Curvature

Abstract

The Riemann curvature tensor is introduced, and the expression of its components in terms of the derivatives of the metric and the structure coefficients is deduced. Tidal forces are discussed in a relativistic context, and it is pointed out that the relativistic gravitational field has both a non-tidal component due to the motion of the reference frame and a tidal component due to spacetime curvature.

Exercises

• 6.1 Parallel transport and curvature

  • (a) A curve $P(\lambda )$ runs through a point $P = P\left( 0 \right)$, and a vector $\vec{A}$ is defined at this point. The vector is parallel transported along the curve so that in each point $P(\lambda )$ there is a well-defined vector $\vec{A}(\lambda )$ . Express the condition that the vectors along the curve are parallel as an equation of the components of the vector $A^{\mu } (\lambda )$. Show that the change of the components of the vector by an infinitesimal displacement ${\text{d}}x^{\mu }$ is

    $$ {\text{d}}A^{\mu } = - {\Gamma}_{\lambda \nu }^{\mu } (x)A^{\lambda } {\text{d}}x^{\nu } . $$

  • (b) A closed curve has the shape of a parallelogram with the sides ${\text{d}}\vec{a}$ and ${\text{d}}\vec{b}$. The corners of the parallelogram are denoted by A, B, C and D, respectively. A vector $\vec{A}$ is parallel transported from A and C along the two curves ABC and ADC. Show that the result in these two cases is in general not the same. Then use this fact to show that the change of $\vec{A}$, by parallel transporting it along the closed curve ABCDA, is

    $$ \delta A^{\alpha } = - R_{\beta \gamma \delta }^{\alpha } A^{\beta } {\text{d}}a^{\gamma } {\text{d}}b^{\delta } , $$

where $R_{\beta \gamma \delta }^{\alpha }$ is the Riemann curvature tensor .

• 6.2 Curvature of the simultaneity space in a rotating reference frame

Calculate the curvature scalar R of a 2-dimensional simultaneity space in a rotating reference frame with the line-element

$$ {\text{d}}l^{2} = {\text{d}}r^{2} + \frac{{r^{2} {\text{d}}\theta^{2} }}{{1 - r^{2} \omega^{2} /c^{2} }}. $$

• 6.3 The tidal force pendulum and the curvature of space

We will again consider the tidal force pendulum. Here we shall use the equation for geodesic deviation to find the period of the pendulum.

• (a) Why can the equation for geodesic equation be used to find the period of the pendulum in spite of the fact that the particles do not move along geodesics? Assume that the centre of the pendulum is fixed at a distance R from the centre of mass of the Earth. Introduce an orthonormal basis $\{ {\mathbf{e}}_{{\hat{a}}} \}$ with the origin at the centre of the pendulum (see Fig. 6.7).

  Fig. 6.7

  

  Spacetime curvature and the tidal force pendulum. Here $a = R_{{\hat{\phi }\hat{t}\hat{\phi }\hat{t}}} \ell^{{\hat{\phi }}} \cos \theta - R_{{\hat{r}\,t\,\hat{r}\,\hat{t}}} \ell^{{\hat{r}}} \sin \theta ,\quad \vec{\ell } = \ell \sin \theta \,\vec{e}_{{\hat{\phi }}} + \ell \cos \theta \,\vec{e}_{{\hat{r}}}$

• (b) Show to first order in $v/c$ and $\phi /c^{2}$, where v is the 3-velocity of the masses and $\phi$ the gravitational potential at the position of the pendulum, that the equation of geodesic deviation takes the form

  $$ \frac{{{\text{d}}^{2} \ell^{{\hat{i}}} }}{{{\text{d}}t^{2} }} + R^{{\hat{i}}}_{{\hat{0}\hat{j}\hat{0}}} \ell^{{\hat{j}}} = 0. $$

• (c) Find the period of the pendulum expressed in terms of the components of Riemann’s curvature tensor.