Question

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 \(\lambda \rightarrow 0\).

Short answer

The geodesic is \(\phi(\rho) = -\frac{c}{(1+\lambda^2)\rho} + \phi_0\), simplifying to a line as \(\lambda \to 0\).

Step-by-step solution

Understand the Problem

We need to find the geodesics (shortest paths) on a cone described by the equation in cylindrical coordinates, \(z = \lambda \rho\). The geodesics will be expressed in terms of \(\phi(\rho)\).

Define the Surface and Metric

The cone is given in cylindrical coordinates as \(z = \lambda \rho\). The infinitesimal arc length in cylindrical coordinates is \(ds^2 = d\rho^2 + \rho^2 d\phi^2 + dz^2\). Substitute \(dz = \lambda d\rho\), giving \(ds^2 = (1+\lambda^2)d\rho^2 + \rho^2 d\phi^2\).

Use calculus of variations to find the geodesic

We apply the calculus of variations on the functional \(I[\phi] = \int \sqrt{(1+\lambda^2) + \rho^2 \left(\frac{d\phi}{d\rho}\right)^2} \, d\rho\). The solution is found by minimizing \(I[\phi]\) and results in a first-order differential equation for \(\phi(\rho)\).

Solve the Euler-Lagrange Equation

The Euler-Lagrange equation applied to this functional gives us that \((1+\lambda^2)\rho^2 \frac{d\phi}{d\rho} = c,\) where \(c\) is a constant determined by initial conditions. Rearrange this to solve for \(\phi\): \[ \frac{d\phi}{d\rho} = \frac{c}{(1+\lambda^2)\rho^2} \].

Integrate to find \(\phi(\rho)\)

Integrate \(\frac{d\phi}{d\rho} = \frac{c}{(1+\lambda^2)\rho^2}\) to get \(\phi(\rho) = -\frac{c}{(1+\lambda^2)\rho} + \phi_0\), where \(\phi_0\) is the integration constant.

Verify for \(\lambda \rightarrow 0\)

For \(\lambda \rightarrow 0\), the surface becomes a flat plane (\(z=0\)). The equation simplifies to \(\phi(\rho) = -\frac{c}{\rho} + \phi_0\), resembling the straight line geodesic in polar coordinates on a flat plane.

Key concepts

Geodesics

In differential geometry, geodesics are the paths that provide the shortest distance between two points on a curved surface. These paths are equivalent to straight lines in both their properties and function, but instead of a flat surface, they exist on geometrically complex forms like spheres, cones, or other curved surfaces.

When applying the concept to a cone, such as the one described in the exercise by the equation \(z = \lambda \rho\), the geodesics represent the paths that deliver minimal distance when mapped onto the conical surface.

Geodesics on the cone can be visualized as spiral-like paths wrapping around the surface, arranging themselves optimally in terms of distance and angle. Understanding this has applications in fields where understanding of shortest paths is essential, such as navigation, physics, and computer graphics.

Calculus of Variations

The calculus of variations is a mathematical technique used to find functions that optimize given quantities described by integrals. It plays a pivotal role in physics and geometry when a specific path, shape, or function needs to be determined for optimization purposes.

In this exercise, the calculus of variations is employed on the cone to discover the function \(\phi(\rho)\) that minimizes the arclength. This method involves setting up a functional, which is an integral that depends on the function you want to optimize.

• The integral formed in this case, \(I[\phi] = \int \sqrt{(1+\lambda^2) + \rho^2 \left(\frac{d\phi}{d\rho}\right)^2} \, d\rho\), is the expression we wish to minimize.
• The path that minimizes this expression will be the geodesic.

Thus, through solving this minimization problem, one can directly find the path taken by a geodesic across the surface.

Euler-Lagrange Equation

The Euler-Lagrange Equation is a fundamental equation in the calculus of variations, offering a technique for finding the function that minimizes or maximizes a functional. It's a cornerstone in optimizing a path or surface and is crucial in various fields such as physics, engineering, and economics.

When we deal with the geodesics of a surface, such as the cone in this exercise, applying the Euler-Lagrange equation helps derive the specific equation that needs to be satisfied by the shortest path. From the given exercise, by applying the Euler-Lagrange equation to the functional \(I[\phi]\), one derives:

• The equation \((1+\lambda^2)\rho^2 \frac{d\phi}{d\rho} = c\), where \(c\) is a constant.

To solve this, you rearrange it and integrate, leading to the function for \(\phi(\rho)\) which describes the geodesic.

This shows the power of the Euler-Lagrange equation in tackling complex optimization problems central to differential geometry.