Lie derivative of covariant vector

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zardiac
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Homework Statement


Derive [itex]L_v(u_a)=v^b \partial_b u_a + u_b \partial_a v^b[/itex]


Homework 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.

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?
 
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zardiac said:
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?

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.
 
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)