Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 10

Find the geodesics on the parabolic cylinder \(y=x^{2}\).

Short Answer

Expert verified
Geodesics largely follow u,z parameters curved in initial parametric input.

Step by step solution

01

Title - Parametrize the Surface

A good starting point is to parametrize the surface of the parabolic cylinder. Use the surface equation and a parameter. Let the parameters be: \( x = u \) \( y = u^2 \) \( z = v \)
02

Title - Find the Metric Tensor

Next, compute the metric tensor. Start by finding the partial derivatives of the parametric equations with respect to the parameters \(u\) and \(v\):\( \mathbf{r}_u = (1, 2u, 0) \)\( \mathbf{r}_v = (0, 0, 1) \)The metric tensor \(g_{ij}\) is then:\[ g_{ij} = \begin{pmatrix} \mathbf{r}_u \ \mathbf{r}_v \end{pmatrix} \begin{pmatrix} \mathbf{r}_u^T & \mathbf{r}_v^T \end{pmatrix} \]Computing the inner products:\[ g_{11} = \mathbf{r}_u \cdot \mathbf{r}_u = 1 + 4u^2, \ g_{12} = \mathbf{r}_u \cdot \mathbf{r}_v = 0, \ g_{22} = \mathbf{r}_v \cdot \mathbf{r}_v = 1 \]So, the metric tensor \(g_{ij}\) is:\[ g_{ij} = \begin{pmatrix} 1+4u^2 & 0 \ 0 & 1 \end{pmatrix} \]
03

Title - Write Down the Geodesic Equations

With the metric tensor known, write down the geodesic equations using the Christoffel symbols. The geodesic equations are:\[ \frac{d^2 u^k}{d \tau^2} + \Gamma_{ij}^k \frac{d u^i}{d \tau} \frac{d u^j}{d \tau} = 0 \]where the Christoffel symbols \( \Gamma_{ij}^k \) are calculated from the metric tensor. For the metric tensor \(g_{ij}\), Christoffel symbols can be computed, but in this specific case, it's easier to see the inherent symmetry and deduce these values directly.
04

Title - Solve the Geodesic Equations

Given the simplicity of the parameterization and metric, simpler geodesic paths occur when the complexity of curvature is bounded by parameters. This means lines parallel to the z-axis and slight curvatures due to \(u\). General solutions are combinations of connections on lines projected on z from the initial lines along the parabolic x,y relationship. Numerical methods might be required for more detail.
05

Title - Interpret the Physical Meaning

Translate the mathematical solutions into a physical and visual interpretation. Normally, the shortest path on the cylinder will follow the curves of the parabolic surface but projected, such paths simplify for lines along 'z', thus direct translation happens more simplified.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Parametric Surfaces
To understand geodesics, we start with the concept of parametric surfaces. This is a way to describe a surface using parameters and equations. For the parabolic cylinder, the surface is described by the equation \(y = x^2\). We use parameters \(u\) and \(v\) to express this surface as:
  • \( x = u \)
  • \( y = u^2 \)
  • \( z = v \)
This means any point on the surface can be described using these two variables. This method helps us simplify and break down complex surfaces into manageable pieces.
Metric Tensor
The metric tensor is a key concept in differential geometry as it helps measure distances on surfaces. Once we have our parametric equations, we find partial derivatives with respect to parameters (\(u\) and \(v\)). For the parabolic cylinder:
  • \(\mathbf{r}_u = (1, 2u, 0)\)
  • \(\mathbf{r}_v = (0, 0, 1)\)
The metric tensor \(g_{ij}\) is built using these partial derivatives. We compute inner products:
  • \( g_{11} = \mathbf{r}_u \cdot \mathbf{r}_u = 1 + 4u^2 \)
  • \( g_{12} = \mathbf{r}_u \cdot \mathbf{r}_v = 0 \)
  • \( g_{22} = \mathbf{r}_v \cdot \mathbf{r}_v = 1 \)

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write and solve the Euler equations to make stationary the integrals. $$\int_{a}^{b} \sqrt{\frac{y^{\prime 2}}{y^{2}}+1} d x$$

(a) Consider the case of two dependent variables. Show that if \(F=F\left(x, y, z, y^{\prime}, z^{\prime}\right)\) and we want to find \(y(x)\) and \(z(x)\) to make \(I=\int_{x_{1}}^{x_{2}} F d x\) stationary, then \(y\) and \(z\) should each satisfy an Euler equation as in (5.1). Hint: Construct a formula for a varied path \(Y\) for \(y\) as in Section \(2[Y=y+\epsilon \eta(x) \text { with } \eta(x) \text { arbitrary }]\) and construct a similar formula for \(z\) llet \(Z=z+\epsilon \zeta(x),\) where \(\zeta(x)\) is another arbitrary function]. Carry through the details of differentiating with respect to \(\epsilon,\) putting \(\epsilon=0,\) and integrating by parts as in Section \(2 ;\) then use the fact that both \(\eta(x)\) and \(\zeta(x)\) are arbitrary to get (5.1). (b) Consider the case of two independent variables. You want to find the function \(u(x, y)\) which makes stationary the double integral $$\int_{y_{1}}^{y_{2}} \int_{x_{1}}^{x_{2}} F\left(u, x, y, u_{x}, u_{y}\right) d x d y$$. Hint: Let the varied \(U(x, y)=u(x, y)+\epsilon \eta(x, y)\) where \(\eta(x, y)=0\) at \(x=x_{1}\) \(x=x_{2}, y=y_{1}, y=y_{2},\) but is otherwise arbitrary. As in Section \(2,\) differentiate \(x=y=y=y\), with respect to \(\epsilon,\) set \(\epsilon=0,\) integrate by parts, and use the fact that \(\eta\) is arbitrary. Show that the Euler equation is then $$\frac{\partial}{\partial x} \frac{\partial F}{\partial u_{x}}+\frac{\partial}{\partial y} \frac{\partial F}{\partial u_{y}}-\frac{\partial F}{\partial u}=0$$ (c) Consider the case in which \(F\) depends on \(x, y, y^{\prime},\) and \(y^{\prime \prime} .\) Assuming zero values of the variation \(\eta(x)\) and its derivative at the endpoints \(x_{1}\) and \(x_{2},\) show that then the Euler equation becomes $$\frac{d^{2}}{d x^{2}} \frac{\partial F}{\partial y^{\prime \prime}}-\frac{d}{d x} \frac{\partial F}{\partial y^{\prime}}+\frac{\partial F}{\partial y}=0$$

A yo-yo (as shown) falls under gravity. Assume that it falls straight down, unwinding as it goes. Find the Lagrange equation of motion. Hints: The kinetic energy is the sum of the translational energy \(\frac{1}{2} m \dot{z}^{2}\) and the rotational energy \(\frac{1}{2} I \dot{\theta}^{2}\) where \(I\) is the moment of inertia. What is the relation between \(\dot{z}\) and \(\dot{\theta}\) ? Assume the yo-yo is a solid cylinder with inner radius \(a\) and outer radius \(b\).

Find a first integral of the Euler equation to make stationary the integrals. $$\int_{a}^{b} \frac{y y^{\prime 2} d x}{\sqrt{1+y^{\prime 2}}}$$

A simple pendulum (Problem 4) is suspended from a mass \(M\) which is free to move without friction along the \(x\) axis. The pendulum swings in the \(x z\) plane and gravity acts in the negative \(z\) direction. Find the Lagrangian and Lagrange's equations for the system.
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free