lecture 6 review of vectors physics in more than one dimension...

Physics 227 Lecture 6 1 Autumn 2008

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), e.g., for 3-D there are 3 standard choices,

, , rectilinear , , spherical , , cylindrical .x y z r z (6.1)

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 (i.e., 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 1 to N,

1

.k k NA A

(6.2)

The symbol is being used here to signify the fact that a vector with the “over-

arrow” 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 3.10 in Edition 3 and 3.8 in Edition 2 of Boas).


Let's consider a 3-D example, 1 2 3, , , ,r r r r x y z

. Consider first multiplication

by the constant c,

, , .cr cx cy cz

(6.3)

Next consider two vectors, which we can add or subtract (component by component),

1 2 1,1 2,1 1,2 2,2 1,3 2,3 1 2 1 2 1 2, , , , .r r r r r r r r x x y y z z

(6.4)

Just as in the case of 1-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,

1 2 3 1 2 3 1 2 3

1 2 2 1 1 2 2 1

associative ,

commutative , but .

r r r r r r r r r

r r r r r r r r

(6.5)

To further define our vector space we need to be able to say something about the

lengths of vectors and they relative directions, i.e., 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, i.e., 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 (i.e., a single number that is unchanged by rotations). It is

written as

3

1 2 1 2 1 2 1 2 1, 2,

1

,k kk

r r x x y y z z r r

(6.6)

and is commutative and distributive, but not associative,

1 2 2 1

1 2 3 1 2 1 3

1 2 3 1 2 3

commutative ,

distributive ,

associative .

r r r r

r r r r r r r

r r r r r r not

(6.7)

The scalar product provides a simple way to calculate the length, or norm, of a vector

variously written as


2

, .r r r r r r r

(6.8)

The geometric interpretation of the scalar product is

1 2 1 2 12cos ,r r r r

(6.9)

where 12 is the polar angle between the directions of the two vectors. Thus, for

vectors with nonzero length, the scalar product vanishes, 1 2 0r r

, if and only if

12 2 (or 3/2), i.e., if the vectors are orthogonal. Note that Eq. (6.9), which is

valid in any number of Euclidean dimensions, is guaranteed to satisfy our expectation

that 12cos 1 by the familiar Schwartz Inequality

2 2

1 2 1 2 12 1, 2, 1, 2, 1 2

1 1 1

cos .N N N

k k k k

k k k

r r r r r r r r r r

(6.10)

Also note that, if we consider complex valued vectors, i.e., the components 1,kr 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

1 2 1, 2, 1, 2,

1 1 1

or , .k k k k k kk k k

r r r r r r r r r

(6.11)

ASIDE 1: 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,, i.e.,

a row vector

2,1 1,1

†

2 2,2 1 1 1 1,2 1,1 1,2 1,3

2,3 1,3

, , , .

T

T

r r

r r r r r r r r r

r r

(6.12)

Then the scalar product is in the form of usual matrix multiplication of a row times


a column, element by element,

2,1 3

1 2 1,1 1,2 1,3 2,2 1, 2,

1

2,3

, , .k kk

r

r r r r r r r r

r

(6.13)

ASIDE 2: Strictly speaking, when we define the scalar product, we should consider

the general bilinear form (see Chapter 10 in Boas) defined (in N dimensions) in

terms of an NxN matrix, or metric klg , (we will introduce a little matrix notation

here that will be useful later)

1 2 1, 2,1 1

.N N

k kl l

k l

r r r g r

(6.14)

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,

1 0 0 0

0 1 0 0

.

