2.3.2 The Stress Tensor When thermal motions are taken into account, a pressure force has to be added to the right-hand side of Eq. (2.15). This force arises from the random motion of particles in and out of a fluid element and does not appear in the equation for a single particle. Let a fluid element Ax Ay Az be centered at (xo, Ay, Az) (Fig.3.3). For simplicity, we shall consider only the x component of motion through the faces A and B. The number of particles per second passing through the face A with velocity vx is Any Vx Ay Az (2.16) Z X-4x A Ay Az X X X+Ax Fig.1 Origin of the elements of the stress tensor where An, is the number of particles per m³ with velocity vx. Each particle carries a momentum mvx. The density n and temperature KT in each cube is assumed to have the value associated with the cube's center. The momentum PA+ carried into the element at xo through A is then P₁+ = £ An, mv² Ay Az = Ay Az [m²] Σ 1 mv (2.17) The sum over An, results in the average v over the distribution. The xo-Ax factor comes from the fact that only half the particles in the cube at 2 xo Ax are going toward face A. Similarly, the momentum carried out through face B is - = Ay Az [mv² /n] (2.18) ΧΟ Thus the net gain in x momentum from right-moving particles is PB+ PA+ − PB+ = AyAz!m( [nv²³] - [m]) (2.19) xo-Ax dx = AyAz{m(— Ax) 3 (nr²) [his result will be just doubled by the contribution of left-moving articles, since they carry negative x momentum and also move in he opposite direction relative to the gradient of nv. The total hange of momentum of the fluid element at xo is therefore ə (nmux) Ax Ay Az ə dx (nv²) Ax Ay Az (2.20) dt Let the velocity vx of a particle be decomposed into two parts, Vx = Ux+Vxr Ux = √x (2.21) vhere ux is the fluid velocity and vxr is the random thermal velocity. For a one-dimensional Maxwellian distribution, we have =-m one-dimensional Maxwellian distribution, we have from Eq. (1.7) 1mAv th2 Avth mv² = KT Eav = d dt (nmux): = Equation (2.20) now becomes | mv² = KT 2 mn We can cancel two terms by partial differentiation: dn d (nux) + mux = dt dx conservation dux dt The equation of mass =-m m & [ n(1² + 200 + 15 )] = − 1 ² 3 [" (1²3 +7)] (₁ =-m KT) (2.23) xr dx xr m -mux dn ə di allows us to cancel the Defining the pressure + mnux dux d dx dx (2.22) - (nKT) (2.24) (nux) = 0 (2.25) dx terms nearest the equal sign in Eq. (3.40). p = nKT (2.26) дих (2.27) + Ux dx This is the usual pressure-gradient force. Adding the electromagnetic forces and generalizing to three dimensions, we have the fluid equation we have finally mn dux dt = др dx du mn + (u. V) V)u] =qn(E+u×B) − Vp (2.28) dt What we have derived is only a special case: the transfer of x momentum by motion in the x direction; and we have assumed that the fluid is isotropic, so that the same result holds in the y and z directions. But it is also possible to transfer y momentum by motion in the x direction, for instance. Suppose, in Fig. 3.3, that Uy