Popper's Dreamland: Vector Analysis

Another chapter down! This is my favorite part. It's like someone is reading my mind.

MATH: Were you looking for this?
RYAN: Um. I dunn-
Omigod. ..... Ummmmm. This is totally awesome. Yes. Yes I-
MATH: There's more, if you'd --
RYAN: ...was. wait. What?
MATH ... stop talking.
RYAN: Oh.
MATH: Check this out:

A scalar function φ is said to be harmonic if it is continuous,

has continuous second partial derivatives, and satisfies Laplace's Equation:

$\nabla^2 \phi = 0$

RYAN: How did you know I-
MATH: I'm just getting warmed up. How about this.

$\begin{align*} \textup{If}\;\;\;\;\; \vec{\mathbf{f}}= ( f_1(x,y,z), \; &f_2(x,y,z), \; f_3(x,y,z)) \\ &\nabla^2 \vec{\mathbf {f}} = \;? \end{align*}$

RYAN: But how -?
MATH: ...do you define the gradient of a vector field? Turn the page.
{rustle}

$(\mathbf{f}\cdot \nabla)\mathbf{g} \;=\; f_1\frac{\partial \mathbf{g}}{\partial x} + f_2\frac{\partial \mathbf{g}}{\partial y} + f_3\frac{\partial \mathbf{g}}{\partial z}$

Like that. Now try this:

$\begin{align*} \textup{curl(curl}\; \mathbf{f}) \;=\; \nabla \times (\nabla \times \mathbf{f}) \;&=\; \nabla(\nabla \cdot \mathbf{f}) - \nabla^2\mathbf{f}\\ &= \textup{grad(div}\;\mathbf{f}) - \nabla^2\mathbf{f}\\ \end{align*}$

And you have a formula for computing Del squared of a field:

$\nabla^2\mathbf{f}\;=\; \nabla(\nabla \cdot \mathbf{f})\;-\;\nabla \,\times \, (\nabla \times \mathbf{f})$

You should probably check the partials by hand.
RYAN: Yeah, I-
MATH: Oh, and you're welcome.
RYAN: Hey! I was go-
MATH: Sure you were.
RYAN: This shit is fucking amazing.
MATH: {Yawn}Yeah. {Stretch} I know. So, I see the Indian girl is still at the next table. You know it's not because she wasn't going to drink it, right?
RYAN: Wait. What?
MATH: The beer. You should talk to her. She's taking a long time to put her hair up for you. You should probably ask her out.
RYAN: Look, buddy. I'm not a mind reader. If she wants to talk to me, I'm right here.
MATH: Spare me. And good luck with that.
RYAN: With what?
MATH: ....
....
RYAN: What?
MATH: You are a complete idiot.

____________________

I am a complete idiot. Anyway, Vectors continue to contain more Radness per component of space than any other container. Expressed in terms of BMX,

$\textup{Total Radness} = \oint_C \nabla BMX \;dx$

Where grad(BMX) is the direction of maximum increase in BMX, in mega-wheelie-hours.

I tend to double, triple, sextuple-check identities. That has to stop. The vector algebra identities involving Del take an Awful Lot of Symbols to verify by hand. I did them all. A few results will live here so I can find them. Like

curl (curl f) =

Notes:

1) The proof is just a verification of consistency:

If the operator distributes partials according to

1) the usual product rules for vectors,

2) swapping dot product order to allow Del to operate on something

Then a result is obtained which is algebraically consistent. In other words, it might as well be a definition.

I think it is self-evident that the intermediate, undefined form grad(f) preserves measure if immediately operated on by dot (scalar) product. It is not a proof that grad(f) has any general meaning.

I assume the justification for adoption of the algebra itself lies in physical measurement of space.

2) I have almost found my difficulty with partial derivatives, implicit functions, and multiple integration. The place where all of the differentials come together remains unclear. There is more than one way to integrate multiple variables. It depends on the relationship between all quantities, there are many such quantities, and I cannot convince myself that substitution of a variable is trivial. I think very soon I will have to go back and build diagrams with every little dx, dy, dz, and work out each connection to a partial or substituted variable.

Grrrr.

In the meantime, moving forward is helping, so I will keep doing so. On to Vector Integration.

I haven't forgotten about the schoolgirl problem or Snell's Law. I (finally) have a correct general solution for Kirikman's schoolgirls: a puzzle consisting of simple blocks with which any of the particular solutions can be built.

I intend to finish Applied Vector Analysis in a month. 8 Chapters remain; most of the math I need is here.

