By 苏剑林 | November 02, 2016
The manifestation and application of Riemannian geometry in General Relativity, while perhaps not a household name, is likely something most readers have heard of. When one talks about the application of Riemannian geometry in physics, the first reflex is usually General Relativity. A common view is that the discovery of General Relativity greatly advanced the development of Riemannian geometry. While this is indeed a fact, what most people do not know is that there are traces of Riemannian geometry even in classical Newtonian mechanics. The content of this article discusses how to geometrize mechanics, thereby using the concepts of Riemannian geometry to describe it. This process essentially provides a framework that can incorporate theories from many other fields into the system of Riemannian geometry.
The starting point of Riemannian geometry is the Riemannian metric, from which geodesics can be obtained through variation. In this sense, a Riemannian metric provides a variational principle. Conversely, can a variational principle provide a Riemannian metric? As is well known, the foundational principles of many disciplines can be summarized as an extremum principle, and once you have an extremum principle, it is not difficult to derive a variational principle (extremum of a functional), such as the principle of least action or the principle of minimum potential energy in physics, or the principle of maximum entropy in probability theory, and so on. If there is a method to derive a Riemannian metric from a variational principle, then it can be described in a geometric way. Fortunately, for variational principles of quadratic forms, this is achievable.
From the Principle of Action to Riemannian Geometry
Let us consider the principle of least action in classical mechanics. To illustrate the point more clearly, we will use a two-dimensional system as an example. The trajectory of a two-dimensional conservative system is the extremum curve of the following action:
$$S = \int \left\{\frac{1}{2}\left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right]-U(x,y)\right\}dt\tag{70}$$
Here we have assumed $m=1$. The resulting equations of motion are
$$\frac{d^2 x}{dt^2}=-\frac{\partial U}{\partial x},\quad \frac{d^2 y}{dt^2}=-\frac{\partial U}{\partial y}\tag{71}$$
Since it is a conservative system, energy conservation is satisfied:
$$\frac{1}{2}\left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right] + U=E\tag{72}$$
We can use this to eliminate the parameter $dt$ in Equation (70). Using Equation (72), we get
$$U=E-\frac{1}{2}\left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right]\tag{73}$$
Substituting this into $S$, we get:
$$S = \int \left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2-E\right]dt\tag{74}$$
From a variational perspective, the term $Edt$ is a total differential and will not produce any actual effect; therefore, the equivalent action is
$$S = \int \left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right]dt=\int \frac{dx^2+dy^2}{dt}\tag{75}$$
Using Equation (72) again, we can obtain
$$dt^2 = \frac{dx^2+dy^2}{2(E-U)}\tag{76}$$
Eliminating $dt$, we get
$$S = \int \sqrt{2(E-U)(dx^2+dy^2)}\tag{77}$$
The result of the variation is independent of constant factors, so finally the action is equivalent to
$$S = \int \sqrt{(E-U)(dx^2+dy^2)}\tag{78}$$
The result obtained from the variation of this action is the shape of the moving curve (phase trajectory), and it itself takes the form of a Riemannian metric, namely
$$ds^2 = (E-U)(dx^2+dy^2)\tag{79}$$
The result is an isothermal parameterization.
From Riemannian Geometry to the Equations of Motion
To conversely prove that the geodesics of this metric are indeed the shapes of the motion curves, we perform a variation on Equation (78), obtaining
$$\begin{aligned}\delta S =& \int \delta \sqrt{(E-U)(dx^2+dy^2)}\\
=&\int \frac{\delta[(E-U)(dx^2+dy^2)]}{2\sqrt{(E-U)(dx^2+dy^2)}}
\end{aligned}\tag{80}$$
Conventionally, we would use the natural parameter $ds=\sqrt{(E-U)(dx^2+dy^2)}$ as the parameter, but if we use the natural parameter, we cannot return to classical mechanics. Therefore, here we use the time parameter $dt$ from Equation (76); then
$$\begin{aligned}\delta S =&\int \frac{\delta[(E-U)(dx^2+dy^2)]}{2\sqrt{2}(E-U)dt}\\
=&\int \frac{-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)(dx^2+dy^2)+2(E-U)(dx d\delta x+dy d\delta y)}{2\sqrt{2}(E-U)dt}\\
=&\frac{1}{\sqrt{2}}\int \left[-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)\frac{dx^2+dy^2}{2(E-U)dt}+\left(\frac{dx}{dt} \frac{d\delta x}{dt}+\frac{dy}{dt} \frac{d\delta y}{dt}\right)dt\right]
\end{aligned}\tag{81}$$
Using Equation (76) again and then applying integration by parts, we get
$$\begin{aligned}\delta S \sim& \int \left[-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)dt+\left(\frac{dx}{dt} d\delta x+\frac{dy}{dt} d\delta y\right)\right]\\
=&\int \left[-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)dt-\left(\frac{d^2 x}{dt^2} \delta x+\frac{d^2 y}{dt^2} \delta y\right)dt\right]\\
=&-\int \left[\left(\frac{d^2 x}{dt^2}+\frac{\partial U}{\partial x}\right)\delta x dt+\left(\frac{d^2 y}{dt^2}+\frac{\partial U}{\partial y}\right)\delta y dt\right]
\end{aligned}\tag{82}$$
Therefore $\frac{d^2 x}{dt^2}+\frac{\partial U}{\partial x}=0, \frac{d^2 y}{dt^2}+\frac{\partial U}{\partial y}=0$, which re-derives the equations of motion (71), indicating that the two can indeed be converted into each other.
