Vector Concepts For Fluid Mechanics

Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

Vector Concepts for Fluid Mechanics

1. A vector has magnitude and direction and can be described as an arrow pointing in the assigned direction with a length equal to its magnitude (Figure 1).

2. Vectors can be described in Cartesian coordinates as:     v  ai  bj  ck Eq. 1

y Figure 1: A vector , represented by the bold arrow. The tip of the arrow can be reached from the origin by moving a distance in the direction,  followed by a distance in the direction, followed v c by a distance in the direction. b a x

3. The coefficients a , b , and c are the x , y , and z components of the vector.      i , j, and k i x j 4. are themselves vectors. is a vector pointing in the direction, is a vector pointing in the y direction, and k is a vector pointing in the z direction. See Figure 2.

Figure 2: The , and unit vectors. These are true vectors in the sense that they have both   magnitude and direction. For example, the j v magnitude of is 1 and its direction is along the x axis.  x k  i z

5. The magnitude of a vector is its length, which is given by:

 Eq. 2 v  v  a 2  b2  c2 , which is simply a statement of the Pythagorean theorem.      j j  0i  1 j  0k Problem 1: Given that the vector can be represented as , show, from Eq. 2, that the magnitude of j is 1. Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

Problem 2: Show that the equation for the magnitude of a vector is simply a statement of the Pythagorean theorem as follows. First note that b and c are perpendicular to one another because b is parallel to the y axis and c is parallel to the z axis (refer to Figure 1). Use the Pythagorean theorem to obtain d (in Figure 3) in terms of b and c . Now, since b and c are both perpendicular to the x axis, d is also perpendicular to the x x axis (i.e., it lies in a plane that is perpendicular to the axis. Use this information to v a d d write the length of the vector in terms of and . Then back substitute for to obtain the length of v in terms of a , b and c .

y Figure 3: Geometry to be used to demonstrate that the equation for the magnitude of a vector is a simple statement of the Pythagorean theorem.  v c b a d x

        6. The dot product between two vectors, v1  a1i  b1 j  c1k and v2  a2i  b2 j  c2k is   v1  v2  a1a2  b1b2  c1c2 Eq. 3   Problem 3: Show that the dot product of i and j is zero. Similarly show that the dot           product of i and k is zero. (Hint: write i  1i  0 j  0k and j  0i  1 j  0k and then use Eq. 3 for the dot product.         Problem 4: Show that the dot product of u  1i  1 j  1k and j  1i  0.5 j  0.5k is zero.

7. The dot product of two vectors is a scalar value. It has only magnitude, not direction. 8. The value of the dot product is:   v v cos Eq. 4 1 2 ,

where  is the angle between the two vectors.     Problem 5: Note that the vectors i and v  1i  1 j are in directions that are 45 apart from one another. Calculate their dot products first by Eq. 3 and then by Eq. 4 and show that they are equal. Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

9. The dot product of one vector with another is physically the projection of that vector on the other vector, multiplied by the length of the other vector.

10.The dot product of a vector with itself is the square of the vector’s magnitude. Thus, 2 v  v  v  v 2  v  v  v  v .

11.A unit vector is a vector with magnitude of one.

12.Any vector can be normalized to become a unit vector by simply dividing by its own magnitude.      v ai  bj  ck u    v 2 2 2 a  b  c .

13.Recalling 9 above, the dot product of any vector with a unit vector is the projection of that vector in the direction of the unit vector. (The projection is the shadow cast by the vector by a light source that is in the plane of the two vectors and that is perpendicular to the unit vector).

14.The cross product of two vectors is a vector and can be written as:    i j k      v1  v2  a1 b1 c1  b1c2  c1b2 i  a1c2  c1a2  j  a1b2  c1a2 k

a2 b2 c2

    v  v  v v sin 14. The magnitude of the cross-product has the value of 1 2 1 2 , where  is the angle between the two vectors.

15. The direction of the cross-product is perpendicular to both of the two original vectors, or, equivalently, perpendicular to the plane in which the two original vectors lie.

Problem 6: Find the magnitude and direction of the cross product of the two vectors given in Problem 5.

Problem 7: Show that the dot product of the result from Problem 6 with the two vectors given in Problem 5 is zero. Hence, show that the cross product of these to vectors is perpendicular to the original two vectors.

The above concepts are ones that students should know from their statics course. The students will need this much review, and they will need to be reminded that they already know these concepts from statics. Next come some concepts that will be Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

new in the sense of understanding vector applications. In other words, while students may have seen the gradient operator in their math class, they may not yet understand what it means physically.     16. Define a vector r  xi  yj  zk . (The operator  means “is defined as,” as opposed to “is equal to.” The difference between  and  is subtle, but often  important). Note that this vector may also be represented by r  x, y, z. A function of the three spatial coordinates x, y, z, i.e. f x, y, z, can then be represented as f  f  f  f r   f  i  j  k . The gradient of the function f r  is defined as x y z .   Problem 8: If f r   x sinyz, find f r .    (Answer: f r  sinyzi  xz cosyz j  xy cosyzk )

17. The gradient is a vector function. In other words, for a given set of values of x ,     y , and z , the gradient can be represented by f r   ai  bj  ck .

