The manifestation and application of Riemannian geometry in general relativity, while perhaps not a household name, is something most readers have likely heard of. When discussing the application of Riemannian geometry in physics, the first reaction for most is usually general relativity. A common view is that the discovery of general relativity greatly promoted the development of Riemannian geometry. While this is indeed a fact, what most people do not know is that even in classical Newtonian mechanics, the presence of Riemannian geometry can be found.
The content of this article discusses how to geometrize mechanics, thereby using the concepts of Riemannian geometry to describe them. 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, the Riemannian metric provides a variational principle. Conversely, can a variational principle provide a Riemannian metric? As is well known, the fundamental principles of many disciplines can be reduced to an extremum principle, and from an extremum principle, it is not difficult to derive a variational principle (functional extremum). For example, in physics, there is the principle of least action and the principle of minimum potential energy; in probability theory, there is the principle of maximum entropy, and so on. If there is a method to derive a Riemannian metric from a variational principle, then it can be described geometrically. 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 main point more clearly, we take a two-dimensional system as an example. The trajectory of a two-dimensional conservative system is the extremal 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, it is assumed that m=1. The derived 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, it satisfies the conservation of energy: \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 dt parameter 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 obtain: 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 Edt term is a total differential and will not bring 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 motion curve (phase trajectory), and it itself has the form of a Riemannian metric, namely: ds^2 = (E-U)(dx^2+dy^2) \tag{79} The result is in isothermal coordinates.
From Riemannian Geometry to Equations of Motion
To conversely prove that the geodesics of this metric are indeed the shape of the motion curves, we vary Equation (78) to get: \begin{aligned} \delta S =& \int \sqrt{(E-U)(dx^2+dy^2)}\\ =& \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} Customarily, 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 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} d\delta x+\frac{dy}{dt} d\delta y\right)\right] \end{aligned} \tag{81} Utilizing Equation (76) again, and then using 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 and \frac{d^2 y}{dt^2}+\frac{\partial U}{\partial y}=0. We have re-derived the equations of motion (71), which indicates that the two can indeed be converted into each other.
General Results
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) with 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. It may be surprising that the above results were completed by Jacobi as early as 1837.
The results above tell us that general relativity is no longer the sole synonym for Riemannian geometry in physics; even without general relativity, Riemannian geometry exists in physics. Geometrizing mechanics helps us connect mechanics, field theory, etc., with geometry. Riemannian geometry is actually a framework for geometric research; as long as a corresponding conversion can be made, many conclusions of Riemannian geometry can be directly applied, potentially leading to richer and more comprehensive content.
Solving Geodesics
The aforementioned results possess not only theoretical value but sometimes practical value as well, such as helping us solve 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 these equations 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 equated to a conservative system with potential energy U=-f(x,y) at E=0. That is, the geodesic equations are: \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 our example f(x,y)=\frac{1}{2}(x^2+y^2) are merely two linear differential equations with separated variables, which are completely solvable.
When reposting, please include the original address of
this article:
https://kexue.fm/archives/4046
For more detailed information regarding reposting, please
refer to:
Scientific Space FAQ