Lecture 6 Review of Vectors Physics in more than one dimension (See Chapter 3 in Boas, but we try to take a more general approach and in a slightly different order) Recall that in the previous two lectures we used the two-dimensional vector analog to study complex numbers The concept of a vector is much more general Vectors give us a notation for handling ordered sets of numbers (generally real numbers, but we can also consider complex valued vectors) The rules for handling vectors are rules for manipulating these sets of numbers Making use of our experience with vectors in the real world, we can often make use of explicit geometric interpretations of these manipulations to guide our intuition Note that the numbers, which represent a specific vector, can have different forms depending on our choice of basis vectors (this is very similar to the use of the rectilinear representation versus polar representation of complex numbers), eg, for 3-D there are 3 standard choices, x y z r z,, rectilinear,, spherical,, cylindrical (6) Also note that the quantities preserved during manipulations of vectors are a property of the geometry For example, while the individual components of a vector are changed by a rotation, the length of the vector is not changed (in spherical coordinates r is not changed by rotations, but the other coordinates are changed) Such invariant quantities specify the symmetry properties of the system (ie, the geometry) Symmetries play an essential role in the understanding of physics An N dimensional vector, or N-vector, corresponds to an ordered set of N numbers, arrayed linearly Thus we can use a single index to label the individual elements or components in this linear array with the index running from to N, A A k k N The symbol is being used here to signify the fact that a vector with the overarrow label, A, is associated with an N-tuple of ordinary numbers, and vice versa The two expressions are not strictly equal The left-hand side of this expression is meant to be abstract, while the right-hand side is concrete Once we have defined how to add and subtract these N dimensional arrays and multiply by a constant, we have defined a N-dimensional (represented often by N-D) linear vector space (see 30 in Edition 3 and 38 in Edition of Boas) (6) Physics 7 Lecture 6 Autumn 008
Let's consider a 3-D example, r r, r, r x, y, z by the constant c, cr 3 cx, cy, cz Consider first multiplication (63) Next consider two vectors, which we can add or subtract (component by component), r r r r, r r, r r x x, y y, z z,,,,,3,3 (64) Just as in the case of -D vectors (the familiar scalars, objects unchanged by rotations) addition and subtraction of vectors of higher dimensionality exhibit the properties of being associative and commutative (for addition) - they can be performed using any grouping and in any order, r r r3 r r r3 r r r3 associative, (65) r r r r commutative, but r r r r To further define our vector space we need to be able to say something about the lengths of vectors and they relative directions, ie, we want to say something about the geometry of the space For that purpose we define products of vectors There are two types of multiplication of vectors, ie, two types of products, which find many uses in physics The first is the scalar (or inner or dot) product, which, no surprise, results in a scalar quantity (ie, a single number that is unchanged by rotations) It is written as r r x x y y z z r r 3,, k, k k and is commutative and distributive, but not associative, r r r r r r r3 r r r r3 r r r r r r not commutative, associative 3 3 distributive, (66) (67) The scalar product provides a simple way to calculate the length, or norm, of a vector variously written as Physics 7 Lecture 6 Autumn 008
r r r, r r r r (68) The geometric interpretation of the scalar product is r r r r cos, (69) where is the polar angle between the directions of the two vectors Thus, for vectors with nonzero length, the scalar product vanishes, r r 0, if and only if (or 3/), ie, if the vectors are orthogonal Note that Eq (69), which is valid in any number of Euclidean dimensions, is guaranteed to satisfy our expectation that cos by the familiar Schwartz Inequality r r r r cos r r r r r r N N N, k, k, k, k k k k (60) Also note that, if we consider complex valued vectors, ie, the components r,k are allowed to take complex values, we want to still define the scalar product in such a way that vectors have positive, real lengths Hence in the complex case we define 3 3 3 r r r, kr, k or r, kr, k, r rk rk k k k (6) ASIDE : If we want a picture of what is happening with the scalar product, we can think of the second vector as being (concretely) represented by the column vector of its coordinates (for some set of basis vectors) while the first vector is represented by the complex transpose (Hermitian conjugate) of its coordinates,, ie, a row vector T r, r, T,,,,,3 r,3 r,3 r r, r r r r r, r, r (6) Then the scalar product is in the form of usual matrix multiplication of a row times Physics 7 Lecture 6 3 Autumn 008
