Relationship between metric tensor and position vector

  • #1
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2

Main Question or Discussion Point

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:

$$\textbf{Z}^{k} \left( \frac{\delta^2 \textbf{Z}_{j}}{\delta Z^{i} \delta Z^{n}} - \frac{\delta^2 \textbf{Z}_{i}}{\delta Z^{j} \delta Z^{n}} \right) = \textbf{Z}^{k} \left(\frac{\delta^3 \textbf{R}}{\delta Z^{i}\delta Z^{j} \delta Z^{n}} -\frac{\delta^3 \textbf{R}}{\delta Z^{i}\delta Z^{j} \delta Z^{n}} \right)=0$$
 
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Answers and Replies

  • #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.
 
  • #3
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2
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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