Tangent, unit vector

One complication is that the boundary conditions (4-264)-(4-266) must be applied at the bubble surface, which is both unknown [that is, specified in terms of functions R(t) and fn(9,boundary conditions are also functions of the bubble shape. In this analysis, we use the small-deformation limit s 1 to simplify the problem by using the method of domain perturbations that was introduced earlier in this chapter. First, we note that the unit normal and tangent vectors can be approximated for small e in the forms... 

Sinee this equality must be valid for any orientation of the rectangle and, thus, of the tangent unit vector 1, we finally have... 

If a vector it is a function of a single scalar quantity s, the curve traced as a function of s by its terminus, with respect to a fixed origin, can be represented as shown in Fig. 8. Within the interval As the vector AR = R2 - Ri is in the direction of the secant to the curve, which approaches the tangent in the limit as As - 0. Tins argument corresponds to that presented in Section 2.3 and illustrated in Fig. 4 of that section. In terms of unit vectors in a Cartesian coordinate system... 

The excitation spectrum proves even more useful near the surface. Since anisotropic molecules at the surface of a liquid tend to orient relative to the surface tangent, one might expect the excitation spectrum to be sensitive to such orientation. For example, suppose we take the extreme case in which molecules at the surface are oriented with their transition moments perpendicular to the surface tangent. Then the only field component which can excite these molecules is the radial field at the surface. When one recalls that only the N type vector field has radial components, one expects that a calculation of the excitation spectrum of such a molecular layer will yield half as many resonant features as shown in Figure 8.4. Indeed this is the case. Figure 8.7 shows the calculated surface average of the square modulus of the radial component of the local electric field, < E er 2>J, where sr is the radial unit vector. 

In the second half of this article, we discuss dynamic properties of stiff-chain liquid-crystalline polymers in solution. If the position and orientation of a stiff or semiflexible chain in a solution is specified by its center of mass and end-to-end vector, respectively, the translational and rotational motions of the whole chain can be described in terms of the time-dependent single-particle distribution function f(r, a t), where r and a are the position vector of the center of mass and the unit vector parallel to the end-to-end vector of the chain, respectively, and t is time, (a should be distinguished from the unit tangent vector to the chain contour appearing in the previous sections, except for rodlike polymers.) Since this distribution function cannot describe internal motions of the chain, our discussion below is restricted to such global chain dynamics as translational and rotational diffusion and zero-shear viscosity. 

This derivative is the unit vector tangent to the curve defined by a succession of points r at the point r, and hence it must give the direction of the vector Vp(r) at this point. Thus... 

Although equal in value to the rectilinear coordinate Qx, the parameter /s can be treated as a curvilinear coordinate that follows the infinitesimal displacement of a point on the seam along the local tangent vector to the curve, t(/3). This moving frame is completed by the normal vector, n(/3). At the expansion point (origin of the frame fj, = 0), the normal and tangent vectors to the seam are parallel to xi and X3 (unit vectors), respectively. However, away from that point, these vectors are different and combine xi and X3 because the seam is curved (Fig. 5). 

Differentiating ti -ti = 1, with respect to arc length si we find ti -dti/dsi = 0 which implies that the vector dti/dsi is perpendicular to the tangent vector ti- Therefore, if we divide dti/dsi by the length dti/dsi, we obtain a unit vector which is orthogonal to ti. 

The bi-normal unit vector of a space curve bi = ti x ri/ is perpendicular to both the surface tangent vector ti and the surface normal vector ri/, such that the vectors (ti, n/, bi) form a right handed system. It is noted that the vector Hsj lies in the plane which contains the vectors n/ and bi, yet the direction of the two normal vectors in this plane generally differ by an angle 6. 

To express the force equilibrium condition in a mathematical form, we can now consider a force balance on an arbitrary surface element of a fluid interface, which we denote as A. A sketch of this surface element is shown in Fig. 2-14, as seen when viewed along an axis that is normal to the interface at some arbitrary point within A. We do not imply that the interface is flat (though it could be) - indeed, we shall see that curvature of an interface almost always plays a critical role in the dynamics of two-fluid systems. We denote the unit normal to the interface at any point in A as n (to be definite, we may suppose that n is positive when pointing upward from the page in Fig. 2-14) and let t be the unit vector that is normal to the boundary curve C and tangent to the interface at each point (see... 

To this point, we have considered only the component of the stress balance (2-134), in the direction normal to the interface. There will be, in general, two tangential components of (2-134), which we obtain by taking the inner product with the two orthogonal unit tangent vectors that are normal to n. If we denote these unit vectors as t (with i = 1 or 2), the so-called shear-stress balances can be written symbolically in the form,... 

Here, ps is the pressure inside the bubble, y is the interfacial tension, and t, represents one of the pair of orthogonal unit vectors that are tangent at any point to the bubble surface. The problem is to calculate velocity and pressure fields in the fluid, as well as to determine the functions R(t) and fn(0, bubble surface. In the present context, it is the latter part of the problem that is the focus of our interest. 

One shows that t is indeed a unit vector by using Eq. (1.81). The rate of change of the tangent is related to the curvature, k, and the normal, He, by... 

Another example are spherical coordinates on a unit sphere so that u and v are the angles 9 and respectively. On the surface one defines the two tangent vectors = dr/du and r = dr/dv. These vectors are not necessarily unit vectors, nor are they necessarily orthogonal. The two vectors define a tangent plane. The equation of the plane is given by r n = 0 where h is the normal to the surface at positions (u,v). The normal is given by the cross product ... 

Equation (2.16) is written here for the simplest case where the tangent to the jet axis is inclined at any point to a certain straight line 0i by an acute angle and it is possible to introduce a Cartesian coordinate system Oi rjC with the correspimding unit vectors i, j and k and to describe the jet axis using the following equations... 

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