[A Brief Introduction to Exterior Differentiation] 3. Orthogonal Frames

【外微分浅谈】3. 正交标架 苏剑林 · 2016-11-05 · Mathematics symmetry exterior differentiation

【A Gentle Introduction to Exterior Differentiation】3. Orthogonal Frames

As is well known, mastering Riemannian geometry requires a strong sense of geometric intuition. But beyond that, describing Riemannian geometry in terms of components also demands solid analytical skill to untangle, because there are so many $N$ indices standing in for components and summations that at first glance indices seem to be everywhere. This kind of cumbersome component-based language is not always well liked, and in some circles it has quite a bad reputation.

In component language, we can in principle set up any kind of local coordinate system, that is, adopt an arbitrary basis $\{\boldsymbol{e}_{\mu}\}$, or what is called a natural frame. But it's undeniable that under an orthogonal frame (an orthonormal basis), many equations become considerably simpler, and thanks to our familiarity with Euclidean space, we tend to have a better feel for working under orthogonal frames. So, whenever conditions allow, we should use an orthogonal frame $\{\hat{\boldsymbol{e}}_{\mu}\}$ — even a moving one — and here we denote the orthogonal frame by $\hat{}$.

For instance, suppose we have the line element

$$d\boldsymbol{r} = \boldsymbol{e}_{\mu}dx^{\mu} \tag{12} $$

measured under a general frame; then we obtain the Riemannian metric

$$ds^2 = \langle d\boldsymbol{r}, d\boldsymbol{r}\rangle= g_{\mu\nu}dx^{\mu} dx^{\nu} \tag{13} $$

where

$$g_{\mu\nu} = \langle \boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\rangle \tag{14} $$

might be a matrix with complicated function entries. Writing the Riemannian metric in matrix form,

$$g_{\mu\nu}dx^{\mu} dx^{\nu}=d\boldsymbol{x}^T \boldsymbol{g}d\boldsymbol{x} \tag{15} $$

we then try to perform the following decomposition

$$\boldsymbol{g}=\boldsymbol{h}^T \boldsymbol{\eta}\boldsymbol{h} \tag{16} $$

where $\boldsymbol{h},\boldsymbol{\eta}$ are matrices of the same shape as $\boldsymbol{g}$, so that

$$ds^2 = (\boldsymbol{h}d\boldsymbol{x})^T\boldsymbol{\eta}(\boldsymbol{h}d\boldsymbol{x}) \tag{17} $$

Written in component form this reads

$$ds^2 = \eta_{\mu\nu}(h_{\alpha}^{\mu} dx^{\alpha} )(h_{\beta}^{\nu} dx^{\beta}) \tag{18} $$

Let us denote

$$\omega^{\mu} = h_{\alpha}^{\mu} dx^{\alpha} \tag{19} $$

In fact $h_{\alpha}^{\mu}$ is exactly a transformation matrix that carries the original arbitrary frame $\{\boldsymbol{e}_{\mu}\}$ over to the (moving) orthogonal frame $\{\hat{\boldsymbol{e}}_{\mu}\}$, i.e.

$$\hat{\boldsymbol{e}}_{\mu} = \boldsymbol{e}_{\alpha}(h^{-1})^{\alpha}_{\mu} , \quad \boldsymbol{e}_{\mu} = \hat{\boldsymbol{e}}_{\alpha} h^{\alpha}_{\mu} \tag{20} $$

We then have

$$d\boldsymbol{r} = \boldsymbol{e}_{\mu} dx^{\mu} =\hat{\boldsymbol{e}}_{\mu} \omega^{\mu} \tag{21} $$

which shows that $\omega^{\mu}$ in the orthogonal frame plays the role of $dx^{\mu}$ in the general frame, together with

$$ds^2 = \eta_{\mu\nu} \omega^{\mu} \omega^{\nu} \tag{22} $$

The above shows that an orthogonal frame helps simplify the Riemannian metric, since the metric tensor is now the simpler $\eta_{\mu\nu}$. It should be pointed out that the ideal decomposition would have $\boldsymbol{\eta}$ equal to the identity matrix, but if we are considering a general Riemannian metric (especially one from general relativity) and restrict ourselves to real numbers, this ideal cannot always be achieved — for example, even the simple case $\boldsymbol{g}=\begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}$ fails to admit it. So we only ask that $\boldsymbol{\eta}$ be as simple as possible, say a constant diagonal matrix, without requiring it to be exactly the identity. Such a decomposition can always be arranged, and in many practical situations $\boldsymbol{g}$ is already a diagonal matrix, in which case this is quite easy to achieve. We therefore assume here that $\eta_{\mu\nu}$ is a diagonal matrix whose diagonal entries are 1 or -1.

Next we write

$$ds^2 = \langle d\boldsymbol{r},d\boldsymbol{r}\rangle=\langle \hat{\boldsymbol{e}}_{\mu},\hat{\boldsymbol{e}}_{\nu}\rangle \omega^{\mu}\omega^{\nu} \tag{23} $$

that is,

$$\eta_{\mu\nu} = \langle \hat{\boldsymbol{e}}_{\mu},\hat{\boldsymbol{e}}_{\nu}\rangle \tag{24} $$

Here $\hat{\boldsymbol{e}}$ is precisely the "orthogonal frame" in the sense of the equation above.