【Understanding Riemannian Geometry】8. Geometry Everywhere (Geometrizing Mechanics)

【理解黎曼几何】8. 处处皆几何 (力学几何化) 苏剑林 · 2016-11-02 · Phy-chem mechanics geometry action riemannian geometry

Riemannian geometry's role and applications in general relativity, while perhaps not exactly common knowledge, are something most readers will have heard of. Whenever Riemannian geometry's application in physics comes up, the first reaction most people have is probably "general relativity." A common view holds that the discovery of general relativity greatly spurred the development of Riemannian geometry. That is indeed true — but what most people don't know is that even in classical Newtonian mechanics, traces of Riemannian geometry can already be found.

What this post will discuss is precisely how to geometrize mechanics so that Riemannian geometric concepts can be used to describe it. In fact, the entire process provides a framework that lets us fold quite a few other theories from other fields into the Riemannian geometric system.

The starting point of Riemannian geometry is the Riemannian metric, from which geodesics can be obtained via variational calculus. 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 reduced to an extremal principle, and once we have an extremal principle it is not hard to derive a variational principle (an extremum of a functional) — for example, in physics there is the principle of least action and the principle of minimum potential energy, and in probability theory there is the principle of maximum entropy, and so on. If there were a way to derive a Riemannian metric from a variational principle, then we could describe it geometrically. Fortunately, for quadratic-form variational principles, this can indeed be done. more

From the Action Principle to Riemannian Geometry

Let us consider the principle of least action in classical mechanics; to state the essential idea more clearly, we use 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 we have already assumed $m=1$, and the resulting equation of motion is

$$\frac{d^2 x}{dt^2}=-\frac{\partial U}{\partial x},\quad \frac{d^2 y}{dt^2}=-\frac{\partial U}{\partial y}\tag{71} $$

Since the system is conservative, energy conservation holds:

$$\frac{1}{2}\left[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2\right] + U=E\tag{72} $$

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 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 the point of view of variation, the term $Edt$ is a total differential and contributes no actual effect, so 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 $(72)$ once more, 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 does not depend on constant factors, so ultimately the action is equivalent to

$$ S = \int \sqrt{(E-U)(dx^2+dy^2)}\tag{78} $$

The result obtained by varying this action gives the shape of the motion curve (the 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 an isothermal parameter.

From Riemannian Geometry to the Equation of Motion

To conversely prove that the geodesics of this metric really are the shape of the motion curve, we vary equation $(78)$ and obtain

$$\begin{aligned}\delta S =& \int \sqrt{(E-U)(dx^2+dy^2)}\\ =&\int\delta \sqrt{(E-U)(dx^2+dy^2)}\\ =&\int \frac{\delta[(E-U)(dx^2+dy^2)]}{2\sqrt{(E-U)(dx^2+dy^2)}} \end{aligned}\tag{80} $$

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, there is no way to get back to classical mechanics. Therefore, here we use the time parameter from equation $(76)$, so that

$$\begin{aligned}\delta S =&\int \frac{\delta[(E-U)(dx^2+dy^2)]}{2\sqrt{2}(E-U)dt}\\ =&\int \frac{-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)(dx^2+dy^2)+2(E-U)(dx d\delta x+dy d\delta y)]}{2\sqrt{2}(E-U)dt}\\ =&\frac{1}{\sqrt{2}}\int \left[-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)\frac{dx^2+dy^2}{2(E-U)dt}+\left(\frac{dx}{dt} d\delta x+\frac{dy}{dt} d\delta y\right)\right] \end{aligned}\tag{81} $$

Using equation $(76)$ once again, and then applying integration by parts, we get

$$\begin{aligned}\delta S \sim& \int \left[-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)dt+\left(\frac{dx}{dt} d\delta x+\frac{dy}{dt} d\delta y\right)\right]\\ =&\int \left[-\left(\frac{\partial U}{\partial x}\delta x+\frac{\partial U}{\partial y}\delta y\right)dt-\left(\frac{d^2 x}{dt^2} \delta x+\frac{d^2 y}{dt^2} \delta y\right)dt\right]\\ =&-\int \left[\left(\frac{d^2 x}{dt^2}+\frac{\partial U}{\partial x}\right)\delta x dt+\left(\frac{d^2 y}{dt^2}+\frac{\partial U}{\partial y}\right)\delta y dt\right] \end{aligned}\tag{82} $$

Hence $\frac{d^2 x}{dt^2}+\frac{\partial U}{\partial x}=0,\frac{d^2 y}{dt^2}+\frac{\partial U}{\partial y}=0$, and we recover the equation of motion $(71)$, which also shows that the two really can be converted into each other.

General Result

The above result can be generalized: 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 a geodesic under the Riemannian metric

$$ds^2=[E-U(\boldsymbol{x})]g_{\mu\nu}dx^{\mu}dx^{\nu}\tag{84} $$

the derivation proceeds in a similar way. In this way we have geometrized the problem of mechanics — or in other words, geometrized the quadratic-form variational problem. What may come as a surprise is that this result was already established by Jacobi back in 1837.

The above result tells us that general relativity is no longer synonymous with "Riemannian geometry in physics" — even without general relativity, Riemannian geometry already appears in physics. Geometrizing mechanics helps us connect mechanics, field theory, and other subjects with geometry. Riemannian geometry is really just a research framework for geometry; as long as we can find the corresponding translation, we can directly apply many of its results, and this may lead to richer and more comprehensive content.

Solving for Geodesics

The above result is not merely of theoretical value — sometimes it also has practical value, for instance in helping us solve for geodesic equations. Let us continue considering 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 equation is:

$$\begin{aligned}\frac{d^2 x}{ds^2} =& -\frac{1}{2f}\frac{\partial f}{\partial x}\left(\frac{dx}{ds}\right)^2+\frac{1}{2f}\frac{\partial f}{\partial x}\left(\frac{dy}{ds}\right)^2-\frac{1}{f}\frac{\partial f}{\partial y}\frac{dx}{ds}\frac{dy}{ds}\\ \frac{d^2 y}{ds^2} =& -\frac{1}{2f}\frac{\partial f}{\partial y}\left(\frac{dy}{ds}\right)^2+\frac{1}{2f}\frac{\partial f}{\partial y}\left(\frac{dx}{ds}\right)^2-\frac{1}{f}\frac{\partial f}{\partial x}\frac{dx}{ds}\frac{dy}{ds}\end{aligned}\tag{84}$$

Except in some very special cases, solving this equation is not easy, even for a special case as simple as $f(x,y)=\frac{1}{2}(x^2+y^2)$. However, based on what we explored above, we know we can use the time parameter

$$dt=\sqrt{\frac{dx^2+dy^2}{2f(x,y)}}\tag{85}$$

to make the system equivalent to the phase trajectory, at energy $E=0$, of a conservative system with potential energy $U=-f(x,y)$, so that 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 equation. In this case, the geodesic for the example $f(x,y)=\frac{1}{2}(x^2+y^2)$ that we chose turns out to be nothing more than two already-separated, fully solvable linear differential equations.