General Result
The above results can be generalized; that is, for a conservative system with the following action:
$$S = \int \left[\frac{1}{2}g_{\mu\nu} \frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}-U(\boldsymbol{x})\right]dt\tag{83}$$
the shape of the motion curve (phase trajectory) at energy $E$ is equivalent to the geodesic under the Riemannian metric
$$ds^2=[E-U(\boldsymbol{x})]g_{\mu\nu}dx^{\mu}dx^{\nu}\tag{84}$$
The derivation process is similar. In this way, we have achieved the geometrization of mechanical problems, or rather, the geometrization of variational problems of quadratic forms. What might be surprising is that the above results were completed by Jacobi as early as 1837.
The result above tells us that General Relativity is no longer the sole representative of Riemannian geometry in physics. Even without General Relativity, Riemannian geometry exists in physics. Geometrizing mechanics helps us link mechanics, field theory, etc., with geometry. Riemannian geometry is essentially a framework for geometric research; as long as a corresponding conversion can be made, many conclusions of Riemannian geometry can be applied directly, potentially leading to richer and more comprehensive content.
Solving for Geodesics
The aforementioned results possess not only theoretical value but sometimes practical value as well, such as helping us find geodesic equations. Let us continue to consider the isothermal parameter $ds^2 = f(x,y)(dx^2+dy^2)$. Under the natural parameter $ds=\sqrt{f(x,y)(dx^2+dy^2)}$, its geodesic equations are:
$$\begin{aligned}\frac{d^2 x}{ds^2} =& -\frac{1}{2f}\frac{\partial f}{\partial x}\left(\frac{dx}{ds}\right)^2+\frac{1}{2f}\frac{\partial f}{\partial x}\left(\frac{dy}{ds}\right)^2-\frac{1}{f}\frac{\partial f}{\partial y}\frac{dx}{ds}\frac{dy}{ds}\\
\frac{d^2 y}{ds^2} =& -\frac{1}{2f}\frac{\partial f}{\partial y}\left(\frac{dy}{ds}\right)^2+\frac{1}{2f}\frac{\partial f}{\partial y}\left(\frac{dx}{ds}\right)^2-\frac{1}{f}\frac{\partial f}{\partial x}\frac{dx}{ds}\frac{dy}{ds}\end{aligned}\tag{84}$$
Except for some very special cases, solving this equation is not an easy task, even for a special case like $f(x,y)=\frac{1}{2}(x^2+y^2)$. However, according to the results we explored earlier, we know that by using the time parameter
$$dt=\sqrt{\frac{dx^2+dy^2}{2f(x,y)}}\tag{85}$$
the system can be made equivalent to a conservative system with potential energy $U=-f(x,y)$ at $E=0$. Thus the phase trajectory, which is the geodesic equation, becomes
$$\frac{d^2 x}{dt^2}=\frac{\partial f}{\partial x},\quad \frac{d^2 y}{dt^2}=\frac{\partial f}{\partial y}\tag{86}$$
This greatly simplifies the form of the geodesic equations. At this point, the geodesics for the example $f(x,y)=\frac{1}{2}(x^2+y^2)$ are merely two linear differential equations with separated variables, which are completely solvable.