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Lie derivative vector fields, show Leibniz rule holds

  1. Jul 21, 2017 #1
    1. The problem statement, all variables and given/known data

    liecomm.png
    2. Relevant 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


    3. The attempt at a solution

    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
     
  2. jcsd
  3. Jul 21, 2017 #2

    andrewkirk

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    First note the definition of the Lie derivative of a vector field (for brevity I'll use the lower comma notation to indicate differentiation of a scalar in a coordinate direction). In a local coordinate system, the ##a##th component of ##\mathcal L_v w## is:
    $$(\mathcal L_vu)^a=[v,u]^a=v^c\partial_cu^a-u^c\partial_c v^a = v^cu^a_{,c}-u^cv^a_{,c}$$
    Then, substituting the vector field ##fw## for ##u## in the above, we have
    \begin{align*}
    \left(\mathcal L_v(fw) - f\mathcal L_vw - w \mathcal L_vf\right)^a
    &= [v^c(fw)^a_{,c}-(fw)^cv^a_{,c}]
    - f[v^cw^a_{,c}-w^cv^a_{,c}]
    - w^a \left(v^c f_{,c}\right)
    \\
    &= [v^c(f_{,c}w^a+fw^a_{,c}-fw^cv^a_{,c}]
    - f[v^cw^a_{,c}-w^cv^a_{,c}]
    - w^a v^c f_{,c}
    \end{align*}

    and collecting terms, we see that this cancels out to zero.

    I think where your attempt went off-course is here:
    The LHS of this equation is a scalar, being a coordinate of a vector. But the RHS is a vector, being a linear sum of the items ##\partial_\alpha##, which are vectors.
    The correct version of this is in my first equation above, usaing ##a## instead of ##u## for the index, and ##u## instead of ##W## for the vector.

    Also, to avoid confusion, either always use upper case, or always use lower case, for vectors. By mixing the two, as happens in that quote, confusion can arise as to what is a vector, what is a scalar coordinate and what is an index.
     
    Last edited: Jul 21, 2017
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