Hassani S. Mathematical Methods: For Students of Physics and Related Fields

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234 Finite-Dimensional Vector Spaces

Now divide the second equation by −3,

+ x

=0,

+2x

= −1, (7.41)

−3x

+0·x

=2.

Multiply the second equation of (7.41) successively by −3 and +3 and add it to the

third and last equations. This will yield

+ x

=0,

= −3, (7.42)

=4.

Subtract the third equation from the last to get

+ x

=0,

= −3, (7.43)

0=7.

In this case, we have an equation of the form 0 = d

for which d

=7. So,the

system is incompatible, i.e., it has no solution.



A system of linear equations (7.30) is homogeneous if the constants b

homogeneous

system of linear

equations

on the RHS are all zero. Such a system always has a trivial solution with

all the unknowns equal to zero. There may be no further solutions, but if

the number of variables exceeds the number of equations, the last equation

of (7.32) will always contain more than one variable at least one of which can

be chosen at will. Furthermore, the inconsistent equations 0 = d

can never

arise for such homogeneous equations. Hence,

Box 7.6.2. Asystemofm homogeneous linear equations in n unknowns,

with n>m, always has a solution in which not all the unknowns are zero.

7.7 Problems

7.1. Show that Equation (7.4) is a linear transformation.

7.2. Verify that the operation of diﬀerentiation of any order is a linear trans-

formation on P

[t].

7.3. Show that

L ≡ p

(x)

+ p

(x)

+ p

(x)

is a linear operator on the space of diﬀerentiable functions.

7.7 Problems 235

7.4. Show that the coeﬃcients in P

[t] of the second derivative of an arbitrary

polynomial f(t)=α

+ α

t + α

+ α

can be obtained by the product of

the matrix of the second derivative obtained in Example 7.1.1, and the 4 × 1

column vector representing f(t).

7.5. Express the element in the ith row and jth column of a unit matrix in

terms of the Kronecker delta.

7.6. Suppose x is an eigenvector of T with eigenvalue λ. Show that, for any

constant α, αx is also an eigenvector of T with the same eigenvalue.

7.7. Find the length of a

of Example 7.4.1 in terms of x

. Now show that

/|a

| =

7.8. Show that the rotation of the plane aﬀects all vectors in the plane. Hint:

Try to ﬁnd an eigenvector of the 2 × 2 rotation matrix (6.24).

7.9. Find the eigenvalues and normalized (unit length) eigenvectors of the

following matrices. In cases where the matrix is symmetric, verify directly

that its eigenvectors corresponding to diﬀerent eigenvalues are orthogonal.

(a)



2 −2



. (b)





. (c)





(d)

⎛

⎝

110

101

011

⎞

⎠

. (e)

⎛

⎝

200

011

⎞

⎠

. (f)

⎛

⎝

111

⎞

⎠

7.10. Show that Box 7.4.1 is not necessarily true for a general inner product

with matrix G. However, if G and T commute (i.e., if GT = TG), then Box

7.4.1 holds. Hint: Follow the argument after Box 7.4.1 and see how far you

can proceed.

7.11. Show that the inner product deﬁned in Equation (7.21) is indeed a

positive deﬁnite inner product.

7.12. Find the fourth Legendre polynomial using the results of Example 7.5.1.

7.13. Find the ﬁrst three Hermite polynomials using the standardization (or

normalization) Equation (7.27).

7.14. The volume element of a four-dimensional Euclidean space with Carte-

sian coordinates x, y, z,andw is dxdydzdw. In any other coordinate system,

it is given by a 4-dimensional generalization of the Jacobian (6.66)

(a) Write this Jacobian for a general transformation to coordinates s, t, u

and v where x, y, z,andw are functions of these new coordinates.

(b) Now consider the 4-dimensional spherical coordinates:

x = r sin μ sin θ cos ϕ

y = r sin μ sin θ sin ϕ

z = r sin μ cos θ

w = r cos μ

236 Finite-Dimensional Vector Spaces

and calculate the 4-dimensional Jacobian to ﬁnd the volume element of a 4-

dimensional sphere.

radius a.

