Solve Geodesic Problem for f:[a,b] to R

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In summary: Since U is a surface in M, the vector perpendicular to U is also in M. The vector is (u,v) and it has the magnitude (f(u), f(v))cos(c), (f(u), f(v))sin(c). Therefore, the vector is also perpendicular to the surface.
  • #1
wii
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If [itex]f:[a,b] \to R[/itex] is a positive real function and[itex]\gamma(u,v) = ( f(u)\cos (v), f(u) \sin (v), u)[/itex] then show that

[itex]\gamma(t) = \sigma(u(t), c)[/itex] is a geodesic in [itex]M[/itex]where [itex]c[/itex] is a constant between 0 and[itex]2\pi[/itex] and
[itex]M=\sigma(U)[/itex] where [itex]U= \{ (u,v)| a<u<b and 0<v< 2\pi \}[/itex]

Actually , I tried to calculate the second derivative of
[itex]\sigma(t)[/itex] but that did not work and also I still have u in the first derivative

which means it is not constant

any suggestion? :\

Thanx
 
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  • #2
:confused: no help?
 
  • #3
wii said:
:confused: no help?

i don't know what u(t) is and M seems incorrectly defined
 
  • #4
wii said:
If [itex]f:[a,b] \to R[/itex] is a positive real function and[itex]\gamma(u,v) = ( f(u)\cos (v), f(u) \sin (v), u)[/itex] then show that

[itex]\gamma(t) = \sigma(u(t), c)[/itex] is a geodesic in [itex]M[/itex]where [itex]c[/itex] is a constant between 0 and[itex]2\pi[/itex] and
[itex]M=\sigma(U)[/itex] where [itex]U= \{ (u,v)| a<u<b and 0<v< 2\pi \}[/itex]

Actually , I tried to calculate the second derivative of
[itex]\sigma(t)[/itex] but that did not work and also I still have u in the first derivative

which means it is not constant

any suggestion? :\

Thanx

OK. Now I think I understand your question.

Choose u(t) so that the curve [itex]\sigma(u(t), c)[/itex] is parameterized by arclength.

This curve may be written as (u(t),f(u(t))cos(c),f(u(t))sin(c))

Its tangent has length one which is expressed in the equation

u'(t)^2 + (f'(u)u'(t))^2 = 1

Differentiate this equation with respect to t. The left hand side is the inner product of the tangent vector with the second derivative. The right hand side is zero. Thus the acceleration of the curve is perpendicular to its tangent. Now you need to show that it is perpendicular to the surface, U.
 
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  • #5
for your question. I can provide a possible solution to your problem. First, let's define what a geodesic is. A geodesic is the shortest path between two points on a curved surface. In this case, the curved surface is represented by the function f:[a,b] \to R. The geodesic we are trying to find is represented by \gamma(u,v) = ( f(u)\cos (v), f(u) \sin (v), u).

To show that \gamma(t) = \sigma(u(t), c) is a geodesic, we need to show that it satisfies the geodesic equation. The geodesic equation is given by:

\frac{d^2\sigma}{dt^2} + \Gamma^k_{ij}\frac{d\sigma^i}{dt}\frac{d\sigma^j}{dt} = 0

where \Gamma^k_{ij} is the Christoffel symbol. In this case, since we are working with a 2-dimensional surface, the Christoffel symbol is given by:

\Gamma^k_{ij} = \frac{1}{2} \frac{\partial g_{jk}}{\partial x^i}

where g_{jk} is the metric tensor. The metric tensor for our surface is given by:

g_{jk} = \begin{pmatrix} f(u)^2 & 0 \\ 0 & 1 \end{pmatrix}

Therefore, the Christoffel symbol becomes:

\Gamma^k_{ij} = \begin{pmatrix} 0 & 0 \\ 0 & -\frac{f(u)f'(u)}{2} \end{pmatrix}

Now, let's calculate the derivatives of \sigma(t):

\frac{d\sigma^i}{dt} = \begin{pmatrix} u'(t) \\ 0 \end{pmatrix}

\frac{d^2\sigma^i}{dt^2} = \begin{pmatrix} u''(t) \\ 0 \end{pmatrix}

Substituting these into the geodesic equation, we get:

\begin{pmatrix} u''(t) \\ 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & -\frac{f(u)f'(u)}
 

1. What is a geodesic problem?

A geodesic problem is a mathematical problem that involves finding the shortest path or distance between two points on a curved surface, such as a sphere or a curved plane. In this case, the problem is to find the shortest path between two points on a function f, which maps from a closed interval [a,b] to the real numbers.

2. How is a geodesic problem solved?

The geodesic problem for f:[a,b] to R is typically solved using the calculus of variations. This involves finding a function, known as a geodesic, that minimizes the distance between two points on f. The solution involves setting up and solving a differential equation known as the Euler-Lagrange equation.

3. What is the importance of solving a geodesic problem?

Solving a geodesic problem has many practical applications in fields such as physics, engineering, and computer graphics. It allows us to find the most efficient paths between two points on a curved surface, which can be useful in optimization problems or in determining the trajectory of moving objects.

4. Are there any limitations to solving a geodesic problem?

Yes, there are some limitations to solving a geodesic problem. One limitation is that it can only be applied to functions that are continuous and differentiable on the interval [a,b]. Additionally, the solution may not always exist or may not be unique, depending on the function and the given points.

5. Can the geodesic problem be extended to higher dimensions?

Yes, the geodesic problem can be extended to higher dimensions, such as three-dimensional space or even higher-dimensional spaces. This is known as the general geodesic problem and involves finding the shortest path between two points on a curved surface in a higher-dimensional space.

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