Lie derivative of covariant vector

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SUMMARY

The discussion focuses on deriving the Lie derivative of a covariant vector, specifically L_v(u_a), using the formula L_v(u_a) = v^b ∂_b u_a + u_b ∂_a v^b. The participants clarify the application of Einstein notation, emphasizing that the indices used in the expressions can be interchanged without altering the mathematical meaning. The conversation highlights the importance of distinguishing between free and dummy indices in tensor calculus, which is crucial for correctly manipulating expressions involving Lie derivatives.

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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)
 

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