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The mystery of geodesics .

  1. Apr 9, 2008 #1
    Hi , everyone
    I have a problem with geodesic equation .
    I know the method of solving it , but I can't understand the solutions .
    When I tried to solve it for a torus I arrived at :
    [tex]\dot{u}
    = \frac{k}
    {{\left( {c + a\cos v} \right)^2 }}
    \][/tex]

    [tex]\[
    \dot v = \pm \sqrt { - \frac{{k^2 }}
    {{a^2 \left( {c + a\cos v} \right)^2 }} + l}
    \][/tex]
    c is the major raduis and a in the minor raduis .
    What dose this mean ?
    Thanks .
     
  2. jcsd
  3. Apr 9, 2008 #2
    geodesics on the orus

    Hi Shadow

    I am not sure what you are asking. Do you want to know what the geodesic differential equation means in general or do you want to understand the torus specifically?

    In any case, for a surface in 3 space a curve is a geodesic if its acceleration vector is normal to the surface. This means that the tangential part of the acceleration is zero so a person living on the surface would think that the geodesic is a straight line. He would see no acceleration.

    (One must require here that the curve has constant speed so that you don't get a tangential acceleration component from the bumpiness of the ride. )

    The simplest example I know is a ball at the end of a string swung in a circle. The acceleration vector always points inwards towards the center of the circle. The circles on the torus that you found are exactly like this. Their accleration vectors point to their centers and are perpendicular to the torus. So torus man would see no acceleration and would think these circles are straight lines. On the other hand, the circles at the top and bottom of the torus are not geodesics because their acceration vectors are actually tangent to the torus.

    Generally the geodesics on a torus are not circles and are difficult to find ( depending on which torus you look at.) But they all have acceleration vectors that are perpendicular to the torus and so are invisible to torus man.

    In 4 space one can produce a flat torus i.e. a torus which is exactly like a double cylinder.
    On it the geodesics are actual straight lines that wind around on it. If you want the equation for a flat torus in 4 space I can give it to you.
     
    Last edited: Apr 9, 2008
  4. Apr 10, 2008 #3
    Hi , wosfy
    Let's forget the tours for now .
    We all know that the geodesics on a 2 - sphere are circles .
    When solved the geodesic equation for it I couldn't undrestand how the solutions representes circles .
    So could you expalin in general what the solutions of the geodesic equation means ?
    and in the special case of a 2 - sphere .
    Thanks .
     
  5. Apr 10, 2008 #4
    Shadow

    I gave not only a tour but an exact geometric criterion for a geodesic.

    The geodesic equation for a hypersurface exactly means that when you parametrize it at constant speed, its accleration vector is perpendicular to the surface. For a sphere, the geodesics are great circles (not any circle) and this is because they are the only curves on the sphere whoses acceleration is perpendicular to the sphere.

    In terms of the connection the geodesic equation says that the curve's constant speed tangent vector is parallel along the curve. The equations expresses this fact in terms of the metric.

    If you want to do the sphere send me the parametrization that you are using and I will illustrate this in detail.


    wofsy

    P.S. Try solving the geodesic equation for the surface (cos(a),sin(a),cos(b),sin(b))

    (a and and b run from zero to 2 pi)
     
    Last edited: Apr 10, 2008
  6. Apr 11, 2008 #5
    Hi , wofsy
    The sphere parametrization that I use is :
    [tex]\[
    x\left( {u,v} \right) = \left( {\cos u\cos v,\sin u\cos v,\sin v} \right)
    \][/tex]
    Are you sure that this is a surface ?
    There is four coordinates .
    Thanks
     
  7. Apr 11, 2008 #6
    It is a surface in 4 dimensional space
     
  8. Apr 11, 2008 #7
    Here is my stupid solution :
    [tex]\[
    \begin{gathered}
    X\left( {a,b} \right) = \left( {\cos a,\sin a,\cos b,\sin b} \right) \hfill \\
    g_{ij} = \left\langle {\frac{{\partial X}}
    {{\partial x^i }},\frac{{\partial X}}
    {{\partial x^j }}} \right\rangle \hfill \\
    \frac{{\partial X}}
    {{\partial a}} = \left( { - \sin a,\cos a,0,0} \right),\frac{{\partial X}}
    {{\partial b}} = \left( {0,0, - \sin b,\cos b} \right) \hfill \\
    g_{11} = \left\langle {\frac{{\partial X}}
    {{\partial a}},\frac{{\partial X}}
    {{\partial a}}} \right\rangle = \sin ^2 a + \cos ^2 a + 0 + 0 = 1 \hfill \\
    ,g_{22} = \left\langle {\frac{{\partial X}}
    {{\partial b}},\frac{{\partial X}}
    {{\partial b}}} \right\rangle = 0 + 0 + \sin ^2 b + \cos ^2 b = 1 \hfill \\
    ,\text{ }g_{12} = g_{21} = \left\langle {\frac{{\partial X}}
    {{\partial a}},\frac{{\partial X}}
    {{\partial b}}} \right\rangle = \left\langle {\frac{{\partial X}}
    {{\partial b}},\frac{{\partial X}}
    {{\partial a}}} \right\rangle = 0 + 0 + 0 + 0 = 0 \hfill \\
    \therefore ds^2 = da^2 + db^2 \hfill \\
    \therefore \Gamma _{jk}^i = 0 \hfill \\
    \frac{{d^2 a}}
    {{ds^2 }} = 0 \Rightarrow a = cs \hfill \\
    \frac{{d^2 b}}
    {{ds^2 }} = 0 \Rightarrow b = ls \hfill \\
    \end{gathered}
    \][/tex]
    I think that this solution is completely wrong .
    Thanks
     