a column, element by element, r, 3 r r r,, r,, r,3 r, r, kr, k (63) k r,3 ASIDE : Strictly speaking, when we define the scalar product, we should consider the general bilinear form (see Chapter 0 in Boas) defined (in N dimensions) in terms of an NxN matrix, or metric g kl, (we will introduce a little matrix notation here that will be useful later) N N r r r, k gkl r, l k l (64) We should think of the metric as defining the geometry of the vector space For the familiar rectilinear coordinates defined by fixed, orthogonal unit basis vectors (more on this later) the metric is just the unit matrix and we obtain the earlier expressions The geometry defined by the unit matrix is called a Euclidean geometry, g kl 0 0 0 0 0 0 0 0 0 0 0 0 kl (65) As you may know from your studies of special relativity, the metric describing 4-D space-time has minus signs (either minus and 3 plus signs, or plus and 3 minus signs, ie, there are different conventions in broad usage), and the corresponding geometry is called a Minkowski geometry In general relativity the metric has even more structure, and gravity arises as a result of the response of the metric to the presence of a nonzero energy-momentum density, ie, the metric is treated as a dynamically varying quantity The second type of useful product is the cross or vector product and results in a new (pseudo) vector (ie, the resulting quantity is another N-tuple that changes under Physics 7 Lecture 6 4 Autumn 008
rotations like the usual vector but does not change under reflection as we will discuss later, and is special to 3-D; also called an axial vector), r r y z y z, z x z x, x y x y (66) The vector product is not commutative, r r r r r r, (67) (ie, it is anti-symmetry) However, it is associative and distributive in the sense that r r r3 r r r3 r r3 r, (68) r r r r r r r 3 3 Note the essential feature that the,,3 ordering is maintained in each expression ASIDE: A particularly handy notation for the vector product of 3-vectors employs the unique completely antisymmetric 3-tensor (the Levi-Cevita symbol in 3-D), which is defined by,, klm lkm kml mlk 3 r r r r r r 3 klm, k, l 3, m k, l, m 3 3 r r r r r r 3 k klm, l 3, m klm, l 3, m lm,, (69) The final expression introduces the common notation that repeated indices are understood as being summed over The geometric interpretation of the cross product in Eq (67) is that it defines a vector with magnitude given by (compare Eq (69)) r r r r sin, (60) Physics 7 Lecture 6 5 Autumn 008
where again is the polar angle between the directions of the two vectors The vector direction of r r is orthogonal to both r and r, ie, orthogonal to the plane defined by r and r (Recall that any, non-parallel, vectors define a plane in which they both lie) The sense of the vector along this direction is given by the infamous right-hand-rule, ie, the direction a right handed screw would advance if we turned it in the direction defined by the cross product, rotating r into r The cross product of two (non-null) vectors vanishes if, and only if, the two vectors are parallel, ie, 0 or The cross product of a vector with itself vanishes, r r 0, as is clear from the antisymmetric expression in Eq (69) Note that the fact that this cross product defines a unique 3rd vector is special to N = 3 In higher dimensions there will be more than one direction orthogonal to the plane defined by the original vectors The so-called scalar triple product in the first line of Eq (68), r r r 3, has a particularly simple geometric interpretation as the volume of the parallelepiped which has the three vectors as contiguous edges This geometric interpretation makes clear why the triple product involving only vectors, eg, r r r, must vanish (because the parallelepiped lies in a plane and has no volume) The vector r r is orthogonal to both r and r This reminds us of the essential connection between vectors as N-tuples of