Lecture 6 Review of Vectors Physics in more than one ...
Transcript of Lecture 6 Review of Vectors Physics in more than one ...
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).
Physics 227 Lecture 6 2 Autumn 2008
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 k
k
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
Physics 227 Lecture 6 3 Autumn 2008
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 k
k 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
Physics 227 Lecture 6 4 Autumn 2008
a column, element by element,
2,1 3
1 2 1,1 1,2 1,3 2,2 1, 2,
1
2,3
, , .k k
k
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
Physics 227 Lecture 6 5 Autumn 2008
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)
Physics 227 Lecture 6 6 Autumn 2008
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
Physics 227 Lecture 6 7 Autumn 2008
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
Physics 227 Lecture 6 8 Autumn 2008
(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)
Physics 227 Lecture 6 9 Autumn 2008
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
Physics 227 Lecture 6 10 Autumn 2008
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 .
Physics 227 Lecture 6 11 Autumn 2008
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
Physics 227 Lecture 6 12 Autumn 2008
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
Physics 227 Lecture 6 13 Autumn 2008
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
Physics 227 Lecture 6 14 Autumn 2008
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).
Physics 227 Lecture 6 15 Autumn 2008
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 1ˆ 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.