7.15. Determine the r of Equation (7.34) for each of the following systems of

linear equations and whether or not the system is compatible. If the system

is compatible, ﬁnd a solution for it.

(a)

2x −y − 4z =1,

x +2y +2z =0,

−x −y +6z =3.

(b)

x + y + z = −1,

2x − y +2z = −5,

3x +3y + z =1.

(c)

x + y + z =2,

2x − y +2z = −2,

+ y − z =4.

(d)

2x + y − 2z =2,

3x −y − 4z = −1,

3x +4y − 2z =7.

(e)

3x +2y =7,

x + y + z =6,

5x +4y +2z =19,

x − 2y = −5.

(f)

x +5y − z =2,

2x + y +3z = −1,

−x +3y +2z = −3,

3x +2y − z =4

Chapter 8

Vectors in Relativity

One of the most rewarding applications of vectors is to relativity. The special

theory of relativity (STR) was a direct consequence of Maxwell’s equations,

which summarize the entire theory of electromagnetism (see Section 15.4).

These equations predict mathematically that there must exist electromagnetic

(EM) waves which travel at the speed of light in empty space. This speed c is

found in terms of purely electric and magnetic measurements:

c =

√





(4π × 10

−7

)(8.854 ×10

−12

)

=2.998 × 10

m/s,

where 

=1/4πk

and μ

=4πk

,withk

and k

the electric and magnetic

constants introduced in Chapter 1.

Imagine two laboratories on two spaceships, S

and S

,withS

behind

(and moving towards) S

at 0.9c relative to S

. The physicists on S

perform

electric and magnetic experiments, measure 

and μ

, and conclude that

EM waves travel at 300,000 km/s in empty space. The physicists on S

also perform electric and magnetic experiments, measure 

and μ

,andalso

conclude that EM waves travel at 300,000 km/s in empty space. Now a

physicist on S

takes a ﬂashlight and sends a beam of light in the forward

direction in empty space. The consequence of Maxwell’s equations is that the

physicists on S

, although seeing S

moving towards them at 0.9c and the

light beam moving away from S

at c, conclude that the speed of the light

beam is c and not 1.9c, as expected from the Newtonian law of addition of

velocities.

To appreciate the strange consequence of Maxwell’s equations, consider

the following example: A train moving at 30 m/s and a passenger throwing

law of addition

of velocities

a ball in the forward direction with a speed of 20 m/s. A ground observer

measures the speed of the ball to be 30 + 20 = 50 m/s: velocities add. Here is

another familiar example: A car moves at 75 mph on a highway on which your

carismovingat50mph. Thespeedofthefastcarrelative to you is 25 mph.

You speed up to 70 mph. Then the other car appears to have “slowed down,”

because, now you measure its speed relative to you to be only 5 mph. Go to

238 Vectors in Relativity

outer space, let someone in your spaceship ﬁre a bullet moving at 500 mph.

Increase your speed to 450 mph, the bullet appears to be moving at 50 mph

away from you. Increase your speed by another 100 mph. You catch up with

the bullet, and if you decrease your speed by 50 mph, the bullet appears

stationary relative to you.

Now shoot a beam of light forward, and once the beam leaves your ﬂash-

light, accelerate your spaceship to a speed of 299,000 km/s. Measure the

speed of the light beam. It is still 300,000 km/s, and not 1000 km/s, as in-

tuitively expected! Maxwell’s equations defy intuition, and the (STR), which

is entirely based on these equations is extremely counter-intuitive. Let us

summarize these observations:

Box 8.0.1. (Principle of Relativity) Every time you detect an electro-

magnetic wave, it moves at the rate of 300,000 km per second in vacuum,

regardless of the motion of its source or its detector. Speed of light in

vacuum is a universal constant.

