Geodesic Sphere Homework: Prove Great Circle Path is Shortest

In summary, a geodesic sphere is a spherical shape made up of interconnected straight lines that represent the shortest path between two points on its surface. It is important to prove that the great circle path is the shortest as it has practical applications in navigation, aviation, and cartography. This proof involves the use of calculus and the Pythagorean theorem. However, there are exceptions to this rule, such as when two points are antipodal. Geodesic spheres are used in various real-life applications, including navigation systems, flight planning, and structural design.
  • #1
roeb
107
1

Homework Statement


[tex]L = R \int_{\theta_1}^{\theta_2} \sqrt{1 + sin^2(\theta ) \phi ' ^ 2} d\theta [/tex]

Use the result to prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f(phi,phi_prime,theta) in the result is independent of phi so the Euler-Lagrange equation reduces to partial_f/partial_phi_prime = c, a constant. This gives you phi_prime as a function of theta. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero and describe the corresponding geodesics.

Homework Equations



[tex]\partial f / \partial x = d/du \partial f / \partial x'[/tex]

The Attempt at a Solution



I am having a bit of difficulty interpreting this problem. Using the Euler-Lagrange equation I get the following:
[tex]\phi ' ^ 2 = C^2 / ( sin^4 (\theta ) - C^2 sin^2(\theta ) [/tex]
The hint kind of confusing me. How can 'c' be zero? If I were to have the z-axis at point 1, I would think that [tex]\theta_1[/tex] = 0 if anything.

Can anyone help guide me on the right path for this problem?
 
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  • #2
roeb said:
I am having a bit of difficulty interpreting this problem. Using the Euler-Lagrange equation I get the following:
[tex]\phi ' ^ 2 = C^2 / ( sin^4 (\theta ) - C^2 sin^2(\theta ) [/tex]

It might be more useful to write this as:

[tex]\left( sin^4 (\theta ) - C^2 sin^2(\theta ) \right) \phi ' ^ 2 = C^2[/tex]

I think the point of the hint is that the above relation holds for all allowable values of [itex]\theta[/itex] and that [itex]\theta=0[/itex] is one such allowable value. What does [itex]C^2[/itex] equal when [itex]\theta=0[/itex] assuming that [itex]\phi'[/itex] is bounded (i.e. not infinite)? Since [itex]C[/itex] is a constant, it must have this value for all [itex]\theta[/itex].
 
  • #3
Thanks for your reply.

Now I can see why C = 0. If I am doing my math correctly that means [tex]\phi (\theta ) = A[/tex] (another constant).

If [tex]\phi (\theta )[/tex] is just some constant, does that mean this constant A will be a segment of the great circle between two locations on the sphere?
 
  • #4
As the choice of axis is arbitrary, you can choose one such that [tex]\theta[/tex] is zero. Then varying [tex]\phi[/tex] (which as you point out should have no [tex]\theta[/tex] dependence) is just tracing out a great circle.
 
  • #5


Sure, let's break down the problem step by step. First, let's define some variables:

- R: radius of the sphere
- (theta_1, phi_1): coordinates of the starting point
- (theta_2, phi_2): coordinates of the ending point
- L: length of the geodesic (shortest path)

Now, we can rewrite the given equation as:

L = R \int_{\theta_1}^{\theta_2} \sqrt{1 + \sin^2(\theta ) \phi ' ^ 2} d\theta = R \int_{\theta_1}^{\theta_2} \sqrt{1 + \frac{C^2}{\sin^4(\theta) - C^2\sin^2(\theta)}} d\theta

where C is a constant.

Next, let's use the Euler-Lagrange equation to find the function phi' as a function of theta. Remember that the Euler-Lagrange equation is given by:

\frac{\partial f}{\partial \phi} - \frac{d}{du}\frac{\partial f}{\partial \phi'} = 0

In our case, f = \sqrt{1 + \frac{C^2}{\sin^4(\theta) - C^2\sin^2(\theta)}} and \frac{\partial f}{\partial \phi} = 0 (since the integrand is independent of phi). Therefore, we have:

\frac{d}{du} \frac{\partial f}{\partial \phi'} = \frac{d}{du} \frac{C^2\sin^2(\theta)}{(sin^4(\theta) - C^2\sin^2(\theta))^{3/2}} = 0

Solving for phi', we get:

\phi' = \frac{C}{\sqrt{\sin^4(\theta) - C^2\sin^2(\theta)}}

Now, let's use the hint provided in the problem. It says that there is no loss of generality in choosing the z-axis to pass through the point 1. This means that we can choose our coordinate system such that the starting point (theta_1, phi_1) lies on the z-axis. In other words, we can set phi_1 = 0.

Using this
 

Related to Geodesic Sphere Homework: Prove Great Circle Path is Shortest

1. What is a geodesic sphere?

A geodesic sphere is a sphere made up of interconnected straight lines, also known as geodesics, that follow the shortest path between two points on the surface of the sphere. It is a mathematical concept used to understand the shortest distance between two points on a curved surface.

2. Why is it important to prove that the great circle path is the shortest?

Knowing that the great circle path is the shortest distance between two points on a geodesic sphere is important in many fields, such as navigation, aviation, and cartography. It allows for more accurate calculations and efficient travel routes.

3. How do you prove that the great circle path is the shortest?

To prove that the great circle path is the shortest, we use the concept of calculus and the Pythagorean theorem. By finding the derivative of the distance formula on a geodesic sphere, we can show that the derivative is equal to zero at the point where the great circle path intersects the sphere, indicating that it is the shortest path.

4. Are there any exceptions to the great circle path being the shortest?

Yes, there are exceptions to the great circle path being the shortest. One exception is when two points on the sphere are antipodal, meaning they are on opposite sides of the sphere. In this case, all paths between the two points would have the same length, and the great circle would not be the shortest.

5. How is the concept of geodesic spheres used in real life?

The concept of geodesic spheres is used in various applications, including navigation systems, flight planning, and map projections. It is also used in the design of structures, such as domes and geodesic domed stadiums, as it allows for the most efficient use of materials in creating a spherical shape.

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