numbers and the associated N-dimensional linear vector space Pursuing this idea of an N- dimensional vector space, we note that the strict definition of such a space requires, along with Eqs (63) and (64), also the existence of three (intuitively reasonable) quantities ) The null vector, 0, such that A0 A for all vectors A in the vector space ) The unit constant,, such that A A for all vectors A in the vector space 3) The opposite (or inverse) vector, A the vector space, such that A A 0 for all vectors A in With these definitions we can define the extremely important concept of linearly independent vectors Consider, for example, three (non-null) vectors, A, B, C, in our Physics 7 Lecture 6 6 Autumn 008
N-dimensional vector space These three vectors are said to be linearly independent if, and only if, the equation c A c B c C 3 0, (6) allows only the trivial solution c c c 3 0 Note that, once we pick basis vectors, Eq (6) is really a set of N equations, one for each of the components, c A c B c C for k,,, N k k 3 k 0 ASIDE: As a basic first example assume that we can find orthogonal vectors, A and B, such that AB0, ie, the geometric notion of orthogonality (being at right-angles) means that the scalar product vanishes Then it should be clear that the equation c A c B 0 has only the solution c c 0 To see this explicitly first dot A into the equation yielding c A 0 c 0 Then dot B in yielding c B 0 c 0 In fact, all we really need for linear independence of the vectors is that they not be parallel, AB A B Now the really interesting question is, for a given vector space, what is the maximal number of linearly independent vectors? The general approach is to start with some minimal set of vectors that allow only the trivial solution to Eq (6) and then keep adding vectors until a nontrivial solution is unavoidable For any finite dimensional vector space this will happen eventually and, in fact, serves to define the dimensionality of the space In particular, if the set of vectors, vk, k,, N, are a maximal linearly independent set in a given vector space, then for any vector r in the space we must be able to find a nontrivial solution to N k c v k k ar 0 N N r c v r v a k k k k k k (6) In other words we must be able to write any vector in the space as a linear combination of the vectors in this maximal linearly independent set of vectors Thus we can associate the N-tuple r k with the vector r and this must serve to uniquely Physics 7 Lecture 6 7 Autumn 008
(and concretely) define the vector Hence the vectors vk, k,, N serve as basis vectors for the space, ie, they can be used to define all vectors in the space Another way to express this fact is to say that the vectors vk, k,, N span the space It is conventional for a simple geometric picture, but not necessary, to choose the basic vectors of the N-dimensional vector space to be orthogonal and of unit length, ie, the unit basis vectors you used in introductory physics, v 0, k l v, k l k l kl (63) This last quantity, kl, carries the name Kronecker Delta Function and is just the NxN unit matrix ( s on the diagonal, zeros elsewhere) of Eq (65), kl 0 0 0 0 0 0 0 0 0 0 0 0 kl (64) The matrix in Eq (64), which defines the scalar products of the basis vectors and thus the geometry, is usually called the metric as noted earlier Orthogonal basis vectors can always be found using, for example, the Gram-Schmidt method (see Section 30 in Boas) It is conventional to define special symbols for the orthonormal (orthogonal and normalized) basis vectors For example, in 3-D with a rectilinear basis we define iˆ ˆj iˆ iˆ xˆ, yˆ, zˆ or iˆ, ˆj, kˆ ˆj kˆ 0, ˆj ˆj, iˆ kˆ kˆ kˆ iˆ ˆj kˆ ˆj iˆ kˆ ˆj kˆ iˆ, kˆ ˆj iˆ kˆ iˆ ˆj iˆ kˆ ˆj (65) Physics 7 Lecture 6 8 Autumn 008
The words labeling the choice of the signs in the cross product are that iˆˆ, j, k ˆ (or xˆ, yˆ, z ˆ ) form a righted-handed set of unit basis vectors In terms of the corresponding 3-tuples that we first discussed we have ˆ ˆ ˆ ˆ ˆ iˆ x,0,0, j y 0,,0, k z 0,0, (66) The 3-tuple, ie, the coordinates, corresponding to any other vector can be found from the appropriate scalar product, x r xˆ r iˆ r x, y, z y r yˆ r ˆj z r zˆ r kˆ (67) In terms of these explicit unit basis vectors and components we can express the 3-D cross product in terms of the (familiar?) determinant of 3x3 matrices We have r x xˆ y yˆ z zˆ, r x xˆ y yˆ z zˆ xˆ yˆ zˆ r r x y z x y z y z x z x y xˆ yˆ zˆ y z x z x y y z z y xˆ z x x z yˆ x y y x zˆ (68) The x sub-matrices are referred to as the cofactors (of the unit vectors) This form clearly agrees with Eq (66) ASIDE: For completeness we recall that the determinant of a 3 x 3 matrix can be expressed as Physics 7 Lecture 6 9 Autumn 008
