Let us consider an empty region of the world, and try to create in it one or more particles of small mass~$\delta m$ by variations of the~$g_{\mu\nu}$ within the region. By \Eq{(60.12)} and~\Eq{(60.2)}, \[ \delta \int G\sqrt{-g}\, d\tau = 8\pi \sum \delta m · ds, \Tag{(63.1)} \] and by~\Eq{(60.42)} the left-hand side is zero because the space is initially empty. In the actual world particles for which $\delta m · ds$ is negative do not exist; hence it is impossible to create any particles in an empty region, so long as we adhere to the condition that the~$g_{\mu\nu}$ and their first derivatives must not be varied on the boundary. To permit the creation of particles we must give up this restriction and accordingly resurrect the term \[ \delta \int G\sqrt{-g}\, d\tau = \int \frac{\dd}{\dd x_{\alpha}} \left(\mf{g}^{\mu\nu}\, \delta\left(\frac{\dd\mf{L}}{\dd\mf{g}_{\alpha}^{\mu\nu}}\right)\right) d\tau, \Tag{(63.2)} \] which was discarded from~\Eq{(60.3)}. On performing the first integration, \Eq{(63.2)}~gives the flux of the normal component of \[ \mf{g}^{\mu\nu}\, \delta\left(\frac{\dd\mf{L}}{\dd\mf{g}_{\alpha}^{\mu\nu}}\right) = g^{\mu\nu} \sqrt{-g}\, \delta\bigl[-\{\mu\nu, \alpha\} + g_{\mu}^{\alpha} \{\nu\beta, \beta\}\bigr] \Tag{(63.3)} \] across the three-dimensional surface of the region. The close connection of this expression with the value of~$\mf{t}_{\mu}^{\nu}$ in~\Eq{(59.6)} should be noticed. Take the region in the form of a long tube and create a particle of gravitational mass~$\delta m$ along its axis. The flux~\Eq{(63.3)} is an invariant, since $\delta m · ds$ is invariant, so we may choose the special coordinates of \SecRef{38} for which the particle is at rest. Take the tube to be of radius~$r$ and calculate the flux for a length of tube $dt = ds$. The normal component of~\Eq{(63.3)} is given by $\alpha = 1$ and accordingly the flux is \begin{multline*} \int g^{\mu\nu} \sqrt{-g}\, \delta\bigl[-\{\mu\nu, 1\} + g_{\mu}^{1} \{\nu\beta, \beta\}\bigr]\, d\theta\, d\phi\, dt \\ = 4\pi r^{2}\, ds · \left[ -g^{\mu\nu}\, \delta\{\mu\nu, 1\} + g^{\mu1}\, \delta \left(\frac{\dd}{\dd x_{\nu}} \log \sqrt{-g}\right)\right], \Tag{(63.4)} \end{multline*} \PageSep{145} which by~\Eq{(38.5)} %[** TN: Re-breaking] \begin{multline*} = 4\pi r^{2}\, ds\, \biggl\{e^{-\lambda}\, \delta(\tfrac{1}{2}\lambda') - \frac{1}{r^{2}}\, \delta(re^{-\lambda}) - \frac{1}{r^{2} \sin^{2}\theta}\, \delta(r\sin^{2}\theta\, e^{-\lambda}) \\ - e^{-\nu}\, \delta(\tfrac{1}{2} e^{\nu - \lambda} \nu') - e^{-\lambda}\, \delta\left(\frac{2}{r}\right)\biggr\}. \Tag{(63.5)} \end{multline*} Remembering that the variations involve only~$\delta m$, this reduces to \begin{align*} &4\pi r^{2}\, ds \left(-\delta\gamma' - \frac{2}{r}\, \delta\gamma\right) \\ &\qquad= 8\pi\, \delta m · ds. \Tag{(63.6)} \end{align*} We have ignored the flux across the two ends of the tube. It is clear that these will counterbalance one another.