Mechanics 1: Coordinates

Transcription

1 Mechanics 1: Coordinates We have emphasized that our development of the definition and properties of vectors did not utilize at all the notion of a coordinate system, i.e., an agreed upon set of locations with respect to which the location of a vector in space is described. In this sense a vector is a coordinate free concept. This is important to understand because many physical quantities should be independent of the particular system of coordinates in which they may be considered. Nevertheless, you may have felt a certain vagueness with the geometrical arguments given to prove the various laws of vector algebra. Coordinates are very useful for calculating certain quantities and making certain notions more concrete. Many of the laws of vector algebra, as well as the properties of the dot and cross products, are easy to verify using coordinates (but, these laws and properties are independent of the particular choice of coordinates). In this lecture we will consider different types of coordinate systems and how we represent a vector in a given coordinate system. We begin with a coordinate system with which you should already have some familiarity. Rectangular, or Cartesian, Coordinates in Three-Dimensions. Consider three unit vectors in threedimensional space, i, j, and k emanating from the same point (which we call 0, for origin) having the property that each unit vector is perpendicular to the other two, i.e., i j = 0, j k = 0, i k = 0. (1) These three vectors define three coordinate axes, which are referred to as the x, y, and z axes, respectively. There is a slight technical point that we need to address here, which is the same technical point that arose in the previous lecture when we defined the direction of A B. We will choose i, j, and k in such a way that our coordinate system is a right-handed coordinate system. Here is the rule for doing that (the same as for choosing the direction of the cross product). Position your right hand so that your thumb is perpendicular to your fingers. Then put your thumb in the direction of i and your fingers in the direction of j. Now there are two choices for the direction of k (remember, these three unit vectors are taken to be perpendicular). Our coordinate system will be right-handed if we choose k to be pointing out of the palm of our hand, see Fig. 1. z i k j y x Figure 1: The i, j, and k unit vectors in the direction of the x, y, and z axes, respectively. Note that it also follows, from the definition of the scalar product as well as the definition given for i, j, and k, that: 1

2 i i = 1, j j = 1, k k = 1. (2) Expressing a Vector in Coordinates. Now consider an arbitrary vector A emanating from the origin of this coordinate system, as shown in Fig. 2. z P ( A. i) i i k 0 j A ( A.k) k y x R ( A. j) j Q Figure 2: The projection of A onto i, j, and k. From this figure we see geometrically that A can be expressed as: A = (A i)i + (A j)j + (A k)k, and the numbers A i, A j, and A k are called the components of A with respect to this coordinate system. They are the x, y, and z coordinates of the tip of the vector A. For notational convenience we may write: or A 1 A i, A 2 A j, A 3 A k, A x A i, A y A j, A z A k. Using Coordinates to Compute the Magnitude of a Vector. We can also compute the length (or, magnitude) of A in terms of its components (and this is one place where coordinates are very useful). Refer to Fig. 2. Let P denote the point at the tip of the vector A, let Q denote its projection onto the plane defined by i and j (projection along the k direction), and let R denote the endpoint of (A i)i. Let OP denote the line segment from O to P, QP denote the line segment from Q to P, OQ denote the line segment from O to Q, OR denote the line segment from O to R, and RQ denote the line segment from R to Q. Let OP, QP, OQ, OR, and RQ denote the lengths of these respective segments. Then the length of A is OP and we will derive an expression for it in terms of the components of A. From the Pythagorean theorem we have: and OP 2 = OQ 2 + QP 2, (3) 2

