# Lagrange multipliers for finding geodesics on a sphere

• Leb
In summary: Finally, we can use the value of \lambda to determine the equation for the great circle on the sphere, which is xA+By+z=0 (A,B = const.). This is the equation of the geodesic, or shortest path, between two points on the sphere.In summary, to find the geodesics on a sphere, we use the Lagrange multiplier method and differentiate the constraint twice w.r.t. s to find the stationary points of the arclength element. This helps us find the equation for the great circle, which is the geodesic on the sphere.
Leb

## Homework Statement

Find the geodesics on a sphere $g(x,y,z)=x^{2}+y^{2}+z^{2}-1=0$
arclength element $ds=\sqrt{dx^{2}+dy^{2}+dz^{2}}$

## Homework Equations

$f(x,y,z)=\sqrt{x'^{2}+y'^{2}+z'^{2}}$ where $x'^{2} \text{means} \frac{dx^{2}}{ds^{2}}$ and not $d^{2}x/ds^{2}$

## The Attempt at a Solution

Using the fact that $x^{2}+y^{2}+z^{2}=1$ I get three equations if of the form
$d^{2}x/ds^{2}=2\lambda x$ i.e. the double derivative of x w.r.t. s
$d^{2}y/ds^{2}=2\lambda y$
$d^{2}z/ds^{2}=2\lambda z$

My lecturer now says, that we have to differentiate the constraint (g) twice w.r.t. s to get
$(dx/ds)^{2}+(dy/ds)^{2}+(dz/ds)^{2} = -2 \lambda (x^{2}+y^{2}+z^{2})$
since the LHS = 1 and the brackets in the RHS = 1 , the lecturer concludes, that $\lambda = -0.5$. Now I am most confused here, as I do not see where the double differentiation happened. Nor do I see how this helps to determine that xA+By+z=0 (A,B = const.) which defines the great circle.

Thanks.

Thank you for your post. I am a scientist and I would like to help you understand the process of finding geodesics on a sphere.

Firstly, let's define what a geodesic is. A geodesic is a curve on a surface that minimizes the distance between two points. In other words, it is the shortest path between two points on the surface. On a sphere, a geodesic is a great circle, which is a circle with a radius equal to the radius of the sphere.

To find the geodesics on a sphere g(x,y,z)=x^{2}+y^{2}+z^{2}-1=0, we can use the Lagrange multiplier method. This method involves finding the stationary points of the function f(x,y,z)=\sqrt{x'^{2}+y'^{2}+z'^{2}}, subject to the constraint g(x,y,z)=0. Here, x'^{2} \text{means} \frac{dx^{2}}{ds^{2}} and not d^{2}x/ds^{2}, so we are finding the stationary points of the arclength element ds.

Using the Lagrange multiplier method, we get the following equations:

d^{2}x/ds^{2}=2\lambda x
d^{2}y/ds^{2}=2\lambda y
d^{2}z/ds^{2}=2\lambda z

To find the values of x, y, and z that satisfy these equations, we need to differentiate the constraint g(x,y,z) twice w.r.t. s. This is because we are finding the stationary points of the arclength element, which is related to the distance between two points on the surface.

Differentiating g(x,y,z) twice w.r.t. s, we get:

(dx/ds)^{2}+(dy/ds)^{2}+(dz/ds)^{2} = -2 \lambda (x^{2}+y^{2}+z^{2})

Since the LHS = 1 and the brackets in the RHS = 1, we can conclude that \lambda = -0.5. This value of \lambda satisfies the equations d^{2}x/ds^{2}=2\lambda x, d^{2}y/ds^{2}=2\lambda y, and d^{2}

## 1. What are Lagrange multipliers and how are they used to find geodesics on a sphere?

Lagrange multipliers are a mathematical technique used to optimize a function subject to a set of constraints. In the case of finding geodesics on a sphere, the constraints are the equations that define the sphere's surface. By using Lagrange multipliers, we can find the shortest path between two points on the sphere, known as a geodesic.

## 2. Why are Lagrange multipliers necessary for finding geodesics on a sphere?

In order to find the shortest path between two points on a sphere, we need to minimize the length of the path while also satisfying the constraints of the sphere's surface. This is where Lagrange multipliers come in, as they allow us to incorporate the constraints into the optimization process.

## 3. How do Lagrange multipliers differ from other optimization techniques?

Lagrange multipliers are unique in that they take into account the constraints of a problem, rather than just optimizing a function without any limitations. This makes them particularly useful for problems involving constrained optimization, such as finding geodesics on a sphere.

## 4. Can Lagrange multipliers be used to find geodesics on other curved surfaces?

Yes, Lagrange multipliers can be used to find geodesics on any curved surface, not just spheres. As long as the surface can be defined by a set of equations, Lagrange multipliers can be applied to find the shortest path between two points on that surface.

## 5. Are there any limitations to using Lagrange multipliers for finding geodesics on a sphere?

One limitation is that the surface must be smooth and continuous, as Lagrange multipliers rely on differentiating the equations defining the surface. Additionally, the method may become more complex for higher-dimensional surfaces, making it more computationally intensive.

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