[6.2] Geodesic บนผิวทรงกระบอก

เกียรติศักดิ์

Administrator
neutrino

Offline

Posts: 296

[6.2] Geodesic บนผิวทรงกระบอก

« on: November 15, 2005, 06:41:06 AM »

Find the length $L$ of a [geodesic] path [from $(\phi_1, z_1)$ to $(\phi_2, z_2)$ ] on a cylinder of radius $R$ , using cylindrical polar coordinates $( \rho, \phi, z )$ . Assume that the path is specified in the form $\phi = \phi (z)$ .


« Last Edit: November 21, 2005, 09:34:38 PM by kiattisak »	Logged

Scientia gaza inaestimabilis est.

เกียรติศักดิ์

Administrator
neutrino

Offline

Posts: 296

solution

« Reply #1 on: November 15, 2005, 09:50:41 AM »

The first part can be skipped; one can just write out the length element straightaway.

Find the length element in cylindrical coordinate system $( \rho, \phi, z )$ .

$\begin{array}{rcl} x &=& \rho \cos \phi \\ y &=& \rho \sin \phi \\ z &=& z \\ \mathbf{r} &=& x \mathbf{\hat{i}} + y \mathbf{\hat{j}} + z \mathbf{\hat{k}} \\ &=& \rho \cos \phi \mathbf{\hat{i}} + \rho \sin \phi \mathbf{\hat{j}} + z \mathbf{\hat{k}} \end{array}$

(Notice that it is given that $\phi$ is a function of $z$ so this is just a parametric equation of a helix with $z$ as a parameter, provided that $\phi (z)$ must be a linear function.)

$\begin{array}{rcl} \mathrm{d} \mathbf{r} &=& \frac{\partial}{\partial \rho} \mathbf{r} \mathrm{d}\rho + \frac{\partial}{\partial \phi} \mathbf{r} \mathrm{d} \phi + \frac{\partial}{\partial z} \mathbf{r} \mathrm{d} z \\ &=& \mathrm{d} \rho (\cos \phi \mathbf{\hat{i}} + \sin \phi \mathbf{\hat{j}}) + \mathrm{d} \phi (-\rho \sin \phi \mathbf{\hat{i}} + \rho \cos \phi \mathbf{\hat{j}}) + \mathrm{d} z(\mathbf{\hat{k}}) \\ &=& \mathrm{d} \rho \sqrt{\cos^2 \phi + sin^2 \phi} \frac{\cos \phi \mathbf{\hat{i}} + \sin \phi \mathbf{\hat{j}}}{\sqrt{\cos^2 \phi + sin^2 \phi}} \\ && + \mathrm{d} \phi \sqrt{\rho^2 \sin^2 \phi + \rho^2 \cos^2 \phi} \frac{-\rho \sin \phi \mathbf{\hat{i}} + \rho \cos \phi \mathbf{\hat{j}}}{\sqrt{\rho^2 \sin^2 \phi + \rho^2 \cos^2 \phi}} \\ && + \mathrm{d} z (\mathbf{\hat{k}}) \\ &=& \mathrm{d} \rho \mathbf{\hat{e}_\rho} + \rho \mathrm{d} \phi \mathbf{\hat{e}_\phi} + \mathrm{d} \mathbf{\hat{e}_z} \end{array}$

$\begin{array}{rcl} \mathrm{d} s^2 &=& \mathrm{d} \mathbf{r} \cdot \mathrm{d} \mathbf{r} \\ &=& \mathrm{d} \rho^2 + \rho^2 \mathrm{d} \phi^2 + \mathrm{d} z^2 \\ \mathrm{d} s &=& \sqrt{\mathrm{d} \rho^2 + \rho^2 \mathrm{d} \phi^2 + \mathrm{d} z^2} \\ \\ \mathrm{d}\rho &=& 0 \qquad \text{ (because we are on the surface) } \\ \rho &=& R \qquad \quad \text{ (...of the cylinder whose radius is } R \text{) } \end{array}$

Given that the path is expressed as $\phi (z)$ , its length is found by integrating over $z$ which is the independent variable.

$\begin{array}{rcl} \mathrm{d} s &=& \sqrt{R^2 \mathrm{d} \phi^2 + \mathrm{d} z^2} = \sqrt{R^2 \left( \frac{\mathrm{d}}{\mathrm{d} z} \phi \right)^2 + 1} \; \mathrm{d} z = \sqrt{R^2 (\phi^\prime)^2 + 1} \; \mathrm{d} z \\ L &=& \int_{z_1}^{z_2} \sqrt{R^2 (\phi^\prime)^2 + 1} \; \mathrm{d} z \end{array}$

We see that for this problem, $f[\phi , \phi^\prime, z]$ is

$\begin{array}{rcl} f = \sqrt{R^2 (\phi^\prime)^2 + 1} \end{array}$

Since this is independent of $z$ , $\frac{\partial}{\partial z} f = 0$ and the second form of the Euler equation (see more below) should be used. (It's easier!)

$\begin{array}{rcl} f - \phi^\prime \frac{\partial}{\partial \phi^\prime} f &=& \text{constant} \equiv \sqrt{\frac{1}{C}} \end{array}$

Solve for $\phi (z)$ .

$\begin{array}{rcl} \sqrt{R^2 (\phi^\prime)^2 + 1} - \phi^\prime \frac{R^2 \phi^\prime}{\sqrt{R^2 (\phi^\prime)^2 + 1}} &=& \sqrt{\frac{1}{C}} \\ \frac{R^2 (\phi^\prime)^2 + 1 - R^2 (\phi^\prime)^2}{\sqrt{R^2 (\phi^\prime)^2 + 1}} &=& \sqrt{\frac{1}{C}} \\ \frac{1}{\sqrt{R^2 (\phi^\prime)^2 + 1}} &=& \sqrt{\frac{1}{C}} \\ \frac{1}{R^2 (\phi^\prime)^2 + 1} &=& \frac{1}{C} \\ R^2 (\phi^\prime)^2 &=& C - 1 \\ (\phi^\prime)^2 &=& \frac{C - 1}{R^2} \\ \phi^\prime &=& \frac{\sqrt{C - 1}}{R} \\ \phi &=& \frac{\sqrt{C - 1}}{R} z + \text{constant} \end{array}$

The path $\phi (z)$ is now easily pictured as a helix. As $z$ increases, $\phi$ linearly increases, and the path traces around a cylinder helically.

Now knowing it is just a straight line if the surface is unwrapt and formed into a plane, when it is given that the end points are $(\phi_1, z_1)$ and $(\phi_2, z_2)$ then the length is, using Pythagorean theorem, $\displaystyle \sqrt{[R (\phi_2 - \phi_1)]^2 + (z_2 - z_1)^2} = \sqrt{C}(z_2 - z_1)$ , and $C$ can be evaluated.

[Unparseable or potentially dangerous latex formula. Error 6 ]

Plug $(\phi^\prime)^2$ into $L$ to obtain the geodesic's length.

$\begin{array}{rcl} L &=& \int_{z_1}^{z_2} \sqrt{R^2 \left( \frac{C - 1}{R^2} \right) + 1} \; \mathrm{d} z \\ &=& \sqrt{C} \int_{z_1}^{z_2} \mathrm{d} z \\ &=& \sqrt{C}(z_2 - z_1) \\ &=& \sqrt{\frac{[R (\phi_2 - \phi_1)]^2}{(z_2 - z_1)^2} + 1}(z_2 - z_1) \\ &=& \sqrt{[R (\phi_2 - \phi_1)]^2 + (z_2 - z_1)^2} \end{array}$


« Last Edit: November 27, 2005, 07:16:07 AM by kiattisak »	Logged

Scientia gaza inaestimabilis est.

เกียรติศักดิ์

Administrator
neutrino

Offline

Posts: 296

The second form of the Euler equation

« Reply #2 on: November 15, 2005, 11:22:12 AM »

Notice that if $\frac{\partial}{\partial x} f = 0$ then

$\begin{array}{rcl} \frac{\mathrm{d}}{\mathrm{d} x} f[y, y^\prime, x] &=& y^\prime \frac{\partial}{\partial y} f + y^{\prime \prime} \frac{\partial}{\partial y^\prime} f + \frac{\partial}{\partial x} f \\ &=& y^\prime \frac{\partial}{\partial y} f + y^{\prime \prime} \frac{\partial}{\partial y^\prime} f \qquad \text{or} \\ y^{\prime \prime} \frac{\partial}{\partial y^\prime} f &=& \frac{\mathrm{d}}{\mathrm{d} x} f - y^\prime \frac{\partial}{\partial y} f \end{array}$

