Euler L. Principles of the motion of fluids (794385), страница 4
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Second part39. Having presented what pertains to all possible motions,let us now investigate the nature of the motion which can reallyoccur in the fluid. Here, besides the continuity of the fluid andthe constancy of its density, we will also have to consider theforces which act on every element of the fluid. When the motionof any element is either non-uniform or varying in its direction,the change of motion must be in accordance with the forcesacting on this element. The change of motion becomes knownfrom known forces, and the preceding formulas contain thischange; we will now deduce new conditions13 which single outthe actual motion among all those possible up to this point.40.
Let us arrange this investigation in two parts as well; atfirst let us consider all motions being performed in the sameplane. Let AL = x, Ll = y be, as before, the definingcoordinates of the position of an arbitrary point l; now, afterthe elapsed time t, the two velocities of the point l parallel tothe axes AL and AB are u and v: since the variability of timehas to be taken into account, u and v will be functions of x, yand t themselves. In respect of which we putdu = Ld x + ldy + Ldtand dv = Md x + mdy + Mdt13 Here Euler probably has in mind the condition of potentiality, which hewill obtain in §§.
47 and 54 for the two-dimensional case and in §. 60 for thethree-dimensional case.and we have established above that because of the formercondition encountered above, we have L + m = 0.41. After an elapsed small time interval dt the point l iscarried to p, and it has travelled a distance udt parallel to theaxis AL, a distance vdt parallel to the other axis AB.
Hence,to obtain the increments in velocities u and v of the point lwhich are induced during time dt, for d x and dy we mustwrite the distance udt and vdt, from which will arise these trueincrements of the velocitiesdu = Ludt + lvdt + Ldt and dv = Mudt + mvdt + Mdt.Therefore the accelerating forces, which produce theseaccelerations areAccel. force w.r.t.
AL = 2(Lu + lv + L)Accel. force w.r.t. AB = 2(Mu + mv + M)to which therefore the forces acting upon the particle of waterought to be equal.1442. Among the forces which in fact act upon the particles ofwater, the first to be considered is gravity; its effect, however,if the plane of motion is horizontal, amounts to nothing. Yet ifthe plane is inclined, the axis AL following the inclination, theother being horizontal, gravity generates a constant acceleratingforce parallel to the axis AL, let it be α. Next we must notneglect friction, which often hinders the motion of water, andnot a little. Although its laws have not yet been exploredsufficiently, nevertheless, following the law of friction for solidbodies, probably we shall not wander too far astray if we set thefriction everywhere proportional to the pressure with which theparticles of water press upon one another.1543.
First, must be brought into the calculation the pressurewith which the particles of water everywhere mutually act uponeach other, by means of which every particle is pressed togetheron all sides by its neighbours; and in so far as this pressure isnot everywhere equal, to that extent motion is communicatedto that particle.16 The water simply will be put everywhereinto a state of compression similar to that which still waterexperiences when stagnating at a certain depth. This depth ismost conveniently employed for representing the pressure at anarbitrary point l of the fluid.
Therefore let that height, or depth,expressing the state of compression at l, be p, a certain functionof the coordinates x and y, and should the pressure at l vary alsowith the time, the time will also enter into the function p.44. Thus let us set d p = Rd x + r dy + Rdt, and let usconsider a rectangular element of water, lmno, whose sides arelm = no = d x and ln = mo = dy, whose area is d xdy(Fig. 3).
The pressure at l is p, the pressure at m is p + Rd x,at n it is p + r dy and at o it is p + Rd x + r dy. Thus the sidelm is pressed by a force = d x( p + 12 Rd x), while the oppositeside no will be pressed by a force = d x( p + 12 Rd x + r dy);14 The unusual factors of 2 in the previous equations have to do with a choiceof units which soon became obsolete; cf.
Truesdell, 1954; Mikhailov, 1999.15 It is actually not clear why Euler takes the friction force proportional to thepressure.16 Here Euler makes full use of the concept of internal pressure, cf. Darrigoland Frisch, 2008.1847L. Euler / Physica D 237 (2008) 1840–1854after the substitution of which values we have the followingequation(L + m)(l − M) +dl − dMdl − dMdl − dMu+v+= 0.dxdydtPlainly, this is satisfied if l = M: so thatdudyFig. 3.therefore by these two forces the element lmno will be impelledin the direction ln by a force = −r d xdy. Moreover, in a similarmanner from the forces dy( p+ 21 r dy) and dy( p+Rd x + 12 r dy),which act on the sides ln and mo will result a force = −Rd xdyimpelling the element in the direction lm.45. Thus will originate an accelerating force parallel to lm =−R and an accelerating force parallel to ln = −r , of which theone directed along the force of gravity α gives α − R.
