Here we will demonstrate how powerful the method from the previous section is for calculating the Riemann curvature tensor! We list all the formulas we obtained once again. First, the conceptual ones: \begin{aligned} &\omega^{\mu}=h_{\alpha}^{\mu}dx^{\alpha}\\ &d\boldsymbol{r}=\hat{\boldsymbol{e}}_{\mu} \omega^{\mu}\\ &ds^2 = \eta_{\mu\nu} \omega^{\mu}\omega^{\nu}\\ &\langle \hat{\boldsymbol{e}}_{\mu}, \hat{\boldsymbol{e}}_{\nu}\rangle = \eta_{\mu\nu} \end{aligned} \tag{65} Then: \begin{aligned} &d\eta_{\mu\nu}=\omega_{\nu\mu}+\omega_{\mu\nu}=\eta_{\nu\alpha}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}\\ &d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0 \end{aligned} \tag{66} These two help us determine \omega_{\nu}^{\mu}. Next is: \mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha} \tag{67} Finally, if you want the components \hat{R}^{\mu}_{\nu\beta\gamma} in the orthonormal frame, you write: \mathscr{R}_{\nu}^{\mu}=\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma} \tag{68} If you want the components R^{\mu}_{\nu\beta\gamma} in the original frame, you write: (h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'} = \sum_{\beta < \gamma} R^{\mu}_{\nu\beta\gamma}dx^{\beta}\land dx^{\gamma} \tag{69} Then read off R^{\mu}_{\nu\beta\gamma} in sequence, as if filling out a table.
A Two-Dimensional Example: The Sphere
Let’s warm up with a two-dimensional example. We will calculate the Riemann curvature tensor for a sphere with ds^2 = d\theta^2 + \sin^2 \theta d\phi^2.
We choose: \omega^1 = d\theta, \quad \omega^2 = \sin\theta d\phi \tag{70} Which means: \boldsymbol{h}=\begin{pmatrix}1&0\\0&\sin\theta\end{pmatrix},\quad \boldsymbol{\eta}=\begin{pmatrix}1&0\\0&1\end{pmatrix} \tag{71} Since \boldsymbol{\eta} is the identity matrix, d\eta_{\mu\nu}=\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha} tells us that \omega_{\nu}^{\mu} is an antisymmetric matrix. We write d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0 in matrix form: \begin{pmatrix} 0 & \omega_2^1 \\ -\omega_2^1 & 0 \end{pmatrix}\land \begin{pmatrix} d\theta \\ \sin\theta d\phi \end{pmatrix}=-d\begin{pmatrix} d\theta \\ \sin\theta d\phi \end{pmatrix}=-\begin{pmatrix} 0 \\ \cos\theta d\theta\land d\phi \end{pmatrix} \tag{72} Due to antisymmetry, \omega_{\nu}^{\mu} has only one independent component. It is not difficult to find that \omega_2^1=-\cos\theta d\phi. This solving process can be done by inspection and trial. Next, we find \mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha}, i.e.: \begin{aligned}\mathscr{R}_{\nu}^{\mu} = &d\begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix}\\ &+\begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix}\land \begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix} \end{aligned} \tag{73} The matrix multiplication term is clearly 0. In fact, it can be proven that in any 2-dimensional space, \omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha} is always 0. Therefore: \mathscr{R}_{\nu}^{\mu} = d\begin{pmatrix} 0 & -\cos\theta d\phi \\ \cos\theta d\phi & 0 \end{pmatrix}=\begin{pmatrix} 0 & \sin\theta d\theta\land d\phi \\ -\sin\theta d\theta\land d\phi & 0 \end{pmatrix} \tag{74} And since: \mathscr{R}_{\nu}^{\mu}=\sum_{\beta < \gamma} \hat{R}^{\mu}_{\nu\beta\gamma}\omega^{\beta}\land \omega^{\gamma}=\hat{R}^{\mu}_{\nu 1 2 }\sin\theta d\theta \land d\phi \tag{75} By comparison, we have: \hat{R}^{\mu}_{\nu 1 2 } = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \tag{76} That is, in the orthonormal frame, we have \hat{R}^{1}_{2 1 2 } = 1, \hat{R}^{2}_{1 1 2 } = -1. Then, according to (h^{-1})_{\mu'}^{\mu}\mathscr{R}^{\mu'}_{\nu'}h_{\nu}^{\nu'} = \sum_{\beta < \gamma} R^{\mu}_{\nu\beta\gamma}dx^{\beta}\land dx^{\gamma}, we calculate R^{\mu}_{\nu\beta\gamma}: \begin{aligned}&\begin{pmatrix}1&0\\0&\sin\theta\end{pmatrix}^{-1}\begin{pmatrix} 0 & \sin\theta d\theta\land d\phi \\ -\sin\theta d\theta\land d\phi & 0 \end{pmatrix}\begin{pmatrix}1&0\\0&\sin\theta\end{pmatrix}\\ =&R^{\mu}_{\nu 12}d\theta\land d\phi\end{aligned} \tag{77} Which is: R^{\mu}_{\nu 1 2 } = \begin{pmatrix} 0 & \sin^2\theta \\ -1 & 0 \end{pmatrix} \tag{78} That is, R^{1}_{2 1 2 } = \sin^2\theta, R^{2}_{1 1 2 } = -1. The entire process only involves matrix multiplication, which we are familiar with, and it saves a tremendous amount of effort compared to the tedious summation over multiple indices. Comparing the forms of \hat{R}^{\mu}_{\nu\beta\gamma} and R^{\mu}_{\nu\beta\gamma}, one can also see that the orthonormal frame indeed plays a simplifying role.
