which we also might have deduced, by a much simpler calculation, from the formula p = 2=Nm(km/=)3 { " dw w1[TM" ds sin s cos2s e-km(~2+a2-20= cos #) or its equivalent 0 0 p = Nm(km/=)3[°_dus°_dv[___ dwu2e ̄*m {(w−a>' + v2 + w° } . We have obtained this formula in § 35 (p. 72) in the form P = p(}G2 + a2), and from it have drawn conclusions respecting the momentum and force of reaction of a stream of air, and also respecting the resistance of air. The magnitude of the resistance is calculated in another way in some memoirs. It has been thought that the formulæ for Q1 and Q2 may be interpreted as if 2Q, represents the pressure which the forward face of a body moving with speed a experiences in air, while 202 represents the pressure of the air against the hinder face. Then the difference of these two magnitudes would give the resistance per unit area, and this reduces to 2(Q1 + Q2) = 2Nman(1 + 4a2/3πQ3) on neglect of higher powers of a. The resistance would thus consist of two parts, of which one would be proportional to the first, and the other to the third, power of the speed a. This mode of interpreting the formula was first employed by Hirn, and the contradiction between his formula and experiment led him to raise objections to the validity of the kinetic theory, which were, however, answered by Clausius.3 It is sufficient here to point out that the deduction of the expressions for Q, and Q2 are not valid for a rigid bounding surface, but only for a hypothetical plane in the interior of the gas. 1 W. B. Smith, Zur Molecular-Kinematik, Göttingen 1879; E. Toepler, Zur Ermittlung des Luftwiderstandes nach der kinetischen Theorie, Wien 1886; G. Sussloff, Journ. russ. phys.-chem. Ges. xviii. p. 79, 1887. 2 Hirn, Recherches sur la Résistance de l'Air en Fonction de la Température,' Mém. de l'Acad. de Belgique, xliii. (2) 1882. 3 Clausius, 'Examen des Objections faites par M. Hirn,' Bull. de l'Acad. de Belgique [3] xi. p. 173, 1886. 439 APPENDIX IV VISCOSITY OF GASES 44*. General Remarks on Viscosity THE viscosity or internal friction of gases is, according to this theory, nothing else than the transference from one place to another of the momentum of translation or flow of the medium by means of the heat-motions of its particles. In order to find the magnitude of the force exerted by one layer upon another by reason of viscosity, we have to determine the amount of momentum which is carried over in unit time by the molecules as they move backwards and forwards across the separating plane or surface of friction. Starting from this conception of the action, which has been explained more at length in § 73, I published, in a memoir 1 that appeared in 1865, a theory of viscosity which I will here first reproduce. However, since for easiness of calculation I then made the not strictly accurate assumption that all the molecules move with the same speed, I shall follow up this calculation, made in accordance with Clausius' assumption, by another which I shall found on Maxwell's law of the distribution of speeds. 45*. Theory of Viscosity on the Assumption of Equal Speeds for all Molecules While I now ascribe at first to all molecules of the gas equal molecular or heat motions that occur equally in all directions, I assume, further, a forward movement of given magnitude and direction; of this I assume the direction to be the same at all Ueber die innere Reibung der Gase,' erste Abhandlung. 'Ueber den Einfluss der Luft auf Pendelschwingungen,' Pogg. Ann. 1865, cxxv. p. 586. points of the gas, but the magnitude to vary continuously from layer to layer. I consider, further, that this motion may be looked upon as vanishingly small in comparison with the heatmotions, though not actually small in itself; for, since the mean molecular speed which will be ascribed to all the molecules is very great, amounting to several hundred metres per second, the forward motion of even a tolerably quick flow, such as occurs with a speed of 10 metres per second, will seem of but little importance in comparison. Consider a system of rectangular coordinates x, y, z such that the y-axis is parallel to the direction of the forward motion, and take the surface of friction, or the plane for which the friction between the gaseous layers on either side of it is to be determined, as perpendicular to the x-axis, and therefore parallel to the yz-plane, and let this plane pass through any arbitrary point in the medium with coordinates