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Math Help - Vector Calculus

  1. #1
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    Vector Calculus

    hi this is about vector calculus subscript notation:

    i generally understand v.c.s.n but can't do something like this:
    Show : (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C)

    my work so far:
     (AXB)i = \varepsilon_{ijk}A_jB_k

     (CXD)i = \varepsilon_{ijk}C_jD_k

    so
    (AXB) . (CXD) =  (\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i

    = \varepsilon_{ijk}\varepsilon_{ijk}A_jC_jB_kD_k

    = ( \delta_{ij} \delta_{jk} - \delta_{ik}\delta_{jj})(A.C)_j(B.D)k

    =  [ \delta_{ij}(A.C)_j ][ \delta_{jk}(B.D)_k ] - [ \delta_{ik}(B.D)_k ][ \delta_{jj}(A.C)_j ]

    = (A.C)_i(B.D)j - (B.D)_i(A.C)_j

    =(A.C)(B.D) - (B.D)(A.C)

    the first part of the answer (in red) i got right.. but the 2nd part is wrong as you can see

    how am i meant to get -(A.D)(B.C)???
    thanks guys please help out
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by physics4life View Post
    hi this is about vector calculus subscript notation:

    i generally understand v.c.s.n but can't do something like this:
    Show : (AXB) . (CXD) = (A.C)(B.D) - (A.D)(B.C)

    my work so far:
     (AXB)i = \varepsilon_{ijk}A_jB_k

     (CXD)i = \varepsilon_{ijk}C_jD_k

    so
    (AXB) . (CXD) =  (\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i
    Here's the problem: you are using jk indices for the second sum. And you have the epsilon-epsilon identity wrong. Here's what you should do:
    (A \times B) \cdot (C \times D)

    \to (\varepsilon_{ijk}A_jB_k)(\varepsilon_{imn}C_mD_n)

    = \varepsilon_{ijk} \varepsilon_{imn}A_jB_kC_mD_n

    = (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km})A_jB_kC_mD_n

    = \delta_{jm} \delta_{kn}A_jB_kC_mD_n - \delta_{jn} \delta_{km}A_jB_kC_mD_n

    = A_jB_kC_jD_k - A_jB_kC_kD_j

    \to (A \cdot C)(B \cdot D) - (A \cdot D)(B \cdot C)

    -Dan
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  3. #3
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    cheers dan, your method is easy to understand. it's frustrating trying to learn all this calculus on your own when there is just a huge pile of books and you're forever researching. so i appreciate your help.
    just a question though, on the 2nd epsilon identity you used: \epsilon_{imn}
    what i wanted to ask is: do you have to use the imn notation.. or can you use any preferred notation.. such as ixy....
    i understand the final result will still be the same as long as the notation is consistent throughout the solution, but i just wanted to confirm whether imn has to be used as a standard for vector calculus. thanks once again .
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by physics4life View Post
    cheers dan, your method is easy to understand. it's frustrating trying to learn all this calculus on your own when there is just a huge pile of books and you're forever researching. so i appreciate your help.
    just a question though, on the 2nd epsilon identity you used: \epsilon_{imn}
    what i wanted to ask is: do you have to use the imn notation.. or can you use any preferred notation.. such as ixy....
    i understand the final result will still be the same as long as the notation is consistent throughout the solution, but i just wanted to confirm whether imn has to be used as a standard for vector calculus. thanks once again .
    You can use any indices you like. (You can even make up your own, but they're much harder to type. ) The point is that they are different from the ones you used previously since they have no relation to the indices in the other cross product.

    By the way, a nice way to remember that epsilon-epsilon identity:
    \varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}
    Note that the "i" index doesn't appear anywhere in the deltas. Now look at the first delta of each pair. They both have a "j," the first index after the "i" in the first epsilon. The second index of those first deltas is in sequence, m and n... the second and third indices of the second epsilon.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    By the way, a nice way to remember that epsilon-epsilon identity:
    \varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}
    Note that the "i" index doesn't appear anywhere in the deltas. Now look at the first delta of each pair. They both have a "j," the first index after the "i" in the first epsilon. The second index of those first deltas is in sequence, m and n... the second and third indices of the second epsilon.
    -Dan
    hold on a minute.
    if we have a standard epsilon-epsilon identity.. like
    \varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl},
    then surely the identity you used would be: \varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}?
    but yours is different. is there something i missed while learning this topic?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by physics4life View Post
    hold on a minute.
    if we have a standard epsilon-epsilon identity.. like
    \varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl},
    then surely the identity you used would be: \varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}?
    but yours is different. is there something i missed while learning this topic?
    The two are the same. Note that in my statement the common index in the epsilons is the first. Rearranging your expression by noting that \varepsilon_{kij} = \varepsilon_{ijk} you will find we have the same expressions.

    -Dan
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