18.5 Equation of geodesic deviation

中文

We now give a geometrical interpretation of the curvature tensor, which will subsequently be used in the measurement of the gravitational field (see Section 18.8). Let $\gamma :I\times J\to M.$ where $I=[t_1,t_2]$ and $J=[{\lambda }_1,{\lambda }_2]$ are closed intervals of the real line, be a one-parameter family of curves on $M$. We will assume that the restriction of the map $\gamma$ to $I^{\prime }\times J^{\prime };$ , where $I^{\prime }$ and $J^{\prime }$ are the open intervals $(t_1,t_2)$ and $({\lambda }_1,{\lambda }_2)$ respectively, is an embedded twodimensional submanifold of $M$. We think of each map ${\gamma }_{\lambda }:I\to M$ defined by ${\gamma }_{\lambda }(t)=\gamma (t,\lambda )$ as being the curve represented by $\lambda =\mathrm{const}$ . and $t$ as the parameter along the curve. The one-parameter family of curves $\gamma$ will be said to be from $p$ to $q$ if

$$p=\gamma (t_1,\lambda ),q=\gamma (t_2,\lambda )$$

for all ${\lambda }_1\le \lambda \le {\lambda }_2$

The tangent vectors to the curves of a one-parameter family constitute a vector field $X$ on the two-dimensional submanifold $\gamma (I^{\prime }\times J^{\prime })$ . If the curves are all covered by a single coordinate chart $(U;x^i)$ , then

$$X=\frac{\partial {\gamma }^i(t,\lambda )}{\partial t}\frac{\partial }{\partial x^i}\text{ where }{\gamma }^i(t,\lambda )=x^i(\gamma (t,\lambda )).$$

The connection vector field $Y$ defined by

$$Y=\frac{\partial {\gamma }^i(t,\lambda )}{\partial \lambda }\frac{\partial }{\partial x^i}$$

is the tangent vector field to the curves connecting points having the same parameter value, $t=\mathrm{const}.$ . (see Fig. 18.1). The covariant derivative of the vector field $Y$ along the curves ${\gamma }_{\lambda }$ is given by

$$\begin{array}{rl}& \frac{D{Y}^i}{\partial t}\equiv \frac{\partial Y^i}{\partial t}+{\Gamma }_{jk}^i{Y}^j{X}^k\\ & =\frac{{\partial }^2{\gamma }^i}{\partial t\partial \lambda }+{\Gamma }_{jk}^i\frac{\partial {\gamma }^j}{\partial \lambda }\frac{\partial {\gamma }^k}{\partial t}\\ & =\frac{{\partial }^2{\gamma }^i}{\partial \lambda \partial t}+{\Gamma }_{jk}^i\frac{\partial {\gamma }^k}{\partial t}\frac{\partial {\gamma }^j}{\partial \lambda }.\end{array}$$

Figure 18.1 Tangent and connection vectors of a one-parameter family of geodesics

Hence

$$\frac{D{Y}^i}{\partial t}=\frac{D{X}^i}{\partial \lambda }.\text{\hspace{1em}(18.44)}$$

Alternatively, we can write

$$D_{X}Y=D_{Y}X.\text{\hspace{1em}(18.45)}$$

If $A=A^i{\partial }_{x^i}$ is any vector field on $U$ then

$$\begin{array}{c}(\frac{D}{\partial t}\frac{D}{\partial \lambda }-\frac{D}{\partial \lambda }\frac{D}{\partial t})A^i=\frac{D}{\partial t}\left(A_{;j}^i{Y}^j\right)-\frac{D}{\partial \lambda }\left(A_{;j}^i{X}^j\right)\\ =A_{;jk}^i{Y}^j{X}^k+A_{;j}^i\frac{D{Y}^j}{\partial t}-A_{;jk}^i{X}^j{Y}^k-A_{;j}^i\frac{D{X}^j}{\partial \lambda }.\end{array}$$

From Eq. (18.44) and the Ricci identity Eq. (18.28)

$$(\frac{D}{\partial t}\frac{D}{\partial \lambda }-\frac{D}{\partial \lambda }\frac{D}{\partial t})A^i=R_{ajk}^i{A}^a{X}^j{Y}^k.\text{\hspace{1em}(18.46)}$$

Let $M$ be a pseudo-Riemannian manifold and $\gamma (\lambda ,t)$ a one-parameter family of geodesics, such that the geodesics $\lambda =\mathrm{const}$ . all have $t$ as an affine parameter,

$$D_{X}X=\frac{D{X}^i}{\partial t}=0,$$

the parametrization chosen to have the same normalization on all geodesics, $g(X,X)=\pm 1$ or 0. It then follows that $g(X,Y)$ is constant along each geodesic, since

$$\begin{array}{l}\frac{\partial }{\partial t}(g(X,Y))=D_X(g(X,Y))=(D_{X}g)(X,Y)+g(D_{X}X,Y)+g(X,D_{X}Y)\\ =g(X,D_{Y}X)\text{ by (18.45) }\\ =\frac{1}{2}{D}_Y(g(X,X))=\frac{1}{2}\frac{\partial }{\partial \lambda }g(X,X)=0.\end{array}$$

Thus, if the tangent and connection vector are initially orthogonal on a geodesic of the one-parameter family, $g(X_p,Y_p)=0$ , then they are orthogonal all along the geodesic.

In Eq. (18.46) set $A^i=X^i$ – this is possible since it is only necessary to have $A^i$ defined in terms of $t$ and $\lambda$ (see Problem 18.14). With the help of Eq. (18.44) we have

$$\frac{D}{\partial t}\frac{D{Y}^i}{\partial t}=R_{ajk}^i{X}^a{X}^j{Y}^k,\text{\hspace{1em}(18.47)}$$

known as the equation of geodesic deviation. For two geodesics, labelled by constants $\lambda$ and $\lambda +\delta \lambda$ , let $\delta {\mathbf{x}}^i$ be the tangent vector

$$\delta x^i\frac{\partial {\gamma }^i}{\partial \lambda }\delta \lambda =Y^i\delta \lambda .$$

For vanishingly small $\Delta \lambda$ it is usual to think of $\delta {\mathbf{x}}^i$ as an ‘infinitesimal separation vector’. Since $\delta \lambda$ is constant along the geodesic we have

$$\delta {\ddot{x}}^i=R_{ajk}^i{X}^a{X}^j\delta x^k\text{\hspace{1em}(18.48)}$$

where $\cdot \equiv D/\partial t$ . Thus ${R^i}_{jkl}$ measures the relative ‘acceleration’ between geodesics.

Problem

Problem 18.14

Equation (18.46) has strictly only been proved for a vector field $A$. Show that it holds equally for a vector field whose components $A^i(t,\lambda )$ ) are only defined on the one-parameter family of curves $\gamma$.