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Thread: 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:
    $\displaystyle (AXB)i = \varepsilon_{ijk}A_jB_k$

    $\displaystyle (CXD)i = \varepsilon_{ijk}C_jD_k$

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

    = $\displaystyle \varepsilon_{ijk}\varepsilon_{ijk}A_jC_jB_kD_k$

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

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

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

    =(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:
    $\displaystyle (AXB)i = \varepsilon_{ijk}A_jB_k$

    $\displaystyle (CXD)i = \varepsilon_{ijk}C_jD_k$

    so
    (AXB) . (CXD) =$\displaystyle (\varepsilon_{ijk}A_jB_k)_i . (\varepsilon_{ijk}C_jD_k)_i$
    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:
    $\displaystyle (A \times B) \cdot (C \times D)$

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

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

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

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

    $\displaystyle = A_jB_kC_jD_k - A_jB_kC_kD_j$

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

    -Dan
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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: $\displaystyle \epsilon_{imn}$
    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: $\displaystyle \epsilon_{imn}$
    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:
    $\displaystyle \varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}$
    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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    Quote Originally Posted by topsquark View Post
    By the way, a nice way to remember that epsilon-epsilon identity:
    $\displaystyle \varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm} \delta_{jm} - \delta_{jn} \delta_{km}$
    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
    $\displaystyle \varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$,
    then surely the identity you used would be: $\displaystyle \varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$?
    but yours is different. is there something i missed while learning this topic?
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    Quote Originally Posted by physics4life View Post
    hold on a minute.
    if we have a standard epsilon-epsilon identity.. like
    $\displaystyle \varepsilon_{ijk} \varepsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$,
    then surely the identity you used would be: $\displaystyle \varepsilon_{ijk} \varepsilon_{imn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}$?
    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 $\displaystyle \varepsilon_{kij} = \varepsilon_{ijk}$ you will find we have the same expressions.

    -Dan
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