Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Definition of the extrinsic-curvature tensor.

  1. Sep 11, 2013 #1
    Some define the extrinsic curvature tensor as

    $$K_{\mu \nu} = h^{\ \ \ \sigma}_\nu h^{\ \ \ \lambda}_\nu \nabla_\sigma n_\lambda.$$

    From the expression it seems like the index of the covariant derivative in can be any spacetime index. However, does it makes sence to ask what the covariant derivative of the normal vector n to a hypersurface, when n is only defined at the surface? How does one make sense of the change in n away from the hypersurface?
     
  2. jcsd
  3. Sep 11, 2013 #2

    WannabeNewton

    User Avatar
    Science Advisor

    That's the whole point of projecting the space-time indices of ##\nabla_{a}n_{b}## onto spatial indices relative to ##h_{ab}##. ##\nabla_{a}n_{b}## by itself may not be well-defined since we won't necessarily know how ##n^{a}## varies off of the hypersurface ##\Sigma## but ##h_{a}{}{}^{c}h_{b}{}{}^{d}\nabla_{c}n_{d}## has only spatial indices and as such lies entirely within ##\Sigma## thus this derivative only requires information of how ##n^{a}## varies on ##\Sigma## itself.

    In other words, ##h_{ab}## and ##\nabla_a## induce a derivative operator on ##\Sigma## itself, denoted ##\tilde{\nabla}_a##, defined by ##\tilde{\nabla}_c T^{a_1...a_k}{}{}_{b_1...b_l} = h_{d_1}{}{}^{a_1}...h_{b_l}{}{}^{e_l}h_{c}{}{}^{f}\nabla_{f}T^{d_1...d_k}{}{}_{e_1...e_l}##.
     
    Last edited: Sep 11, 2013
  4. Sep 11, 2013 #3
    But from the definition it seems like the derivative is acting first and then it's projected. Does the projection of a not necessarily defined vector then become well defined?

    Another author defined the extrinsic curvature as

    $$K_{ab} = \vec{e}_b \cdot \nabla_a \vec{n}$$

    where latin indicies are associated with the hypersurface and greek indicies are associated with the ambient manifold. He states that the that the covariant derivative is to be taken in the full manifold. Are these two definitions somehow equivalent?

    (BTW: Do you have any suggestions on where to read about these things? :) )
     
  5. Sep 11, 2013 #4

    WannabeNewton

    User Avatar
    Science Advisor

    You act on the tensor field with ##h_{a}{}{}^{b}\nabla_{b}## so the projection of the derivative operator onto ##\Sigma## has already been done.

    Also keep in mind that usually ##n^a## is defined as the unit normal field in space-time to a one-parameter family ##\Sigma_t## of space-like hypersurfaces foliating space-time, in which case ##\nabla_a n_b## is well defined on all of space-time and ##h_{a}{}{}^{b}\nabla_{b}n_{c}## is simply the tangential component of ##\nabla_a n_b## along ##\Sigma_t## for any ##t##.

    In such a case, the extrinsic curvature would simply be defined as ##K_{ab} = h_{a}{}{}^{c}\nabla_c n_b##. This has a very clear geometric interpretation as follows. First note that ##n^a## is hypersurface orthogonal i.e. ##n_{[a}\nabla_{b}n_{c]} = 0##; also, since ##n^a## has length unity, we have that ##n^b \nabla_a n_b = 0##. Thus ##\mathcal{L}_{n}h_{ab} = n^{c}\nabla_{c}h_{ab} + h_{cb}\nabla_a n^c + h_{ac}\nabla_{b}n^c\\ = n^{c}n_b\nabla_{c}n_a + n^{c}n_a\nabla_{c}n_b + \nabla_a n_b + \nabla_b n_a\\ = 2(n_a n^c \nabla_c n_b + \nabla_a n_b )\\ = 2h_{a}{}{}^{c}\nabla_c n_b = 2K_{ab}##.

    Now choose Gaussian normal coordinates (i.e. coordinates adapted to ##\Sigma_t##) so that the components of the unit normal field become ##n^{\mu} = \delta^{\mu}_t##. Then ##K_{\mu\nu} = \frac{1}{2}\partial_{t}h_{\mu\nu}##.

    What is ##\vec{e}_b##?

    See chapters 9 and 10 of Wald (particularly chapter 10).
     
  6. Sep 17, 2013 #5

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    To take a covariant derivative one only needs to know the vector field along a curve. One knows the unit normal along any curve on the hypersurface.

    Away from the hypersurface you would need to extend the normal to a neighborhood of the hypersurface. For small distance it is possible to do this without creating singularities( if the hypersurface is compact)
     
    Last edited: Sep 17, 2013
  7. Sep 17, 2013 #6

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    The covariant derivative of the unit normal with respect to a tangent vector to the hypersurface is itself tangent to the hypersurface. This follows because the connection is compatible with the metric.

    Specifically, if X is a tangent vector then X.<N,N> = 0 (since <N,N> = 1) = 2<∇[itex]_{X}N[/itex],N>

    The extrinsic curvature can be expressed in two ways

    K(X,Y) = <∇[itex]_{X}N[/itex],Y> and

    K(X,Y) = -<∇[itex]_{X}Y[/itex],N>

    Equality of these two expressions follows by expanding the derivative X.<Y,N> in terms of the connection.
     
    Last edited: Sep 17, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Definition of the extrinsic-curvature tensor.
Loading...