1.1 Non-Relativistic Particles

Let’s start by forgetting about special relativity and spacetime and focus instead on the non-relativistic motion of a particle in curved space. Mathematically, these spaces are known as manifolds, and the study of curved manifolds is known as Riemannian geometry. However, for much of this section we will dispense with any formal mathematical definitions and instead focus attention on the physics.

1.1.1 The Geodesic Equation

We begin with something very familiar: the non-relativistic motion of a particle of mass m in flat Euclidean space ${\mathbf{R}}^3$ . For once, the coordinates ${\mathbf{x}}^i=(x,y,z)$ actually are the usual Cartesian coordinates. The Lagrangian that describes the motion is simply the kinetic energy,

$$L=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)\text{\hspace{1em}(1.3)}$$

The Euler-Lagrange equations (1.2) applied to this Lagrangian simply tell us that ${\ddot{x}}^i=0$ , which is the statement that free particles move at constant velocity in straight lines.

Now we want to generalise this discussion to particles moving on a curved space. First, we need a way to describe curved space. We will develop the relevant mathematics in Sections 2 and 3 but here we offer a simple perspective. We describe curved spaces by specifying the infinitesimal distance between any two points, $x^i$ and $x^i+d{x}^i$ , known as the line element. The most general form is

$$d{s}^2=g_{ij}(x)d{x}^{i}d{x}^j\text{\hspace{1em}(1.4)}$$

where the $3\times 3$ matrix $g_{ij}$ is called the metric. The metric is symmetric: $g_{ij}=g_{ji}$ since the anti-symmetric part drops out of the distance when contracted with $d{x}^{i}d{x}^j$ We further assume that the metric is positive definite and non-degenerate, so that its inverse exists. The fact that $g_{ij}$ is a function of the coordinates x simply tells us that the distance between the two points $x^i$ and $x^i+d{x}^i$ depends on where you are.

Before we proceed, a quick comment: it matters in this subject whether the indices $i,j$ are up or down. We’ll understand this better in Section 2 but, for now, remember that coordinates have superscripts while the metric has two subscripts.

We’ll see plenty of examples of metrics in this course. Before we introduce some of the simpler metrics, let’s first push on and understand how a particle moves in the presence of a metric. The Lagrangian governing the motion of the particle is the obvious generalization of (1.3)

$$L=\frac{m}{2}{g}_{ij}(x){\dot{x}}^i{\dot{x}}^j\text{\hspace{1em}(1.5)}$$

It is a simple matter to compute the Euler-Lagrange equations (1.2) that arise from this action. It is really just an exercise in index notation and, in particular, making sure that we don’t inadvertently use the same index twice. Since it’s important, we proceed slowly. We have

$$\frac{\partial L}{\partial x^i}=\frac{m}{2}\frac{\partial g_{jk}}{\partial x^i}{\dot{x}}^j{\dot{x}}^k$$

where we’ve been careful to relabel the indices on the metric so that the i index matches on both sides. Similarly, we have

$$\frac{\partial L}{\partial {\dot{x}}^i}=m{g}_{ik}{\dot{x}}^k\Rightarrow \frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{x}}^i}\right)=m\frac{\partial g_{ik}}{\partial x^j}{\dot{x}}^j{\dot{x}}^k+m{g}_{ik}{\ddot{x}}^k$$

Putting these together, the Euler-Lagrange equation (1.2) becomes

$$g_{ik}{\ddot{x}}^k+\left(\frac{\partial g_{ik}}{\partial x^j}-\frac{1}{2}\frac{\partial g_{jk}}{\partial x^i}\right){\dot{x}}^j{\dot{x}}^k=0$$

Because the term in brackets is contracted with ${\dot{x}}^j{\dot{x}}^k$ , only the symmetric part contributes. We can make this obvious by rewriting this equation as

$$g_{ik}{\ddot{x}}^k+\frac{1}{2}\left(\frac{\partial g_{ik}}{\partial x^j}+\frac{\partial g_{ij}}{\partial x^k}-\frac{\partial g_{jk}}{\partial x^i}\right){\dot{x}}^j{\dot{x}}^k=0\text{\hspace{1em}(1.6)}$$

Finally, there’s one last manoeuvre: we multiply the whole equation by the inverse metric, $g^{-1}$ , so that we get an equation of the form ${\ddot{x}}^k=\dots$ We denote the inverse metric $g^{-1}$ simply by raising the indices on the metric, from subscripts to superscripts. This means that the inverse metric is denoted $g^{ij}$ . By definition, it satisfies

$$g^{ij}{g}_{jk}={\delta }_k^i$$

Finally, taking the opportunity to relabel some of the indices, the equation of motion for the particle is written as

$${\ddot{x}}^i+{\Gamma }_{jk}^i{\dot{x}}^j{\dot{x}}^k=0\text{\hspace{1em}(1.7)}$$

where

$${\Gamma }_{jk}^i(x)=\frac{1}{2}{g}^{il}\left(\frac{\partial g_{lj}}{\partial x^k}+\frac{\partial g_{lk}}{\partial x^j}-\frac{\partial g_{jk}}{\partial x^l}\right)\text{\hspace{1em}(1.8)}$$

These coefficients are called the Christoffel symbols. By construction, they are symmetric in their lower indices: ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ . They will play a very important role in everything that follows. The equation of motion (1.7) is the geodesic equation and solutions to this equation are known as geodesics.

