Wikipedia:Reference desk/Archives/Mathematics/2021 December 17

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December 17

line integral and length

I am a bit confused regrading line integral and its relation with a length of a curve. On some places, the line integral over a curve is used to measure it's length (for example via integrating over 1). Which clearly isn't 0. But this seems to contradict the result in complex analysis that over a closed curve with analytical function, the integral value is exactly 0. 1 is analytical, and for example, a unit circle has a length of 2pi. but integrating over a circle should be 0 . What do I miss-understand. --Exx8 (talk) 16:25, 17 December 2021 (UTC)

You're conflating two different types of line integrals: arc length integrals and contour integrals. From a theoretical standpoint, the difference is how you measure the base of the rectangles in the Riemann sum. From a practical standpoint, if $r(t)$ nicely parameterizes a curve, the arc length integral is calculated as $\int _{a}^{b}f(r(t))|r'(t)|\,dt$, while the contour integral is $\int _{a}^{b}f(r(t))r'(t)\,dt$.--108.36.85.111 (talk) 17:03, 17 December 2021 (UTC)

Well, it depends what you're integrating. Not just the part before the d, but the part after the d as well.

When you write $\oint 1dz$, you can think of evaluating the function 1 at many many places that are very close together. Two of the places might be z and z+dz, where dz is a very small complex number equal to the difference between these two points close together on the circle. At all these places, you're adding up $1dz$; that is, $dz$. If you think about it you can maybe see why those cancel out.

On the other hand, say you integrate instead $\oint 1ds$ around the circle, where $ds$ is the "element of length". That means that at two nearby points on the circle, you're not adding up their difference as complex numbers, but the Euclidean distance between them. In that case, nothing cancels out, and you wind up getting the circumference of the circle. --Trovatore (talk) 17:05, 17 December 2021 (UTC)

More generally, if $\mathbf {r} :[a,b]\rightarrow C$ describes a path in $\mathbf {R} ^{n}$, then $\int _{a}^{b}1\,\mathbf {r} '(t)\,dt$ is simply the vector from $\mathbf {r} (a)$ to $\mathbf {r} (b)$, and if $C$ is closed then the integral is the 0 vector. On the other hand $\int _{a}^{b}1\,|\mathbf {r} '(t)|\,dt$ is the length of $C$. The difference is that in one integral you're adding vectors and in the other you're adding lengths of vectors.

Btw, it seems like an expression like $\int _{C}f(\mathbf {r} )\,d\mathbf {r}$ ought to be "a thing", but it looks like we don't cover it in line integral. It's been a long time since I took this stuff in school, so is this because it reduces trivially to 1-dimensional integrals? --RDBury (talk) 06:24, 19 December 2021 (UTC)

The notation is used here, where it appears to be a homework assignment.  --Lambiam 08:56, 19 December 2021 (UTC)

I don't think that's the same; the integrand in the SE post is a dot product and the result is a scalar. I'm thinking of scalar multiplication and the result would be a vector. The case where the integrand is a dot product is covered in our article under "Line integral of a vector field". --RDBury (talk) 14:09, 19 December 2021 (UTC)

That post has the identity

$\int _{C}\mathbf {F} \cdot {\rm {d}}\mathbf {r} =\int _{a}^{b}\mathbf {F} (\mathbf {r} (t))\mathbf {r} '(t)dt$.

In the context, the implied product operation between $\mathbf {F} (\mathbf {r} (t))$ and $\mathbf {r} '(t)$ in the rhs is the dot product, which does not do what we want. However, the identity is – in principle – meaningful for any product defined between the types of the operands. Assuming that scalar multiplication applies with $1\,\mathbf {v} =\mathbf {v} ,$ we have $\textstyle {\int _{C}1}\,{\rm {{d}\mathbf {r} =}}$ $\textstyle {\int _{a}^{b}}\mathbf {r} '(t)dt=$ $\mathbf {r} (b)-\mathbf {r} (a).$  --Lambiam 16:21, 19 December 2021 (UTC)

Until one gets into differential forms, we generally only integrate scalar integrands. Hence $\mathbf {F} \cdot {\rm {{d}\mathbf {r} }}$, since dot product gives a scalar. In the case of contour integrals, a real 2-vector is recast as a complex scalar. Integrating a vector integrand would just be integrating the individual coordinates separately, so I'm not surprised we don't see it much.108.36.85.111 (talk) 17:33, 19 December 2021 (UTC)

