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Lie derivative and vector field notion.

  1. Jun 26, 2010 #1
    Here is an approach for lie derivative. And i would like to know how wrong is it.
    Assuming lie derivative of a vector field measures change of a vector field along a vector field, take a coordinate system, xi , and the vector field fi along which Ti is being changed. I go this way, i take the vector 'T' at point xi, transform the coordinate as- xi -> xi + efi= x'i, e nearly zero. Now Ti(x) is transformed to T'i(x').
    Now I subtract T'(x') with T(x).
    We would do the same thing to find directional derivative of a scalar function- S( x + ef)- S(x)= S(x')-S(x)
    Now back to vector T. T'j(x')= Ti(x)dx'j/dxi.
    Now take Ti(x') = Ti(x) + efk[tex]\nabla[/tex]kTi. Since T(x) is a vector field. But I don't ask for a definition of connection. We use arbitrary connection and try to see if T(x') is completely specified by T(x) by requiring specification of connection [tex]\nabla[/tex].
    So T'j(x')= Ti(x') - e( fk[tex]\nabla[/tex]kTi(x) - Tk(x')dfi /dxk
    To first approximation.
    This finally gives, considering first order in e,
    T'j(x')= Ti(x) - e( fk[tex]\nabla[/tex]kTi(x) - Tk(x')[tex]\nabla[/tex]kfi)
    = Ti(x) + e(lie derivative required) + e2(terms..)
    As is seen here, the final form does not include the notion of a definite vector field for T. Even connection [tex]\nabla[/tex] can be removed from lie derivative for vector. I need to know if it is correct, the assumption that lie derivative can be considered as change in tensor, not necessarily a tensor field. Or is a precisely defined tensor field property already used here?
     
    Last edited: Jun 26, 2010
  2. jcsd
  3. Jun 28, 2010 #2
    I really need to know if its correct. Its simple maths though looks long. Could someone help? Experts' comments?
     
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