DifferentiationSpace curves
Surfaces
Tangent, curvature, torsion

Divergence, Gradient, Curl

Integration
Line, surface and volume integrals
Gauss' (Divergence) Theorem
Green's Theorem
Volume-to-surface; surface-to-line transformations
Irrotational Fields
Orthogonal Curvilinear Coordinates
~~Cripple~~ Special coordinate systems

Fluid Mechanics:
Equations of continuity, motion, and energy
Fluid States
Steady flow and streamlines
Vortex flow and circulation

Electromagnetic Theory:
Electromagnetic field
Maxwell's equations
Potential
Energy and the Poynting vector
Static fields

Rad.
I'll take notes here.

PARAMETRIC SURFACES: NORMAL VECTOR

The normal N to any surface represented by a vector function r(x,y) is given by

$N= \frac {\boldsymbol{r}\!_x \times \boldsymbol{r}\!_y } {\left| \boldsymbol{r}\!_x \times \boldsymbol{r}\!_y \right|}$
where
$\boldsymbol{r}\!_x = \frac {\partial \boldsymbol{r} } {\partial x}, \;\;\;\;\; \boldsymbol{r}\!_y = \frac {\partial \boldsymbol{r} } {\partial y}$

I. Explicit function of x, y
Let a surface S be given by the equations
$x = x, \;\;\; y=y,\;\;\; z = z(x, y)$
where x, y, z are the coordinate axes. Then the position vector of any point on the surface is given by the vector function
$r(x,y)= x \boldsymbol{\hat\imath} + y\boldsymbol{\hat \jmath} + z(x,y)\boldsymbol{\hat k}$
Compute the partial derivatives,
$\begin{align*}\boldsymbol{r}\!_x = \boldsymbol{\hat\imath} \;+ \frac{\partial z}{\partial x}\boldsymbol{\hat k} \\ \boldsymbol{r}\!_y = \boldsymbol{\hat \jmath} \;+ \frac{\partial z}{\partial y}\boldsymbol{\hat k} \end{align*}$

Compute the cross product:
$\boldsymbol{r}\!_x \times \boldsymbol{r}\!_y \;\;=\;\; \begin{vmatrix} \boldsymbol{\hat \imath} &\boldsymbol{\hat \jmath} &\boldsymbol{\hat k} \\ 1 &0 &\dfrac{\partial z}{\partial x} \\ 0 &1 &\dfrac{\partial z}{\partial y} \end{vmatrix} \;\; = \;\; -\frac{\partial z}{\partial x} \boldsymbol{\hat\imath}\; - \frac{\partial z}{\partial y} \boldsymbol{\hat\jmath}\; + \boldsymbol{\hat k}$

And the magnitude, for normalization:
$\begin{align*} |\boldsymbol{r}\!_x \times \boldsymbol{r}\!_y| &\;\;=\;\; \sqrt{ (\boldsymbol {r}\!_x \times \boldsymbol {r}\!_y)\cdot (\boldsymbol {r}\!_x \times \boldsymbol {r}\!_y)} &\;\;=\;\; \sqrt{ \left( \frac{\partial z}{\partial x}\right)^2 + \left( \frac{\partial z}{\partial y}\right)^2 +1 } \end{align*}$

The normal vector to any point on the surface is then
$N \;=\; \frac{ -\dfrac{\partial z}{\partial x} \boldsymbol{\hat\imath}\; - \dfrac{\partial z}{\partial y} \boldsymbol{\hat\jmath}\; + \boldsymbol{\hat k}} {\sqrt{ \left (\dfrac{\partial z}{\partial x}\right)^2 + \left( \dfrac{\partial z}{\partial y} \right)^2 +1}}$

II. Implicit function of x, y.
I'm still not comfortable with partial derivatives of implicit functions. I lose the intuition quickly, and return to the definitions often.
Fix it, Ryan.
Okay. The following result makes me happy.

"

Problem 4-11 Find a unit normal vector n to a surface S that is represented by $\phi (x, y ,z) = 0$

Go one step at a time. This is a surface. Any surface can be represented by a (series of) function(s) of two variables. So, the given function has two independent variables. Phi implicitly defines a function of two variables. By convention, I assume the independent variables are listed first, and are therefore x and y.

Phi defines z implicitly as a function of x and y.