A Trivial Example: Flat Space Again

Let’s start by considering flat space ${\mathbf{R}}^3$ . Pythagoras taught us how to measure distances using his friend, Descartes’ coordinates,

$$d{s}^2=d{x}^2+d{y}^2+d{z}^2\text{\hspace{1em}(1.9)}$$

Suppose that we work in polar coordinates rather than Cartesian coordinates. The relationship between the two is given by

$$\begin{array}{rl}& x=r\sin \theta \cos \varphi \\ & y=r\sin \theta \sin \varphi \\ & z=r\cos \theta \end{array}$$

In polar coordinates, the infinitesimal distance between two points can be simply derived by substituting the above relations into (1.9). A little algebra yields,

$$d{s}^2=d{r}^2+r^{2}d{\theta }^2+r^2{\sin }^2\theta d{\varphi }^2$$

In this case, the metric (and therefore also its inverse) are diagonal. They are

$$g_{ij}=\left(\begin{array}{ccc}1& 0& 0\\ 0& r^2& 0\\ 0& 0& r^2{\sin }^2\theta \end{array}\right)\text{and}{g}^{ij}=\left(\begin{array}{ccc}1& 0& 0\\ 0& r^{-2}& 0\\ 0& 0& (r^2{\sin }^2\theta {)}^{-1}\end{array}\right)$$

where the matrix components run over $i,j=r,\theta ,\varphi$ . From this we can easily compute the Christoffel symbols. The non-vanishing components are

$${\Gamma }_{\theta \theta }^r=-r,{\Gamma }_{\varphi \varphi }^r=-r{\sin }^2\theta ,{\Gamma }_{\theta r}^{\theta }={\Gamma }_{r\theta }^{\theta }=\frac{1}{r}$$

$${\Gamma }_{\varphi \varphi }^{\theta }=-\sin \theta \cos \theta ,{\Gamma }_{\varphi r}^{\varphi }={\Gamma }_{r\varphi }^{\varphi }=\frac{1}{r},{\Gamma }_{\theta \varphi }^{\varphi }={\Gamma }_{\varphi \theta }^{\varphi }=\frac{\cos \theta }{\sin \theta }\text{\hspace{1em}(1.10)}$$

There are some important lessons here. First, $\Gamma \ne 0$ does not necessarily mean that the space is curved. Non-vanishing Christoffel symbols can arise, as here, simply from a change of coordinates. As the course progresses, we will develop a diagnostic to determine whether space is really curved or whether it’s an artefact of the coordinates we’re using.

The second lesson is that it’s often a royal pain to compute the Christoffel symbols using (1.8). If we wished, we could substitute the Christoffel symbols into the geodesic equation (1.7) to determine the equations of motion. However, it’s typically easier to revert back to the original action and determine the equations of motion directly. In the present case, we have

$$S=\frac{m}{2}\int dt\left({\dot{r}}^2+r^2{\dot{\theta }}^2+r^2{\sin }^2\theta {\dot{\varphi }}^2\right)\text{\hspace{1em}(1.11)}$$

and the resulting Euler-Lagrange equations are

$$\ddot{r}=r{\dot{\theta }}^2+r{\sin }^2\theta {\dot{\varphi }}^2,\frac{d}{dt}(r^2\dot{\theta })=r^2\sin \theta \cos \theta {\dot{\varphi }}^2,\frac{d}{dt}(r^2{\sin }^2\theta \dot{\varphi })=0\text{\hspace{1em}(1.12)}$$

These are nothing more than the equations for a straight line described in polar coordinates. The quickest way to extract the Christoffel symbols is usually to compute the equations of motion from the action, and then compare them to the geodesic equation (1.7), taking care of the symmetry properties along the way.

A Slightly Less Trivial Example: ${\mathbf{S}}^2$

The above description of ${\mathbf{R}}^3$ in polar coordinates allows us to immediately describe a situation in which the space is truly curved: motion on the two-dimensional sphere ${\mathbf{S}}^2$ This is achieved simply by setting the radial coordinate r to some constant value, say $r=R$ . We can substitute this constraint into the action (1.11) to get the action for a particle moving on the sphere,

$$S=\frac{m{R}^2}{2}\int dt\left({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2\right)$$

Similarly, the equations of motion are given by (1.12), with the restriction $r=R$ and $\dot{r}=0$ . The solutions are great circles, which are geodesics on the sphere. To see this in general is a little complicated, but we can use the rotational invariance to aid us. We rotate the sphere to ensure that the starting point is ${\theta }_0=\pi /2$ and the initial velocity is $\dot{\theta }=0$ . In this case, it is simple to check that solutions take the form $\theta =\pi /2$ and $\varphi =\Omega t$ for some Ω, which are great circles running around the equator.

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正文：英文 · OCR 机器稿 · 待校对

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