By 苏剑林 | October 18, 2016
Now we turn our attention to Riemann curvature. Overall, Riemann curvature provides a scheme that allows people living inside a space to calculate the degree of curvature of the space they inhabit. As the saying goes, "One does not know the true face of Mount Lu, only because one is inside the mountain," and "The observer sees clearly, while the participant is mired in confusion." Therefore, being able to discover the curvature of a space while being inside it is a remarkable feat, as if we have transcended our existing space and gone to a higher-dimensional space to look down from above. Truly, "As far as the heart can reach, that is how far the road and the world extend."
If we look from the perspective of a higher-dimensional space, it is easy to spot the curvature of a space. For example, a geodesic in a curved space, when viewed from a higher-dimensional space, is a curve for which curvature can be calculated. However, in the original space, it is "straight"—the geodesic is a generalization of the concept of a straight line. Thus, it is impossible to discover the curvature of space through this path directly; some indirect routes must be taken. It might not be immediately obvious, but once various paths converge to the same result, it feels inevitable.
How can we better derive the Riemann curvature so that it clearly reflects the fundamental difference between curved space and flat space? I have pondered this for a long time and consulted many reference books (Gravity and Spacetime, Field Theory, Gravitation, etc.), comparing several ways to derive the Riemann curvature, which are summarized below.
In general tensor analysis or Riemannian geometry textbooks, the way Riemann curvature is derived is by considering the difference in the order of second-order covariant derivatives: $$A^{\mu}_{;\alpha;\beta}-A^{\mu}_{;\beta;\alpha}=-R^{\mu}_{\nu\alpha\beta}A^{\nu} \tag{40} $$ From this, the Riemann curvature tensor $R^{\mu}_{\nu\alpha\beta}$ can be isolated. This is indeed a straightforward path, but its geometric meaning is not very apparent; it is difficult to see how it reflects whether a space is curved or flat. Moreover, we have not yet defined the second-order covariant derivative (a single covariant derivative adds an index, essentially making it a matrix rather than a vector; defining higher-order covariant derivatives requires further detail). Since this definition is basically pure algebraic calculus and holds little significance for us at the moment, we will not define it here. Readers can refer directly to textbooks, so we will not discuss this scheme further.
Alternatively, Riemann curvature can be derived through geodesic deviation (which corresponds to tidal forces in General Relativity). This is a scheme with clear geometric and physical meaning, but the calculations involved are quite cumbersome. The main idea is: consider the geodesic equation $$\frac{d^2 x^{\mu} }{ds^2}+\Gamma_{\alpha\beta}^{\mu}(x) \frac{d x^{\alpha} }{ds}\frac{d x^{\beta} }{ds}=0 \tag{41} $$ Suppose there is another geodesic $x(s)+\delta x(s)$, which satisfies the equation $$\frac{d^2 (x^{\mu} + \delta x^{\mu}) }{ds^2}+\Gamma_{\alpha\beta}^{\mu}(x+\delta x) \frac{d (x^{\alpha}+\delta x^{\alpha}) }{ds}\frac{d (x^{\beta}+\delta x^{\beta}) }{ds}=0 \tag{42} $$ Assuming $\delta x$ and $d\delta x/ds$ are infinitesimal quantities, subtracting the two equations yields $$\frac{d^2 \delta x^{\mu}}{ds^2}+\frac{\partial \Gamma_{\alpha\beta}^{\mu}}{\partial x^{\nu}}\delta x^{\nu} \frac{d x^{\alpha} }{ds}\frac{d x^{\beta}}{ds}+2\Gamma_{\alpha\beta}^{\mu}\frac{d \delta x^{\alpha} }{ds}\frac{d x^{\beta}}{ds}=0 \tag{43} $$ Where $\delta x$ is called the geodesic deviation, or in Riemannian geometry, it is called the "Jacobi vector field." The above form is simple enough; however, we tend to write it in terms of covariant derivatives, as the covariant derivative is the proper derivative in curved space. We previously defined the derivative along a geodesic $\frac{DA^{\mu}}{Ds}$. Repeating this once allows us to obtain the second-order derivative along the geodesic $\frac{D^2 A^{\mu}}{Ds^2}=\frac{D}{Ds}\left(\frac{DA^{\mu}}{Ds}\right)$. This is easily achievable, but since we are not particularly focused on this scheme here, we will not write out the specific form of $\frac{D^2 A^{\mu}}{Ds^2}$; readers can derive it themselves. After calculation, it is found that: $$\frac{D^2 \delta x^{\mu}}{Ds^2}=-R^{\mu}_{\nu\alpha\beta}\delta x^{\alpha}\frac{dx^{\nu}}{ds}\frac{dx^{\beta}}{ds} \tag{44} $$ This brings us to the curvature tensor $R^{\mu}_{\nu\alpha\beta}$. Mathematically, a non-zero $R^{\mu}_{\nu\alpha\beta}$ actually indicates an uneven distribution of geodesics, which is one manifestation of curved space.
This scheme reminds me of Jimmy Liao's comic Turn Left, Turn Right, which tells of a man and a woman who are used to walking left and walking right respectively, so it seems they will never meet. But one day they meet at a circular fountain—walking away from each other at one end of the circle, they eventually meet at the other end. In curved space, such as on a sphere, even two parallel lines have the chance to intersect. This actually shows that "curvature" is more profound and interesting; it gives our world more possibilities.
Finally, there is a scheme that analyzes the changes in a vector after parallel transport along a closed curve, which we will analyze in detail here. In fact, it is equivalent to geodesic deviation, but its geometric meaning is clearer and helps derive more profound results. It shows that if a vector "strolls" around a loop and returns, it is not necessarily the same as the original vector. The following example clearly illustrates this point.
