Hence in Galilean coordinates \[ T^{\mu\nu} = \rho\, \frac{dx_{\mu}}{dt}\, \frac{dx_{\nu}}{dt}, \Tag{(53.3)} \] so that if $u$, $v$, $w$ are the components of velocity \[ T^{\mu\nu} = \begin{array}[t]{@{}c@{,\quad}c@{,\quad}c@{,\quad}c@{}} \rho u^{2} & \rho vu & \rho wu & \rho u\Add{,} \\ \rho uv & \rho v^{2} & \rho wv & \rho v\Add{,} \\ \rho uw & \rho vw & \rho w^{2} & \rho w\Add{,} \\ \rho u & \rho v & \rho w & \rho\Add{.} \\ \end{array} \Tag{(53.4)} \] In matter atomically constituted, a volume which is regarded as small for macroscopic treatment contains particles with widely divergent motions. Thus the terms in~\Eq{(53.4)} should be summed for varying motions of the particles. For macroscopic treatment we express the summation in the following way.--- %[** TN: No indent in the original] {\Loosen Let $(u, v, w)$ refer to the motion of the centre of mass of the element, and \index{Stress-system}% $(u_{1}, v_{1}, w_{1})$ be the internal motion of the particles relative\footnote {In the sense of elementary mechanics, i.e.\ the simple difference of the velocities.} to the centre of mass. Then in a term of our tensor such as $\sum \rho(u + u_{1})(v + v_{1})$, the cross-products will vanish, leaving $\sum \rho uv + \sum \rho u_{1}v_{1}$. Now $\sum \rho u_{1}v_{1}$ represents the rate of transfer of $u$-momentum by particles crossing a plane perpendicular to the $y$-axis, and is therefore equal to the internal stress usually denoted by~$p_{xy}$. We have therefore to add to~\Eq{(53.4)} the tensor formed by the internal stresses, bordered by zeroes. The summation can now be omitted, $\rho$~referring to the whole density, and $u$, $v$, $w$ to the average or mass-motion of macroscopic elements. Accordingly} \[ T^{\mu\nu} = \begin{array}[t]{@{}c@{,\quad}c@{,\quad}c@{,\quad}c@{}} p_{xx} + \rho u^{2} & p_{yx} + \rho vu & p_{zx} + \rho wu & \rho u\Add{,} \\ p_{xy} + \rho uv & p_{yy} + \rho v^{2} & p_{yz} + \rho wv & \rho v\Add{,} \\ p_{xz} + \rho uw & p_{yz} + \rho vw & p_{zz} + \rho w^{2} & \rho w\Add{,} \\ \rho u & \rho v & \rho w & \rho\Add{.} \\ \end{array} \Tag{(53.5)} \] Consider the equations \[ \frac{\dd T^{\mu\nu}}{\dd x_{\nu}} = 0. \Tag{(53.6)} \] Taking first $\mu = 4$, this gives by~\Eq{(53.5)} \[ \frac{\dd(\rho u)}{\dd x} + \frac{\dd(\rho v)}{\dd y} + \frac{\dd(\rho w)}{\dd z} + \frac{\dd\rho}{\dd t} = 0, \Tag{(53.71)} \] which is the usual ``equation of continuity'' in hydrodynamics. \index{Continuity, equation of}% \index{Hydrodynamics, equations of}% \PageSep{118}