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Vector analysis problem

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data

    How does the cross product (i.e. A X B) of two vectors transform under inversion? [The cross product of two vectors is properly called a pseudovector because of this "anomalous behavior]. Is the cross product of two pseudo vectors, a vector, and a pseudovector? Name two psuedovector quantities in classical mechanics.

    2. Relevant equations

    A=A_x*x-hat+A_y*y_hat+A_z*z_hat

    B=B_x*x-hat+B_y*y_hat+B_z*z_hat

    3. The attempt at a solution
    should I show/prove that
    A X B=-B X A
     
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  3. Aug 28, 2009 #2

    gabbagabbahey

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    Do you understand what inversion means in this context?

    If so, what happens to an actual vector [itex]\textbf{C}[/itex] under inversion?
     
  4. Aug 28, 2009 #3
    Under inversion , C=C-1
     
  5. Aug 28, 2009 #4

    gabbagabbahey

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    Huh?! How exactly does one compute [itex]\textbf{C}^{-1}[/itex]???:confused:

    No. In this context the word inverse has nothing to do with compositional/functional inverses (such as [itex]\tan^{-1}x[/itex]). Rather, when you talk about vectors and inversion, you want to find the effect of inverting the coordinate axes (i.e. [itex]x\yo-x[/itex], [itex]y\to-y[/itex] and [itex]z\to-z[/itex])....What effect does this kind of inversion have on [itex]\textbf{C}[/itex]?
     
  6. Aug 28, 2009 #5
    I was thinking that C was a matrix rather than a vector because I assumed the C u were referring to was C=A XB; If C is a vector, then C=-C
     
  7. Aug 28, 2009 #6

    gabbagabbahey

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    Why do you think [itex]\textbf{A}\times\textbf{B}[/itex] is a matrix and not a vector?

    Yes, [itex]\textbf{C}\to-\textbf{C}[/itex] under inversion (Strictly speaking, the equal sign is incorrect, a transformation like this is represented by an arrow sign))....so [itex]\textbf{A}\to[/itex]___? And [itex]\textbf{B}\to[/itex]?....Which makes [itex]\textbf{A}\times\textbf{B}\to[/itex]____?
     
  8. Aug 28, 2009 #7
    Because A X B can be written in matrix form


    A X B =(A_y*B_z-B_y*A_z)x-hat+(A_x*B_z-B_x*A_z)y-hat+(A_x*B_y-A_y*B_x)z-hat


    A-> -A, B-> -B => A X B => -A X -B
     
  9. Aug 28, 2009 #8

    gabbagabbahey

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    But that's not matrix form, that's just component form for a vector....matrix form is more like [itex]C_{xx}\hat{\textbf{x}}\wedge\hat{\textbf{x}}+C_{xy}\hat{\textbf{x}}\wedge\hat{\textbf{y}}+C_{xz}\hat{\textbf{x}}\wedge\hat{\textbf{z}}+C_{yx}\hat{\textbf{y}}\wedge\hat{\textbf{x}}\ldots[/itex]

    You can write the cross product as a determinant of a matrix

    [tex]\textbf{A}\times\textbf{B}=\begin{vmatrix}\hat{\textbf{x}} & \hat{\textbf{y}} & \hat{\textbf{z}} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}[/tex]

    But, a determinant is not a matrix.


    Keep going...the product of two negative numbers is a _______ number?
     
  10. Aug 29, 2009 #9
    -Ax -B=AxB
     
  11. Aug 29, 2009 #10

    gabbagabbahey

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    Right (again, you need to use an arrow instead of an equal sign, otherwise you are telling whoever is marking your paper that the two vectors are equal, when you want to tell the marker that the first vector transforms into the second), but if [itex]\textbf{A}\times\textbf{B}[/itex] was an actual vector (instead of a pseudo vector) then you would expect it to transform as [itex]\textbf{A}\times\textbf{B}\to-(\textbf{A}\times\textbf{B})[/itex] wouldn't you?....Therefor, [itex]\textbf{A}\times\textbf{B}[/itex] is a _________.
     
  12. Aug 29, 2009 #11
    a pseudovector since the expression AXB does not change when AXB -> -AX-B
     
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