  1 Problem 9: For the answer to Problem 8, what is the vector f r  at r  2, , 4 . 1  1   (Answer: i  j   2 k ). 2 2 2

18. However, the function itself is not a vector.

  1 Problem 10: What is the value of the function f r  at r  2, , 4 .

19.The gradient is a function of the spatial coordinates.

20. Each term of the gradient is the rate of change of the function as one moves in the given coordinate direction, assuming that there is no change in position with respect to the other two coordinate directions. To illustrate the gradient, consider the situation shown in Figure 4. A paper mill produces a sulfurous odor that diminishes in pungency with distance from the mill. A person at point A experiences a relatively large rate of reduction of the odor as he moves in the x direction. The same person experiences a relatively small rate of reduction of odor as he moves in the y direction. Thus, the x component of the gradient is large and the y y component of the gradient is small.

Figure 4: The concept of gradient illustrated by the rate y at which the odor from a paper mill decreases as one moves in a particular direction. A x

Paper Mill Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

Problem 11: Assume in the paper mill example that the odor drops off with distance 2 from the mill according to the equation O  A r . Give the equation for the gradient at any location x, y from the mill, assuming that the origin of the x and y axes is r  x 2  y 2 at the center of the mill. Hint: You should write  and then take the xi  yj  O  2A 4 derivatives with respect to x and y . Answer: r .   Problem 12: For the same paper mill, what are the i and j components of the gradient found in Problem 3 at a point x, y  2,0? How do they agree with the above statement that the odor should drop off more quickly along the x direction than along the y direction at this location?

21. Velocity is a vector quantity. It’s magnitude is the speed of an object. It’s direction is the direction in which the object is moving.         i  j  k 22. The gradient operator can be described as a vector, x y z . Thus, the gradient of a function is the gradient operator applied to the function. 23. The divergence of a vector v (such as velocity) is the dot product of the gradient operator with the vector.

  v v v div(v)    v  1  2  3 x y z

24. The divergence is a scalar quantity because it is a dot product.

25. One may take the divergence of the gradient of a function. This scalar quantity is known as the Laplacian operator.

          f  f  f   2 f  2 f  2 f f r 2 f (r) i j k i j k                   2  2  2  x y z   x y z  x y z

2 26. One may also consider the divergence to be an operator  defined as:

               2  2  2 2 i j k i j k                 2  2  2  x y z   x y z  x y z Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

  2  2  2  2 f      f  x 2 y 2 z 2  27. Thus,  

28.The curl of a vector is the cross-product of the gradient operator with the vector.

      v2 v3   v1 v3   v1 v2  curl(v)    v    i     j    k  z y   z x   y x 

29. The curl of a vector is another vector.

30.The curl of a vector can be written as:    i j k      curlv    v  x y z

v1 v2 v3

and this form provides one with an easy way to remember where the “–“ signs must go.

Problem 13: Which of these are not legitimate vector operations. Why? Assume f is a scalar function.   2 a.   v c.   f    v  f   e. g. b.   f     v f  f  2 d. f. h.    f 

Problem 14: For each legitimate operation in Problem 5, state what kind of entity (vector or scalar) is the result.

31. The equation f r   c0 , where c0 is a constant, represents a surface in three- dimensional space. A vector normal to this surface is f r  .

32. f r  is also the direction of maximum rate of change in f r  .

33.A curve in space is the intersection of two surfaces.

Problem 15: What is the curve represented by the intersection of the following two surfaces:

x 2  y 2  z 2  25 ; x 2  y 2  z  0 ? Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

Solution: First notice that the first curve is a sphere of radius 5 and the second curve is a parabola of rotation. If we substitute z  x 2  y 2 from the second curve into the first curve, the result is:

z 2  z  25  0 , 1  1  100 which has two solutions, z  , one of which is negative and therefore 2 irrelevant. It follows that the curve is in the plane represented by z   101 1 2 and on the curve in that plane x 2  y 2   101 1 2 .

31.There are several vector identities that should be recognized by the student (though not necessarily memorized).

a. a  b c  b  c a  c  a b  a  c b  b  a c  c  b a

Proof of the first equality:

a  b  a1 a2 a3  a2b3  a3b2 i  a1b3  a3b1 j  a1b2  a2b1 k

b1 b2 b3

a  b c  a2b3  a3b2 c1  a1b3  a3b1 c2  a1b2  a2b1 c3

 b2c3  b3c2 a1  b1c3  b3c1 a2  b1c2  c1b2 a3 i j k

 b1 b2 b3  a  b  c a

c1 c2 c3

The other equalities follow similarly. Steven A. Jones BIEN 501, Physiological Modeling March 8, 2006

b. a  b c  a  cb  b  ca

Proof: a  b is a vector perpendicular to both a and b . Any vector that is perpendicular to a  b must therefore be in the plane of both a and b . In particular a  b c must be in that plane. Any vector in that plane can be described by a linear combination of a and b . Thus, a  b c  ma  nb . Furthermore, since c must be perpendicular to a  b c , it follows that a  b c c  0 so that ma  nb c  0 , i.e. ma  c  nb  c . Therefore, a  b c  ma  nb