Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contravariant and covariant vectors

  1. Dec 19, 2015 #1
    I know if the number of coordinates are same in both the old and new frame then A.B=A`.B` . But if the number of coordinates are not same in both old and new frame then A.B=0 means that both the vectors A and B are perpendicular. Why is it so that if the number of coordinates of both the frames are not same, then both the vectors must be perpendicular.
     
  2. jcsd
  3. Dec 19, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The choice of frame has no effect on the dot product. It is always zero for two perpendicular vectors because the definition of perpendicularity is for the dot product to be zero.
    The dot product is not defined for vectors with different dimensionality. It is necessary to embed the lower dimensional vector in the higher-dimensional space of the other vector (or to embed both in a third space) in order to obtain a dot product. The embedding may not necessarily be unique, so the dot product may not necessarily be unique.
     
  4. Dec 22, 2015 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What, exactly, do you mean by "number of coordinates"? In any frame, the "number of coordinates" should be equal to the dimension of the space which is independent of the choice of frame. I don't see how it is possible to have "the number of coordinates not the same in both old and new frame".
     
  5. Dec 22, 2015 #4

    BiGyElLoWhAt

    User Avatar
    Gold Member

    I think he means something to the effect of ##\vec{A}\cdot \vec{B}## with ##\vec{A} \in \mathbb{R}^2## and ##\vec{B} \in \mathbb{R}^3##
     
  6. Dec 22, 2015 #5

    BiGyElLoWhAt

    User Avatar
    Gold Member

    So replace frames with coordinate systems.
     
  7. Dec 22, 2015 #6

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Maybe. But then the dot product of ##\vec A## and ##\vec B## is undefined.
     
  8. Dec 22, 2015 #7

    BiGyElLoWhAt

    User Avatar
    Gold Member

    True. But that's the best way that I can interpret what the OP means.
     
  9. Dec 22, 2015 #8

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Just to add that , if the dimension of the embedded object or subobject is lower than that of the ambient space of dimension n, the coordinates can be parametrized with fewer than n variables, e.g., a curve in space being parametrized by a single variable.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Contravariant and covariant vectors
Loading...