An immediate consequence of the principle of relativity is the fact that time

is observer-dependent. As Einstein said “Time is something that is measured

by clocks.” So, let us look at the eﬀect of motion on clocks. The clock best

suited for this investigation is the “arm” of the Michelson–Morley apparatus

shown in Figure 8.1. It consists of a source S of light, or electromagnetic

waves, and a mirror M. The distance between S and M is L. Therefore, it

takes light Δτ

tick

≡ 2L/c to go from S to M and back. If we place a lightthe

Michelson–Morley

clock

sensitive “ticker” at S, the clock will tick every Δτ

tick

second. We call such

a clock a Michelson–Morley clock, or an MM clock,andΔτ

tick

the proper

tick of the MM clock. Δτ

tick

is the tick measured by an observer for whom

the clock is at rest, or for whom the beginning and the end of a tick occur at

thesamelocation.

Figure 8.1: A Michelson–Morley clock. A “tick” of this clock occurs when the light

signal makes a round trip along the length L.

8.1 Proper and Coordinate Time 239

8.1 Proper and Coordinate Time

An MM clock is placed on a train and observed by two observers, O (on the

ground) and O



(on the train) moving to the right of O. Consider three events:

The emission of a light beam at S, its reﬂection at M, and its reception at

S. These three events constitute one tick. Let us denote them by E

, E

and E

, respectively. How does O



see the ticking of the clock? The clock

is sitting right beside her, and she observes the whole process of ticking as

the light going straight up and coming straight down. She concludes that her

clock’s ticks are Δτ

tick

long.

Now, let us see how O perceives the succession of these three events. Since

the clock is moving to the right, the light signal that leaves S will reach M

only after M has moved to the right. Thus, to O, the events E

and E

are

separated not only by a vertical distance, but also by a horizontal distance (see

Figure 8.2). Since the speed of light is the same for all observers, O concludes

that it takes light more than 2L/c to travel

and E

. Therefore, he

concludes that the clock on the train must tick slower!

moving clocks

slow down.

We can quantify the above statement by referring to the triangle E

of Figure 8.2. Pythagoras’ theorem implies

Let the speed of the train be v and the light beam’s travel time from S to M

be δt according to O.Then

A = vδt and E

= cδt with c the (universal)

speed of light. Putting all of this in the above equation gives

(cδt)

=(vδt)

+ L

⇒ c

(δt)

= v

(δt)

+ L

, (8.1)

(δt)

−

(δt)

, ⇒ (δt)



1 −



Figure 8.2: A moving Michelson–Morley clock. The path of light (represented by a

black dot) is not a vertical line but a slanted one due to the motion of M.

240 Vectors in Relativity

This yields

(δt)

1 − v

⇒ δt =

L/c



1 − (v/c)

Let us denote by Δt

tick

the duration of the light’s round trip as seen by O.

Then

Δt

tick

2L/c



1 − (v/c)

Δτ

tick



1 − (v/c)

. (8.2)

In deriving this equation, we have tacitly assumed that motion does not aﬀect

transverse lengths. Thus the length of the MM clock does not change because it

motion does not

aﬀect transverse

lengths.

is perpendicular to the direction of motion. To see this, consider the distance

between two wheels of a train, and suppose that this distance shrinks

due to

its motion as seen by a ground observer. This means that the wheels will fall

between the rails. On the other hand, the engineer of the train sees the rail

moving and concludes that the distance between the rails shrink; i.e., that the

wheels fall outside the rails. This contradicts the previous conclusion. Thus,

the length perpendicular to the direction of motion must not change.

Although Equation (8.2) is derived for a single tick, it really applies to all

time intervals, because any such interval is a multiple of a single tick. We now

rewrite Equation (8.2) without the subscript “tick,” realizing that Δτ is the

proper time between any two events, i.e., the time interval between the two

events measured by a clock that is present at both events:

relation between

proper time and

coordinate time

Δt =

Δτ



1 − (v/c)

. (8.3)

Δτ can also be deﬁned as the time measured by an observer for whom the two

events occur at the same spatial point. Δt, called the coordinate time,is

the time measured by another observer, moving relative to the ﬁrst one with

speed v, for whom the two events occur at two diﬀerent spatial points.

