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Permutation Index

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm taking a Magnetic Fields class, and the professor taught us doing cross and dot products using the permutation index. But I don't quite understand how it works completely.

    I have these problems:

    Given:

    [tex]\vec A=\hat x + 2\hat y - 3\hat z[/tex]
    [tex]\vec B=3\hat x - 4\hat y[/tex]
    [tex]\vec C=3\hat y - 4\hat z[/tex]

    Find:

    1) [tex]\vec A \times \vec C[/tex]

    2) [tex]\hat x \times \vec B[/tex]

    2. Relevant equations



    3. The attempt at a solution

    1) Using what I know about the permitivity constant:

    [tex](\vec A \times \vec C)=[/tex]

    [tex]\varepsilon_{xyz}\vec A_y \vec C_z=[/tex]

    But I don't know where to go from here. All I know is that [tex]\varepsilon_{xyz} = 1[tex] because indices are a cyclic permutation, but I don't know what to do next.

    2) For this one I don't even know where to begin.

    Please someone help, any help at all would be great.
     
  2. jcsd
  3. Feb 3, 2008 #2

    Avodyne

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    Science Advisor

    You can do these with rules for the cross products of the unit vectors:

    [tex]\hat x\times\hat x = 0\quad\quad\hat x\times\hat y = \hat z\quad\quad\hat x\times\hat z = -\hat y[/tex]

    [tex]\hat y\times\hat x = -\hat z\quad\quad\hat y\times\hat y = 0\quad\quad\hat y\times\hat z = \hat x[/tex]

    [tex]\hat z\times\hat x = \hat y\quad\quad\hat z\times\hat y = -\hat x\quad\quad\hat z\times\hat z = 0[/tex]
     
  4. Feb 4, 2008 #3
    Hi!

    I'm new to the forum so I'm still getting used to the formatting techniques... instead of using the fancy mathematical symbols, I'm going to use easier assignments. In this case, E will be my permutation symbol.

    Remember that the Einstein notation for vectors is incredibly useful, but it operates with an implied summation. E(xyz) is summed as x goes from 0 to 3, y goes from 0 to 3, and z goes from 0 to 3. These represent nothing more than component indexing numbers, with 1=i'th component, 2=j'th component, and 3=k'th component.

    Your representation of the cross product is accurate, but don't include the vector line above each letter. E(xyz)A(y)C(z)=AxC. Using the summation, let's expand this, we get...

    =E(123)A(2)C(3)1+E(132)A(3)C(2)1+E(213)A(1)C(3)2+E(231)A(3)C(1)2+
    E(312)A(1)C(3)3+E(321)A(2)C(1)3

    Then, for all cyclic arrangements of E -> E(123)=E(312)=E(231)=1
    For all non-clyclic arrangements of E -> E(132)=E(213)=E(321)=-1

    Now simply plug in the corresponding values for each component index.

    Part B in your problem is solved in almost the exact same way, just remember that the x-unit vector does not have a y or z component, so all those components will be zero.

    Hope this helps!

    Steve
     
  5. Feb 4, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    As nevetsman said, this should be
    [tex]\varepsilon_{xyz} A_y C_z=[/tex] where Ax and Cz are the x component of A and the z component of C, not vectors themselves.
    For any set of indices [itex]\varepsilon_{ijklm}[/itex] is defined to be "1 if ijklm is an even permutation of 12345, -1 if an odd permutation, 0 otherwise". There are 3!= 6 permutions of 123. 3 are even: 123, 231, and 312, 3 are odd: 132, 213, and 321.So [itex]\varepsilon_{123}= \varepsilon_{231}= \varepsilon_{321}= 1[/itex] while [itex]\varepsilon_{132}= \varepsilon_{213}= \varepsilon_{321}= -1[/itex].

    Do you know how to find a 3 by 3 determinant? That's another mnenonic that might be simpler.
     
    Last edited: Feb 4, 2008
  6. Feb 4, 2008 #5
    Thanks for the replies. I accidentally left A and C as vectors, didn't mean to. And now I see what I have to do, thank you very much!
     
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