  9. Apr 11, 2008 #8
    your answer is correct and is not stupid.

    It says that the manifold has zero curvature - is flat - and that geodesics are just the images of straight lines in the parameter domain . This is totally correct.

    If you think about it you will see that the manifold is a torus. For if you imagine the parameter domain as a square with edges a distance of 2 pi apart then opposite edges are identified by the sines and cosines and make a two way cylinder i.e. a torus.

    In 3 space you can not have a torus with zero curvature.

    Notice also that the points in the manifold have a constant distance of the square root of 2 to the the origin. So the whole torus lies in the sphere of radius square root of 2 in 4 -space
     
  10. Apr 12, 2008 #9
    Hi , wofsy
    In the the solution that I gave , I arrived at that the geodesics are straight lines .
    In the case of the 2 - sphere ( a sphereical surface in a 3D space ) I can't understand how dose the solutions tell us that the geodesics are great circles .
    Could you explain that ?
    Thanks .
     
  11. Apr 14, 2008 #10
    Hi Shadow

    Here is the first installment on your question. You might have to correct my computations. They worry me a little.

    Anyway I used the other parameterization by accident - that is

    (sinucosv,sinusinv,cosu)

    The geodesic equations I got were two ODE's

    The second said |v'| = K|sinu| K is a constant
    At the origin, the initiial condition is |v'| = Ksin(0) = 0 so K = 0 for the entire curve.
    This means v is constant so the solution is a radial line beginning at the origin. This is a great circle.

    If you take the initial condition, u = pi/2 you again get K = 0 and the curve is the equator, again a great circle.

    For the other points I'm not sure yet how to solve the equations directly but am working on it. But it suffices to observe that any point on the sphere can serve as the origin of this coordinate system by just first rotating that point to the north pole. This is a perfectly rigorous argument because rotation is an isometry and so preserves geodescis.

    Note also by uniqueness of solutions for ODEs for a given initial condition that these radial lines are the only geodesics passing through the origin up to initial velocity.

    By the way I saw a different ODE on the web which is why I worry that I have made a computational mistake. However the exact same argument works for that ODE.
    Please correct my equation if it is wrong.
     