A Four-Dimensional Example: The Schwarzschild Metric
The first exact solution to Einstein’s field equations is the Schwarzschild metric, which is obtained by solving for a metric of the following form: ds^2= -e^{2\Phi}dt^2 + e^{2\Lambda} dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2 \tag{79} Its starting point is to consider an isotropic metric, so \Phi and \Lambda are assumed to be functions of r only. Let’s calculate the Riemann curvature tensor for this case (Intermediate-Advanced difficulty).
Naturally, we choose: \omega^1 = e^{\Phi}dt, \quad \omega^2 = e^{\Lambda }dr, \quad \omega^3 = rd\theta, \quad \omega^4 = r\sin\theta d\phi \tag{80} In this case: \boldsymbol{h}=\begin{pmatrix}e^{\Phi}&0&0&0\\0&e^{\Lambda }&0&0\\0&0&r&0\\0&0&0&r\sin\theta\end{pmatrix},\quad \boldsymbol{\eta}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} \tag{81} From d\eta_{\mu\nu}=\eta_{\alpha \nu}\omega_{\mu}^{\alpha}+\eta_{\mu \alpha}\omega_{\nu}^{\alpha}, we know that \omega_{\mu\nu}=\omega_{\mu}^{\alpha}\eta_{\alpha \nu} is antisymmetric. Combined with the form of \boldsymbol{\eta}, a brief analysis shows that \omega_{\nu}^{\mu} has the following form: \begin{pmatrix}0&\omega_2^1&\omega_3^1&\omega_4^1\\ \omega_2^1&0&\omega_3^2&\omega_4^2\\ \omega_3^1&-\omega_3^2&0&\omega_4^3\\ \omega_4^1&-\omega_4^2&-\omega_4^3&0\end{pmatrix} \tag{82} Its characteristic is that, viewed as a block matrix: \left(\begin{array}{c:ccc}0&\omega_2^1&\omega_3^1&\omega_4^1\\ \hdashline \omega_2^1&0&\omega_3^2&\omega_4^2\\ \omega_3^1&-\omega_3^2&0&\omega_4^3\\ \omega_4^1&-\omega_4^2&-\omega_4^3&0\end{array}\right) =\left(\begin{array}{c:c}E & F\\ \hdashline G&H\end{array}\right) \tag{83} it is symmetric as a block matrix, but the diagonal blocks E and H are antisymmetric. The specific partitioning depends on how we partition \boldsymbol{\eta} into \left(\begin{array}{c:c}-I & 0\\ \hdashline 0&I\end{array}\right), where I represents the identity matrix.