x, y, z. In this plane take an infinitely small rectangle with edges dy and dz, and find the number of particles which pass through it and the amounts of momentum, which I will denote by Q, and Q2, carried over it in both directions by these particles. For this purpose consider an infinitely small volume-element dx'dy'dz' at another point (x', y', z') of the gas, and first determine the number of particles which, starting from it in a straight course, meet the surface dy dz and pass through it. If N is the number of molecules contained in unit volume, there are N dx'dy'dz' particles in this volume-element at any moment; and if T denotes the average interval between two successive collisions of a particle with others, the number of straight paths commenced in unit time by this group of molecules is NT-'dx'dy'dz'; this is also the number of particles which issue from the element in unit time in all directions. Of these a portion, whose number is NT-e-dx'dy'dz', traverse a path of length r without a collision; herein ẞL = 1, or B is the reciprocal of the mean free path L which, on the assumption of equal speeds for all molecules, we have to put equal to the value found by Clausius', so that B = gs - s and A denoting 1 § 67 of the text, or § 33* of Appendix III. as before the radius of the sphere of action and the mean distance between neighbouring particles. Putting for r, which is still undetermined, the distance of the point (x, y, z) from the point (x', y', z'), or p2 = (x′ − x)2 + (y' − y)2 + (z' — z)2, we may interpret the magnitude NT-1e-dx'dy'dz' as the number of particles which start from the element dx'dy'dz' in unit time and traverse a sphere of radius r described about the element as centre, so as to cut the surface-element dy dz. From the number of particles traversing the whole spherical surface we deduce the number of those crossing the element dy dz by comparing the projection of the element on the spherical surface with the area of the whole sphere. The latter amounts to 4πr2, and the former to dy dz cos s, where s denotes the acute angle which the direction of makes with the x-axis. The number of particles, therefore, which in unit time reach and pass through the element dy dz, having started from the volume-element dx'dy'dz', is NT-le-(42)1 cos s dy dz dx'dy'dz'. The next question is, How much momentum is carried over by these particles? Since the molecular motion, of which heat consists, is taken to be the same throughout the medium, its transference causes no change, and it may therefore be left out of account, and we have to consider only the forward motion of the layers. Let this occur with velocity v at the point (x, y, z), and let v' be the corresponding value of this function at the point (x', y', z'). Then the momentum leaving the element dx'dy'dz' and crossing the element dy dz in unit time is dQ= (m/4)NT-v'er-2 cos s dy dz dx'dy'dz'. From this, by integration with respect to x', y', z' over one-half of the medium, we obtain the total value of momentum carried over from this half of the medium through the element dy dz of the dividing plane into the other half. If we take the medium as unlimited, this quantity which is carried over in the direction of increasing x is ܣ܂ Q1 = dy dz(m/47)NT − 15°。° da'["_dy'["_dz' e-"v'r2 cos s, 81 while that carried over in the other direction is -1 -2 Q2 = dy dz(m/47)NT-1[ˆ___dx'["__dy'["__ dz' e-"v'r-2 cos s. 81 81 The difference between these magnitudes F= Q1-Q2, or the sum of the gain and loss of one-half, is the friction exerted from the side of increasing x upon the other; and so - F= Q2- Q1 denotes the reaction exerted by the half corresponding to the smaller values of x on the half with the larger values of x. Since, as above assumed, v is a continuous function of x, y, z, and therefore also va continuous function of x', y', z', Taylor's theorem gives the development After substitution of this series the integrations can be carried out, and present no difficulty if the rectilinear coordinates are replaced by polar coordinates whose origin is at the point (x, y, z), i.e. by the coordinates r and s already introduced, and a second angle given by x) = r cos s (y' — y) = r sins cos (z) = r sin s sin o, where the sign must be determined so that the acute angle s may satisfy these relations. Then we obtain Q1 = dy dz(m/4x)NT−1e-v' cos s sin s dr ds dø π 0 Q2 = dy dz(m/4x)NT-1f foe-v' cos s sin s dr ds dø, where for shortness we put 0 |