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Einstein Summation

  1. Apr 7, 2016 #1
    1. The problem statement, all variables and given/known data
    My question is regarding a single step in a solution to a given problem. The step begins at:

    ##\large \frac{\partial \alpha _j}{\partial x ^i}
    \frac{\partial x^i}{y^p}
    \frac{\partial x^j}{\partial y^q} -
    \frac{\partial \alpha _j}{\partial x ^i}
    \frac{\partial x^i}{\partial y^q}
    \frac{x^j}{ \partial y^p}
    ##


    The solution says that the second term has the dummy i and j indices which can be switched in order to factorise which gives us:


    ##\large \big (\frac{\partial \alpha _j}{\partial x ^i}
    -
    \frac{\partial \alpha _i}{\partial x ^j} \big )
    \frac{\partial x^i}{\partial y^p}
    \frac{\partial x^j}{\partial y^q}
    ##


    1. To clarify, the Einstein summation means that changing the index i in the second term has no implication for the index i in the first term?


    2. If the dummies i and j were switched from the first term then I would result in the negative answer. So how does one know beforehand which term for whcih the indices have to be switched?
     
  2. jcsd
  3. Apr 7, 2016 #2

    George Jones

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    In the second term, does replacing index i by index m and replacing index j by index n change anything?
     
  4. Apr 7, 2016 #3
    Well no because the summation will still be the same. But that makes it diffucult to see the factorization that is needed.
     
  5. Apr 7, 2016 #4

    George Jones

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    So, now the second term is

    $$\frac{\partial \alpha _n}{\partial x ^m} \frac{\partial x^m}{\partial y^q} \frac{\partial x^n}{ \partial y^p}.$$

    Does replacing index m by index j and replacing index n by index i change anything?
     
  6. Apr 8, 2016 #5
    But I still don't see how this answers by second question?
     
  7. Apr 8, 2016 #6

    PeroK

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    I might suggest a radical approach of putting the sigmas back in so you can see what's going on. The summation convention is only a shorthand, after all.
     
  8. Apr 9, 2016 #7

    George Jones

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    If you type in the expression that results when this is done, we can discuss the expression.

    Also, this might help you to understand things. I sometimes find that my understanding changes when I actually write an expression down. Unfortunately, it can go both ways! Sometime, I don't understand an expression when I visualize it in my mind, but things become clear when I write the expression down. Other times, I think that I understand something, but I when I write it down, my "understanding" fades.

    I only recommend doing this as a short-term aid for learning how to deal with expressions that omit the Sigmas.
     
  9. Apr 9, 2016 #8
    Your question 1 is true; the Einstein summation convention only applies to terms connected by multiplication . For question 2 there is no way to know before hand and your reasoning must lead to the conclusion that the term in braces equals zero, i.e. ∂αj/∂xi = ∂αi/∂xj
     
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