Also

$\begin{array}{rcl} \frac{\mathrm{d}}{\mathrm{d} x} \left( y^\prime \frac{\partial}{\partial y^\prime} f \right) &=& y^{\prime \prime} \frac{\partial}{\partial y^\prime} f + y^\prime \frac{\mathrm{d}}{\mathrm{d} x} \frac{\partial}{\partial y^\prime} f \\ \end{array}$

Plug the first equation into the second one to get

$\begin{array}{rcl} \frac{\mathrm{d}}{\mathrm{d} x} \left( y^\prime \frac{\partial}{\partial y^\prime} f \right) &=& \frac{\mathrm{d}}{\mathrm{d} x} f - y^\prime \frac{\partial}{\partial y} f + y^\prime \frac{\mathrm{d}}{\mathrm{d} x} \frac{\partial}{\partial y^\prime} f \\ &=& \frac{\mathrm{d}}{\mathrm{d} x} f - y^\prime \left( \frac{\partial}{\partial y} f - \frac{\mathrm{d}}{\mathrm{d} x} \frac{\partial}{\partial y^\prime} f \right) \\ &=& \frac{\mathrm{d}}{\mathrm{d} x} f \end{array}$

The term in parentheses vanishes in view of the Euler equation.

That is,

$\begin{array}{rcl} \frac{\mathrm{d}}{\mathrm{d} x} \left( f - y^\prime \frac{\partial}{\partial y^\prime} f \right) &=& 0 \end{array}$

This equation is called the second form of the Euler equation in the special case when $\frac{\partial}{\partial x} f = 0$ . (Otherwise the term appears.)


« Last Edit: November 15, 2005, 11:29:45 AM by kiattisak »	Logged

Scientia gaza inaestimabilis est.

เกียรติศักดิ์

Administrator
neutrino

Offline

Posts: 296

Re: geodesic บนผิวทรงกระบอก (ข้อ 6.2, Taylor's)

« Reply #3 on: November 16, 2005, 05:18:12 PM »

คิดไปคิดมา ก็ยังไม่รู้ว่าจะแก้อย่างไรดีครับ


	Logged

Scientia gaza inaestimabilis est.

ปิยพงษ์ - Head Admin

Administrator
SuperHelper

Offline

Posts: 5614

มีน้ำใจ ไม่อวดตัว มั่วไม่ทำ

Re: geodesic บนผิวทรงกระบอก (ข้อ 6.2, Taylor's)

« Reply #4 on: November 16, 2005, 09:15:32 PM »

Quote from: kiattisak on November 16, 2005, 05:18:12 PM

คิดไปคิดมา ก็ยังไม่รู้ว่าจะแก้อย่างไรดีครับ

ที่ทำมาไม่ผิดหรอก geodesic บนผิวทรงกระบอกเป็นเส้นเกลียว (helix) (พิสูจน์หน่อย!) ดังนั้นถ้า $z_1 = z_2$ จุดสองจุดบนเส้นทางต้องเป็นจุดเดียวกัน ซึ่งทำให้ระยะทางระหว่างสองจุดนั้นเป็นศูนย์


	Logged

มีน้ำใจ ไม่อวดตัว มั่วไม่ทำ

เกียรติศักดิ์

Administrator
neutrino

Offline

Posts: 296

Re: [6.2] Geodesic บนผิวทรงกระบอก

« Reply #5 on: November 22, 2005, 08:00:52 AM »

คราวนี้ดูน่าจะถูกแล้วรึเปล่าครับ ผลที่ได้เป็นไปอย่างที่ผมคาดหวังตั้งแต่ทีแรกละ

จุดสองจุดบนทรงกระบอกที่เลือกคือ $(\phi_1, z_1)$ กับ $(\phi_2, z_2)$ ถ้าบังเอิญว่าเลือกให้ $z_1 = z_2$ ก็เป็นไปได้ว่า ตำแหน่ง polar ที่เลือกอาจไม่เท่ากันก็ได้ ทำให้ความยาวของ geodesic path ที่อธิบายโดย $\phi (z)$ เท่ากับ $R (\phi_2 - \phi_1)$ ได้ครับ