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Thread: orthogonality

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    orthogonality

    In a universe with positive or negative curvature, parallel lines do not exist, therefore in such a universe, orthogonality is one-sided from a vantage point of view (POV). As shown by the image below, lines AB and CD are not parallel. Therefore, distances lq, mr, and ns are meaningful only from POV of line AB and distances qm, rn, and sk are meaningful only from POV of line CD.
    Attachment 256
    Last edited by AntonioLao; 01-14-2008 at 03:27 PM.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Smile Re: orthogonality

    Quote Originally Posted by AntonioLao
    In a universe with positive or negative curvature, parallel lines do not exist, therefore in such a universe, orthogonality is one-sided from a vantage point of view (POV). As shown by the image below, lines AB and CD are not parallel. Therefore, distances lq, mr, and ns are meaningful only from POV of line AB and distances qm, rn, and sk are meaningful only from POV of line CD.
    Attachment 256
    Great diagram Antonio,but is not the drawing one of straight lines?

    kind regards michael.
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    reveal herself?

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    Re: orthogonality

    Antonio,

    I've noticed that if you extend the lines AB and CD from the points A and C until the lines join at the end, and then you draw a third line from point B to point D, you create a non right angle triangle. But if you continue the sequence of lines within the trinagle, lqmrnsk... Until you cover the whole riangle from side to side, you can use the triangles formed by those lines to calculate the area of the whole triangle. What is the meaning of the area? Could it be the field in which the force acts now that it's not orthogonal?

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    Cool Re: orthogonality

    Antonio, is this in 3 dimensional space?
    The first is only interesting if it is the beginning of something. The first is not interesting if it is the only - Djanet Sears

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    Re: orthogonality

    Quote Originally Posted by harmonygirl
    is this in 3 dimensional space?
    No. It is in a 4 dimensional curvilinear space and time.
    Quote Originally Posted by Guille
    What is the meaning of the area? Could it be the field in which the force acts now that it's not orthogonal?
    No triangles no areas and no areas no volumes. But if triangles exist then inequalities exist. However, equalities exist only if there is no areas and only primary forces exist. Secondary forces can exist only in inequalities.
    Quote Originally Posted by mkirkpatrick
    but is not the drawing one of straight lines?
    Yes. They are straight lines in the strictest sense. However, 90 degrees lines exist only with respect to one straight line not two or more straight lines.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Cool Re: orthogonality

    but Antonio, if one accepts the hypothesis that space is curved (I am not sure how to incorporate time), couldn't AB and CD be parallel (or have they been defined as not parallel?)
    The first is only interesting if it is the beginning of something. The first is not interesting if it is the only - Djanet Sears

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    Re: orthogonality

    Quote Originally Posted by harmonygirl
    if one accepts the hypothesis that space is curved (I am not sure how to incorporate time), couldn't AB and CD be parallel (or have they been defined as not parallel?)
    For one, I don't think space is curved locally but it is very difficult to refute Einstein's special and general relativity even though these theories are global theories. For me, AB and CD are always parallel to each other at the local infinitesimal region of space and time.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

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    Cool Re: orthogonality

    are you saying that the curvature of space may be the way to incorporate time? so AB and CD are parallel now, but over time are not?
    The first is only interesting if it is the beginning of something. The first is not interesting if it is the only - Djanet Sears

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    Re: orthogonality

    Quote Originally Posted by harmonygirl
    are you saying that the curvature of space may be the way to incorporate time? so AB and CD are parallel now, but over time are not?
    Maybe what you have in mind is positive curvature as that found on the surface of a sphere where no parallel lines can be drawn. What I have in mind is negative curvature as that found on hyperboloid surfaces where an infinite number of parallel lines can be drawn.
    Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: a(tr(t)=c²

 

 

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