Parallel Transport
Suppose there is an arbitrary vector $A^{\mu}$ at $x^{\mu}$. Starting from $x^{\mu}$, first shift by an infinitesimal $dx^{\mu}$, then by an infinitesimal $\delta x^{\mu}$, then by $-dx^{\mu}$, and finally by $-\delta x^{\mu}$. This means walking around an infinitesimal parallelogram and returning to the origin: $$x^{\mu} \to x^{\mu}+dx^{\mu} \to x^{\mu}+dx^{\mu}+\delta x^{\mu} \to x^{\mu}+\delta x^{\mu} \to x^{\mu}$$
Parallel Transport
We step-by-step calculate the changes in $A^{\mu}$ during the translation process. From $x^{\mu}$ to $x^{\mu}+dx^{\mu}$, $A^{\mu}$ becomes $$A^{\mu}-\Gamma^{\mu}_{\alpha\beta}(x) A^{\alpha}dx^{\beta} \tag{45} $$ Next, from $x^{\mu}+dx^{\mu}$ to $x^{\mu}+dx^{\mu}+\delta x^{\mu}$, $A^{\mu}$ becomes $$\begin{aligned}&A^{\mu}-\Gamma^{\mu}_{\alpha\beta}(x) A^{\alpha}dx^{\beta}-\Gamma^{\mu}_{\nu\gamma}(x+dx) \left[A^{\nu}-\Gamma^{\nu}_{\alpha\beta}(x) A^{\alpha}dx^{\beta}\right]\delta x^{\gamma}\\ =&A^{\mu}-\Gamma^{\mu}_{\alpha\beta}(x) A^{\alpha}dx^{\beta}-\Gamma^{\mu}_{\nu\gamma}(x) A^{\nu} \delta x^{\gamma} \\ &\quad- \frac{\partial \Gamma^{\mu}_{\nu\gamma}(x)}{\partial x^{\beta}} A^{\nu} dx^{\beta} \delta x^{\gamma} + \Gamma^{\mu}_{\nu\gamma}(x) \Gamma^{\nu}_{\alpha\beta}(x) A^{\alpha} dx^{\beta}\delta x^{\gamma} \end{aligned} \tag{46} $$ Here we only calculate accurately to second-order terms.
Similarly, if we consider the change caused by the path $x^{\mu} \to x^{\mu}+\delta x^{\mu} \to x^{\mu}+dx^{\mu}+\delta x^{\mu}$, we only need to swap $dx$ and $\delta x$: $$\begin{aligned}&A^{\mu}-\Gamma^{\mu}_{\alpha\beta}(x) A^{\alpha}\delta x^{\beta}-\Gamma^{\mu}_{\nu\gamma}(x) A^{\nu} d x^{\gamma} \\ &\quad- \frac{\partial \Gamma^{\mu}_{\nu\gamma}(x)}{\partial x^{\beta}} A^{\nu} \delta x^{\beta} d x^{\gamma} + \Gamma^{\mu}_{\nu\gamma}(x) \Gamma^{\nu}_{\alpha\beta}(x) A^{\alpha} \delta x^{\beta} d x^{\gamma}\end{aligned} \tag{47} $$ Then naturally, the change caused by the path $x^{\mu}+dx^{\mu}+\delta x^{\mu} \to x^{\mu}+\delta x^{\mu} \to x^{\mu}$ is the negative of the above term. Thus, the total change brought by the entire closed path $x^{\mu} \to x^{\mu}+dx^{\mu} \to x^{\mu}+dx^{\mu}+\delta x^{\mu} \to x^{\mu}+\delta x^{\mu} \to x^{\mu}$ is the difference between the two expressions. Adjusting the summation indices and subtracting, it is not difficult to get: \begin{equation}\label{bihelujingbianhua}\begin{aligned}\Delta A^{\mu} =&-\left(\frac{\partial \Gamma^{\mu}_{\alpha\gamma}}{\partial x^{\beta}}-\frac{\partial \Gamma^{\mu}_{\alpha\beta}}{\partial x^{\gamma}}+\Gamma^{\mu}_{\nu\beta}\Gamma^{\nu}_{\alpha\gamma}-\Gamma^{\mu}_{\nu\gamma}\Gamma^{\nu}_{\alpha\beta}\right)A^{\alpha} dx^{\beta}\delta x^{\gamma}\\ =&-R^{\mu}_{\alpha\beta\gamma} A^{\alpha} dx^{\beta}\delta x^{\gamma}\end{aligned} \tag{48} \end{equation} Here $$R^{\mu}_{\alpha\beta\gamma}=\frac{\partial \Gamma^{\mu}_{\alpha\gamma}}{\partial x^{\beta}}-\frac{\partial \Gamma^{\mu}_{\alpha\beta}}{\partial x^{\gamma}}+\Gamma^{\mu}_{\nu\beta}\Gamma^{\nu}_{\alpha\gamma}-\Gamma^{\mu}_{\nu\gamma}\Gamma^{\nu}_{\alpha\beta} \tag{49} $$ is the definition of the Riemann curvature tensor. It has four indices and is a very "grand" quantity.
Three different ways of deriving the Riemann curvature tensor demonstrate the difference between curved and flat space from three perspectives: In flat space, the order of covariant derivatives can be swapped, while in curved space it cannot; in flat space, the distribution of geodesics is uniform and linear, while in curved space it is not; in flat space, a vector returns unchanged after "strolling" in a loop, whereas in curved space, the vector may no longer be the same after its "stroll."