Lecture 17

Vector Fields & Path Integrals

APMA E2000

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Drew Youngren dcy2@columbia.edu

$\gdef\RR{\mathbb{R}}$ $\gdef\vec{\mathbf}$ $\gdef\bv#1{\begin{bmatrix} #1 \end{bmatrix}}$ $\gdef\proj{\operatorname{proj}}$ $\gdef\comp{\operatorname{comp}}$

Announcements

  • HW 9 due Thurs 11/7.
  • Quiz 6 this week.
    • Double/triple integrals
    • Coordinate systems
  • Exam 2 Tues, 11/12
    • Through HW9.
    • Same format as last time
    • New seat assignments
    • Not cumulative in the narrow sense but all math is cumulative.

1-minute review

MoI Integrals

Suppose the same mass $M$ were distributed evenly over the following regions in space:

  • $B$ a solid ball of radius $R$.
  • $S$ a (hollow) sphere of radius $R$.
  • $C$ a ring (circle) of radius $R$.

Which has the greatest moment of inertia about its center?

Other Integrals

Framework

Every integral has three essential parts, summarized as the where, the what, and the how.

\[ \int\limits_D F\,d\omega \]
  • Where? the domain of integration $D$
  • What? the integrand $F$
  • How? the differential (form) $d\omega$
We Need a New Integral

Let $C$ be the (1-dimensional) ring above object \[\iiint_C f\,dV = 0\] for any $f$.

Assume a constant linear density of $\frac{M}{2\pi R} \frac{\text{unit of mass}}{\text{unit of length}}$.

\[ \int_C (\text{distance to axis})^2 \times (\text{linear density})\,d(\text{length}). \] This requires a new integral called a line integral or path integral.

Path Integrals

Taxonomy

A line integral, also called a path integral, is an integral whose domain of integration is $C$, the image of a curve parametrized by $\vec r(t)$.

They come in different "flavors" ...

Go to 3Demos → Story → Path Integrals for a demo of the flavors.

w/ respect to arc length

\[ \int\limits_{\mathcal C} f\,ds \]
  • Where? path in space $\mathcal C$
    • given by $\vec r(t) = \langle x(t), y(t),\ldots \rangle$, $a \leq t \leq b$
  • What? scalar field $f(x,y,\ldots)$
  • How? w.r.t. arc length: $ds$
Formula
\[ \int\limits_{\mathcal C} f\,ds \approx \sum_{i= 1}^N f(x_i^*, y_i^*) \sqrt{\Delta x_i^2 + \Delta y_i^2} \]
Formula
\[ \int\limits_{\mathcal C} f\,ds \approx \sum_{i= 1}^N f(x(t_i^*), y(t_i^*)) \sqrt{\frac{\Delta x_i^2}{\Delta t^2} + \frac{\Delta y_i^2}{\Delta t^2}}\, \Delta t \]
\[ \longrightarrow \int_a^b f(x(t), y(t)) \sqrt{x'(t)^2 + y'(t)^2}\,dt \]
Formula
\[ \int\limits_{\mathcal C} f\,ds = \int_a^b f(x(t), y(t)) \sqrt{x'(t)^2 + y'(t)^2}\,dt \]
\[ = \int_a^b f(\vec r(t)) |\vec r'(t)|\,dt \]
Interpretation

If $f$ is the height of a wall whose base runs along $\mathcal C$, then $\int_{\mathcal C} f\,ds$ would be the area of the wall.

N.B. Though $\int_C ds$ is always positive, $f$ could yield negative values, so $\int_C f\,ds$ can be negative.

w/ respect to a coordinate

\[ \int\limits_{\mathcal C} f\,dx \]
  • Where? oriented path in space $\mathcal C$
    • given by $\vec r(t) = \langle x(t), y(t),\ldots \rangle$, $a \leq t \leq b$
  • What? scalar field $f(x,y,\ldots)$
  • How? w.r.t. the change in the $x$-direction: $dx$.
Formula
\[ \int\limits_{\mathcal C} f\,dx \approx \sum_{i= 1}^N f(x_i^*, y_i^*) \Delta x_i \]
Formula
\[ \int\limits_{\mathcal C} f\,dx \approx \sum_{i= 1}^N f(x(t_i^*), y(t_i^*)) \frac{\Delta x_i}{\Delta t} \, \Delta t \]
\[ \longrightarrow \int_a^b f(x(t), y(t)) \,x'(t)\,dt \]
Formula
\[ \int\limits_{\mathcal C} f\,dx = \int_a^b f(x(t), y(t))\, x'(t)\,dt \]
\[ = \int_a^b f(\vec r(t)) \,x'(t) \,dt \]
Interpretation

These integrals are certainly more abstract. They are most often the building blocks of vector field integrals.

