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Urgent & Immediate-Proof of Vector Identity Using Tensors

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello everyone, can anyone help me prove this using tensors?
    Given three arbitrary vectors not on the same line, A, B, C, any other vector D can be expressed in terms of these as:
    539fd77825ca4b49399a84cf91b85b53.png
    where [A, B, C] is the scalar triple product A · (B × C)

    2. Relevant equations
    I know that scalar triple product A · (B × C) in tensor notation is

    NumberedEquation1.gif

    and triple product can be written as:

    Inline5.gif Inline6.gif Inline7.gif
    Inline8.gif Inline9.gif Inline10.gif
    Inline11.gif Inline12.gif Inline13.gif
    Inline14.gif Inline15.gif Inline16.gif
    Inline17.gif Inline18.gif Inline19.gif

    3. The attempt at a solution

    I just tried to prove as normal, and after simplifying numerator of fractions with their repective denominator in tensor form, finally I arrived to D=3D !!
    please help!
    thanks
     
  2. jcsd
  3. Sep 20, 2015 #2

    Orodruin

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    You need to show us your actual attempt and not just describe it and give your result. If you do not show us what you did, how are we supposed to see where you went wrong and help you correct it?
     
  4. Sep 20, 2015 #3
    We assume vectors have the following tensor notation:
    6.png
    After simplifying numerators with denominators we have
    1.png
    And we know that the permutation symbol has the following characteristic (I also mentioned on vector algebra):
    2.png
    So we have
    3.png
    And I assumed that (from permutation symbol characteristics):
    4.png
    So I arrived to
    5.png
    Which is WRONG!
    please help
     
    Last edited: Sep 20, 2015
  5. Sep 20, 2015 #4

    Orodruin

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    You cannot simply cancel terms in the denominator and numerator like you are doing. The repeated indices represent sums (Einstein's summation convention) and both the denominator and numerator are therefore containing sums where all of ##A_i## (for example) represent different values for each term of the sum. In other words, what you are doing is the equivalent of
    $$
    \frac{x_1 y_1 + x_2 y_2}{x_1 z_1 + x_2 z_2} = \frac{y_1 + y_2}{z_1 + z_2}.
    $$
    This is obviously not true unless you have very particular values for the numbers included.

    This is utter nonsense. The different ##\varepsilon##s do not even have the same indices so you cannot possibly expect this relation to hold.

    You are also doing something obscure when you are trying to identify a vector with its components through ##A = A_i##. This cannot possibly hold as the left hand side is a vector and the right hand side is a number. You can take the component of a vector in a certain direction, but then you need to do that in the same direction for all terms of the vector. What you have done here
    between the second and third line does not make any sense. You should be taking the ##i##th component of the vector ##D##, which in the first term would be proportional to ##A_i##, in the second proportional to ##B_i##, and in the third to ##C_i## (note the same index ##i## on all of these). Until you have these basics down, you will have no chance of solving this problem.
     
  6. Sep 20, 2015 #5
    Hello
    the first note is true,
    $$
    \frac{x_1 y_1 + x_2 y_2}{x_1 z_1 + x_2 z_2} = \frac{y_1 + y_2}{z_1 + z_2}.
    $$
    is totally wrong and now I understood.
    For your second note, I used the following:
    ca96fade4343202b0b4c317f1f4ce469.png
    220px-Permutation_indices_3d_numerical.svg.png
    and I just assumed m is replaced insted of k, j, i respectively.
    whats the problem with it?
    And for the third, I think I understood what you mean. I try to rewrite the answer and post it again.
    thanks so much
    I am beginner and the way I tried to prove is what my instructor used. I will try to correct it
     
    Last edited: Sep 20, 2015
  7. Sep 20, 2015 #6

    Ray Vickson

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    You are doing it the hard way. If I were doing it I would just express ##\vec{D}## in the form ##\vec{D} = a \vec{A} + b \vec{B} + c \vec{C}##, then solve the equations for ##a,b,c##.
     
  8. Sep 20, 2015 #7

    Orodruin

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    All equations on index form must have the same indices on both sides, you cannot just assume that one index is replacable by another. ##\varepsilon_{ijk}## is only equal to ##\varepsilon_{mjk}## if ##i=m##, which is not always the case in your equations.
     
  9. Sep 20, 2015 #8

    Orodruin

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    To OP: This does seem like a much better way to do it for your purposes. You do not even need to be familiar with tensor notation to take it from here.
     
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