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About polar vectors and pseudo vectors

  1. Apr 5, 2008 #1
    "Polar vector or real vector is a vector which possesses direction inherently (eg. displacement), the direction of polar vector remains unchanged irrespective of the coordinate system chosen. If the components of a polar vector are reversed, the vector obtained is different from the original vector. Components of polar vector change sign when the coordinate system is inverted but components of pseudo vector does not change in such a case.

    Psedo vector remains unchanged even if its components are reversed (eg. angular velocity)

    Are the above statements I read about pseudo vector correct in general? Angular velocity vector remains unchanged even if its components are reversed? I am unclear about this, could some one kindly elaborate?
     
  2. jcsd
  3. Apr 6, 2008 #2

    pam

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    You have to be more careful to distinguish between the components of a vector (with respect to a coordinate system) and the vector itself. If x, y, and z of a coordinate system are changed into -x, -y, -z, the components of a polar vector will change into their negatives, which will keep the vector unchanged. The components of a pseudovector, will not change, which means that the pseudovector will now point in the opposite direction.
     
  4. Apr 18, 2008 #3
    Is torque T a polar vector or a pseudo vector?
     
  5. Apr 18, 2008 #4

    Hootenanny

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    In general, the cross product of two vectors produces a pseudovector. A torque is defined as a cross product and therefore is a pseudovector.
     
  6. Apr 18, 2008 #5

    Shooting Star

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    Don't want to split hairs, but since the word vectors cover both polar(or proper or true) and pseudo, "the cross product of two polar vectors produces a pseudo vector".

    Also, the curl of a proper vector field is a pseudo-vector field.
     
  7. Apr 18, 2008 #6

    pam

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    The cross product of two psuedo vectors also produces a pseudo vector.
    The word vector, with no adjective, usually denotes polar vector.
     
  8. Apr 18, 2008 #7

    Shooting Star

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    Yes. I forgot to mention that. Thanks for reminding.

    This made me remember that the cross product of a polar vector and a pseudo vector is also a polar vector. I mean, of course we all know it, but don't think about it separately, but use it in things like F = q(vXB), v = ΩXR etc. Thinking about these things, I ran into a sort of a puzzle, which took me some time to figure out. I thought I must share it.

    Suppose B and C are pseudo vectors. If A is a polar vector, then,

    AX(BXC) = (A.C)B - (A.B)C.

    The LHS is a polar vector, but the RHS is a linear combination of a two pseudo vectors.
     
  9. Apr 19, 2008 #8
    As we did not own a car, my mother was fond of referring to her
    wash machine as our "pseudo-automobile" because it was not a car
    but had four wheels. My baby brother grew up speaking in this way
    and became a Great Physicist, since he grasped the notion of
    pseudovector at once: it is not a vector but has three components.
    The rest of us were damned to lifelong
    confusion about axial and polar vectors.

    But there is Good News: forget about axial and polar vectors!
    You don't need them! Here a quote:

    "Books on vector algebra commonly make a distinction between polar vectors
    and axial vectors, with a x b identified as an axial vector if a
    and b are polar vectors. This confusing practice of admitting two kinds of
    vectors is wholly unnecessary. An "axial vector" is nothing more than a
    bivector disguised as a vector. So with bivectors at our disposal, we can do
    without axial vectors. As we have defined it, the quantity a x b is a
    vector in exactly the same sense that a and b are vectors."

    from David Hestenes "New Foundations for Classical Mechanics"
    (second edition) p. 61

    A somewhat longer essay on this topic is found in the attachment.
     

    Attached Files:

  10. Apr 19, 2008 #9
    Some sense can be made of pseudo vectors when they are replaced with the wedge product and another operator, which is valid aswell, in spaces other than three dimensions, whereas the cross product is unique to three.wedge product.

    Accepting an equation like this:
    [tex]\nabla \times B - \mu \epsilon (dE/dt) = \mu J[/tex]
    requires a leap of faith, that I'd forgotten I'd taken along with many others. One is required to accept that each term in this equation represents a vector in the three spatial dimensions. Fortunately all the hard work has been done a long time ago in recasting these sort of equations in a logical form.
     
    Last edited: Apr 19, 2008
  11. Apr 24, 2008 #10

    robphy

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    I think the coefficient, A.C, is a pseudoscalar.... so (A.C)B is actually a polar vector. Similarly for the other vector.
     
  12. Mar 7, 2010 #11
    force and R are vector(polar vector),so their vector product is pseudo vector.

    if u exert a force on something from right to left in front of a mirror,the force will exert from left to right on the image of that.

    have a good time:smile:
     
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