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Math Help - Lie derivatives

  1. #1
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    Lie derivatives

    Say you have a scalar field \Phi(x) defined on a manifold M. Then imagine evaluating the field at a point an infitesimal distance away from x at x'=x+\epsilon X where X is a illing vector of the manifold. Then

    \Phi(x')=\Phi(x)+\epsilon X^{\mu}\partial_{\mu}\Phi(x) to first order. Now the second term on the rhs is the Lie derivative of a scalar, so

    \Phi(x')=\Phi(x)+\epsilon L_{X}\Phi(x)

    Say now that \Phi is not a scalar, say maybe a vector of spinor or tensor. Does the same thing follow, that

    \Phi^{\nu}(x')=\Phi^{\nu}(x)+\epsilon L_{X}\Phi^{\nu}(x) Where now L_{X} is not the appropriate Lie derivative of whatever object \Phi is.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ppyvabw View Post
    Say you have a scalar field \Phi(x) defined on a manifold M. Then imagine evaluating the field at a point an infitesimal distance away from x at x'=x+\epsilon X where X is a illing vector of the manifold. Then

    \Phi(x')=\Phi(x)+\epsilon X^{\mu}\partial_{\mu}\Phi(x) to first order. Now the second term on the rhs is the Lie derivative of a scalar, so

    \Phi(x')=\Phi(x)+\epsilon L_{X}\Phi(x)

    Say now that \Phi is not a scalar, say maybe a vector of spinor or tensor. Does the same thing follow, that

    \Phi^{\nu}(x')=\Phi^{\nu}(x)+\epsilon L_{X}\Phi^{\nu}(x) Where now L_{X} is not the appropriate Lie derivative of whatever object \Phi is.
    Yes, though you might wind up with something uglier like
    A_{\nu}(x')=A_{\nu}(x)+\epsilon g_{\nu \lambda}X^{\mu}\partial_{\mu}A^{\lambda}(x)
    for your Lie derivative.

    (That would be from General Relativity where  [g_{\mu \nu} ] is a metric. I'm not sure of the general structure if the embedding in the manifold isn't a space-time.)

    -Dan
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  3. #3
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    Thanks very much for your quick reply.

    The manifold is a spacetime, I forgot to mention.

    Yes, though you might wind up with something uglier like

    for your Lie derivative.
    Maybe I didn't explain very well. That expression is exactly the same as the expression I wrote, because that derivative should be a covariant derivative when operating on a vector and so the metric commutes through it and lowers the index on A_{\lambda} Is that right? That expression won't work for say a vector because we are comparing vectors in two different tangent spaces.

    I meant would it still be the same expression, for say a vector

    A_{\nu}(x')=A_{\nu}+\epsilon L_{X} A_{\nu}(x)

    where now the expression for the Lie derivatie on a covariant vector is

    L_{X}A_{\nu}=X^{\mu}\nabla_{\mu}A_{\nu}+A_{\mu}\na  bla_{\nu}X^{\mu} if I got the indices right.

    I mean roughly can the Lie derivative be thought of as a generalisation of the directional derivative in vector calculus to more general settings.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ppyvabw View Post
    That expression won't work for say a vector because we are comparing vectors in two different tangent spaces.
    Well, it's been (regrettably) about 10 months since I even looked at my QFT books, and I came up with that expression on the fly, so I'm not particularly surprised that it wasn't quite right for your needs.

    Quote Originally Posted by ppyvabw View Post
    I meant would it still be the same expression, for say a vector

    A_{\nu}(x')=A_{\nu}+\epsilon L_{X} A_{\nu}(x)

    where now the expression for the Lie derivatie on a covariant vector is

    L_{X}A_{\nu}=X^{\mu}\nabla_{\mu}A_{\nu}+A_{\mu}\na  bla_{\nu}X^{\mu} if I got the indices right.

    I mean roughly can the Lie derivative be thought of as a generalisation of the directional derivative in vector calculus to more general settings.
    Yes, as far as I know.

    -Dan
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  5. #5
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    Well, it's been (regrettably) about 10 months since I even looked at my QFT books, and I came up with that expression on the fly, so I'm not particularly surprised that it wasn't quite right for your needs.
    lol, I know. In the office last week we were discussing the CMB background and whether it violates relativity by defining a prefered reference frame. We decided it didn't, but later we were discussing the area of a paralelogram and the determinant of a two by two matrix, and also how to factorise cubics. We were totally lost.

    So on the same day we saved general relativity, but failed to remember some A-level algebra lol.

    Thanks for your help. I'll just go with it.
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