# Geodesic of Sphere in Spherical Polar Coordinates (Taylor's Classical Mechanics)

• fehilz
In summary, the shortest path between two points on a curved surface, such as a sphere, is called a geodesic. To find a geodesic, one must set up an integral to determine the length of the path on the surface. This integral will be similar to (6.2) but may involve different coordinates. For a path on a sphere of radius R, using spherical polar coordinates, the length is given by the integral of sqrt(1+sin^2 theta * phi'(theta)^2) with respect to theta, between the two specified points. The distance element on a line of longitude and a line of latitude can be used to derive this formula. Expanding and simplifying the terms results in a total of
fehilz

## Homework Statement

"The shortest path between two point on a curved surface, such as the surface of a sphere is called a geodesic. To find a geodesic, one has to first set up an integral that gives the length of a path on the surface in question. This will always be similar to the integral (6.2) but may be more complicated (depending on the nature of the surface) and may involve different coordinates than x and y. To illustrate this, use spherical polar coordinates (r, $\theta$,$\phi$ ) to show that the length of a path joining two points on a sphere of radius R is

L=R$\int$$\sqrt{1+sin^2\theta\phi'(\theta)^2}$d$\theta$ (Don't know how to do it on latex but the integral is between $\theta$1 and $\theta$2)

if ($\theta$1,$\phi$1) and ($\theta$2,$\phi$2) specify two points and we assume that the path is expressed as $\phi$=$\phi$($\theta$)."

## Homework Equations

x=rsin(ϕ)cos(θ)
y=rsin(ϕ)sin(θ)
z=rcos(ϕ)

ds=$\sqrt{dx^2+dy^2+dz^2}$

## The Attempt at a Solution

I'm unsure of how much this question is asking for. I was able to quickly work out the solution after looking up the line element for the surface of a sphere in spherical polar and using that in place of the Cartesian form of ds.

But then I was wondering if whether the question was asking me to derive the line element. It doesn't seem likely since this is one of the * questions which are supposed to be the easiest and I've already completed the ** questions with no difficulty.

In any case, I worked on deriving the line element for the sake of it and got stuck. I found the differentials for x,y and z (dx, dy and dz) and put them into the equation for ds. What do I do from here? Do I tediously expand the brackets involving three terms or is there something that I'm missing?

Probably you can just do this intuitively instead of diving into the algebra - just write down the formula for a distance element along a line of longitude on the sphere, and then along a line of latitude. Use Pythagoras to put them together and you're home and dry

I think you need to adjust your expectations of what constitutes "tedious." When you expand, you'll get only 7 terms. Two of them will obviously cancel, and you can simplify the rest with one or two lines of algebra.

your distance formula will not hold here.

I would like to commend you for your efforts in attempting to solve this problem. It is important to always try and derive equations and understand the underlying concepts in order to truly grasp a topic.

To answer your question, the homework statement is not asking you to derive the line element. It is simply asking you to use spherical polar coordinates to find the length of a path on the surface of a sphere. Your solution using the line element for a sphere in spherical polar coordinates is correct.

As for your attempt at deriving the line element, you are on the right track. You have correctly found the differentials for x, y, and z and substituted them into the equation for ds. To continue, you can expand the brackets and simplify the expression by using trigonometric identities. This will eventually lead you to the same line element that you found by looking it up.

In conclusion, it is important to understand the underlying concepts and try to derive equations, but in this case, it is not necessary to do so. Your solution using the line element for a sphere in spherical polar coordinates is sufficient. Keep up the good work!

## 1. What is a geodesic of a sphere in spherical polar coordinates?

A geodesic is the shortest path between two points on a curved surface, such as a sphere. In spherical polar coordinates, the geodesic is the path that follows the curve of the sphere and maintains a constant radius and angle from the center of the sphere.

## 2. How is the geodesic of a sphere in spherical polar coordinates derived?

The geodesic of a sphere in spherical polar coordinates can be derived using Taylor's Classical Mechanics. This involves applying the equations of motion and the principle of least action to find the path that minimizes the action, which is the integral of the Lagrangian function.

## 3. What is the significance of the geodesic of a sphere in spherical polar coordinates?

The geodesic of a sphere in spherical polar coordinates is important in understanding the dynamics of objects moving on a spherical surface, such as planets orbiting a star. It also has applications in navigation and geodesy, as it can be used to calculate the shortest path between two points on Earth's surface.

## 4. How does the geodesic of a sphere in spherical polar coordinates differ from a straight line?

While a straight line is the shortest path between two points in Euclidean space, the geodesic of a sphere in spherical polar coordinates takes into account the curvature of the sphere and follows the natural curve of the surface. This means that the geodesic may appear to be curved, even though it is the shortest distance between the two points on the sphere.

## 5. Can the geodesic of a sphere in spherical polar coordinates be generalized to other curved surfaces?

Yes, the concept of a geodesic can be applied to any curved surface, not just a sphere. In fact, the geodesic equation can be used to find the shortest path on any surface, whether it is a sphere, a cylinder, or a more complex shape. However, the specific equations and coordinates used may differ depending on the surface.

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