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Consider a right circular cylinder of radius \(R\) centered on the \(z\) axis. Find the equation giving \(\phi\) as a function of \(z\) for the geodesic (shortest path) on the cylinder between two points with cylindrical polar coordinates \(\left(R, \phi_{1}, z_{1}\right)\) and \(\left(R, \phi_{2}, z_{2}\right) .\) Describe the geodesic. Is it unique? By imagining the surface of the cylinder unwrapped and laid out flat, explain why the geodesic has the form it does.

Short Answer

Expert verified
The geodesic is \(\phi(z) = \phi_1 + \frac{z-z_1}{z_2 - z_1}(\phi_2 - \phi_1)\) and is unique as a straight line in the unwrapped plane.

Step by step solution

01

Unwrap the Cylinder

Imagine the cylinder being unwrapped into a flat rectangle. The height of the cylinder is the length of the rectangle, and its circumference, which is \(2\pi R\), becomes the width of the rectangle. The points \( (R, \phi_1, z_1) \) and \( (R, \phi_2, z_2) \) now lie on this rectangle.
02

Map Polar Coordinates onto the Rectangle

Convert the cylindrical coordinates \((R, \phi, z)\) to the unwrapped rectangular coordinates \((x, z)\), where \(x = R\phi\). The points become \((R\phi_1, z_1)\) and \((R\phi_2, z_2)\) on the rectangle.
03

Determine the Line Equation

In the rectangle, the shortest path or geodesic between two points is a straight line. The equation of the line can be determined using the point-slope form. The slope \(m\) is \(\frac{z_2 - z_1}{R\phi_2 - R\phi_1}\). Thus, the line equation is \(z = mz + c\), where \(c\) can be determined using the point \( (R\phi_1, z_1) \).
04

Convert Back to Cylindrical Coordinates

Convert the line equation back to cylindrical coordinates. Since \(x = R\phi\), the equation becomes \(z = m(R\phi) + c\), where \(m = \frac{z_2 - z_1}{R(\phi_2 - \phi_1)}\). Rearrange to find \(\phi\) in terms of \(z\): \[ \phi = \frac{z - c}{R \cdot m} \] Substitute \(c\) and \(m\) to get \[ \phi(z) = \phi_1 + \frac{z - z_1}{z_2 - z_1}(\phi_2 - \phi_1) \].
05

Describe and Check the Uniqueness of the Geodesic

The equation \(\phi(z) = \phi_1 + \frac{z-z_1}{z_2 - z_1}(\phi_2 - \phi_1)\) is a linear function describing the geodesic. It confirms that the shortest path on the unwrapped cylinder is a straight line, which corresponds to a helical path when rewrapped onto the cylinder. This geodesic is unique assuming \(\phi_1\) and \(\phi_2\) differ by less than \(\pi\), preventing overlap on the cylinder's multiple revolutions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Coordinates
When dealing with problems involving cylinders, cylindrical coordinates are extremely helpful. Cylindrical coordinates are a three-dimensional coordinate system, where a point in space is defined by its distance from a central axis (radius or \(R\)), its angular position around that axis (angle denoted by \(\phi\)), and its height along the axis (\(z\)).
  • Radius \(R\): This is the distance from the point to the axis of the cylinder, and it remains constant on the surface of the cylinder.
  • Angle \(\phi\): This is measured from a chosen reference direction on the plane perpendicular to the axis. It is important in describing positions around the cylinder and is measured in radians.
  • Height \(z\): This coordinate measures the position along the cylinder's axis, which in many cases coincides with the vertical direction.
These coordinates are particularly suited for circular and cylindrical shapes because they align naturally with the geometry, making calculations simpler. In the context of geodesic problems on cylinders, they simplify the description of paths and aid in the transformation to other coordinate systems for easier visualization.
Unwrapped Cylinders
A great way to imagine paths on a cylinder is to "unwrap" it. By unwrapping a cylinder, you can visualize its surface as a flat rectangle. Here's how this works:
  • Width of the rectangle is the same as the circumference of the cylinder, which is \(2\pi R\).
  • Height of the rectangle is equal to the height of the cylinder, maintaining the original height dimension \(z\).
When looking at the unwrapped rectangle, the problem transforms from finding a path on a cylindrical surface to finding a path on a flat surface.
This perspective allows you to use straight-line geometry to determine the geodesic. The points on the cylinder described by \((R, \phi_1, z_1)\) and \((R, \phi_2, z_2)\) can now be plotted on this flat rectangle as \((R\phi_1, z_1)\) and \((R\phi_2, z_2)\). This approach simplifies the later steps because lines on flat surfaces are simple to handle compared to curves on curved surfaces.
Shortest Path
The concept of a shortest path, especially on a cylinder, ties back into the idea of geodesics. A geodesic is the shortest curve between two points on a given surface.
When you transform the cylindrical surface into a flat rectangle, the geodesic becomes a straight line across this rectangle. You determine this line by:
  • Using the slope formula: \(m = \frac{z_2 - z_1}{R\phi_2 - R\phi_1}\), indicating how much \(z\) changes with respect to \(x = R\phi\).
  • Writing the equation: The line's equation in the rectangle is \(z = m(R\phi) + c\).
This line, when translated back into cylindrical terms, represents a helical path around the cylinder.
The equation for this helical curve in terms of the angle \(\phi\) against height \(z\) is the linear function: \[\phi(z) = \phi_1 + \frac{z - z_1}{z_2 - z_1}(\phi_2 - \phi_1)\]This formula ensures that as \(z\) changes from \(z_1\) to \(z_2\), \(\phi\) transitions smoothly from \(\phi_1\) to \(\phi_2\), drawing the shortest path on the cylindrical surface. This geodesic path is unique as long as the angular distance between \(\phi_1\) and \(\phi_2\) does not exceed \(\pi\) radians, which prevents overlapping paths on the cylindrical surface.

