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Lie derivative of covariant vector

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Derive [itex]L_v(u_a)=v^b \partial_b u_a + u_b \partial_a v^b[/itex]


    2. Relevant equations
    [itex]L_v(w^a)=v^b \partial_b w^a - w^b \partial_b v^a[/itex]

    [itex]L_v(f)=v^a \partial_a f[/itex] where f is a scalar.

    3. The attempt at a solution
    In the end I get stuck with something like this,
    [itex]L_v(u_a)w^a=v^b u_a \partial_b w^a -u_a v^b \partial_b w^a +v^b w ^a \partial_b u_a +u_a w^b \partial_b v^a [/itex]
    Which makes me think I am on the right way, but I end up with
    [itex]L_v(u_a)w^a=v^b w ^a \partial_b u_a +u_a w^b \partial_b v^a [/itex]

    Which is what I want if I can change place of a and b in the last term only. But I am not allowed to do that just like that right?
     
  2. jcsd
  3. Nov 5, 2012 #2
    Of course you are. There's no significance to what letter you use to represent the sums. [itex] u^a v_a = u^b v_b = u^\Upsilon v_\Upsilon [/itex] all mean exactly the same thing.
     
  4. Nov 5, 2012 #3
    Really? I guess I am not used to the einstein notation contention. How can you see what indexes are free and which ones are dummy?
    I thought since I was to determine [itex]L_v(u_a)[/itex] then a was a free index, and thus if I change in one of the terms I have to change in all of the terms. (Which would not lead to the correct result in this case)
     
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