    Last edited: Apr 14, 2008
  12. Apr 15, 2008 #11
    Hi , wofsy
    look , I'll solve the equation and you tell me the meanning of the solutions :
    [tex]\[
    \begin{gathered}
    X\left( {u,v} \right) = \left( {\cos u\cos v,\sin u\cos v,\sin v} \right) \hfill \\
    g_{ij} = \left\langle {\frac{{\partial X}}
    {{\partial x^i }},\frac{{\partial X}}
    {{\partial x^j }}} \right\rangle \hfill \\
    \frac{{\partial X}}
    {{\partial u}} = \left( { - \sin u\cos v,\cos u\cos v,0} \right) \hfill \\
    \frac{{\partial X}}
    {{\partial v}} = \left( { - \cos u\sin v, - \sin u\sin v,\cos v} \right) \hfill \\
    g_{11} = \left\langle {\frac{{\partial X}}
    {{\partial u}},\frac{{\partial X}}
    {{\partial u}}} \right\rangle = \sin ^2 u\cos ^2 v + \cos ^2 u\cos ^2 v = \cos ^2 v\left( {\sin ^2 u + \cos ^2 u} \right) = \cos ^2 v \hfill \\
    g_{22} = \left\langle {\frac{{\partial X}}
    {{\partial v}},\frac{{\partial X}}
    {{\partial v}}} \right\rangle = \cos ^2 u\sin {}^2v + \sin ^2 u\sin ^2 v + \cos ^2 v \hfill \\
    = \sin ^2 v\left( {\cos ^2 u + \sin ^2 u} \right) + \cos ^2 v = 1 \hfill \\
    \therefore ds^2 = \cos ^2 vdu^2 + dv^2 \hfill \\
    \end{gathered}
    \][/tex]
    The non zeor [tex]\Gamma _{jk}^i[/tex] are :
    [tex]\[
    \begin{gathered}
    \Gamma _{12}^1 = \Gamma _{21}^1 = - \tan v \hfill \\
    \Gamma _{11}^2 = \cos v\sin v \hfill \\
    \therefore u'' - 2\tan vu'v' = 0\text{ }.....i \hfill \\
    ,v'' + \cos v\sin vu'^2 = 0\text{ }......ii \hfill \\
    \left( i \right): \hfill \\
    u'' = 2\tan vu'v' \hfill \\
    \therefore \frac{{u''}}
    {{u'}} = 2\tan vv' \Rightarrow \int {\frac{{u''}}
    {{u'}} = \int {2\tan vv'} } \hfill \\
    \therefore \ln u' = - 2\ln \cos v + C \hfill \\
    \therefore u' = \frac{C}
    {{\cos ^2 v}} \hfill \\
    \end{gathered}
    \][/tex]
    [tex]\[
    \begin{gathered}
    \therefore \frac{{u'}}
    {{v'}} = \frac{C}
    {{\cos v\sqrt {\cos ^2 v - C^2 } }} \hfill \\
    \therefore u = \int {\frac{{C\sec ^2 v}}
    {{\sqrt {1 - C^2 - C^2 \tan ^2 v} }}} .dv \hfill \\
    = \sqrt {\frac{1}
    {{1 - C^2 }}} \int {\frac{{C\sec ^2 v}}
    {{\sqrt {\frac{{1 - C^2 - C^2 \tan ^2 v}}
    {{1 - C^2 }}} }}} .dv \hfill \\
    let:w = \frac{{C\tan v}}
    {{\sqrt {1 - C^2 } }} \hfill \\
    \therefore u = \int {\frac{{dw}}
    {{\sqrt {1 - w^2 } }}} = \sin ^{ - 1} w + d \hfill \\
    \end{gathered}
    \][/tex]
    in some lectures There is a continue :
    [tex]\[
    \begin{gathered}
    \therefore \sin \left( {u - d} \right) = w \hfill \\
    \therefore \sin \left( {u - d} \right) = \lambda \tan v \hfill \\
    :\lambda = \frac{C}
    {{\sqrt {1 - C^2 } }} \hfill \\
    \because\sin \left( {u - d} \right) = \sin u\cos d - \sin d\cos u \hfill \\
    \therefore \sin u\cos d - \sin d\cos u - \lambda \tan v = 0 \hfill \\
    \therefore \frac{{\sin u\cos v}}
    {{\cos v}}\cos d - \frac{{\cos u\cos v}}
    {{\cos v}}\sin d - \lambda \frac{{\sin v}}
    {{\cos v}} = 0 \hfill \\
    \therefore \sin u\cos v\cos d - \cos u\cos v\sin d - \lambda \sin v = 0 \hfill \\
    ,x = \cos u\cos v,y = \sin u\cos v,z = \sin v \hfill \\
    \therefore - \left( {\sin d} \right)x + \left( {\cos d} \right)y - \lambda z = 0 \hfill \\
    \end{gathered}
    \][/tex]
    these lectures say about the last equation that :
    What dose this mean ??
     
  13. Apr 15, 2008 #12
    Without reading your proof yet, what the conclusion means is that a goedesic is the intersection of the sphere with a plane throught the origin. But a plane through the origin slices the sphere in a great circle. Notice also that a circle paramtereized by arclength centered at the origin will have acceleration pointing to the origin i.e. perpendicular to the sphere.

    My proof was also correct. At the risk of saying something you may already know, I'd like to say that mathematicians try to avoid calculation and will typically use isometries of a manifold to account for geodesics. For spaces which are uniform like the sphere the isometries carry a single geodesic into all others. So one actually only needs to sove the equation in one case and know the isometry group.

    Another case is the hyperbolic plane. You may also like to try this on higher dimensional spheres. Here is the sphere in 4 space.

    (sinusinvcosw,sinusinvsinw,sinucosv,cosu)

    By the way, how do you get thoses cool mathematical symbols. What program do you use? I have nothing.
     
  14. Apr 15, 2008 #13
    Now that I read your proof I see where I made my mistake. I lost a minus sine and so got the sine on the wrong side of the equation. If you look at you equation u' = C/cos_2v and
    rewrite your equation as cos_2vu' = C and take the initial condition that the curve's coordinates pass through the origin then the solution must have C = 0 along it, so u is constant. Thus the coordinates of this curve lie on a radial line throught the origin. This line maps to a great circle throught the point (1,0,0)
     
    Last edited: Apr 15, 2008
  15. Apr 15, 2008 #14
    oops. Not through the origin but through (0,pi/2)
     
  16. Apr 15, 2008 #15
    Thanks man .
    I understand now .
    And I'll try to solve the 4D sphere's equation , but it'll take a bit of time .
    I use a program called (TeXaide) ( you can download it from here) and here is the way to use it .
    P.S. Using TeXaide is easier than using Latex Reference ( this segma symbol in the tools )
     
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