When account is taken of the stresses in continuous matter, or of the \index{Invariant density (proper-density)}% molecular motions in discontinuous matter, the proper-density of the matter \index{Density!definitions of proper-}% requires rather careful definition. There are at least three possible definitions which can be justified; and we shall denote the corresponding quantities by $\rho_{0}$, $\rho_{00}$,~$\rho_{000}$. (1) We define \[ \rho_{0} = T. \] By reference to~\Eq{(54.6)} it will be seen that this represents the sum of the densities of the particles with different motions, \emph{each particle being referred to axes with respect to which it is itself at rest}. (2) We can sum the densities for the different particles referring them all to axes which are at rest in the matter as a whole. The result is denoted by~$\rho_{00}$. Accordingly \[ \text{$\rho_{00} = T_{44}$ referred to axes at rest in the matter as a whole.} \] (3) If a perfect fluid is referred to axes with respect to which it is at rest, the stresses $p_{xx}$, $p_{yy}$, $p_{zz}$ are each equal to the hydrostatic pressure~$p$. The \index{Hydrostatic pressure}% \index{Pressure!hydrostatic}% energy-tensor~\Eq{(53.5)} accordingly becomes \[ T^{\mu\nu} = \begin{array}[t]{@{}c@{\Add{,}\quad}c@{\Add{,}\quad}c@{\Add{,}\quad}c@{}} p & 0 & 0 & 0\Add{,} \\ 0 & p & 0 & 0\Add{,} \\ 0 & 0 & p & 0\Add{,} \\ 0 & 0 & 0 & \rho_{00}\Add{.} \\ \end{array} \] \PageSep{122} Writing $\rho_{00} = \rho_{000} - p$, the pressure-terms give a tensor~$-g^{\mu\nu} p$. Accordingly we have have the tensor equation applicable to any coordinate-system \[ T^{\mu\nu} = \rho_{000}\, \frac{dx_{\mu}}{ds}\, \frac{dx_{\nu}}{ds} - g^{\mu\nu} p. \Tag{(54.81)} \] Thus if the energy-tensor is analysed into two terms depending respectively on two invariants specifying the state of the fluid, we must take these invariants to be $p$ and~$\rho_{000}$. The three quantities are related by \[ \rho_{0} = \rho_{00} - 3p = \rho_{000} - 4p. \Tag{(54.82)} \] If a fluid is \emph{incompressible}, i.e.\ if the closeness of packing of the particles is independent of~$p$, the condition must be that $\rho_{0}$~is constant\footnotemark.\footnotetext {Many writers seem to have defined incompressibility by the condition $\rho_{00} = \text{constant}$. This \index{Incompressibility}% is surely a most misleading definition.} Incompressibility is concerned with constancy not of mass-density but of particle-density, so that no account should be taken of increases of mass of the particles due to motion relative to the centre of mass of the matter as a whole.