# The Unapologetic Mathematician

## Line Integrals

We now define some particular kinds of integrals as special cases of our theory of integrals over manifolds. And the first such special case is that of a line integral.

Consider an oriented curve $c$ in the manifold $M$. We know that this is a singular $1$-cube, and so we can pair it off with a $1$-form $\alpha$. Specifically, we pull back $\alpha$ to $c^*\alpha$ on the interval $[0,1]$ and integrate.

More explicitly, the pullback $c^*\alpha$ is evaluated as

$\displaystyle\left[c^*\alpha\left(\frac{d}{dt}\right)\right](t_0)=\left[\alpha_{c(t_0)}\right]\left(c_{*t_0}\frac{d}{dt}\bigg\vert_{t_0}\right)$

That is, for a $t_0\in[0,1]$, we take the value $\alpha_{c(t_0)}\in\mathcal{T}^*_{c(t_0)}M$ of the $1$-form $\alpha$ at the point $c(t_0)\in M$ and the tangent vector $c'(t_0)\in\mathcal{T}_{c(t_0)}M$ and pair them off. This gives us a real-valued function which we can integrate over the interval.

So, why do we care about this particularly? In the presence of a metric, we have an equivalence between $1$-forms $\alpha$ and vector fields $F$. And specifically we know that the pairing $\alpha_{c(t)}\left(c'(t)\right)$ is equal to the inner product $\langle F(c(t)),c'(t)\rangle$ — this is how the equivalence is defined, after all. And thus the line integral looks like

$\displaystyle\int\limits_c\alpha=\int\limits_{[0,1]}\langle F(c(t)),c'(t)\rangle\,dt$

Often the inner product is written with a dot — usually called the “dot product” of vectors — in which case this takes the form

$\displaystyle\int\limits_{[0,1]}F(c(t))\cdot c'(t)\,dt$

We also often write $ds=c'(t)\,dt$ as a “vector differential-valued function”, in which case we can write

$\displaystyle\int\limits_{[0,1]}F\cdot ds$

Of course, we often parameterize a curve by a more general interval $I$ than $[0,1]$, in which case we write

$\displaystyle\int\limits_IF\cdot ds$

This expression may look familiar from multivariable calculus, where we first defined line integrals. We can now see how this definition is a special case of a much more general construction.

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Posted by John Armstrong | Differential Geometry, Geometry

## 5 Comments	»

1. [...] any -form on the image of — in particular, given an defined on — we can define the line integral of over . We already have a way of evaluating line integrals: pull the -form back to the parameter [...]

Pingback by | October 24, 2011 | Reply

2. [...] flip side of the line integral is the surface [...]

Pingback by | October 27, 2011 | Reply

3. [...] Stokes’ theorem only works in dimension , where the differential can take us straight from a line integral over a -dimensional region to a surface integral over an -dimensional [...]

Pingback by | November 23, 2011 | Reply

4. [...] if we have a curve that starts at a point and ends at a point then fundamental theorem of line integrals makes it easy to [...]

Pingback by | December 15, 2011 | Reply

5. [...] We calculate the work done by a force in moving a particle along a path is given by the line integral [...]

Pingback by | January 13, 2012 | Reply

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