The Unapologetic Mathematician

Mathematics for the interested outsider

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.

October 21, 2011 - Posted by | 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 […]

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

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  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 […]

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  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 […]

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

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