- #1
binbagsss
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Homework Statement
Homework Equations
##V=V^u \partial_u ##
I am a bit confused with the notation used for the Lie Derivative of a vector field written as the commutator expression:
Not using the commutator expression I have:
## (L_vU)^u = V^u \partial_u U^v - U^u\partial_u V^v## (1)
When using the commutator expression however I sometimes see it written as :
simply, without an index, indicating a tensor or rank zero ## L_v w = [v,w]##
and sometimes with an index : ## (L_v w)^u = [v,w]^u ##
Due to this I am confused as to what the commutator should be once expanded out.
Going with the notation my guess is that :
##(L_v W)^u= V^u w^{\alpha} \partial_{\alpha} - W^u v^{\alpha} \partial_{\alpha}##
##L_v W = (v^u \partial_u w^v - w^u\partial_u v^v) \partial_v ##
So I see the expression multiplying the ##\partial_v## is a tensor of rank 1 and the expression agrees with (1 ) ). Is this what the notation means , or could someone please fill me in (non-sexually)?
Since the question specifies no index, I am going to assume the expression
The Attempt at a Solution
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So for the LHS I get:
##v^u\partial_u f w^v \partial_v - f w^u \partial_u v^v \partial_v ##
RHS:
## f(v^u\partial_u w^v \partial_v - w^u \partial_u v^v\partial_v) + w^v\partial_v v^u\partial_uf ##(where the last term comes from the expression given for the lie derivative acting on a scalar)
so the first term from the LHS and the third term from the RHS agree, but not the rest...
Many thanks in advance