With the specific form of \omega_{\nu}^{\mu}, we can write d\omega^{\mu}+\omega_{\nu}^{\mu}\land \omega^{\nu}=0: \begin{aligned} &\begin{pmatrix}0&\omega_2^1&\omega_3^1&\omega_4^1\\ \omega_2^1&0&\omega_3^2&\omega_4^2\\ \omega_3^1&-\omega_3^2&0&\omega_4^3\\ \omega_4^1&-\omega_4^2&-\omega_4^3&0\end{pmatrix}\land \begin{pmatrix}e^{\Phi}dt\\ e^{\Lambda }dr\\rd\theta\\r\sin\theta d\phi\end{pmatrix}\\ =&-d\begin{pmatrix}e^{\Phi}dt\\ e^{\Lambda }dr\\rd\theta\\r\sin\theta d\phi\end{pmatrix}=-\begin{pmatrix}e^{\Phi} \dot{\Phi} dr\land dt\\ 0 \\dr\land d\theta\\ \sin\theta dr\land d\phi+ r\cos\theta d\theta\land d\phi \end{pmatrix} \end{aligned} \tag{84} Here \dot{} denotes the derivative with respect to r. By inspection, we can quickly determine the components. For example, since the second row is identically zero, we can conclude that \omega_2^1, \omega_3^2, \omega_4^2 are related only to dt, d\theta, d\phi respectively. Combined with the first row, we find \omega_2^1 = e^{\Phi-\Lambda}\dot{\Phi} dt, and that \omega_3^1, \omega_4^1 are related only to d\theta, d\phi. Then, from the third row, we find \omega_3^1=0 and \omega_3^2=-e^{-\Lambda}d\theta, and that \omega_4^3 is related only to d\phi. Finally, from the fourth row, we quickly find \omega_4^1=0, \omega_4^2 = -\sin\theta e^{-\Lambda} d\phi, \omega_4^3 = -\cos\theta d\phi. Thus: \omega_{\nu}^{\mu}=\begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix} \tag{85} Now we can calculate \mathscr{R}_{\nu}^{\mu} = d\omega_{\nu}^{\mu}+\omega_{\alpha}^{\mu} \land \omega_{\nu}^{\alpha}: \begin{aligned}\mathscr{R}_{\nu}^{\mu} =&d\begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix}\\ &+\begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi&\cos\theta d\phi&0\end{pmatrix}\\ &\land \begin{pmatrix}0& e^{\Phi-\Lambda}\dot{\Phi} dt & 0 & 0 \\ e^{\Phi-\Lambda}\dot{\Phi} dt &0&-e^{-\Lambda}d\theta&-\sin\theta e^{-\Lambda} d\phi\\ 0 &e^{-\Lambda}d\theta&0&-\cos\theta d\phi\\ 0&\sin\theta e^{-\Lambda} d\phi& \cos\theta d\phi&0\end{pmatrix} \end{aligned} \tag{86} Calculating these gives: \begin{aligned} &\mathscr{R}_1^1=\mathscr{R}_2^2=\mathscr{R}_3^3=\mathscr{R}_4^4=0\\ &\mathscr{R}^1_2=\mathscr{R}^2_1=-e^{\Phi-\Lambda}(\ddot{\Phi}+\dot{\Phi}^2-\dot{\Phi}\dot{\Lambda}) dt\land dr\\ &\mathscr{R}^1_3=\mathscr{R}^3_1=-e^{\Phi-2\Lambda} \dot{\Phi} dt\land d\theta\\ &\mathscr{R}^1_4=\mathscr{R}^4_1=-e^{\Phi-2\Lambda} \dot{\Phi} \sin\theta dt\land d\phi\\ &\mathscr{R}^2_3=-\mathscr{R}^3_2=e^{-\Lambda}\dot{\Lambda} dr\land d\theta\\ &\mathscr{R}^2_4=-\mathscr{R}^4_2=e^{-\Lambda}\dot{\Lambda}\sin\theta dr\land d\phi\\ &\mathscr{R}^3_4=-\mathscr{R}^4_3=(1-e^{2\Lambda})\sin\theta d\theta\land d\phi \end{aligned} \tag{87} From these, we can read off the components \hat{R}^{\mu}_{\nu\beta\gamma} in sequence, for example: \hat{R}^{1}_{212}=-e^{-2\Lambda}(\ddot{\Phi}+\dot{\Phi}^2-\dot{\Phi}\dot{\Lambda}), \quad\hat{R}^{1}_{312}=-\frac{1}{r}e^{-2\Lambda}\dot{\Phi} \tag{88} and so on. If you wish, you can continue to find R^{\mu}_{\nu\beta\gamma}. Since \boldsymbol{h} is a diagonal matrix, this does not add much work.
If readers perform the calculations themselves, they might still complain about the time required and feel that there is no simplification. However, by following the steps above one by one, it is manageable even by hand. This is at least a method that can be calculated manually; the process above was calculated by the author by hand, without using software like Mathematica. I believe no one has calculated the Riemann curvature tensor for more than three dimensions from the original expression R^{\mu}_{\nu\beta\gamma}=\frac{\partial \Gamma^{\mu}_{\nu\gamma}}{\partial x^{\beta}}-\frac{\partial \Gamma^{\mu}_{\nu\beta}}{\partial x^{\gamma}}+\Gamma^{\mu}_{\alpha\beta}\Gamma^{\alpha}_{\nu\gamma}-\Gamma^{\mu}_{\alpha\gamma}\Gamma^{\alpha}_{\nu\beta}? Not to mention the calculation itself, even keeping track of the summation indices is not easy. In comparison, the techniques of exterior calculus are much more effective. Of course, regardless of the method, it first requires some training to become familiar; second, even after becoming familiar, it takes some time and thought to work it out—it is impossible to see the answer at a glance, unless you are a computer.
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