a b c a b c e f d f d e det d e f d e f a b c h i g i g h g h i g h i a ei fh b di fg c dh eg Note that, given a choice of specific basis vectors and the corresponding components, we move from the discussion of relatively abstract vectors to the more concrete discussion of components We can rewrite various properties of vectors in terms of ordinary equations involving their components The statement r r 0 is the statement that the vectors are orthogonal without reference to any specific basis vectors Once we have chosen a basis set (ie, we have both the vector space and a metric in that space) we can write for orthogonal vectors (the symbol stands for perpendicular or orthogonal) r r 0 r r x x y y z z 0 Likewise parallel vectors are defined by a vanishing cross product, which we can rewrite in terms of components (in 3-D) as r r 0 r r y z z y z x x z x y y x 0 x y z x y z r C r C a constant We can pictorially represent a vector as the line from the origin, 0,0,0 to the point defined by the components x, y, z, where r x y z is the distance from the origin to the point x, y, z Thus there is a one-toone mapping between points and vectors (from the origin to the point, once we have chosen an origin) Both are specified by an N-tuple of numbers More generally we can think of this vector as the difference between two (other) vectors, one from an arbitrary point to the point x, y, z and the second from the same arbitrary point to the origin, ie, the point 0,0,0 (69) (630) Physics 7 Lecture 6 0 Autumn 008
Next let s think about how we define various geometric figures in a 3-dimensional space using the quantities above The simplest as a 0-D object, the point, defined by x, y, z, which also defines a vector once we choose an origin Next consider a -D object, the straight line We can uniquely define a straight line by specifying (ordered) points as above, or by specifying both the direction of the line (modulo the sense) and a point through which the line passes For example, a straight line passing through the point x0, y0, z0 r 0 and parallel to a vector v In general we expect a -D object to involve a single continuous parameter, which we will call t here and which identifies the continuum of points along the line We know from Eq (630) that all vectors of the form vt are parallel to the vector v We can thus define a line passing through the point x0, y0, z 0 by defining a vector as a function of the parameter t in the form r t r vt 0 As the parameter t varies from to the points defined by the vector rt out the desired straight line Note that, in general, neither the vectorrt nor the (63) trace vector r 0 are along the line, but rather describe the location of points on the line with respect to the origin The obvious physical interpretation is the trajectory of a free r t 0 r r t 0 r t v constant (the particle with initial conditions, 0 linear dependence on the product vt is what guarantees a straight line) Note that this definition in terms of vectors works in any number of dimensions In terms of components in 3-D we represent a straight line by the equations (compare to Eq (630), the form here, ie, the number of equations, depends on the number of dimensions) x x0 y y0 z z0 t (63) v v v x y z For example, with,,0 r 0 and v xˆ zˆ we want to define a line in terms of two points, r 0 and r, we can just substitute r r in the above equations so that the analog of Eq (63) is 0 we have r t t xˆ yˆ tzˆ If Physics 7 Lecture 6 Autumn 008
x x0 y y0 z z0 t x x y y z z 0 0 0 (633) The next most interesting (but simple) object is a (flat) -D surface or plane In 3-D we can define this object in terms of passing through a point x0, y0, z0 r 0 and being orthogonal to a vector N, the normal vector to the plane Note that this latter constraint produces a -D surface only in a 3-D space In N-dimensions such a constraint produces an N--dimensional subspace From Eq (69) it follows that the required surface in 3-D is given by the equation r r N 0 x x N y y N z z N 0 0 0 x 0 y 0 z (634) Since this equation is a single constraint on the 3 parameters x, y, z, it clearly defines a -D (-parameter) surface (or subspace) Since it is a linear constraint, the surface is a flat plane (think about that, the normal vector N is the same vector