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A Relationship between metric tensor and position vector

  1. Jul 23, 2016 #1
    Given the definition of the covariant basis (##Z_{i}##) as follows:

    $$Z_{i} = \frac{\delta \textbf{R}}{\delta Z^{i}}$$

    Then, the derivative of the covariant basis is as follows:

    $$\frac{\delta Z_{i}}{\delta Z^{j}} = \frac{\delta^2 \textbf{R}}{\delta Z^{i} \delta Z^{j}}$$

    Which is also equal to the following:

    $$\frac{\delta Z_{i}}{\delta Z^{j}} = \Gamma^{k}_{i j} Z_{k}$$

    Such that:

    $$\frac{\delta^2 \textbf{R}}{\delta Z^{i} \delta Z^{j}}=\Gamma^{k}_{i j} Z_{k}$$

    (which is one way we know that ##i## and ##j## commute)

    Then if we take the derivative of ##\frac{\delta Z_{i}}{\delta Z^{j}}## with respect to ##Z^{k}##, is the following equation true?:

    $$\frac{\delta^2 Z_{i}}{\delta Z^{j} \delta Z^{k}} = \frac{\delta^3 \textbf{R}}{\delta Z^{i} \delta Z^{j} \delta Z^{k}}$$

    The reason I ask is to see if the following is true:

    $$\textbf{Z}^{k} \left( \frac{\delta^2 \textbf{Z}_{j}}{\delta Z^{i} \delta Z^{n}} - \frac{\delta^2 \textbf{Z}_{i}}{\delta Z^{j} \delta Z^{n}} \right) = \textbf{Z}^{k} \left(\frac{\delta^3 \textbf{R}}{\delta Z^{i}\delta Z^{j} \delta Z^{n}} -\frac{\delta^3 \textbf{R}}{\delta Z^{i}\delta Z^{j} \delta Z^{n}} \right)=0$$
    Last edited: Jul 23, 2016
  2. jcsd
  3. Jul 24, 2016 #2


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    I don't understand your notation. In an affine space you can define a position vector (which implies that this concept doesn't apply in General Relativity, where the spacetime is pseudo-Riemannian manifold but not an affine space) and parametrize a certain part of the affine space with general coordinates ##q^j##. Then at each point in this domain you can define a corresponding basis of tangent vectors of the coordinate lines
    $$\vec{b}_j=\frac{\vec{R}}{\partial q^j}.$$
    The Christoffel symbols are then defined as (Einstein summation convention implied)
    $$\frac{\partial \vec{b}_j}{\partial q^k}=\Gamma^{i}_{jk} \vec{b}_i.$$
    $$\frac{\partial \vec{b}_j}{\partial q^k}=\frac{\partial^2 \vec{R}}{\partial q^j \partial q^k} = \frac{\partial^2 \vec{R}}{\partial q^k \partial q^j}$$
    this definition implies
    i.e., an affine space is tortion free.

    Of course, you can now go on and take further derivatives to define all kinds of relations.
  4. Jul 24, 2016 #3
    Notation from Introduction to Tensor Analysis.... (Pavel Grinfeld).
    Got it; makes sense.

    One other question:

    The Riemann-Christoffel Tensor (##R^{k}_{\cdot n i j}##) is defined as:

    R^{k}_{\cdot n i j}= \frac{\delta \Gamma^{k}_{j n}}{\delta Z^{i}} - \frac{\delta \Gamma^{k}_{i n}}{\delta Z^{j}}+ \Gamma^{k}_{i l} \Gamma^{l}_{j n}- \Gamma^{k}_{j l} \Gamma^{l}_{i n}

    Given following equation for the Christoffel symbol (##\Gamma^{k}_{i j}##):

    \Gamma^{k}_{i j} = \textbf{Z}^{k} \frac{\delta \textbf{Z}_{i}}{\delta Z^{j}}

    Based on this equation, we consider the following term in the Riemann curvature tensor equation



    \Gamma^{k}_{il}\Gamma^{l}_{jn} &= \textbf{Z}^{k} \frac{\delta \textbf{Z}_i}{\delta Z^{l}} \textbf{Z}^{l} \frac{\delta \textbf{Z}_j}{\delta Z^{n}}


    &=\textbf{Z}^{k l} \frac{\delta \textbf{Z}_i}{\delta Z^{l}} \frac{\delta \textbf{Z}_j}{\delta Z^{n}}






    \Gamma^{k}_{j l}\Gamma^{l}_{i n} &= \textbf{Z}^{k} \frac{\delta \textbf{Z}_j}{\delta Z^{l}} \textbf{Z}^{l} \frac{\delta \textbf{Z}_i}{\delta Z^{n}}


    &=\textbf{Z}^{k l} \frac{\delta \textbf{Z}_j}{\delta Z^{l}} \frac{\delta \textbf{Z}_i}{\delta Z^{n}}


    &=\textbf{Z}^{k l} \frac{\delta \textbf{Z}_i}{\delta Z^{l}} \frac{\delta \textbf{Z}_j}{\delta Z^{n}}




    \Gamma^{k}_{i l} \Gamma^{l}_{j n}- \Gamma^{k}_{j l} \Gamma^{l}_{i n}=0

    If this is true, the Riemann curvature tensor can be simply written as follows:

    R^{k}_{\cdot n i j}= \frac{\delta \Gamma^{k}_{j n}}{\delta Z^{i}} - \frac{\delta \Gamma^{k}_{i n}}{\delta Z^{j}}

    Where is my mistake? I'm not sure.
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