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Most popular questions from this chapter

Consider a medium in which the refractive index \(n\) is inversely proportional to \(r^{2}\); that is, \(n=a / r^{2},\) where \(r\) is the distance from the origin. Use Fermat's principle, that the integral (6.3) is stationary, to find the path of a ray of light travelling in a plane containing the origin. [Hint: Use twodimensional polar coordinates and write the path as \(\phi=\phi(r) .\) The Fermat integral should have the form \(\int f\left(\phi, \phi^{\prime}, r\right) d r,\) where \(f\left(\phi, \phi^{\prime}, r\right)\) is actually independent of \(\phi .\) The Euler-Lagrange equation therefore reduces to \(\partial f / \partial \phi^{\prime}=\) const. You can solve this for \(\phi^{\prime}\) and then integrate to give \(\phi\) as a function of \(r .\) Rewrite this to give \(r\) as a function of \(\phi\) and show that the resulting path is a circle through the origin. Discuss the progress of the light around the circle.]

Find the equation of the path joining the origin \(O\) to the point \(P(1,1)\) in the \(x y\) plane that makes the integral \(\int_{o}^{P}\left(y^{2}+y y^{\prime}+y^{2}\right) d x\) stationary.

A ray of light travels from point \(P_{1}\) in a medium of refractive index \(n_{1}\) to \(P_{2}\) in a medium of index \(n_{2},\) by way of the point \(Q\) on the plane interface between the two media, as in Figure \(6.9 .\) Show that Fermat's principle implies that, on the actual path followed, \(Q\) lies in the same vertical plane as \(P_{1}\) and \(P_{2}\) and obeys Snell's law, that \(n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2} .\) [Hints: Let the interface be the \(x z\) plane, and let \(P_{1}\) lie on the \(y\) axis at \(\left(0, h_{1}, 0\right)\) and \(P_{2}\) in the \(x, y\) plane at \(\left(x_{2},-h_{2}, 0\right) .\) Finally let \(Q=(x, 0, z)\) Calculate the time for the light to traverse the path \(P_{1} Q P_{2}\) and show that it is minimum when \(Q\) has \(z=0\) and satisfies Snell's law.]

Find the geodesics on the cone whose equation in cylindrical polar coordinates is \(z=\lambda \rho .\) [Let the required curve have the form \(\phi=\phi(\rho) .\) ] Check your result for the case that \(\lambda \rightarrow 0\).

An aircraft whose airspeed is \(v_{\mathrm{o}}\) has to fly from town \(O\) (at the origin) to town \(P\), which is a distance \(D\) due east. There is a steady gentle wind shear, such that \(\mathbf{v}_{\text {wind }}=V y \hat{\mathbf{x}},\) where \(x\) and \(y\) are measured east and north respectively. Find the path, \(y=y(x),\) which the plane should follow to minimize its flight time, as follows: (a) Find the plane's ground speed in terms of \(v_{o}, V, \phi\) (the angle by which the plane heads to the north of east), and the plane's position. (b) Write down the time of flight as an integral of the form \(\int_{0}^{D} f d x .\) Show that if we assume that \(y^{\prime}\) and \(\phi\) both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand \(f\) takes the approximate form \(f=\left(1+\frac{1}{2} y^{2}\right) /(1+k y)\) (times an uninteresting constant) where \(k=V / v_{\mathrm{o}}\). (c) Write down the Euler-Lagrange equation that determines the best path. To solve it, make the intelligent guess that \(y(x)=\lambda x(D-x),\) which clearly passes through the two towns. Show that it satisfies the EulerLagrange equation, provided \(\lambda=(\sqrt{4+2 k^{2} D^{2}}-2) /\left(k D^{2}\right) .\) How far north does this path take the plane, if \(D=2000\) miles, \(v_{\mathrm{o}}=500 \mathrm{mph},\) and the wind shear is \(V=0.5 \mathrm{mph} / \mathrm{mi} ?\) How much time does the plane save by following this path? [You'll probably want to use a computer to do this integral.]

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