0 0 1 0

0 0 0 1

kl

kl

g

(6.15)

As you may know from your studies of special relativity, the metric describing 4-D

space-time has minus signs (either 1 minus and 3 plus signs, or 1 plus and 3 minus

signs, i.e., there are 2 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, i.e., 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 (i.e., the resulting quantity is another N-tuple that changes under


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),

1 2 1 2 2 1 1 2 2 1 1 2 2 1, , .r r y z y z z x z x x y x y

(6.16)

The vector product is not commutative,

1 2 2 1 2 1,r r r r r r

(6.17)

(i.e., it is anti-symmetry). However, it is associative and distributive in the sense that

1 2 3 1 2 3 2 3 1

1 2 3 1 2 1 3

,

.

r r r r r r r r r

r r r r r r r

(6.18)

Note the essential feature that the 1,2,3 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

123

3

1 2 3 1, 2, 3,

, , 1

3

2 3 2, 3, 2, 3,

, 1

, 1,

,

.

klm lkm kml mlk

klm k l m

k l m

klm l m klm l mkl m

r r r r r r

r r r r r r

(6.19)

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. (6.17) is that it defines a

vector with magnitude given by (compare Eq. (6.9))

1 2 1 2 12sin ,r r r r

(6.20)


where again 12 is the polar angle between the directions of the two vectors. The

vector direction of 1 2r r

is orthogonal to both 1r

and 2r

, i.e., orthogonal to the plane

defined by 1r

and 2r

. (Recall that any 2, 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, i.e., the direction a right handed screw would advance if we turned it

in the direction defined by the cross product, rotating 1r

into 2r

. The cross product of

two (non-null) vectors vanishes if, and only if, the two vectors are parallel, i.e.,

12 0 or . The cross product of a vector with itself vanishes, 1 1 0r r

, as is

clear from the antisymmetric expression in Eq. (6.19). 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 2

vectors.

The so-called scalar triple product in the first line of Eq. (6.18), 1 2 3r r r

, 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 2 vectors, e.g., 1 1 2r r r

, must vanish

(because the parallelepiped lies in a plane and has no volume). The vector 1 2r r

is

orthogonal to both 1r

and 2r

.

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. (6.3) and (6.4), also the existence of three (intuitively reasonable)

quantities.

1) The null vector, 0

, such that 0A A

for all vectors A

in the vector space.

2) The unit constant, 1 , such that 1A A

for all vectors A

in the vector space.

3) The opposite (or inverse) vector, A

, such that 0A A

for all vectors A

in

the vector space.

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


N-dimensional vector space. These three vectors are said to be linearly independent

if, and only if, the equation

1 2 3 0,c A c B c C

(6.21)

allows only the trivial solution 1 2 3 0c c c . Note that, once we pick basis vectors,

Eq. (6.21) is really a set of N equations, one for each of the components,

1 2 3 0k k kc A c B c C for 1,2, ,k N .

ASIDE: As a basic first example assume that we can find 2 orthogonal vectors, A

and B

, such that 0A B

, i.e., the geometric notion of orthogonality (being at

right-angles) means that the scalar product vanishes. Then it should be clear that

the equation 1 2 0c A c B

has only the solution 1 2 0c c . To see this explicitly

first dot A

into the equation yielding 2

1 10 0c A c

. Then dot B

in yielding

2

2 20 0c B c

. In fact, all we really need for linear independence of the 2

vectors is that they not be parallel, A B 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. (6.21) 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, , 1, ,kv 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

1

1 1

0

1.

N

k k

k

N N

k k k k

k k

c v ar

r c v r va

(6.22)

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 kr with the vector r

and this must serve to uniquely


(and concretely) define the vector. Hence the vectors , 1, ,kv k N

serve as basis

vectors for the space, i.e., they can be used to define all vectors in the space. Another

way to express this fact is to say that the vectors , 1, ,kv 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, i.e., the

unit basis vectors you used in introductory physics,

0,

.1,

k l kl

k lv v

k l

(6.23)

This last quantity, kl , carries the name Kronecker Delta Function and is just the NxN

unit matrix (1’s on the diagonal, zeros elsewhere) of Eq. (6.15),

1 0 0 0

0 1 0 0

.

0 0 1 0

0 0 0 1

kl

kl

(6.24)

The matrix in Eq. (6.24), 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 3.10 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

ˆ ˆ ˆ ˆ

ˆ ˆˆ ˆ ˆ ˆ ˆˆ ˆ ˆ, , or , , 0, 1,

ˆ ˆ ˆˆ

ˆ ˆˆ ˆ ˆ ˆ

ˆ ˆˆ ˆ ˆ ˆ, .

