of matter on the two sides of any plane area is equal and parallel to the mutual force across any equal, similar, and parallel plane area, the stress is said to be homogeneous through that space. In other words, the stress experienced by the matter is homogeneous through any space if all equal similar and similarly turned portions of matter within this space are similarly and equally influenced by force." (Thomson and Tait, $659.) Zx Fig. 162. Consider a unit cube (Fig. 162) subject to homogeneous internal stresses and in equilibrium. The stress on each of the six sides may be decomposed into three along the coordinate axes, but as, from the definition of a homogeneous stress, the forces acting in the same direction across opposite faces must be equal, we need only consider three faces of the cube. We denote by X, Y, Z, the three components of force acting on the face yz, the index a indicating that the face is normal to the axis of x. Similarly X, Y, Zy, and Xz, Yz, Z2, indicate the components acting on the faces normal to the axes of y and z respectively. If we consider the force which acts on the cube from the outside, two stresses X act in opposite directions on the two faces normal to OX. If we take X to be positive the two forces tend to produce elongation. Similarly Y, and Z are stresses tending to produce elongations along the axes of y and z respectively. Xz The force X (Fig. 163) is a tangential force acting in opposite directions on two opposite faces, but not along the same line, so that a couple of moment X2 is formed. We take X to be positive when, as drawn in the figure, the force acting on a surface parallel to ay from below is along the negative axis of y, the axis of being positive upwards. But the two forces Z also form a couple, which however tends to produce rotation round OY in the opposite direction, hence for equilibrium, it follows that Χι Zx Fig. 163. The two equal couples Y and Z form together a simple shearing It may be proved in the same manner that stress. Xx, Yy, Zz; Y2=Zy; Zx= X2; Xy= Yx, completely define a homogeneous stress. We shall introduce the notation of Thomson and Tait, and write for these six components of stress M HB K C P, Q, R, S, T, U. 130. Shearing stress produced by combined tension and pressure at right angles. Let ABCD be a section of a cube, which is subject to a uniform tension P at right angles to BC, and a uniform pressure at right angles to CD. No stress is supposed to act at right angles to the plane of the paper. Let H, K, L, M be the middle points of the sides of the square ABCD, and draw the square HKLM. If the part HBK is in equilibrium, a force must act on the plane which is at right angles to the plane of the paper, and passes through HK. The elementary laws of Statics show that this force must be in the plane, and that its value per unit surface is P. The rectangular volume of HKLM is therefore acted on by tangential stresses of the nature of shearing stresses, or: Fig. 164. A longitudinal traction (or negative pressure) parallel to one line and an equal longitudinal positive pressure parallel to any line at right angles to it, is equivalent to a shearing stress of tangential tractions parallel to the planes which cut those lines at 45°. And the numerical pressure of this shearing stress, being the amount of the tangential traction in either set of planes, is equal to the amount of the positive or negative normal pressure, not doubled." (Thomson and Tait, § 681.) The caution at the end of the quotation is necessitated by the fact that in the analogous proposition referring to shears, the amount of the shear is obtained by doubling the elongation, as has been proved in Art. 127. 131. Connexion between Strains and Stresses. If a simple shearing stress, as defined in Art. 129, act on a homogeneous body, it produces a shearing strain, and the ratio of the stress to the strain is the resistance to change of shape or the "Rigidity" of the substance. Calling the rigidity n, it follows that we may put in isotropic bodies. S=na; T=nb; U=nc .(3) The three stresses P, Q, R produce elongations e, f, g, and there must be a linear relationship between them. Also by symmetry a stress along OX must produce the same contraction in all directions at right angles to itself. Hence A and B being constants, we may write down at once the equations It remains to prove how A and B are connected with the rigidity and the bulk modulus. If e, f, g are equal But the cubical dilatation being 3e and the bulk modulus being equal to the ratio of the uniform stress P to the cubical dilatation, it follows that As a second special case take R=0, and Q-P, which conditions indicate a shearing stress in planes equally inclined to the axis of X and Y, and these will cause a shearing strain equal in amount to Pn. This shearing strain is equivalent by Art. 127 to an elongation in the direction of P of P/2n, and an equal contraction in the direction of Q. Substituting e-f= P/2n into the first of the equations (4), we find if g=0 Combining (5) and (6), it follows that A=k+n, B=k-n. (6). In place of the components of strain, we may introduce their equivalents in terms of the displacement (Arts. 126 and 127). Equations 132. Equations of Motion in a disturbed medium. Returning to the stresses acting on the cube in Art. 129, we consider the case where these stresses are not constant through the volume, but alter slowly from place to place. If the distance between the two faces of the cube which are at right angles to the axis of X is da, there will be a force Xady dz acting in the negative direction on the face which is coincident with the coordinate plane and a force on the opposite face equal to (X+dx) Similarly the force Xdady acting on the plane xy in the direction of a together with the force combine to a resultant ( X 2 + dx = dz dz and the forces in that same direction are complete when we have added the resultant dX of the two forces which act on the faces which are normal to the axis of y. If p be the density of the substance, so that pdxdydz be the mass of the volume considered, and if a be the displacement in the direction, the equations of motion may be written down by the laws of dynamics, leaving out the factor dadydz on both sides, da_dXx dxy + = + dy dz + Re-introducing the notation of Thomson and Tait, the equations become dy dT Pat2 = da + + dy dz' ds + dy dz + + dy dz To eliminate the stresses use equations (7) and (8). Substituting the values of A and B from Art. 131, and rearranging the terms, we obtain These equations govern wave propagation in all elastic media. We may obtain from them the characteristic equations for the longitudinal waves of sound by putting the rigidity n of the medium equal to zero. When applied to light, the medium is taken as incompressible, so that but k at the same time becomes infinitely large. Writing These equations, together with certain relations which must hold at the surfaces of the elastic body, constitute the elastic solid theory of light. For plane waves, the displacements are the same at all points of the wave-front, which we may imagine to be at right angles to the axis of z. The differential coefficient of a, ẞ, y with respect to x and y must therefore vanish. The equations (9) then reduce to The last equation represents a longitudinal wave propagated with infinite velocity and having no relation to any observed phenomenon of light. Each of the first two equations represents a rectilinear wave propagated with velocity √n/p, a result already deduced by the simpler but less general methods of Art. 12. The investigation of wave propagation in crystalline media presents great difficulties. The simplest hypothesis from a mathematical point of view is that of assuming that the inertia of the medium may differ for displacements in different directions. By substituting P1, P2, P3, respectively, for p on the left-hand side of equations (9), we obtain equations which lead to a wave surface which is similar to, but not identical with, Fresnel's wave surface. A theory of double refraction |