Chapter 9: Problem 17

Find the geodesics on the cone \(x^{2}+y^{2}=z^{2}\),

Short Answer

Expert verified

The geodesics on the cone are given by \( r = \sqrt{kt + b} \) and \( \theta = \frac{c}{2k} \, \text{ln} (r^{2}) + C \).

Step by step solution

Express the cone equation in cylindrical coordinates

In cylindrical coordinates, we have the following relations: \(x = r \, \text{cos} \, \theta\), \(y = r \, \text{sin} \, \theta\), and \(z = r\). The cone equation \(x^{2} + y^{2} = z^{2}\) transforms into \(r^{2} = r^{2}\), which is automatically satisfied.

Write the metric in cylindrical coordinates

In cylindrical coordinates, the line element (metric) on the cone is given by \( ds^{2} = dr^{2} + r^{2} d\theta^{2}\).

Use the Euler-Lagrange equations

The geodesics can be found by minimizing the integral of the line element. Use the Lagrangian \( L = \frac{1}{2} \, (\dot{r}^{2} + r^{2} \dot{\theta}^{2}) \) where \( \dot{r} \) and \( \dot{\theta} \) are derivatives with respect to some parameter \( t \). The Euler-Lagrange equations for \(r\) and \(\theta\) are: \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0 \] \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \]

Solve the Euler-Lagrange equations

First, solve the equation for \(\theta\): Since \( \frac{\partial L}{\partial \theta} = 0 \), we find that \( \frac{d}{dt} (r^{2} \dot{\theta}) = 0 \), implying that \( r^{2} \dot{\theta} \) is a constant, say \(c\). Thus, \( \dot{\theta} = \frac{c}{r^{2}} \). Now substitute \( \dot{\theta} \) into the equation for \(r\): \[ \ddot{r} - r \left( \frac{c}{r^{2}} \right)^{2} = 0 \] \[ \ddot{r} = \frac{c^{2}}{r^{3}} \]

Solve for \( r(t) \) and \( \theta(t) \)

To solve \( \ddot{r} = \frac{c^{2}}{r^{3}} \), use multiplication by \( \dot{r} \): \[ \dot{r} \ddot{r} = \frac{c^{2} \dot{r}}{r^{3}} \] This can be integrated to give: \[ \dot{r}^{2} = \frac{c^{2}}{r^{2}} + C \] Solve for \(r\) and \(\theta\) by integrating once more: \[ r = \sqrt{kt + b} \] \[ \theta = \frac{c}{2k} \, \text{ln} (r^{2}) + C \]

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Key Concepts

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

cylindrical coordinates

In the realm of geometry and physics, cylindrical coordinates offer a unique way to describe a point in a 3D space. Instead of using Cartesian coordinates \(x, y, z\), cylindrical coordinates use \(r, \theta, z\). This system is especially handy when dealing with problems exhibiting cylindrical symmetry, such as a cone.

Euler-Lagrange equations

The Euler-Lagrange equations are fundamental in the calculus of variations. They help us find functions that extremize a given functional. In simpler terms, they help determine the path, curve, surface, etc., for which a particular quantity is optimal (minimum or maximum). The general form of the Euler-Lagrange equation is:
\[ \frac{\text{d}}{\text{dt}}\bigg(\frac{\text{d}L}{\text{d}\text{{\text{dot}}{\text{q}}}}\bigg) - \frac{\text{d}L}{\text{d}q} = 0 \]
Here, \(L\) represents the Lagrangian, often expressed as the difference between kinetic and potential energy, and \(q\) represents generalized coordinates.

metric in differential geometry

The metric is a fundamental concept in differential geometry, providing a way to measure distances and angles. In cylindrical coordinates, the metric for a cone can be derived from the line element \(ds^2\). For the cone \(x^2 + y^2 = z^2\), in cylindrical coordinates we have:
\[ ds^2 = dr^2 + r^2 d\theta^2 \]
This expression helps us compute the distance between two points on the cone's surface and forms the basis for finding geodesics through the Euler-Lagrange equations.