ˆ ˆˆ ˆ ˆ ˆ

i j i i

x y z i j k j k j j

i k k k

i j k j i k

j k i k j i

k i j i k j

(6.25)


The words labeling the choice of the signs in the cross product are that ˆˆ ˆ, ,i j k (or

ˆ ˆ ˆ, ,x y z ) form a “righted-handed” set of unit basis vectors. In terms of the

corresponding 3-tuples that we first discussed we have

ˆˆ ˆˆ ˆ ˆ1,0,0 , 0,1,0 , 0,0,1 .i x j y k z (6.26)

The 3-tuple, i.e., the coordinates, corresponding to any other vector can be found

from the appropriate scalar product,

ˆˆ

ˆˆ, , .

ˆˆ

x r x r i

r x y z y r y r j

z r z r k

(6.27)

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

1 1 1 1 2 2 2 2

1 2 1 1 1

2 2 2

1 1 1 1 1 1

2 2 2 2 2 2

1 2 1 2 1 2 1 2 1 2 1 2

ˆ ˆ ˆ ˆˆ ˆ,

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ ˆ ˆ.

r x x y y z z r x x y y z z

x y z

r r x y z

x y z

y z x z x yx y z

y z x z x y

y z z y x z x x z y x y y x z

(6.28)

The 2x2 sub-matrices are referred to as the cofactors (of the unit vectors). This form

clearly agrees with Eq. (6.16).

ASIDE: For completeness we recall that the determinant of a 3 x 3 matrix can be

expressed as


det

.

a b c a b ce f d f d e

d e f d e f a b ch 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 1 2 0r r

is the

statement that the 2 vectors are orthogonal without reference to any specific basis

vectors. Once we have chosen a basis set (i.e., 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)

1 2 1 2 1 2 1 2 1 20 0.r r r r x x y y z z

(6.29)

Likewise 2 parallel vectors are defined by a vanishing cross product, which we can

rewrite in terms of components (in 3-D) as

1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

1 1 1

2 2 2

1 2

0 || 0

a constant

.

r r r r y z z y z x x z x y y x

x y zC

x y z

r C r

(6.30)

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 2 2 2r x y z

is the distance from

the origin to the point , ,x y z . Thus there is a one-to-one 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, i.e., the point

0,0,0 .


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 1-D object, the straight line. We can uniquely define a straight line by specifying 2

(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 0 0 0 0, ,x y z r

and parallel to a

vector v

. In general we expect a 1-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. (6.30) that all

vectors of the form vt

are parallel to the vector v

. We

can thus define a line passing through the point 0 0 0, ,x y z by defining a vector as a function of the parameter t in the form

0 .r t r vt

(6.31)

As the parameter t varies from to the points defined by the vector r t

trace

out the desired straight line. Note that, in general, neither the vector r t

nor the

vector 0r

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

particle with initial conditions 00r t r

, 0r t r t v constant (the

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.

(6.30), the form here, i.e., the number of equations, depends on the number of

dimensions)

0 0 0 .x y z

x x y y z zt

v v v

(6.32)

For example, with 01,2,0 r

and ˆ ˆ2v x z

we have ˆ ˆ ˆ1 2 2r t t x y tz

. If

we want to define a line in terms of two points, 0r

and 1r

, we can just substitute

1 0r r

in the above equations so that the analog of Eq. (6.32) is


0 0 0

1 0 1 0 1 0

x x y y z zt

x x y y z z

(6.33)

The next most interesting (but simple) object is a (flat) 2-D surface or plane. In 3-D

we can define this object in terms of passing through a point 0 0 0 0, ,x y z r

and

being orthogonal to a vector N

, the normal vector to the plane. Note that this latter

constraint produces a 2-D surface only in a 3-D space. In N-dimensions such a

constraint produces an N-1-dimensional subspace. From Eq. (6.29) it follows that the

required surface in 3-D is given by the equation

0 0 0 00 0.x y zr r N x x N y y N z z N

(6.34)

Since this equation is a single constraint on the 3 parameters , ,x y z , it clearly defines

a 2-D (2-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).

