Lie derivative of covariant vector

1. Nov 4, 2012

zardiac

1. The problem statement, all variables and given/known data
Derive $L_v(u_a)=v^b \partial_b u_a + u_b \partial_a v^b$

2. Relevant equations
$L_v(w^a)=v^b \partial_b w^a - w^b \partial_b v^a$

$L_v(f)=v^a \partial_a f$ where f is a scalar.

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

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. Nov 5, 2012

clamtrox

Of course you are. There's no significance to what letter you use to represent the sums. $u^a v_a = u^b v_b = u^\Upsilon v_\Upsilon$ all mean exactly the same thing.

3. Nov 5, 2012

zardiac

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 $L_v(u_a)$ 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)