Differentiate with respect to x:
$\frac {\partial \phi}{\partial x} + \frac{\partial \phi}{\partial z} \frac{\partial z}{\partial x} \;\;=\;\; \phi_x +\; \phi_z \frac{\partial z}{\partial x} \;\;=0$

Solve for the partial of z with respect to x:
$(1)\;\;\;\;\;\frac{\partial z}{\partial x} \;=\;-\frac{\phi_x}{\phi_z}$

Differentiate with respect to y, and solve for the partial of z with respect to y:
$\begin{align*} &\frac {\partial \phi}{\partial y}\; + \frac{\partial \phi}{\partial z} \frac{\partial z}{\partial y} \;\;=\;\; \phi_y \, +\, \phi_z \frac{\partial z}{\partial y} \;\;=\;0 \\ &(2)\;\;\;\;\; \frac{\partial z}{\partial y} \;\;=- \frac{\phi_y}{\phi_z} \end{align*}$

Now, we have implicitly defined z as a function having the same form as the explicit equation in section I. Substitute the conspicuously Jacobian values (1) and (2) into the equations from I, and we now have the desired partial derivatives:
$\begin{align*} \boldsymbol{r}\!_x = \boldsymbol{\hat\imath} \; - \frac{\phi_x}{\phi_z}\boldsymbol{\hat k} \\ \boldsymbol{r}\!_y = \boldsymbol{\hat \jmath} \; - \frac{\phi_y}{\phi_z}\boldsymbol{\hat k} \end{align*}$

Again, the cross product:
$\boldsymbol{r}\!_x \times \boldsymbol{r}\!_y \;\;=\;\; \begin{vmatrix} \boldsymbol{\hat \imath} &\boldsymbol{\hat \jmath} &\boldsymbol{\hat k} \\ 1 &0 &-\dfrac{\phi_x}{\phi_z} \\ 0 &1 &-\dfrac{\phi_y}{\phi_z} \end{vmatrix} \;\; = \;\; \frac{\phi_x}{\phi_z} \boldsymbol{\hat\imath}\; + \frac{\phi_y}{\phi_z} \boldsymbol{\hat\jmath}\; + \boldsymbol{\hat k}$
The relationship is clearer, and normalization simplified, if this is rewritten as:
$\frac{1}{\phi_z}\left(\phi_x \boldsymbol{\hat\imath}\; + \phi_y \boldsymbol{\hat\jmath}\; + \phi_z\boldsymbol{\hat k}\right)$

Normalize!
$N = \frac{\boldsymbol {r}_x\times \boldsymbol{r}_y} {|\boldsymbol{r}\!_x \times \boldsymbol{r}\!_y|} &\;\;=\;\; \frac {\phi_x \boldsymbol{\hat\imath}\; + \phi_y \boldsymbol{\hat\jmath}\; + \phi_z\boldsymbol{\hat k}} {\sqrt{(\phi_x)^2 +(\phi_y)^2 + (\phi_z)^2}}$

Goodbye, extra z-partial.

$= \frac{ \dfrac{\partial \phi}{\partial x}\boldsymbol{\hat\imath}\; + \dfrac{\partial \phi}{\partial y}\boldsymbol{\hat\jmath}\; + \dfrac{\partial \phi}{\partial z}\boldsymbol{\hat k}} { \sqrt{ \left(\dfrac{\partial \phi}{\partial x}\right)^2 + \left(\dfrac{\partial \phi}{\partial y}\right)^2 + \left(\dfrac{\partial \phi}{\partial z}\right)^2 }}$

Notes:
1. It is not clear to me why the implicit function is differentiated without reference to the unit vectors i, j and k. "Because it's implicit". Yeah. I get it. But rendering z = z(x,y) as a vector function is independent of the function's status as implicit or explicit. The answer is about measure, and I need to find it.

2. In the implicit function, it doesn't matter what x, y and z are. One of the three partial derivatives of φ is nonzero, or it does not meet the general definition of a surface. This nonzero partial corresponds to the dependent variable. I assume the partial derivative of φ with respect to z is nonzero. If it turns out to be x or y, the result can be computed in the same fashion.
The convention is for simplicity: list the dependent variables last. It saves the following step: suppose I intend to solve problem 4-11 and I require the dependent variable in the third position. Then I will solve the system

$\phi (u, v ,w) = 0$

instead, where w is the dependent variable. Map the variables x, y, z in problem 4-11 to u, v, w accordingly as the surface function is known. I do not have to change a letter of the solution.

Popper's Dreamland

Pages

Wednesday, March 14, 2012

Vectors: Grad, Div and Curl

Thursday, March 8, 2012

Vector surfaces

Labels

Followers

Blog Archive

About Me