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

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

• unit tangent: $\TT = \frac{\ct'(t)}{|\ct'(t)|}$
• unit normal (2D): $\NN = \TT^\perp$
• total displacement, total arc length  Two different curves $\ct(t)$ and $\cs(u)$ 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.

\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 $\{(x,x^2,x^3) \st x \in [1,4]\}$. 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 $C$ is the Curve, equipped with an orientation, corresponding to the curve $\ct$.

velocity displacement function $\frac{d\ct}{dt} = \ct'(t) \quad \Rightarrow \quad d\ct = \ct'(t) dt$ $\ct(t)-\ct(a) = \int_a^t \ct'(u) du = \int_C d\ct$ $\frac{ds}{dt} = \|\ct'(t)\| \quad \Rightarrow \quad ds = \|\ct'(t)\| dt$ $s(t) = \int_a^t \|\ct'(u)\| du = \int_C ds$

As mentioned before, the total displacement and total arc length are parametrization-independent (up to orientation). %

Parametrization-independent integration

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

Verify:

• $\cs'(s) = \TT$ and thus also $|\cs'(s)| = 1$ (parametrization independent!)
• $\TT ds = d\ct$ and $\TT \cdot d\ct = ds$

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

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

Differential geometry perspective

Recall from differential geometry that we had the following properties between the standard basis $U_1, U_2, U_3$ for vector fields and their dual 1-forms $dx_1, dx_2, dx_3$. Specifically, $dx_i$ is defined so that \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 $dx_i$ offer some insights that will help in subsequent discussion.

• $dx_i(\vv)$ gives the coefficient of $\vv$ in the basis $\{U_1, U_2, U_3\}$.
• $dx_1,dx_2,dx_3$ act like row vectors of the inverse of the basis matrix $A = [U_1 \; U_2 \; U_3]$. In this special case, since the basis is orthonormal, the inverse is simply the transpose $A^T$. However, for a more general basis as we will soon see, this is not the case.

Curve differentials as 1-forms

Goal: The description of line integrals seems to use two differentials $dt$ and $ds$, as well as a vector $d\ct$ 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 $dt(\vv_{\ct(t)}) = 0$ for any vector orthogonal to $\ct'(t)$.)

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

Exercises

• Definition implications: With $\ct(t) = \vvec{x(t),y(t),z(t)}$, write $dt$ in the standard 1-form basis; i.e. figure out the coefficients in $dt = a\,dx + b\,dy + c\,dz$ using the second definition above. What is the magnitude/norm of the 1-form $dt$? How is this consistent with the magnitude of $\ct'(t)$?
$dt = \frac{x'(t) dx+y'(t) dy+z'(t) dz}{\|\ct'(t)\|^2}$, so that $dt(\vv) = \frac{\ct'(t) \cdot \vv}{\ct'(t) \cdot \ct'(t)}$. The magnitude $\frac{1}{\|\ct'(t)\|}$ is the reciprocal of $\|\ct'(t)\|$ as expected from the definition.
• Arc-length parametrization: Show that in the special case of a unit-speed parametrization $\cs(s) = \vvec{\tilde{x}(t),\tilde{y}(t),\tilde{z}(t)}$, write $ds$ in the 1-form basis $\{dx,dy,dz\}$ as above, leading to the conclusion that $ds(\vv) = \TT \cdot \vv$, or equivalently, $ds(\TT) = 1$. Verify the conversion formula between the two 1-forms $ds = |\ct'(t)|dt$.
$ds = \tilde{x}'(s) dx + \tilde{y}'(s) dy + \tilde{z}'(s) dz = \frac{x'(t) dx+y'(t) dy+z'(t) dz}{\|\ct'(t)\|} = \|\ct'(t)\|dt$. Note that $\vvec{\tilde{x}'(s), \tilde{y}'(s), \tilde{z}'(s)} = \TT$ and the magnitude is 1 as expected.

Note: Try these problems before going into the next section as the solutions would be referred to below.

Integration of an arbitrary 1-form along a curve

Observing that $dt(\ct'(t)) = 1$ and that $ds(\ct'(t)) = |\ct'(t)|$, 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,

• $\int g\, dt = \int g\, dt(\ct'(t))\,dt = \int g\, (1)\, dt = \int g\, dt$ as expected
• $\int f\, ds = \int f\, ds(\ct'(t))\, dt = \int f\, \TT \cdot \ct'(t)\, dt = \int f\, |\ct'(t)| dt$ as expected But this definition applies to arbitrary 1-forms in $\RR^3$ leading to the following observation.

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