It is not suggested that there is anything incorrect in the principle of least action as used in classical mechanics. The break-down occurs when we attempt to generalise it for variations of the state of the system beyond those hitherto contemplated. Indeed it is obvious that the principle must break down if pressed to extreme generality. We may discriminate \Item{(a)}~possible states of the world, \Item{(b)}~states which although impossible are contemplated, \Item{(c)}~impossible states which are not contemplated. Generalisation of the principle consists in transferring states from class~\Item{(c)} to class~\Item{(b)}; there must be some limit to this, for otherwise we should find ourselves asserting that the equation $\delta A \neq 0$ is not merely not a possible equation but also not even an impossible equation. \PageSep{140} \Section{61.}{A property of invariants} Let $K$~be any invariant function of the~$g_{\mu\nu}$ and their derivatives up to any order, so that \[ \int K \sqrt{-g}\, d\tau\quad\text{is an invariant.} \] The small variations $\delta(K \sqrt{-g})$ can be expressed as a linear sum of terms involving $\delta g_{\mu\nu}$, $\delta(\dd g_{\mu\nu}/dx_{\alpha})$, $\delta(\dd^{2} g_{\mu\nu}/dx_{\alpha}\, dx_{\beta})$, etc. By the usual method of partial integration employed in the calculus of variations, these can all be reduced to terms in~$\delta g_{\mu\nu}$, together with complete differentials. Thus for variations which vanish at the boundary of the region, we can write \[ \delta \int K \sqrt{-g}\, d\tau = \int P^{\mu\nu}\, \delta g_{\mu\nu}\, \sqrt{-g}\, d\tau, \Tag{(61.1)} \] where the coefficients, here written~$P^{\mu\nu}$, can be evaluated when the analytical expression for~$K$ is given. The complete differentials yield surface-integrals over the boundary, so that they do not contribute to the variations. In accordance with our previous notation \Eq{(60.43)}, we have \[ P^{\mu\nu} = \frac{\Ham K}{\Ham g_{\mu\nu}}. \Tag{(61.2)} \] We take $P^{\mu\nu}$ to be symmetrical in $\mu$ and~$\nu$, since any antisymmetrical part would be meaningless owing to the inner multiplication by~$\delta g_{\mu\nu}$. Also since $\delta g_{\mu\nu}$ is an arbitrary tensor $P^{\mu\nu}$~must be a tensor. Consider the case in which the $\delta g_{\mu\nu}$ arise merely from a transformation of coordinates. Then \Eq{(61.1)}~vanishes, not from any stationary property, but because of the invariance of~$K$. The $\delta g_{\mu\nu}$ are not now arbitrary independent variations, so that it does not follow that $P^{\mu\nu}$~vanishes.