Differential Geometry – Integrals

\gdef\pp{\mathbf{p}} \gdef\uu{\mathbf{u}} \gdef\vv{\mathbf{v}} \gdef\ct{\mathbf{\mathbf{r}}} \gdef\cs{\mathbf{\mathbf{q}}} \gdef\TT{\mathbf{T}} \gdef\NN{\mathbf{N}} \gdef\FF{\mathbf{F}} \gdef\GG{\mathbf{G}} \gdef\RR{\mathbb{R}} \gdef\vvec#1{\left<#1\right>} \gdef\st{\;\mid\;} \gdef\cross{\times} \gdef\dot{\cdot} \gdef\grad{\nabla} \gdef\curl{\grad\cross} \gdef\div{\grad\dot} \gdef\tgrad{\text{grad}\;} \gdef\tcurl{\text{curl}\;} \gdef\tdiv{\text{div}\;} \gdef\pdiff#1#2{\frac{\partial #1}{\partial #2}}

Curves

A smooth, oriented curve (trajectory) r\ct is a map from an interval [a,b] [a,b] into Rn\RR^n such that r(t)\ct'(t) exists and r(t)0|\ct'(t)| \ne 0. The image r([a,b])\ct([a,b]) (i.e. the collection of corresponding points in Rn\RR^n is called a Curve CC.

Parametrization independent quantities: Those that depend only on CC and possibly the orientation, but not on the specific trajectory r\ct.

Two different curves r(t)\ct(t) and q(u)\cs(u) are reparametrizations of each other if their images are equal. These two curves correspond to the traversal of the same Curve with different speeds. E.g.

r(t)=<t,t2,t3>,t[4,1]q(u)=<eu,e2u,e3u>,u[0,ln2] \begin{aligned} \ct(t) = \vvec{-t, t^2, -t^3}, t \in [-4, -1] \\ \cs(u) = \vvec{e^u, e^{2u}, e^{3u}}, u \in [0, \ln 2] \end{aligned}

are different parametrizations/traversals of the Curve {(x,x2,x3)    x[1,4]}\{(x,x^2,x^3) \st x \in [1,4]\}. Observe that these two parametrizations have opposite orientation.

Our goal is to compute useful quantities with curves that are independent of parametrization, but may be dependent on orientation. To do this, we need the following concepts.

Velocity and speed

Below, assume that CC is the Curve, equipped with an orientation, corresponding to the curve r\ct.

velocity drdt=r(t)dr=r(t)dt\frac{d\ct}{dt} = \ct'(t) \quad \Rightarrow \quad d\ct = \ct'(t) dt displacement function r(t)r(a)=atr(u)du=Cdr\ct(t)-\ct(a) = \int_a^t \ct'(u) du = \int_C d\ct
speed dsdt=r(t)ds=r(t)dt\frac{ds}{dt} = \|\ct'(t)\| \quad \Rightarrow \quad ds = \|\ct'(t)\| dt arc length s(t)=atr(u)du=Cdss(t) = \int_a^t \|\ct'(u)\| du = \int_C ds

As mentioned before, the total displacement and total arc length are parametrization-independent (up to orientation). \(\begin{aligned} \text{displacement}(C) & = & \ct(b)-\ct(a) & = & \int_a^b \ct'(t) dt & = \int_C d\ct \\ \text{arc-length}(C) & = & s(b) & = & \int_a^b \|\ct'(t)\| dt & = \int_C ds \end{aligned}\)

Parametrization-independent integration

Since s(t)s(t) is strictly increasing, its inverse t(s)t(s) exists so that the unit-speed parametrization q(s)=r(t(s))\cs(s) = \ct(t(s)) with the same orientation as r\ct is well-defined.

Verify:

Can now integrate functions along oriented curve CC in a parametrization-independent way. These are called line integrals.

scalar value Cfds=Cfr(t)dt\int_C f ds = \int_C f |\ct(t)'| dt arc length when f=1f=1
vector field along curve CFTds=CFdα\int_C \FF \cdot \TT ds \quad = \quad \int_C \FF \cdot d\alpha how much F\FF is flowing along CC
vector flux through curve (2D) CFNds\int_C \FF \cdot \NN ds how much F\FF is flowing through CC

Differential geometry perspective

Recall from differential geometry that we had the following properties between the standard basis U1,U2,U3U_1, U_2, U_3 for vector fields and their dual 1-forms dx1,dx2,dx3dx_1, dx_2, dx_3. Specifically, dxidx_i is defined so that dxi(Uj)=δij={1 if i=j0 if ij[dx1dx2dx3][U1U2U3]=I \begin{aligned} dx_i(U_j) = \delta_{ij} = \left\{\begin{matrix}1 & \text{ if } i = j\\0 & \text{ if } i\ne j\end{matrix}\right. \qquad \Rightarrow \qquad \left[\begin{matrix} dx_1\\dx_2\\dx_3 \end{matrix}\right] \left[\begin{matrix} U_1 & U_2 & U_3 \end{matrix}\right] = I \end{aligned}

The following equivalent interpretations for the dual 1-forms dxidx_i offer some insights that will help in subsequent discussion.

Curve differentials as 1-forms

Goal: The description of line integrals seems to use two differentials dtdt and dsds, as well as a vector drd\ct of differentials. Understanding the 1-form interpretation of these differentials allows for a formalization of 1-form integration.

Recall that differential 1-forms, by definition, are linear scalar-valued functions on tangent vectors at each point and thus have the same dimension as the tangent space. Since at each point on a curve, the tangent space is a 1-dimensional, so is the space of 1-forms. So, we need to define just a single basis 1-form of the tangent space of a curve.

(One can, however, extend the 1-form to apply to tangent vectors of the embedding space by letting dt(vr(t))=0dt(\vv_{\ct(t)}) = 0 for any vector orthogonal to r(t)\ct'(t).)

Definition of the dtdt 1-form: Follow the pattern above from differential geometry, define dtdt to be the unique 1-form along the curve such that dt(rt)=dt(r(t))=1dt(\pdiff{\ct}{t}) = dt(\ct'(t)) = 1 and dt(v)=0dt(\vv) = 0 for all vr(t)\vv \perp \ct'(t). In other words, dtdt is the dual 1-form of the vector field rt\pdiff{\ct}{t} which spans the tangent space along the curve.

Exercises

Integration of an arbitrary 1-form along a curve

Observing that dt(r(t))=1dt(\ct'(t)) = 1 and that ds(r(t))=r(t)ds(\ct'(t)) = |\ct'(t)|, one can define the integration of a 1-form along the curve via \(\int_C \phi = \int_C \phi(\ct'(t))\, dt = \int_C \phi(\TT(t))\, ds\) Thus,

Line integrals of vector fields: One can represent a vector field in R3\RR^3 directly as F=<P,Q,R>\FF = \vvec{P,Q,R} (where each coordinate is a function), or alternatively, in its 1-form representation in the ambient space: ϕ=Pdx+Qdy+Rdz\phi = P dx + Q dy + R dz. Noting that drd\ct (=Tds=\TT ds) is a vector <dx,dy,dz>\vvec{dx,dy,dz} of differentials, here are equivalent ways of integrating the tangential component FT\FF\cdot\TT of vector field. \(\int_C \FF\cdot\TT ds = \int_C \FF\cdot d\ct = \int_C \vvec{P,Q,R} \cdot \vvec{dx,dy,dz} = \int_C Pdx +Q dy+R dz = \int_C \phi\)