The reader may feel that there is still some doubt as to the rigour of this justification of the neglect of~$m^{2}$\footnotemark.\footnotetext {To illustrate the difficulty, what exactly does $\rho_{0}$ mean, assuming that it is not defined by \Eq{(46.6)} and~\Eq{(46.7)}? If the particles do not interfere with each other's fields, $\rho_{0}$~is $\sum m$~per unit volume; but if we take account of the interference, $m$~is undefined---it is the constant of integration of an equation which does not apply. Mathematically, we cannot say what $m$~would have been if the other particles had been removed; the question is nonsensical. Physically we could no doubt say what would have been the masses of the atoms if widely separated from one another, and compare them with the gravitational power of the atoms under actual conditions; but that involves laws of atomic structure which are quite outside the scope of the argument.} Lest he attach too great importance to the matter, we may state at once that the subsequent developments will not be based on this investigation. In the next chapter we shall arrive at the same formulae by a different line of argument, and proceed in the reverse direction from the laws of continuous matter to the particular case of an isolated particle. The equation~\Eq{(46.2)} is a useful expression for the gravitational field due \index{Stress-system!gravitational field due to}% to a static distribution of mass. It is only a first approximation correct to the order~$m/r$, but \emph{no second approximation exists} except in the case of a solitary particle. This is because when more than one particle is present accelerations necessarily occur, so that there cannot be an exact solution of Einstein's equations corresponding to a number of particles continually at rest. It follows that any constraint which could keep them at rest must necessarily be of such a nature as to contribute a gravitational field on its own account.