Prove that a great circle is a geodesic

In summary, the Euler-Lagrange equation reduces to \partial f / \partial \phi' = c, where f(\phi, \phi', \theta) is independent of phi. By choosing the z-axis to pass through one of the given points, the constant c is necessarily zero and the geodesic is described by the equations \sin{(\theta)} = \frac{C_{1}}{\sin{p}} and \phi' = \frac{\sin^{2}{p}}{C_{1} \, \cos{p}}. This identifies the equation of a great circle in spherical coordinates.
  • #1
miew
27
0

Homework Statement



L = R [tex]\int \sqrt{1+ sin^2 \theta \phi ' ^ 2} d\theta
[/tex] from theta 1 to theta 2

Using this result, prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f([tex]\phi,\phi',\theta[/tex]) in the result is independent of phi so the Euler-Lagrange equation reduces to [tex]\partial f [/tex]/[tex]\partial \phi ' [/tex] = c, a constant. This gives you [tex]\phi ' [/tex] 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





The Attempt at a Solution


Doing some calculations, I ended up getting

[tex]\phi ' ^ 2[/tex] - A sin^2 [tex]\theta[/tex] [tex]\phi ' ^ 2 [/tex]= A^2. The hint tells me that I can take theta as 0, and so I get that phi_prime = A.

I integrate that, but I don't get anything that looks like a circle.
 
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  • #2
your integrand function does not depend on [itex]\phi[/itex] explicitly. That is why you can immediately perform one integration of the Euler's equation:

[tex]
\frac{\partial F}{\partial \phi'} = C_{1}
[/tex]

Finding this partial derivative gives:

[tex]
\frac{R \, \sin^{2}{(\theta)} \, \phi'}{\sqrt{1 + \sin^{2}{(\theta)} (\phi')^{2}}} = C_{1}
[/tex]

You can solve this equation by a parametric trick:

[tex]
\sin{(\theta)} \, \phi' = \tan{p}
[/tex]

which reduces the equation to:

[tex]
\sin^{2}{(\theta)} \, \phi' \, \cos{p} = C_{1}
[/tex]

where I redefined the arbitrary integrating constant [itex]C_{1} \rightarrow R \, C_{1}[/itex].

These equations give:

[tex]
\sin{(\theta)} \, \cos{p} = C_{1} \, \cot{p} \Rightarrow \sin{(\theta)} = \frac{C_{1}}{\sin{p}}
[/tex]

and

[tex]
\phi' \eqiuv \frac{d \phi}{d \theta} = \frac{\sin^{2}{p}}{C_{1} \, \cos{p}}
[/tex]

You need to find [itex]\phi(p)[/itex] and identify the equation of a great circle in spherical coordinates.
 

1. What is a great circle?

A great circle is a circle on the surface of a sphere that has the same center as the sphere and divides it into two equal halves.

2. How is a great circle different from other circles?

A great circle has the largest possible circumference and is the most symmetrical circle on a sphere. It also has the property that any two points on the circle are at the same distance from the center of the sphere.

3. What is a geodesic?

A geodesic is the shortest path between two points on a curved surface, such as a sphere. It is the equivalent of a straight line on a flat surface.

4. How do you prove that a great circle is a geodesic?

To prove that a great circle is a geodesic, we can use the concept of spherical triangles. By drawing a triangle on the surface of a sphere with one side being a portion of a great circle, we can show that the shortest distance between any two points on a sphere is along the great circle that connects them.

5. Why is it important to prove that a great circle is a geodesic?

Understanding the properties of great circles and geodesics is crucial in many areas of science and engineering. For example, in navigation and cartography, great circles are used to determine the shortest distance between two points on the Earth's surface. In physics and mathematics, geodesics play a significant role in understanding the curvature of space and time in Einstein's theory of general relativity.

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