N.B. Orientation of path matters here. Reversing the direction negates the integral.

Example

Let $C$ be the piece of the parabola $y=(x-1)^2$ from $(0,1)$ to $(3,4)$, and let $f(x,y) = x+y$.

Compare the line integrals: \[\int_C f\, dx, \int_C f\, dy, \text{ and } \int_C f\, ds \]

Solution

\[ \int_C f\,dx = \int_0^3 (t + (t - 1)^2)\,dt = 7.5\]

\[ \int_C f\,dy = \int_0^3 (t + (t - 1)^2) 2(t-1)\,dt = 16.5\]

\[ \int_C f\,ds = \int_0^3 (t + (t - 1)^2)\sqrt{1 + 4(t-1)^2}\,dt \approx 20.1 \]

Vector Fields

Definition

A vector field is simply a function \[ \vec F: \RR^n \to \RR^n. \]

We plot them by drawing arrows at regularly-spaces points in the domain. Here is:

\[\vec F(x,y) = \bv{\sin y \\ \cos x} \]

Line Integral of a Vector Field

\[ \int\limits_{\mathcal C} \vec F \cdot\,d\vec r \]
  • Where? oriented path in space $\mathcal C$
    • given by $\vec r(t) = \langle x(t), y(t),\ldots \rangle$, $a \leq t \leq b$
  • What? vector field $\vec F$
    • $\vec F(x,y,\ldots) = \langle P(x,y, \ldots), Q(x,y, \ldots), \ldots \rangle$
  • How? along the path: $d\vec r$
Formula
\[ \int\limits_{\mathcal C} \vec F\cdot d\vec r = \int_a^b \langle P(x(t), y(t)), Q(x(t), y(t)) \rangle \cdot \langle x'(t), y'(t) \rangle\,dt \]
\[ = \int_a^b \vec F(\vec r(t)) \cdot \vec r'(t)\,dt \]
Interpretation

$\int_C \vec F\cdot d\vec r$ measures how much a vector field $\vec F$ pushes along a path $C$.

Two paths with common endpoints plotted over a vector field.
Interpretation

$\int_C \vec F\cdot d\vec r$ measures how much a vector field $\vec F$ pushes along a path $C$.

  • If $\vec F$ is a force, and $C$ is the path of an object, the line integral is the work done by the force on the object.
  • Consider the question of whether a marathon run was "wind-assisted."

Work Integrals

Work = Force × Distance
Suppose a particle moves along a path $C$ parametrized by $\vec r(t)$ for $a \leq t \leq b$, and is subject to a force field $\vec F$ at each point of the curve, then the total work done by the force is
\[W =\int\limits_C \vec F\cdot d\vec r \]
\[W =\int\limits_C \vec F\cdot d\vec r = \int_a^b \vec F(\vec r(t))\cdot \vec r'(t)\,dt \]
Work = Force × Distance

Hooke's Law: force of a spring is proportional to displacement. $F = -k x$.

Compute the work done by spring moving an object from position $x_0$ to $x_1$.

$x(t) = x_0 + (x_1 - x_0)t$ for $0\leq t \leq 1$. \[ \int_0^1 -k( x_0 + (x_1 - x_0)t) (x_1 - x_0)\,dt = \frac12kx_0^2 - \frac12 k x_1^2 \]

Example - Rolling rolling rolling

A ball falls down a parabolic ramp $y=x^2$ from $(-2,4)$ to the origin. How much work was done by gravity?

$\vec r(t) = \left\langle t, t^2 \right\rangle$ for $-2\leq t \leq 0$. \[ \int_{-2}^{0} \left\langle 0, -mg \right\rangle \cdot \left\langle 1, 2 t \right\rangle= mg4 \]

Alternate Forms

In Components
\[ \vec F(x,y) = \langle P(x,y), Q(x,y) \rangle \text{ and } \vec r(t) = \langle x(t), y(t) \rangle, a \leq t \leq b\]
\[\int\limits_C \vec F\cdot d\vec r = \int_a^b \langle P(x(t),y(t)), Q(x(t),y(t)) \rangle\cdot \langle x'(t), y'(t)\rangle \,dt\]
\[ = \int_a^b P(x(t),y(t)) \, x'(t)\,dt + Q(x(t),y(t))\, y'(t) \,dt\]
\[ = \int\limits_C P\,dx + Q\,dy\]
Per Arc Length
\[\int\limits_C \vec F\cdot d\vec r = \int_a^b \vec F(\vec r(t))\cdot \vec r'(t) \,dt\]
\[= \int_a^b \vec F(\vec r(t))\cdot \frac{ \vec r'(t) }{|\vec r'(t)|} |\vec r'(t)| \,dt\]
\[ = \int\limits_C \vec F \cdot \vec T\,ds \]