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Thread: Hello, is distance invariant in Euclidean space?

  1. #1
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    Hello, is distance invariant in Euclidean space?

    Hello!
    I need a reliable link to proof someone that distance doesn't depend on a reference frame in Euclidean space.

    For example: if there are two points on the Earth the distance is the same no matter if we use parallels and meridians as coordinates or Cartesian coordinates, associated with the centre of the Earth.

    I have such links, but in other language. I need them in English, but can't find. Please help me.
    Thanks in advance.
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  2. #2
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    Re: Hello, is distance invariant in Euclidean space?

    Why do you need links?

    Surely this works straight out from the galilean transformations.

    x' = x-vt
    y' = y
    z'= z

    distance [between P1, (x1,y1,z1) and (x2,y2,z2)
    is defined as


    d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}


    and in the primed system, substituting the transformations


    d' = \sqrt {{{\left( {{x_2} - vt - \left[ {{x_1} - vt} \right]} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}


    d' = \sqrt {{{\left( {{x_2} - vt - {x_1} + vt} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}

    which equals d
    Last edited by studiot; Apr 26th 2017 at 11:02 AM.
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  3. #3
    MHF Contributor
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    Re: Hello, is distance invariant in Euclidean space?

    Euclidean space doesn't depend on coordinate systems. The distance is a property of the space.

    You can make transformations that don't preserve distance though (think about measuring in feet or metres), but that's a property of the transformation.
    Last edited by Archie; Apr 26th 2017 at 11:08 AM.
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