3 Substituting (4) into (3) gives: or, Therefore, we have: OQ 2 = OR 2 + RQ 2. (4) OP 2 = OR 2 + RQ 2 + QP 2, (5) A 2 = A A2 2 + A2 3. (6) A = A A2 2 + A2 3. (7) Note that in passing from (6) to (7) a choice of sign was made in taking the square root. Why did we choose the positive sign? The choice of sign always needs to be taken into account when taking any root of a number. Taking it for granted will eventually cause some regret. (But in this case we used the general fact that length or distance is always nonnegative.) There is another issue here that is important, but often taken for granted. We chose a particular coordinate system to compute the length of a vector. However, one might imagine that length should be an intrinsic property of a vector and, therefore, be a number that is independent of the particular coordinate system with respect to which it is computed. In the realm of classical mechanics (this course) this is true (but this notion may require reconsideration when you learn about the theory of relativity). Vector Addition and Multiplication by Scalars in Coordinates. Vector addition in coordinates is easy. You just add the components. Let Then A = A 1 i + A 2 j + A 3 k, B = B 1 i + B 2 j + B 3 k. A + B = (A 1 + B 1 )i + (A 2 + B 2 )j + (A 3 + B 3 )k. Multiplication of a vector by a scalar is also straightforward. Let a be a scalar. Then: aa = aa 1 i + aa 2 j + aa 3 k. Identification of a Vector with a 3-Tuple. Once our coordinate system is chosen, then the components of a vector (in the chosen coordinate system) uniquely define the vector. Hence, we could identify any vector with the 3-tuple consisting of the components of the vector, i.e., (A 1, A 2, A 3 ). Vector addition corresponds to adding 3-tuples as follows: (A 1, A 2, A 3 ) + (B 1, B 2, B 3 ) = (A 1 + B 1, A 2 + B 2, A 3 + B 3 ), and scalar multiplication of a vector corresponds to scalar multiplication of the 3-tuple as follows: a(a 1, A 2, A 3 ) = (aa 1, aa 2, aa 3 ). 3

4 The Dot, or Scalar Product in Coordinates. Using (1) and (2), the dot product of vectors is easy to compute when the vectors are represented in these coordinates. Let Then A = A 1 i + A 2 j + A 3 k, B = B 1 i + B 2 j + B 3 k. A B = A 1 B 1 + A 2 B 2 + A 3 B 3. (8) It also follows immediately by comparing (8) (with A = B) and (7) that: A = A A. The Vector, or Cross Product in Coordinates. It is also easy to compute the cross product of two vectors in coordinates. However, first we need to derive some properties of cross products of the unit vectors. First, it should be clear that: i i = 0, j j = 0, k k = 0, (9) since the cross-product of two parallel vectors is zero. Now consider i j. By definition this is: i j = i j sin π 2 n = n = k, The unit vector n is from our original definition of cross product. The only issue here is why did we identify n with k? This is because we have chosen our coordinate system according to the right-hand-rule described above. Reasoning in the same way, we have: i j = k, j k = i, j i = k, k j = i, k i = j, i k = j. (10) With (9) and (10) in hand, we have the means to compute the cross product of two vectors. Let and we wish to compute: A = A 1 i + A 2 j + A 3 k, B = B 1 i + B 2 j + B 3 k. A B = (A 1 i + A 2 j + A 3 k) (B 1 i + B 2 j + B 3 k) = (A 1 i + A 2 j + A 3 k) (B 1 i + B 2 j) + (A 1 i + A 2 j + A 3 k) B 3 k, = (A 1 i + A 2 j + A 3 k) B 1 i 4

