigned a natural length — as in Euclidean space — then this linear map (at some specific point of your space) can be expressed in terms of the dot product with some fixed tangent vector at that point.

For example, for a curve α(s) through the point P (let's say α(0) = P) the differential form ds at P applied to an arbitrary tangent vector v at P would be the dot product

α'(0)⋅v​

of the tangent vector α'(0) to the curve at P, and the tangent vector v.

At first, this definition (which has actually omitted some technical details) probably sounds ridiculously complicated. But it turns out to put differentials on a totally rigorous basis that does not involve infinitesimals. They were first introduced by the French mathematician Élie Cartan in 1899.