8.2 Spacetime Distance

The most elegant way of relating an event’s space and time properties as

described by two observers is to use geometry. We start with the description

of the event itself. An event has a position and an instant of time. Therefore,

it can be represented by a set of four coordinates: three for position and

one for time. It is common to multiply the time t by c (to make a distance

out of it) and put it as the ﬁrst coordinate. Thus in Cartesian coordinate

system, an event is described by (ct,x,y,z). Geometrically, we have added

spacetime

introduced

the extra “dimension” of time to the three-dimensional space to create the

four-dimensional spacetime.

At the heart of any geometry is the distance between two nearby points,

and how it is written in terms of the coordinates of the points. Euclidean

The same argument applies to the case where the distance expands.

8.2 Spacetime Distance 241

geometry started without coordinates, with the notion of the distance be-

tween two points being “evident.” In fact, we use the properties of Euclidean

distance (such as the Pythagoras’ theorem involving three distances corre-

sponding to the three sides of a right triangle) to show that the distance

geometry and

distance formula

between two points whose Cartesian coordinates diﬀer by (Δx, Δy, Δz)is



(Δx)

+(Δy)

+(Δz)

In the case of the spacetime geometry, we have started with coordinates.

Now we have to ﬁnd a distance formula in terms of the diﬀerence between

coordinates of two events. We get some clues from Euclidean distance as

expressed in terms of coordinates. The ﬁrst clue is that distance is observer-

independent: If observer O uses his Cartesian coordinate system to label point

by (x

)andP

by (x

), and ﬁnds

(

)

≡ Δr =



− x

)

+(y

− y

)

+(z

− z

)

and if observer O



uses her Cartesian coordinate system to label point P



)andP

by (x



), and ﬁnds

(

)



≡ Δr







− x



)

+(y



− y



)

+(z



− z



)

then Δr



=Δr. The second clue is that if P

and P

lie along a single axis of

an observer, then the distance is the (absolute value of the) diﬀerence between

the coordinates of P

and P

Now consider two events E

and E

, which occur at the same spatial

location according to O



,withE

happening after E

. This means that O



(his clock) is present at both events, i.e., that E

and E

lie along the time

axis of O



,andthatO



is measuring the proper time interval between the

two events: Δτ = t



− t



. By the second clue above, cΔτ = c(t



− t



)is

the distance we are looking for (again we multiply by c to make a distance

out of it). We introduce the notation Δs ≡ cΔτ and call Δs the spacetime

distance or the invariant interval between the two events.

Another observer O assigns spacetime coordinates (ct

)toE

and (ct

)toE

. Now the spatial separation between E

and E

according to O is



− x

)

+(y

− y

)

+(z

− z

)

and since O



is at E

when it happens and at E

when it happens, this

equation is precisely the distance that O



travels in time t

− t

with respect

to O. Therefore, the speed of O



relative to O is

v =



− x

)

+(y

− y

)

+(z

− z

)

− t

− x

)

+(y

− y

)

+(z

− z

)

− t

)

242 Vectors in Relativity

Up to this point, we have not used any physics (except for the deﬁnition

of speed). Now comes the crucial ﬁnal step. Equation (8.3) (which is a direct

result of Box 8.0.1) can now be used to ﬁnd the expression of Δs in terms of

coordinate diﬀerences. Equation (8.3) implies that

cΔt =

cΔτ



1 − (v/c)

Δs



1 − (v/c)

Δs = cΔt



1 − (v/c)

= c(t

− t

)



1 − v



− t

)

− v

− t

)

Substituting the expression for v

above, we getspacetime distance

Δs =



− t

)

− (x

− x

)

− (y

− y

)

− (z

− z

)

We rewrite this important formula as

(Δs)

