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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

    vanhees71

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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)
    $$\frac{\partial \vec{b}_j}{\partial q^k}=\Gamma^{i}_{jk} \vec{b}_i.$$
    Since
    $$\frac{\partial \vec{b}_j}{\partial q^k}=\frac{\partial^2 \vec{R}}{\partial q^j \partial q^k} = \frac{\partial^2 \vec{R}}{\partial q^k \partial q^j}$$
    this definition implies
    $$\Gamma_{jk}^i=\Gamma_{kj}^i,$$
    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

    $$

    \begin{align}

    \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}}

    \end{align}

    $$


    Similarly:

    $$

    \begin{align}

    \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}}

    \end{align}

    $$


    Thus:

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


    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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