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Homework Help: Einstien Summation Convention

  1. Apr 15, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    Ok so I'm meant to answer:

    To what scalar or vector quantities do the following expressions in suffix notation correspond? (expand and sum where possible).

    1) aibjci
    2) aibjcjdi
    3) dijaiaj
    4) dijdij
    5) eijkaibk
    6)eijkdij


    2. Relevant equations



    3. The attempt at a solution

    Okay so I'm pretty stuck to say the least..any help would be great..Here are some thoughts:

    1)doesnt this just mean we sum that components of a and c and b can take any value? I'm confused..does this mean the result is a vector.?
    2)so we sum a and d and b and c..but how do i simplify this?
    3)ok so this one i think i know....isn't it just ajaj so sum the components of a? a.a?
    4)doesnt this just equal dij?
    5) Not sure
    6) Again,not sure :(

    thanks
     
  2. jcsd
  3. Apr 15, 2010 #2

    Cyosis

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    Does the d_ij represent the Kronecker delta? Are a, b, c and d numbers or can they also be operators. If they are just numbers remember that order doesn't matter. How would you write down the dot product in index notation, what about the cross product?

    As for 4). Your answer is not right. Keep in mind that you're summing over both i and j so when i != j you get 0, which doesn't contribute to the sum. Therefore the only entries that contribute to the sum are the entries where i=j. Try to write it out by letting i and j run from 0 to 1 and note that your answer can't be right.
     
    Last edited: Apr 15, 2010
  4. Apr 15, 2010 #3

    bon

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    yes d_ij is k delta.

    i think abc are just numbers..

    dot product would be a_ib_i
    cross would be e_ijk a_i b_j e_k?

    where the fist e is an epsillon..

    but i dont see how to do the problems..:S
     
  5. Apr 15, 2010 #4

    bon

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    so is it d_ii?
     
  6. Apr 15, 2010 #5

    Cyosis

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    Yes d_ii is correct. However note that d_ii has repeated indices so we are summing all it's diagonal entries together. Can you express d_ii as a number?

    Your dot and cross products are correct. So for 1) we get [itex]a_i b_j c_i=a_ic_ib_j=(\mathbf{a} \cdot \mathbf{c})b_j[/itex]. So what kind of object do we get in the end?
     
  7. Apr 15, 2010 #6

    bon

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    Aha..so is d_ii = 3?

    And so for 1) do we get a vector in the end? so should i write (a.c) (b1, b2, b3)?

    Thanks a lot for your help..starting to make more sense..

    So for 2) is it just a number, namely (a.d)(b.c)?

    3)a.a?
    4)3
    5)still not sure?
    6) ditto..

    thanks again
     
  8. Apr 15, 2010 #7

    bon

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    Also in another question, im asked to find the div and curl of this expression, where r = (x,y,z) and a and b are fixed vectors..

    (a.r)b

    so i took the div, which i found i could rearrange to be (a.r)(del.b) but can i simplify further? what does it mean when it says a and b are fixed vectors? does it mean that all components are constants, in which case, del.b = 0? thanks :)
     
  9. Apr 15, 2010 #8

    cristo

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    This cross product is incorrect. Remember that the cross product of two vectors is itself a vector, and so much have a free index. So, if [itex]{\bm a}\times{\bm b}={\bm c}[/itex], then [itex]c_i=\epsilon_{ijk}a_jb_k[/itex]
     
  10. Apr 15, 2010 #9

    Fredrik

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    I think he meant that e_k is a basis vector, and in that case it's correct.

    Bon, I'm not sure what you find confusing about 1. It's really trivial if you think about what the notation means. For example, what does bj mean? Perhaps it would help if you write aibjci with a summation sigma and some parentheses?
     
  11. Apr 15, 2010 #10

    cristo

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    Oh, well then that's incredibly confusing to use the same symbols to mean different things!
     
  12. Apr 15, 2010 #11

    Cyosis

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    Cristo what you have written is the i-th component of the cross product. What bon has written down is the full vector with e being the unit vector.

    Example:
    [tex]
    \epsilon_{ijk}a_ib_j\mathbf{e_k}=a_2b_3\mathbf{e_1}-a_3b_2\mathbf{e_1}+a_3b_1\mathbf{e_2}-a_1b_3\mathbf{e_2}+a_1b_2\mathbf{e_3}-a_2b_1\mathbf{e_3}=(a_2b_3-a_3b_2,a_3b_1-a_1b_3,a_1b_2-b_2a_1)
    [/tex]
     
  13. Apr 15, 2010 #12

    cristo

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    Yes, if you right things out properly then it is obvious. What was written down above was ambiguous.
     
  14. Apr 15, 2010 #13

    bon

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    Many thanks all..
    so is anyone able to verify the comments i made in post 6? Also any thoughts on post 7?

    Thank you
     
  15. Apr 15, 2010 #14

    Cyosis

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    The first four look good now. As for 5 compare it to your definition of the cross product. For 6 do something similar to 4) and think what it means for the Levi-Civita symbol if two of the indices are the same.
     
  16. Apr 15, 2010 #15

    bon

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    aha ok get them now..

    5) = jth component of a x b

    and 6) = 0

    finally ( i promise) any thoughts on my question in post 7? ty!
     
  17. Apr 15, 2010 #16

    bon

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    dont worry if you arent sure about that.. i have another q in the meantime which is:

    im meant to find div and curl of (a x r) where a is a fixed vector and r = (x,y,z)

    after simplifying the expression for div i get:

    e_ijk d/dx_i x_j a_k

    but isnt this (del x r)times a which is a vector times a vector with no dot/cross = illegal?
     
  18. Apr 15, 2010 #17

    Cyosis

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    6) is correct. 5) is not entirely right remember that the Levi-Civita symbol is antisymmetric.

    Yes that is illegal, but it is not what they are asking.

    We have [itex]\nabla \times (a \times b)[/itex] first compute axb, which yields a vector and then compute the curl of that vector which again yields a vector. For the divergence we have [itex]\nabla \cdot (axb)[/itex]. First compute (a \times b) again which yields a vector then take the divergence which yields a scalar.
     
  19. Apr 15, 2010 #18

    Cyosis

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    Yes it means that all components are constant, but r is not a constant vector so you can't just put it in front of the del operator.
     
  20. Apr 15, 2010 #19

    bon

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    hmm so is it -(axb)_j i.e. -1 times the jth component?
     
  21. Apr 15, 2010 #20

    Cyosis

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    Yes and -(axb) is?
     
  22. Apr 15, 2010 #21

    bon

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    ok so for div (a.r) b my working is:

    d/dx_i e_i . (a_j x_j b_k e_k)

    why cant i just rewrite this as a_i x_i d/dx_j bj?

    how do i deal with it then?

    thanks
     
  23. Apr 15, 2010 #22

    bon

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    aha (bxa)!
     
  24. Apr 15, 2010 #23

    Cyosis

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    Yes that is correct.

    For the same reason you can't write [itex]\frac{d}{dx}f(x)[/itex] as [itex]f(x)\frac{d}{dx}[/itex]. You are differentiating when taking the divergence.
     
  25. Apr 15, 2010 #24

    bon

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    ah ok i see..well im not sure how to deal with it in that case..so far when doing divergence ive just dotted the e_i in the divergence term with the later e_j and simplified to d_ij etc..

    what can i do now? thanks
     
  26. Apr 15, 2010 #25

    Cyosis

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    The handling of the unit vectors are fine, they do not depend on x y or z. Could you compute the divergence of r for me with r=(x,y,z) the 'normal' way (with that I mean don't use index notation)?
     
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