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What is equation for Lie derivative in Riemann curvature?

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data
    (Self study.)
    Several sources give the following for the Riemann Curvature Tensor:
    d1aa621fc7af204a1dc4a2a247129ad5.png
    The above is from Wikipedia.
    My question is what is [itex] \nabla_{[u,v]} [/itex] ?

    2. Relevant equations
    [A,B] as general purpose commutator: AB-BA (where A & B are, possibly, non-commutative operators),
    [A,B] as Lie bracket, which is a directional derivative: [itex] \nabla_A B - \nabla_B A [/itex],
    The directional derivative: [itex] \nabla_A B = \frac {\partial B} {\partial a} [/itex] where a is the parameter for the curve A, and
    Additivity: [itex] \nabla_{gA + hB} C = g \nabla_A C + h \nabla_B C[/itex] (MTW p 252, equation (10.2d)). This shows how MTW "moves" the coefficients of the subscripts "up" to coefficients of the differentials as well as the addition operation. This is just a notational issue.

    3. The attempt at a solution

    According to the Wikipedia page, the [u,v] as subscripts on the last term of Riemann is a Lie derivative. That would make it: [itex] \nabla_{[u,v]} w = \nabla_{(\nabla_u v - \nabla_v u)} w = \nabla_{\nabla_u v} w - \nabla_{\nabla_v u} w [/itex].

    But, what does it mean to take a directional derivative with respect to a directional derivative? I get what it means to take a directional derivative, along one direction, of a directional derivative along a, possibly different, direction, but is a directional derivative always a direction? I would think not. In general, a directional derivative may not, itself, be a direction, so what does the [itex] {\nabla_u v} [/itex] in the bottom mean?

    In an attempt to make sense of this, let's expand the nablas. We get [itex] \nabla_{(\frac {\partial v} {\partial u})} w - \nabla_{(\frac {\partial u} {\partial v})} w [/itex] = [itex] \frac {\partial w} {\partial (\frac {\partial v} {\partial u})} - \frac {\partial w} {\partial (\frac {\partial u} {\partial v})} [/itex].

    I realize we're headed for a second derivative, and I would think it would be [itex] \frac {\partial^2 w} {{\partial v} {\partial u}} - \frac {\partial^2 w} {{\partial u} {\partial v}} [/itex] (noting the order of the bottom factors since the "u" is the "lowest" on the first nabla & the right-most partial on the bottom is the first applied), but I've been unable to find anything in my books or on the inet that would justify such a step.

    So am I completely misunderstanding this? Is this really the correct form of the equation? Does anyone have any references that explain this notation to really, really stupid people?

    Thanks.
     
  2. jcsd
  3. Nov 23, 2014 #2

    stevendaryl

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    Well, if [itex]u[/itex] and [itex]v[/itex] are vector fields, then [itex][u,v][/itex] is another vector field. So if you define [itex]q = [u,v][/itex], then the question is about taking a directional derivative with respect to [itex]q[/itex]: [itex]\nabla_q[/itex].
     
  4. Nov 23, 2014 #3
    might be the same as [itex] \nabla_{[u,v]} [/itex].
    This is not correct. This expression I gave is actually [itex] \nabla_u \nabla_v - \nabla_v \nabla_u [/itex] which is definitely not equal to [itex] \nabla_{[u,v]} [/itex] in general.

    Ok, thanks, but how does that [itex] \nabla_q [/itex] expand out in terms of partial derivatives with respect to what when q=[u,v]? I think that is where I'm having the biggest problem. (I'm having a lot of problems, that's just the biggest one right now. :confused:) If I could just see where all the partials go, I think I could then figure out how to get there.

    Your response has also raised a couple of more questions I need to go investigate.

    Thanks.
     
  5. Mar 1, 2015 #4
    Ok, after 3 months of studying, I think I may have an answer. If anyone even sees this, please let me know if you think I've gotten it wrong. Oh, and just ignore that ridiculous nonsense in that first post by that FreeThinking guy; he's a moron.

    The key to the commutator on the bottom right of the last nabla (also called "del") seems to be equation 2.7 on page 44 of "Geometrical methods of mathematical physics" by Bernard Schutz, which is the equation for the commutator of two vectors, and using equation 2.3 on page 32, which is the definition of an arbitrary (non-coordinate), tangent vector.

    If we let A & B be two arbitrary, tangent vectors, then by eqn 2.3 we can write:
    [itex] \vec A = {{d} \over {d\lambda}} = {{dx^i}\over{d\lambda}}{{\partial}\over{\partial x^i}}[/itex]
    and
    [itex] \vec B = {{d} \over {d\mu }} = {{dx^i}\over{d\mu }}{{\partial}\over{\partial x^i}}[/itex].

    Plugging these into equation 2.7 we get:
    [itex] [\vec A,\vec B]=(A^i{{\partial B^j}\over{\partial x^i}}-B^i{{\partial A^j}\over{\partial x^i}}){\partial \over dx^j}[/itex].

    Taking the del of the commutator gives us:
    [itex] \nabla_{[\vec A,\vec B]}=\nabla_{\left(A^i{{\partial B^j}\over{\partial x^i}}{\partial \over dx^j}\right)}-\nabla_{\left(B^i{{\partial A^j}\over{\partial x^i}}{\partial \over dx^j}\right)}[/itex].

    Now, just for convenience and easier writing, let
    [itex] \vec S = {\left(A^i{{\partial B^j}\over{\partial x^i}}{\partial \over dx^j}\right)}[/itex]
    and
    [itex] \vec T = {\left(B^i{{\partial A^j}\over{\partial x^i}}{\partial \over dx^j}\right)}[/itex],
    then, by equation 11.8 on page 271 of "Gravitation" by Misner, Thorne, and Wheeler, we get:
    [itex] R(\vec A,\vec B)=\left(\nabla_{\vec A}\nabla_{\vec B}-\nabla_{\vec B}\nabla_{\vec A}\right)-\left(\nabla_{\vec S}-\nabla_{\vec T}\right)[/itex].

    Now we have everything in terms of the dels of vectors. From here, everything else should be straight forward to compute. This is what I was looking for (if it's correct, of course).

    (Thanks again to stevendaryl. Your comment "[u,v] is another vector field" triggered a memory that helped me get started on the solution.)
     
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