Can each dimension of space be represented as a vector?

Discussion in 'Physics & Math' started by Richard777, Jun 2, 2019.

  1. Richard777 Registered Member

    Messages:
    17
    Can space be represented as a “structure” of three “dimension vectors”? The vectors have a common origin and are “connected” by sharing common components. The connected vectors represent space.

    Assume that matter and motion deform space. Characteristics of matter and motion are represented as “length scales”. The length scales are associated with mass, charge, and rotation. A ratio of length scale represents distortion. Deformation is represented as a ratio of incremental length. Assume deformation is equal to distortion.

    Five “rules” may govern the structure and deformation “incremental space”.

    If vector components are suitably defined then a “deformal spatial equation” will give the Kerr-Newman metric. If charge and rotation do not apply, then the Kerr-Newman metric will reduce to the Schwarzschild metric. Empty incremental space is represented as the Minkowski metric.

    Can space be represented as a “structure” of three “dimension vectors” ?

    References;

    http://newstuff77.weebly.com 23 The Kerr-Newman Vectors
     
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  3. RainbowSingularity Valued Senior Member

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    7,447
    yet you can look at the 3 dimensions from 6 sides...
    if you have 6 sides, you can locate the point ?
    how do you locate the time ?
     
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  5. QuarkHead Remedial Math Student Valued Senior Member

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    1,740
    Yes, but not for the reason you seem to think.

    Any field can be considered to be a vector space over itself (proof is easy - try it or ask). So obviously \(\mathbb{R}\) is a vector space. Then so is \(\mathbb{R}^3\).

    Now any vector space, in order to make any sense, must be equipped with a set of basis vectors, such that any vector can be represented as the sum of basis vectors, each multiplied by a scalar from the field over which it is defined. Note that the basis vectors are genuine vectors, the number of them required to "cover" the space defines the dimension of the vector space.

    So that the vector space \(\mathbb{R}^3\) (which are calling "space") has 3 basis vectors, which can be taken as coordinates for this space
     
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