Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kronecker delta symbol

  1. Dec 18, 2009 #1

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    [tex]
    \begin{array}{l}
    \delta _{jk} A_k \\
    \\
    \delta _{jk} A_k = \left( {\delta _{1,1} + \delta _{1,2} + \delta _{1,3} + \delta _{2,1} + \delta _{2,2} + \delta _{2,3} + \delta _{3,1} + \delta _{3,2} + \delta _{3,3} } \right)A_k \\
    \,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {1 + 0 + 0 + 0 + 1 + 0 + 0 + 0 + 1} \right)A_k \\
    \,\,\,\,\,\,\,\,\,\,\,\,\, = 3A_k \\
    \end{array}
    [/tex]

    But the answer should be [tex]A_j[/tex]. Where did I go wrong?
     
  2. jcsd
  3. Dec 18, 2009 #2

    Pengwuino

    User Avatar
    Gold Member

    As far as I know, the sum is only over k, not j.
     
  4. Dec 18, 2009 #3

    diazona

    User Avatar
    Homework Helper

    That's true, and you also need to recognize that [itex]A_k[/itex] has different values for different terms in the sum.
     
  5. Dec 18, 2009 #4

    Pengwuino

    User Avatar
    Gold Member

    and to further add, the delta will only be 1 when k=j in the sum.
     
  6. Dec 18, 2009 #5

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Sorry, I don't understand your explanations. We don't have a textbook for this, only class notes. So I don't even know what "the sum is over k" means. Is it possible to work out the example? Thanks!
     
  7. Dec 19, 2009 #6
    The sum is always only over a repeated index, the other indices are fixed. For example suppose A is a 2x2 matrix and x is a 2-vector:

    [tex]A_{ij} x_j = \sum_{j=1}^2 A_{ij} x_j = A_{i1} x_1 + A_{i2} x_2 [/tex]
     
  8. Dec 19, 2009 #7
    really your only problem is that you summed over j
     
  9. Dec 19, 2009 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi tony873004! :smile:

    See http://en.wikipedia.org/wiki/Einstein_summation_convention" [Broken] :wink:
     
    Last edited by a moderator: May 4, 2017
  10. Dec 20, 2009 #9

    HallsofIvy

    User Avatar
    Science Advisor

    No. As others have said, the sum is over k, the repeated index, not i.
    [tex]\delta_{ik}A_k= (A_{i1}A_1+ A_{i2}A_2+ A_{i3}A_3[/itex]
    for every i. That is
    [tex]\delta_{1k}A_k= A_{11}A_1+ A_{12}A_2+ A_{13}A_3= A_1[/tex]
    [tex]\delta_{2k}A_k= A_{21}A_1+ A_{22}A_2+ A_{23}A_3= A_2[/tex]
    [tex]\delta_{3k}A_3= A_{21}A_1+ A_{32}A_2+ A_{33}A_3= A_3[/tex]
    That is, it is the vector [itex]<A_1, A_2, A_3>[/itex] which can be written as [itex]A_i[/itex] or [itex]A_j[/itex] as they mean the same thing.


     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook