In accordance with the definition of force as rate of change of momentum, the quantity on the right will be recognised as the (negative) body-force acting on unit volume, the three components of the force being given by $\mu = 1$, $2$,~$3$. When the velocity of the matter is very small compared with the velocity of light as in most ordinary problems, we need only consider on the \PageSep{125} right the component~$\mf{T}^{44}$ or~$\rho$; and the force is then due to a field of acceleration of the usual type with components $-\frac{1}{2}\dd g_{44}/\dd x_{1}$, $-\frac{1}{2}\dd g_{44}/\dd x_{2}$, $-\frac{1}{2}\dd g_{44}/\dd x_{3}$. The potential~$\Omega$ of the field of acceleration is thus connected with~$g_{44}$ by the relation $g_{44} = 1 - 2\Omega$. When this approximation is not sufficient there is no simple field of acceleration; the acceleration of the matter depends not only on its position but also on its velocity and even on its state of stress. Einstein's law of gravitation for empty space $G_{\mu\nu} = 0$ follows at once from the above identification of~$T_{\mu}^{\nu}$. \Section{56.}{Dynamics of a particle} \index{Dynamics of a particle}% \index{Particle!dynamics of}% \index{Particle!symmetry of}% An isolated particle is a narrow tube in four dimensions containing a non-zero energy-tensor and surrounded by a region where the energy-tensor is zero. The tube is the world-line or track of the particle in space-time. \index{World-line}% The momentum and mass of the particle are obtained by integrating~$\mf{T}_{\mu}^{4}$ over a three-dimensional volume; if the result is written in the form \[ -Mu,\ -Mv,\ -Mw,\ M, \] then $M$~is the mass (relative to the coordinate system), and $(u, v, w)$ is the \emph{dynamical velocity} of the particle, i.e.\ the ratio of the momenta to the mass. \index{Dynamical velocity}% The \emph{kinematical velocity} of the particle is given by the direction of the \index{Kinematical velocity}% tube in four dimensions, viz.\ $\left(\dfrac{dx_{1}}{dx_{4}}, \dfrac{dx_{2}}{dx_{4}}, \dfrac{dx_{3}}{dx_{4}}\right)$ along the tube. For completely continuous matter there is no division of the energy-tensor into tubes and the notion of kinematical velocity does not arise. It does not seem to be possible to deduce without special assumptions that the dynamical velocity of a particle is equal to the kinematical velocity. The law of conservation merely shows that $(Mu, Mv, Mw, M)$ is constant along the tube when no field of force is acting; it does not show that the direction of this vector is the direction of the tube.