Havingignored friction so far, we obtain the following equations17 :α − R = 2Lu + 2lv + 2LorR = α − 2Lu − 2lv − 2L−r= 2Mu + 2mv + 2M and r = −2Mu − 2mv − 2Mdp = αd x − 2(Lu + lv + L)d x − 2(Mu + mv + M)dy + Rdt,a differential which must be complete or integrable.46. Because the term α d x is integrable by itself and nothingis determined for R, from the nature of complete differentialsit is necessary that the following holds in the notation alreadyemployed:d.Mu + mv + Md.Lu + lv + L=.dydxdudx=L,dudy= l;dvdx=M, anddvdy= m it follows thatSince thisdv 18dx ,d p = αd x − 2u(Ld x + ldy) − 2v(ld x + mdy) − 2Ld x(L + m)(l − M) +dldmdL dMdL dMu−+v−+−= 0.dydxdydxdydx47.
In fact, since we knew Ld x + ldy + Ldt and Md x +mdy + Mdt to be complete differentials,dLdl=dydtandMoreover Ld x + ldy = du − Ldt; ld x + mdy = dv − Mdt;hence we haved p = αd x −2udu − 2vdv + 2Ludt + 2Mvdt − 2Ld x − 2Mdy + Rdt.Therefore, if we wish to ascertain for the present time thepressure at each point of the fluid, with no account of itsvariation in time, we shall have to consider this equationd p = αd x − 2udu − 2vdv − 2Ld x − 2Mdy,dudtand M =d p = αd x − 2udu − 2vdv − 2which is reduced to this form:dmdM=;dxdydvdx .=condition requires thatit appears finally that thedifferential formula ud x + vdy must be complete; in this liesthe criterion of actual motion.48. This criterion is independent from the preceding one,which was provided by the continuity of the fluid and itsuniform constant density.
Therefore even if the fluid in motionchanges its density, as happens in the motion of elastic fluidssuch as air, this property will hold nonetheless, namely ud x +vdy has to be a complete differential. In other words, thevelocities u and v must always be functions of the coordinatesx and y, together with time t, in such a way that when the timeis taken constant the formula ud x + vdy admits an integration.49. We shall now determine the pressure p itself, which isabsolutely necessary for perfectly determining the motion ofthe fluid.
Since we have found that M = l we haveand in our notation L =udLvdldLudMLl ++ lm ++= ML +dydydydxvdmdM+ mM +=dxdxdLdl=;dydx=− 2Mdy + Rdt.from which we gather thatSincedudydMdM=dxdt17 Here the so-called Euler equations of incompressible fluid dynamics appearfor the first time, but the notation and the units are not very modern, in contrastto the memoir he will write three years later (Euler, 1755).dv 19dt .Hencedvdud x − 2 dy,dtdtin the integration of which the time is to be taken constant.50. This equation is integrable by hypothesis, and is indeedunderstood as such, if we consider the criterion of the motionwhich, as we have seen, consists in that ud x + vdy be acomplete differential when the time t is taken constant. Lettherefore S be its integral, which consequently will be afunction of x, y and t themselves.
For dt = 0 we obtaindS = ud x + vdy, while assuming the time t variable as well,18 Here there are two problems. The minor problem is a typographical error inthe published version ( ddux instead of ddvx ), which is not present in a 1752 copyof the manuscript (not in Euler’s hand), henceforth referred to as Euler, 1752.A more serious problem is that Euler here repeats the mistake of D’Alembert,1752 who confused a sufficient condition – the vanishing of the vorticity – witha necessary one.19 The printed version has L = dv instead of L = du . Euler, 1752 is correct.dtdt1848L. Euler / Physica D 237 (2008) 1840–1854AB, AC.
Let u, v, w denote these functions, which depend oncoordinates x, y, z, besides t. After a differentiation we obtainlet us writedS = ud x + vdy + Udt,on which account we obtaindudt=dUdxanddvdt=dUdy .Then, infact U = dSdt .51. After inserting these values we will obtaindudvdUdU.d x +.dy =.d x +.dydtdtdxdyand this differential formula is manifestly integrated at constanttime t to give U. For this to become clearer, let us set dU =dUdUdUKd x + kdy; thus dUd x = K and dy = k, so that d x .d x + dy =Kd x + kdy = dU.
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