5 + (A 1 i + A 2 j + A 3 k) B 2 j + (A 1 i + A 2 j + A 3 k) B 3 k, = A 1 i B 1 i + A 2 j B 1 i + A 3 k B 1 i + A 1 i B 2 j + A 2 j B 2 j + A 3 k B 2 j + A 1 i B 3 k + A 2 j B 3 k + A 3 k B 3 k, = (A 2 B 3 A 3 B 2 )i + (A 3 B 1 A 1 B 3 )j + (A 1 B 2 A 2 B 1 )k. A useful mnemonic for remembering the expression for the cross product, A B, in cartesian coordinates is through the determinant, expressed as follows: A B = det i j k A 1 A 2 A 3 B 1 B 2 B 3 = (A 2B 3 A 3 B 2 )i + (A 3 B 1 A 1 B 3 )j + (A 1 B 2 A 2 B 1 )k. (11) Derivatives and Integrals of Vectors Suppose you have a vector, A, that is a function of a scalar variable, t, i.e., A = A(t). What would we mean by the derivative of A(t), and what types of properties would it satisfy? This is a key point for the subject of mechanics because we have already said that many of the quantities of interest will be expressed as vectors, and mechanics is the study of how those quantities change in time (which is the reason we chose to denote the scalar variable by t). Since the mathematical operation of differentiation is concerned with the instantaneous rate of change (of whatever we may be differentiating) it stands to reason that understanding the idea of the derivative of a vector will be important to us. Remember, if a vector changes, it can change in two ways; direction and length, since these are the two characteristics that define a vector. We will assume familiarity with elementary properties of differentiation and integration of functions of one variable. Differentiation of Vectors. The derivative of a vector is defined in terms of the usual difference quotient that you are already familiar with (hopefully) : A(t + t) A(t) (t) = lim, provided the limit exists. (12) t 0 t However, in this course we are only going to differentiate and integrate vectors that are represented in some coordinate system, for now, the coordinate system will be that defined by the unit vectors i, j and k. So letting: then the derivative of A(t) is given by: A(t) = A 1 (t)i + A 2 (t)j + A 3 (t)k, (13) (t) = 1 (t)i + 2 (t)j + 3 (t)k. (14) Is it obvious that (14) follows from the definition (12)? Probably not, it does require a proof, but we won t do that in this course. However, you could try to derive (14) from (12) yourself. A key fact is that the derivatives of i, j and k are zero, i.e., they don t change in length or direction. This follows from their definition, but it is an important point to realize since later we will learn about moving coordinate systems, and in those the unit vectors defining the coordinate axes may change their direction (with respect to...?). 5

6 Example. Let A(t) = e t sinti + lntj + t3 3 k, then (t) = ( e t sin t + e t cost)i + 1 t j + t2 k. So you see that when the vector is represented in cartesian coordinates, differentiation just means differentiating each of the components. Therefore you only need the knowledge of the standard calculus of one variable. Higher order derivatives are done in the same way. For example, the second derivative of A(t) is expressed as: d 2 A 2 (t) = d2 A 1 2 (t)i + d2 A 2 2 (t)j + d2 A 3 2 (t)k. We compute the second derivative of the example given above. Example. Let then A(t) = e t sinti + lntj + t3 3 k, d 2 A 2 (t) = (e t sin t e t cost e t cost e t sin t)i 1 t2j + 2tk. = 2e t costi 1 t2j + 2tk In calculus you learned how to differentiate products of functions. Similar rules exist for vector calculus. However, in vector calculus there are three types of products: the product of a vector function with a scalar function, the scalar product of two vectors, and the cross product of two vectors. For a scalar function a(t), and vector functions A(t), B(t), the rules for differentiating these products are given below. Product Rules for Differentiating Products Involving Vectors d d d (a(t)a(t)) = da a(t)(t) + (t)a(t), (t) + (t) B(t), (A(t) B(t)) = A(t) db (A(t) B(t)) = A(t) db (t) + (t) B(t). Integration of Vectors. Integration of vector functions expressed in coordinates is developed in an analogous way. We have: A(t) = A 1 (t)i + A 2 (t)j + A 3 (t)k (15) 6

7 There is a corresponding fundamental theorem of calculus for integrals of vectors. If there exists a vector function B(t) such that then the indefinite integral of A(t) is given by: A(t) = where c is a constant vector. The definite integral of A(t) is expressed as: A(t) = db (t), (16) db (t) = B(t) + c, (17) We give an example. b a A(t) = b a db (t) = (B(t) + c) b a = B(b) B(a). (18) Example. Let A(t) = e t sinti + lntj + t3 3 k, and we want to compute A(t). Now e t sin t = e t 2 (sin t + cost) + c 1, lnt = t lnt t + c 2, Therefore t 3 3 = t4 12. A(t) = e t t4 (sin t + cost)i + (t lnt t)j k + c 1i + c 2 j + c 3 k 7