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Lie derivative

  1. Dec 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi,

    it's the first time I post here, so apologies if this is not the right place.
    I'm trying to self-study GR, but I'm stuck with Lie Derivatives. The book I'm using (Ludvigsen - General Relativity. A geometric approach) starts with the usual definitions and then gives the formula for a vector field (in abstract index notation) as

    2. Relevant equations

    [tex]L_{v}w^a = v^{b}\nabla_{b}w^a - w^{b}\nabla_{b}v^a[/tex]

    It then moves on to find the Lie derivative for a covector, and states:

    [tex]L_{v}(u_{a}w^a) = (L_{v}u_a)w^a + u_{a}L_{v}w^a[/tex] (1)


    which is fair. Then after pointing out that [tex]u_{a}w^a[/tex] is a scalar, gives the (almost) final formula:

    [tex](L_{v}u_a)w^a = (v^{c}\nabla_{c}u_a + u_{c}\nabla_{a}v^c)w^a[/tex]

    which, simplified by w^a gives the final rule

    3. The attempt at a solution

    I tried to work out the missing passages following this line of thought.

    [tex]u_{a}w^a[/tex] is a scalar, so:

    [tex]L_{v}(u_{a}w^a) = v^{c}\nabla_{c}(u_{a}w^a) = v^{c}w^{a}\nabla_{c}u_a + v^{c}u_{a}\nabla_{c}w^a[/tex] (2)


    On the other hand:

    [tex]u_{a}L_{v}w^a = u_{a}v^{c}\nabla_{c}w^a - u_{a}w^{c}\nabla_{c}v^a[/tex] (3)


    Combining 1, 2 and 3, I get:
    [tex](L_{v}u_a)w^a = v^{c}w^{a}\nabla_{c}u_a + u_{a}w^{c}\nabla_{c}v^a[/tex]

    which is almost like the final formula, but a couple of indexes don't match. Where have I gone wrong?

    Thanks
     
  2. jcsd
  3. Dec 3, 2009 #2
    Welcome to PF!

    Not sure where you went wrong, but it seems to me that if you want to find the Lie derivative of a form (covector) you would just apply http://en.wikipedia.org/wiki/Lie_derivative#The_Lie_derivative_of_differential_forms":

    [tex]
    \mathcal{L}_{X}(\omega)=i_Xd\omega+d\left(i_X\omega\right)
    [/tex]

    where [itex]d\omega[/itex] is the exterior derivative of [itex]\omega[/itex] and [itex]i_X[/itex] is the insert operator: [itex]i_X(\omega)=\omega(X)[/itex].
     
    Last edited by a moderator: Apr 24, 2017
  4. Dec 3, 2009 #3

    Hurkyl

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    Those look the same to me.

    Are you forgetting things like [itex]v_iw^i = v_jw^j[/itex]?
     
  5. Dec 4, 2009 #4
    Hi Hurkyl,

    thanks for your reply. Are you suggesting that:

    [tex]u_{c}\nabla_{a}v^cw^a = u_{a}w^{c}\nabla_{c}v^a[/tex] ?

    I had this idea of swapping a<->c in mind, but I was afraid that it would not apply to such complex expressions, especially if combined with other expressions which use the same indexes. It seemed to me that if I did the swap for [tex]u_{a}w^{c}\nabla_{c}v^a[/tex], I'd be forced to do the same for the other parts of the equation, and I would be back to start...
     
  6. Dec 4, 2009 #5
    Hi jdwood983,

    yes, I've seen the identity trying to find a solution on the web, but the point is that the book I'm using never mentions it, and I wanted to "stick to the rules", so to say.

    Thanks for your suggestion anyway
     
    Last edited by a moderator: Apr 24, 2017
  7. Dec 4, 2009 #6
    I guess now that I've had some time to think about it, your book is showing the proof of Cartan's identity for forms using its own notation rather than other, more common notation:

    [tex]
    \left[i_Xd+di_X\right]df=i_Xd(df)+di_Xdf=0+d[X(f)]=d\mathcal{L}_Xf=\mathcal{L}_Xdf
    [/tex]

    where [itex]df[/itex] is the exterior derivative of a function (which makes a 1-form as the exterior derivative of a [itex]p[/itex]-form makes it a [itex]p+1[/itex]-form). Recall also (for relating this post to your first post) that assuming [itex]\omega[/itex] is a 1-form that [itex]\omega(X)[/itex] is a function

    This is at least the proof that I have seen in Frankel's The Geometry of Physics; there may be other proofs out there, but this is the one that I've seen and used.
     
  8. Dec 4, 2009 #7
    Always keep in mind that these repeated indices are just dummy indices, which you can swap for anything you like. What is implicit in the notation is that you are summing over the index, but only in that term! So the expression you had actually states:

    [tex]u_{a}v^{c}\nabla_{c}w^a - u_{a}w^{c}\nabla_{c}v^a =
    \left[\sum_{a}\sum_c u_{a}v^{c}\nabla_{c}w^a\right] - \left[\sum_{a}\sum_c u_{a}w^{c}\nabla_{c}v^a\right][/tex]

    As you can see there is no harm in replacing the summation indices by another within a term. So switching a<->c in the the second term, but not in the first is perfectly fine. If you ever find yourself having trouble with these indices, it's a good idea to take a step back and use the explicit notation.
     
  9. Dec 4, 2009 #8
    Thanks for the clarification. Very useful!
     
  10. Dec 4, 2009 #9
    I believe the difference in notation comes from the different approach. From what I see, Fraenkel's looks more calculus-oriented, while the book I use has, by its own title, a geometric approach. Thanks for the clarification anyway, I think I'll spend some time looking at this way of dealing with the subject...
     
  11. Dec 4, 2009 #10
    I'm not sure what you mean by this part, Frankel's book is a differential geometry approach to classical mechanics, electrodynamics, special relativity, and to a small part, thermodynamics. Your textbook looks to be the same to me (horray for google books because I don't own this book) in terms of applying differential geometry.
     
  12. Dec 9, 2009 #11
    Well, it's not easy to explain what I mean. But my book (at least so far in my study) is not following the usual differential geometry approach, i.e. talking about how co-/contra-variant vectors transformation, p-forms and similar things. Instead, it's using a (sort of) axiomatic approach, starting from basic definitions (e.g. 4-velocity, metric tensor) and developing the theory from there. I've seen there are free copies downloadable from the web, although I can't judge if they're complete or not (I have my own paper copy, so I didn't go through the process of getting one). If you're curious about that, you may want to try...
     
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