Calculate the arc length between two points over a hyper-sphere

  1. Good morning,

    I'm trying to compute the arclength (geodesic distance) between two n-dimensional points over a n-dimensional sphere (hypersphere). Do you know if it is possible? If yes, please, I'd be very pleased if you, as experts, provide me this knowledge.

    Thank you very much
  2. jcsd
  3. tiny-tim

    tiny-tim 26,054
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    welcome to pf!

    hi 7toni7! welcome to pf! :smile:

    won't it just be the radius times the angle between them?

    (which you can get from the dot-product)
  4. Hello tiny-tim,

    Thank you very much for your answer, and I'm pleased to be in this forum.
    Yes, I think the same.
    In 2D and 3D is just: (arclength = S, radius = R (in radians), angle between points= ω)

    S = R*ω.

    Then, I have 3 doubts:
    1 - when dealing with dimensions greater than 3...the order of hundreds, could we do the same computation than 2 and 3 Dimensions? Could we extend this equation to higher dimensions?

    2 - The dot product in N just the same than 2 and 3 dimensions?

    3 - This formula is in an euclidean space, isn't it?

    Thank you very much,
    Best regards.
  5. tiny-tim

    tiny-tim 26,054
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    hello 7toni7! :smile:
    1. yes

    2. yes: (a1,a2,…an).(b1,b2,…bn) = a1b1 + a2b2 + …anbn

    (don't forget that the dot product gives you R2cosω, so you'll have to divide by R2, and then use the cos-1 button ! :wink:)

    3. yes :smile:
  6. Thank you.
    Then, the arclength on a n-sphere can be computed as follows:

    S = R*acos(a.b/R2).

    I think it is correct. Isn't it?

    A last question, do you know how to compute the intersection point between a n-vector and a n-sphere?

    Thank you so much again.
  7. tiny-tim

    tiny-tim 26,054
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    Homework Helper

    yes :smile:
    (this is from your other thread, isn't it?)

    do you mean an n-vector starting from the origin (the centre of the n-sphere)?

    if not, how are you defining the n-vector and the n-sphere? :confused:
  8. Hello,

    Yes, suppose that we have one n-sphere. Inside it, we have a n-point (this point different of the origin, it is another point named H).

    So, I have to compute the intersection of the line (that goes from the origin of the n-sphere passing from H) with the n-sphere. do you understand? is it possible?

    Thank you in advance again,
  9. tiny-tim

    tiny-tim 26,054
    Science Advisor
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    ah, so the line is a diameter of the n-sphere?

    then yes, it's easy …

    the n-vector to the intersection will be a scalar multiple of the n-vector to H, such that the magnitude of the n-vector (ie, the square-root of the dot-product with itself) equals the radius :wink:
  10. Well,
    This is how I do it in 2 dimensions. See image.

    Now, my question is: could this development be extended to N dimensions?


    Thank you
  11. tiny-tim

    tiny-tim 26,054
    Science Advisor
    Homework Helper

    yes, the same formula (radius times the unit vector in the P direction) works in n dimensions …

    Q = R*(P/|P|) :smile:
  12. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
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    In n-dimensional Euclidean space, the (hyper)sphere with radius R and center at [itex](a_1, a_2, ..., a_n)[/itex] has equation [itex](x_1- a_1)^2+ (x_2- a_2)^2+\cdot\cdot\cdot+ (x_n- a_n)^2= R^2[/itex]. The line through the origin and point [itex](b_1, b_2, ..., b_n)[/itex] is given by the parametric equations [itex]x_1= b_1t[/itex], [itex]x_2= b_2t[/itex], ..., [itex]x_n= b_nt[/itex]. Replacing [itex]x_1[/itex], etc. in the equation of the sphere with those gives a single quadratic equation for t. Finding the two solutions to that equation gives the two points at which the line crosses the sphere.
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