As is well known, mastering Riemannian geometry requires a strong sense of geometric intuition. However, beyond that, Riemannian geometry described in the language of components also requires good analytical skills to sort out, because there are N many indices representing components and summations, making it look like indices are everywhere at first glance. This tedious component language is not always liked, and in many places, it is even notorious.
In the language of components, we can essentially establish any form of coordinate system locally, that is, using any form of basis \{\boldsymbol{e}_{\mu}\}, or a natural frame. But it is undeniable that under an orthogonal frame (orthonormal basis), many equations will be much simpler, and thanks to our familiarity with Euclidean space, our research under orthogonal frames may feel more intuitive. Therefore, if conditions permit, we should use orthogonal frames \{\hat{\boldsymbol{e}}_{\mu}\}, even if they are moving frames; here we use \hat{} to mark the orthogonal frame.
For example, we have the differential element d\boldsymbol{r} = \boldsymbol{e}_{\mu}dx^{\mu} \tag{12} measured in a general frame. Then we can obtain the Riemannian metric ds^2 = \langle d\boldsymbol{r}, d\boldsymbol{r}\rangle= g_{\mu\nu}dx^{\mu} dx^{\nu} \tag{13} where g_{\mu\nu} = \langle \boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\rangle \tag{14} which might be a matrix with complex functions. We write the Riemannian metric in matrix form: g_{\mu\nu}dx^{\mu} dx^{\nu}=d\boldsymbol{x}^T \boldsymbol{g}d\boldsymbol{x} \tag{15} Then we attempt to perform such a decomposition: \boldsymbol{g}=\boldsymbol{h}^T \boldsymbol{\eta}\boldsymbol{h} \tag{16} where \boldsymbol{h} and \boldsymbol{\eta} are matrices of the same shape as \boldsymbol{g}. Then ds^2 = (\boldsymbol{h}d\boldsymbol{x})^T\boldsymbol{\eta}(\boldsymbol{h}d\boldsymbol{x}) \tag{17} Written in component language, this is: ds^2 = \eta_{\mu\nu}(h_{\alpha}^{\mu} dx^{\alpha} )(h_{\beta}^{\nu} dx^{\beta}) \tag{18} We denote: \omega^{\mu} = h_{\alpha}^{\mu} dx^{\alpha} \tag{19} In fact, h_{\alpha}^{\mu} is a transformation matrix that transforms the original arbitrary frame \{\boldsymbol{e}_{\mu}\} into a (moving) orthogonal frame \{\hat{\boldsymbol{e}}_{\mu}\}, namely: \hat{\boldsymbol{e}}_{\mu} = \boldsymbol{e}_{\alpha}(h^{-1})^{\alpha}_{\mu} , \quad \boldsymbol{e}_{\mu} = \hat{\boldsymbol{e}}_{\alpha} h^{\alpha}_{\mu} \tag{20} At this point, we have: d\boldsymbol{r} = \boldsymbol{e}_{\mu} dx^{\mu} =\hat{\boldsymbol{e}}_{\mu} \omega^{\mu} \tag{21} This indicates that \omega^{\mu} in the orthogonal frame is equivalent to dx^{\mu} in the general frame, and: ds^2 = \eta_{\mu\nu} \omega^{\mu} \omega^{\nu} \tag{22}
The above equation shows that orthogonal frames help simplify the Riemannian metric; now the metric tensor is the simpler \eta_{\mu\nu}. It should be pointed out that the ideal decomposition is for \boldsymbol{\eta} to be the identity matrix, but if we consider general Riemannian metrics (especially those in general relativity) and limit ourselves to the real number field, this ideal may not always be achievable. For example, a simple \boldsymbol{g}=\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} cannot achieve this. Therefore, we only hope that \boldsymbol{\eta} is as simple as possible, such as a constant diagonal matrix, but we do not require it to be the identity matrix. Such a decomposition can always be done, especially in many practical cases where \boldsymbol{g} is a diagonal matrix, making it quite easy to implement. Thus, we assume here that \eta_{\mu\nu} is a diagonal matrix whose diagonal elements are 1 or -1.
Next, we write: ds^2 = \langle d\boldsymbol{r},d\boldsymbol{r}\rangle=\langle \hat{\boldsymbol{e}}_{\mu},\hat{\boldsymbol{e}}_{\nu}\rangle \omega^{\mu}\omega^{\nu} \tag{23} which is: \eta_{\mu\nu} = \langle \hat{\boldsymbol{e}}_{\mu},\hat{\boldsymbol{e}}_{\nu}\rangle \tag{24} Here \hat{\boldsymbol{e}} is the “orthogonal frame” in the sense of the above equation.
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