True. However, consider this. Let $V$ is a vector space with basis $\mathbf {v} _{1},...,\mathbf {v} _{n}$, so that we can represent any vector $\mathbf {v}$ as a sum of scaled basis vectors by writing $\mathbf {v} =a_{1}\mathbf {v_{1}} +\cdots +a_{n}\mathbf {v_{n}} ,$ so that we can define projection functions $\pi _{i}:V\to \mathbb {R}$ by $\pi _{i}(\mathbf {v} )=a_{i}.$ Now suppose we have some function $\mathbf {G} :\mathbb {R} \to V.$ Then the obvious adaptation of the Riemann integral to this setting is to interpret the addition of the Riemann sum as addition in the ring $V$. This is the counterpart of the componentwise definition of the fluxion of a vector. Then $\textstyle {\int _{a}^{b}}\mathbf {G} (t)dt$ is equivalent to componentwise integration, giving the value $\textstyle {\sum _{i}}I_{i}\mathbf {v} _{i},$ where $I_{i}=\textstyle {\int _{a}^{b}}\pi _{i}(\mathbf {G} (t))dt.$ Now, if $\mathbf {r}$ is parametrized with $t\in [a,b],$ so it is effectively a function of type $\mathbb {R} \to V,$ the integral

$\int _{a}^{b}F(\mathbf {r} (t))\mathbf {r} '(t)dt,$

(compare the rhs above) is of this form we can handle, provided that $F:V\to \mathbb {R} .$ Just put $\mathbf {G} (t)=F(\mathbf {r} (t))\mathbf {r} '(t).$  --Lambiam 20:26, 19 December 2021 (UTC)

I think all of this stuff about paramaterization is not really all that helpful. It's true that the way you calculate a line or surface integral generally goes through a parameterization, and it's good to know the formulas.

But the paramaterization is irrelevant conceptually. It's best to think of integrals the way Newton and Leibniz did, as sort of continuous summations of infinitesimal quantities. The integrand, including the bit after the d, tells you what that infinitesimal quantity is.

With that understood, integrals like $\oint f(z)ds$ or $\oint f(z)dz$ have clear and different natural meanings. And you can easily work out what $\int {\vec {f}}({\vec {r}})\cdot {\vec {dr}}$ or $\int {\vec {f}}({\vec {r}})\times {\vec {dr}}$ or $\int _{A}f({\vec {r}})\cdot {\vec {dS}}$ should mean (the last ${\vec {dS}}$ being an element of surface area, with a direction normal to the surface).

Then if you want to formalize all this and make it rigorous, you just adapt the definitions of the Riemann–Stieltjes integral or Lebesgue–Stieltjes integral in the obvious way. No paramaterization needed; in principle you can even integrate along a non-rectifiable curve. (I guess you do need an ordering.) --Trovatore (talk) 22:17, 23 December 2021 (UTC)

I agree it is not a big deal. Note, though, that the Riemann-Stieltjes integral is also regularly denoted in shorthand form using parametrization, as seen e.g. here, in a textbook. And the familiar identity $\textstyle {{\int }FdG=FG-{\int }GdF}$ uses parametrization implicitly. Apart from the convenience of the shorthand notation, it is also the case that students may be familiar with the Riemann integral but not yet with any of its generalizations. The explanation as I gave above then does not require the introduction of new concepts, but merely an explanation of the notation. The formal concept of a curve already requires the notion of parametrization.  --Lambiam 09:37, 24 December 2021 (UTC)

Well, it doesn't have to. You could take a curve to be a 1-dimensional submanifold. Admittedly you'd lose the possibility of self-intersection that way.

I think for purposes of defining the Riemann–Stieltjes integral on it, you could take a curve to be an arbitrary subset of the larger space together with a linear order on that set. The integral might not be well-defined very often, but when it is, it should match any of the other ways you could define it, and it shows that the paramaterization is not essential. --Trovatore (talk) 19:33, 24 December 2021 (UTC)

Actually, now that I look closer, I can't figure out what you mean when you say "$\textstyle {{\int }FdG=FG-{\int }GdF}$ uses parametrization implicitly", or that your textbook link uses paramaterization. I don't see any parameterization in either place. --Trovatore (talk) 19:39, 24 December 2021 (UTC)