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Lie derivative of contraction and of differential form

  1. Aug 20, 2014 #1
    Hello. I'm learning about Lie derivatives and one of the exercises in the book I use (Isham) is to prove that given vector fields X,Y and one-form ω identity [itex]L_X\langle \omega , Y \rangle=\langle L_X \omega, Y \rangle + \langle \omega, L_X Y \rangle[/itex] holds, where LX means Lie derivative with respect to field X.
    My idea was to just expand everything in terms of local components. This is where I got (skipping some intermediate steps, because I'm pretty sure I got them right):
    [tex]L_X \langle \omega , Y \rangle = \langle \omega , L_X Y \rangle + Y^{\mu}(\omega_{\nu}X^{\nu}_{\; \; ,\mu}+\omega_{\mu , \nu}X^{\nu})[/tex]
    where Einstein convention is assumed.
    Now if I were able to prove that expression in the brackets is actually equal to [itex](L_X \omega)_{\mu}[/itex] (which is listed as the next exercise in book, by the way) the task would be finished - I have no idea how to do that, though. Alternatively if I knew how to prove the original formula without expanding it in terms of components I could use my results so far in solving the next task. Any tips will be much appreciated.

    Edit: I noticed that I wrote in wrong subforum (for which I'm sorry) after wrting this post and I'm not able to delete it now. If whoever is in charge could move it instead of deleting I would be grateful.
     
    Last edited: Aug 20, 2014
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  3. Aug 20, 2014 #2

    Matterwave

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    So, what are you allowed to use to "prove" that the expression in brackets is equal to the Lie derivative of a one form? The way I've seen it done, usually one derives that result ##(L_X \omega)_\mu=\omega_\nu\partial_\mu X^\nu + X^\nu\partial_\nu \omega_\mu## from requiring the identity you posted at the beginning...which is just requiring the Lie derivative to be Leibniz.
     
  4. Aug 21, 2014 #3
    By contracting a one-form with vector field we get a scalar function. Lie derivative of a scalar field [itex]f[/itex] is defined to be [itex]L_X f = Xf[/itex] in Isham. From that I deduced that
    [tex]L_X \langle \omega , Y \rangle = X^{\mu}\partial_{\mu} (\omega_{\nu} Y^{\nu})=\langle \omega , L_X Y \rangle + Y^{\mu}(\omega_{\nu}X^{\nu}_{\; \; ,\mu}+\omega_{\mu , \nu}X^{\nu})[/tex]
    I have no idea what else can I do with these expressions.
     
  5. Aug 21, 2014 #4

    Matterwave

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    So how has Isham defined ##L_X \omega##? I do not have this book so I don't know which way he is approaching this problem. However, if he asks you to prove the statement in your OP, surely he has defined ##L_X\omega## by this point?
     
  6. Aug 21, 2014 #5
    [itex]L_X \omega := \left. \frac{\mathrm{d}}{\mathrm{d}t}\phi_t^{*}\omega \right|_{t=0}[/itex]
    Where [itex]\phi _t[/itex] is local flow associated with field X.
     
  7. Aug 21, 2014 #6

    Matterwave

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    As it is 3 in the morning, I can't guarantee my answer. I apologize if I got the derivation wrong, but let's examine the expression ##L_X\omega(Y)##.

    We can use the definition to get:

    $$(L_X \omega) (Y) = \lim_{\Delta t\rightarrow 0}\left.\frac{\phi^*_{t+\Delta t}\omega-\phi^*_t \omega}{\Delta t}\right|_{t=0}(Y)$$

    We know that ##\phi^*_0## is the identity transformation, so we can simplify the above to:


    $$(L_X \omega)(Y) = \lim_{\Delta t\rightarrow 0}\left.\frac{(\phi^*_{\Delta t}\omega)(Y)-\omega(Y)}{\Delta t}\right|_{t=0}$$

    Now we use the definition of the pull back to obtain:

    $$(L_X \omega) (Y) = \lim_{\Delta t\rightarrow 0}\frac{\left.\omega([\phi^*_{\Delta t}]^{-1}Y)\right|_{t=\Delta t}-\left.\omega(Y)\right|_{t=0}}{\Delta t}$$

    The difficult part now is to figure out what ##\left.\omega([\phi^*_{\Delta t}]^{-1}Y)\right|_{t=\Delta t}## is equal to. The problem is of course that all the quantities in this expression are based at a distance ##\Delta t## up the congruence of ##X##, whereas we would like to be able to work with quantities based at 0.

    First we examine the push forward of ##Y## term. What this term is doing is it's pushing forward ##Y## along a congruence of ##X##. At the parameter ##\Delta t##, there is already a vector ##Y(\Delta t)## there. This vector is different from the vector ##[\phi^*_{\Delta t}]^{-1}Y(0)## by exactly the Lie derivative ##L_X Y## as ##\Delta t\rightarrow 0##. We can therefore say:

    $$[\phi^*_{\Delta t}]^{-1}Y(0)\approx Y(\Delta t)-L_X Y(0)\Delta t$$

    We can then expand ##Y## because it is presumably an analytic vector field: ##Y(\Delta t)\approx Y(0)+ \Delta t \frac{d}{dt}Y(0)## we keep only first order terms of course. Additionally we can expand the one form field ##\omega## to read ##\omega(\Delta t)\approx \omega(0)+\Delta t\frac{d}{dt}\omega(0)##.

    Now we have expressed everything in terms of objects at 0. We get finally then that:


    $$(L_X \omega) (Y) = \lim_{\Delta t\rightarrow 0}
    \left. \frac{(\omega+\Delta t \frac{d}{dt}\omega)(Y+\Delta t \frac{d}{dt}Y-L_X Y\Delta t)-\omega(Y)}{\Delta t}\right|_{t=0}$$

    Taking this limit (so that we can drop all terms of order ##(\Delta t)^2##) we get:

    $$(L_X \omega) (Y) =\left(\frac{d}{dt}\omega\right)(Y)+\omega(\frac{d}{dt} Y)-\omega([X,Y])$$

    We know of course that if we define ##Y=\frac{d}{ds}## then:

    $$\omega([X,Y])=\omega\left(\frac{d}{dt} Y - \frac{d}{ds} X\right)$$

    We can cancel one of the terms to finally get:

    $$(L_X \omega) (Y) =\left(\frac{d}{dt}\omega\right)(Y)+\omega\left(\frac{d}{ds} X\right)$$

    And in coordinate notation, this is indeed as desired:

    $$(L_X \omega) (Y)=Y^\nu X^\mu \partial_\mu \omega_\nu+\omega_\mu Y^\nu\partial_\nu X^\mu$$

    My derivation might be completely wrong however...hopefully someone better at this can help you out.
     
    Last edited: Aug 21, 2014
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