=(cΔτ)

= c

(Δt)

− (Δx)

− (Δy)

− (Δz)

. (8.4)

Let’s emphasize the signiﬁcance of this equation: If observer O uses his

Cartesian coordinate system to label event E

by (ct

)andE

(ct

), and ﬁnds

(Δs)

= c

− t

)

− (x

− x

)

− (y

− y

)

− (z

− z

)

and if observer O



uses her Cartesian coordinate system to label event E

(ct



)andE

by (ct



), and ﬁnds

(Δs



)

= c



− t



)

− (x



− x



)

− (y



− y



)

− (z



− z



)

then (Δs



)

=(Δs)

. Thus, although events are coordinatized diﬀerently by

diﬀerent observers, the spacetime distance between two events is universal.

In contrast to Newtonian physics, neither the time interval nor the spatial

distance between two events is universal in relativity.

Example 8.2.1.

Observer O spots a light beam (event E

)at(x

)attime

. A little later he ﬁnds the beam (event E

)at(x

)attimet

. What is thezero spacetime

distance for two

diﬀerent events

spacetime interval for this light beam (i.e., for the two events E

and E

Since light travels from (x

)to(x

)withspeedc,wehave



− x

)

+(y

− y

)

+(z

− z

)

= c(t

− t

Therefore,

(Δs)

= c

− t

)

− (x

− x

)

− (y

− y

)

− (z

− z

)

=0,

which holds for any light signal, as the two events above are quite general. Thus the

spacetime distance between two diﬀerent events which can be connected by a light

signal is zero. This is in contrast to the Euclidean case where two diﬀerent points

always have a nonzero distance between them.



8.3 Lorentz Transformation 243

8.3 Lorentz Transformation

Because of the intuitiveness of the concept of distance in Euclidean geometry,

it is not essential to know how the coordinates of a point in one coordinate

system (CS) are related to the coordinates of that same point in another

CS. This transformation was found long after the maturity of the Euclidean

geometry [see Section 6.1.3 and especially Equation (6.22) for a discussion of

the two-dimensional version of coordinate transformation], and it was based

entirely on the expression for the distance between two points in terms of the

coordinates of those points.

In spacetime geometry such a transformation is indispensable due to the

counter-intuitive properties of the invariant interval (see Example 8.2.1 above).

And while in Euclidean geometry, one can picture diﬀerent coordinate systems

and how they relate to one another (see Figure 6.7, for example), spacetime

geometry does not readily allow such a direct pictorial representation without

some preliminary algebraic discussion.

Let r

=(ct

)andr

=(ct

) be the spacetime “po-

sition vectors” of two events E

and E

relative to a coordinate system O.

Construct the diﬀerence

Δr = r

− r

=(ct

− ct

− x

− y

− z

and deﬁne the square of the “length” of this vector to be (Δs)

.Infact,

this is generalized for any four-dimensional vector. But ﬁrst, let’s introduce a

notation.

A spacetime vector has the form a =(a

), which is usually four-vectors

introduced

called a four-vector or a 4-vector.

It is also denoted by (a

,a)where

a ≡ (a

)isthespace part (or the 3-vector part) of the 4-vector. A pri-

mary example of a four-vector is r =(ct,x,y,z) ≡ (ct,r). The generalization

mentioned above deﬁnes the square of the length of a (or the inner product

of a with itself) as

a ·a ≡ a

− a

≡ a

−a ·a = a

−|a|

. (8.5)

Then it is easy (see Problem 8.1) to show that the inner product of any two

vectors must be given by

a ·b ≡ a

− a

= a

−a ·



b. (8.6)

In matrix form this can be written as

a ·b =

⎛

⎜

⎝

10 0 0

0 −10 0

00−10

00 0−1

⎞

⎟

⎠

⎛

⎜

⎝

⎞

⎟

⎠

, (8.7)

Note that the ﬁrst component of a has zero as an index, and is called the time compo-

nent. This is common in relativity.