everywhere on the surface) x y z, x, y, z, x, y, z Similarly we can use three points,,, 3 3 3, to define a plane We just use the 3 points to define vectors (in the plane) by taking differences and then use their cross product to define the normal vector to the plane Substituting into Eq (634) then yields a triple scalar product form Using the first point as the reference point (this is an arbitrary choice) we have the defining equation as r r r r r3 r 0 x x y y z3 z z z y3 y y y z z x3 x x x z3 z z z x x y y y y x x 3 3 (635) The reader is encouraged to verify that this expression is invariant under permutations of the indices,,3 These expressions also illustrate the fact that vector notation is typically much more efficient (ie, simpler to express, if not to use) than the more concrete component notation (recall our motto) As a specific example let x, y, z, x, y, z, x3, y3, z 3 be,,,,3,0, 0,, Thus the normal vector to the desired surface is Physics 7 Lecture 6 Autumn 008
xˆ yˆ zˆ N r r r r 3 0 3 0 xˆ yˆ zˆ xˆ yˆ zˆ 6 9 3 4 (636) The equation defining the plane is then (writing the triple product in determinant form) r r N 0 3 6 x 8 y z x y z 0 3 6x 8y z 0 3x 4y z 6 0 (637) In physics we are often interested in the minimum or perpendicular distance between a point and a (straight) line or between a point and a (flat) surface With the point specified by the vector r (with respect to some arbitrary origin) and the line specified by an equation as in Eq (63) (with r 0 defined with respect to the same arbitrary origin), we can calculate the perpendicular distance by taking a cross product of the direction ˆv with the vector from r to any point rt along the line (recall that the cross product yields the orthogonal component, while scalar product yields the parallel component) In the figure the desired distance is the length of the (unlabeled) dashed line, which is clearly r r sin Thus we have the perpendicular distance from a point to a line as d r r t v r r v (638) v v 0 point-line We take the modulus because the sign is irrelevant here and we divide by v since we only care about the direction of the line and not the magnitude of the vector ˆv Note that, as expected, the distance does not depend on the parameter t that defines a,, r and specific point along the line For example, if the point is defined by Physics 7 Lecture 6 3 Autumn 008
the line is the one above, r t t xˆ yˆ tzˆ a perpendicular distance defined by r r t t xˆ yˆ t zˆ (,,0 r0 point-line, v xˆ zˆ ) we have xˆ yˆ zˆ r r t v t t xˆ t t yˆ zˆ d xˆ yˆ zˆ 0 4 4 3 4 5 (639) Another useful distance is the perpendicular distance from a point to a (flat) plane In this case we obtain the required distance by projecting onto the normal to the plane If r is the vector to the point of interest with respect to an arbitrary origin, r is the vector to any point on the plane (with respect to the same origin) and N is the normal to the plane, the required perpendicular distance is d r r N (640) N point-plane (Note that we have accounted for the fact that the normal may not be defined as a unit vector) Consider, for example, the point above,,, r, and the plane defined in Eq (637) above by the equation 3x 4y z 6 0 with normal N 3xˆ 4yˆ zˆ Since we can choose any point in the plane, ie, any point that r 0,0, 6 Thus we have satisfies the equation for the plane, we can take r r xˆ yˆ 7, zˆ r r N 3 4 7 (64) d point-plane 6 9 6 6 You should convince yourself that you will obtain the same distance starting with any other point on the plane, eg, (-,0,0) Physics 7 Lecture 6 4 Autumn 008
As a final geometric application of these ideas consider the line defined by the intersection of two planes (in 3-D) Since this line must lie in both planes, it must be perpendicular to both normal vectors, ie, the normal vectors to the planes Thus the direction of the line is provided by the cross product of the normal vectors To fully specify the line we need only point common to both planes and then we can apply Eq (63) For example, consider planes defined by the equations x y 3z 4 and x y z 5 with normal vectors N ˆ ˆ ˆ x y 3z and N ˆ ˆ ˆ x y z (the reader is encouraged to verify that these definitions are consistent) By inspection (ie, solving simultaneous equations) we see that the point r0 4 5, 3 5,0 lies in both planes and we have xˆ yˆ zˆ N N 3 xˆ 7yˆ 5zˆ v 4 3 r t t xˆ 7t yˆ 5tzˆ 5 5 (64) where the final expression defines the desired line of intersection Physics 7 Lecture 6 5 Autumn 008