Similarly we can use three points, 1 1 1, ,x y z , 2 2 2, ,x y z , 3 3 3, ,x y z , to define a plane. We just use the 3 points to define 2 vectors (in the plane) by taking differences

and then use their cross product to define the normal vector to the plane. Substituting

into Eq. (6.34) 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

1 2 1 3 1

1 2 1 3 1 2 1 3 1

1 2 1 3 1 2 1 3 1

1 2 1 3 1 2 1 3 1

0

.

r r r r r r

x x y y z z z z y y

y y z z x x x x z z

z z x x y y y y x x

(6.35)

The reader is encouraged to verify that this expression is invariant under

permutations of the indices 1,2,3. These expressions also illustrate the fact that

vector notation is typically much more efficient (i.e., simpler to express, if not to use)

than the more concrete component notation (recall our motto). As a specific example

let 1 1 1 2 2 2 3 3 3, , , , , , , ,x y z x y z x y z be 1,1,1 , 2,3,0 , 0,1, 2 . Thus the normal vector to the desired surface is


2 1 3 1

ˆ ˆ ˆ

2 1 3 1 0 1

0 1 1 1 2 1

ˆ ˆ ˆ ˆˆ ˆ6 1 9 2 2 3 4 .

x y z

N r r r r

x y z x y z

(6.36)

The equation defining the plane is then (writing the triple product in determinant

form)

1

1 1 1

0 3 2 1 6 1 8 1 2 1

1 0 3

6 8 2 12 0 3 4 6 0.

x y z

r r N x y z

x y z x y z

(6.37)

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 1r

(with respect to some arbitrary origin) and the line

specified by an equation as in Eq. (6.31) (with 0r

defined with respect to the same arbitrary origin), we

can calculate the perpendicular distance by taking a

cross product of the direction v̂ with the vector from 1r

to any point r t

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

1 sinr r

. Thus we have the perpendicular distance from a point to a line as

1 1 0

point-line .r r t v r r v

dv v

(6.38)

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 v̂ . Note that, as expected, the distance does not depend on the parameter t that defines a

specific point along the line. For example, if the point is defined by 11,1,1 r

and


the line is the one above, ˆ ˆ ˆ1 2 2r t t x y tz

( 01,2,0 r

, ˆ ˆ2v x z

) we have

a perpendicular distance defined by

1

1

point-line

ˆ ˆ ˆ2 1

ˆ ˆ ˆ

ˆ ˆ ˆ2 1 1 1 2 2 2 2

2 0 1

ˆ ˆ ˆ2 2

1 4 4 3.

1 4 5

r r t t x y t z

x y z

r r t v t t x t t y z

x y z

d

(6.39)

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 1r

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

1

point-plane .r r N

dN

(6.40)

(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, 11,1,1 r

, and the plane defined in

Eq. (6.37) above by the equation 3 4 6 0x y z with normal

ˆ ˆ ˆ2 3 4N x y z

. Since we can choose any point in the plane, i.e., any point that

satisfies the equation for the plane, we can take 0,0, 6r

. Thus we have

1

1

point-plane

ˆ ˆ ˆ7 ,

2 3 4 7 12

12 6.

2 9 16 1 26

r r x y z

r r N

d

(6.41)

You should convince yourself that you will obtain the same distance starting with any

other point on the plane, e.g., (-2,0,0).


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, i.e., the 2 normal vectors to the planes. Thus

the direction of the line is provided by the cross product of the 2 normal vectors. To

fully specify the line we need only 1 point common to both planes and then we can

apply Eq. (6.31). For example, consider 2 planes defined by the equations

2 3 4x y z and 2 5x y z with normal vectors 1 ˆ ˆ ˆ2 3N x y z

and

2ˆ ˆ ˆ2N x y z

(the reader is encouraged to verify that these definitions are

consistent). By inspection (i.e., solving 2 simultaneous equations) we see that the

point 0 14 5, 3 5,0r

lies in both planes and we have

1 2

ˆ ˆ ˆ

ˆ ˆ ˆ1 2 3 7 5

2 1 1

14 3ˆ ˆ ˆ7 5

5 5

x y z

N N x y z v

r t t x t y tz

(6.42)

where the final expression defines the desired line of intersection.

lecture 6 review of vectors physics in more than one dimension...

Documents