space, time and matter

TIME AND MATTERSPACE

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N E W J E R S E Y L O N D O N S I N G A P O R E BE IJ ING S H A N G H A I H O N G K O N G TA I P E I C H E N N A I

World Scientific

Dipak K SenUniversity of Toronto, Canada

TIME AND MATTERSPACE

8897hc_9789814522830_tp.indd 2 21/2/14 5:22 pm

Published by

World Scientific Publishing Co. Pte. Ltd.

5 Toh Tuck Link, Singapore 596224

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British Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.

SPACE, TIME AND MATTER

Copyright 2014 by World Scientific Publishing Co. Pte. Ltd.

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ISBN 978-981-4522-83-0

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Printed in Singapore

February 22, 2014 14:3 9in x 6in Space, Time and Matter b1716-fm

PREFACE

In 1918, Hermann Weyl wrote a book entitled Raum-Zeit-Materie.

This short monograph, with the same title, however, treats the same

subject matter in a somewhat unconventional way.

We present a novel, albeit equivalent, formalism of relativistic kine-

matics and general relativistic field dynamics in which time does not

have the same primary role as space as in conventional relativity.

Finally, we present a theory of formation of fundamental particles

where the fundamental constituents are left-handed and right-handed

2-component Weyl neutrinos.

The material presented here uses the technics of geometry of

manifolds extensively, in particular, that of vector fields on mani-

folds. Readers unfamiliar with this subject should start first with

Appendices A and B.

Dipak K. Sen

Toronto, Canada

December, 2013

v


CONTENTS

PREFACE v

1. SPACE AND TIME 1

1.0. Introduction . . . . . . . . . . . . . . . . . . . . . . . 1

1.1. The Hyperbolic Structure of the Space

of Relative Velocities . . . . . . . . . . . . . . . . . . 4

1.2. Relativistic Kinematics on 3-Manifolds . . . . . . . . 8

1.3. Class of Inertial Observers and Generalized

Lorentz Matrices . . . . . . . . . . . . . . . . . . . . . 11

1.4. Classical Relativistic Field Dynamics . . . . . . . . . 13

1.4.1. One-dimensional Heat equation . . . . . . . . . 13

1.4.2. One-dimensional Wave equation . . . . . . . . 17

1.4.3. GaussEinstein equations . . . . . . . . . . . . 18

1.5. Relativistic Field Dynamics on 3-Manifolds . . . . . . 21

1.5.1. Three-dimensional field equations

and relationship with Einstein equations . . . 21

1.6. Flat-Space Solutions . . . . . . . . . . . . . . . . . . . 24

1.7. The de Sitter Solution . . . . . . . . . . . . . . . . . . 26

1.8. A New Solution of the Vacuum Einstein

Field Equations . . . . . . . . . . . . . . . . . . . . . 27

vii


viii Space, Time and Matter

1.9. A Solution of Mathematical Interest . . . . . . . . . . 34

1.10. The Schwarzschild Solution . . . . . . . . . . . . . . . 38

1.10.1. The Schwarzschild metric . . . . . . . . . . . . 39

1.11. From Space-Time 4-Metric to 3-Metric

and 3-Vector Field . . . . . . . . . . . . . . . . . . . . 44

1.12. The Kerr Solution . . . . . . . . . . . . . . . . . . . . 49

1.13. The Maxwell Equations . . . . . . . . . . . . . . . . . 56

References . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2. MATTER 61

2.0. Introduction . . . . . . . . . . . . . . . . . . . . . . . 61

2.1. The Photon and the Weyl Neutrinos . . . . . . . . . 63

2.2. A Neutrino Theory of Matter . . . . . . . . . . . . . 68

2.2.1. Composite LR system without

interaction . . . . . . . . . . . . . . . . . . . . 68

2.2.2. Composite LR system with interaction . . . 70

2.2.3. A reduced 1-dimensional model . . . . . . . . . 72

2.2.4. Conclusion . . . . . . . . . . . . . . . . . . . . 77

References . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

APPENDIX A. VECTOR FIELDS ON MANIFOLDS 79

A.1 Vector Fields on Manifolds . . . . . . . . . . . . . . . 79

A.2 Example of a Vector Field whose Integral

Curves are Knots . . . . . . . . . . . . . . . . . . . . 83

References . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

APPENDIX B. DYNAMICAL VECTOR FIELDS

OF CLASSICAL MECHANICS 87

B.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 87


Contents ix

B.2 Configuration Spaces of Mechanical Systems

as Manifolds . . . . . . . . . . . . . . . . . . . . . . . 89

B.3 Dynamical Vector Fields

in Classical Mechanics . . . . . . . . . . . . . . . . . . 99

References . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

APPENDIX C. MAPLE PROGRAM 1 129

APPENDIX D. MAPLE PROGRAM 2 131

APPENDIX E. MAPLE PROGRAM 3 133

AUTHOR INDEX 137

SUBJECT INDEX 139

February 22, 2014 14:2 9in x 6in Space, Time and Matter b1716-ch01

1. SPACE AND TIME

1.0. Introduction

It was Minkowski who paraphrased that, . . .Henceforth space by

itself, and time by itself, are doomed to fade away into mere shadows,

and only a kind of union of the two will preserve an independent

reality. And ever since relativists are accustomed to always think in

the framework of a 4-dimensional space-time, it has almost become

second nature to them.

Strictly speaking, however, the Minkowski signature ( + + +)makes a clear distinction as to which coordinates can be interpreted

as spatial and which ones as temporal, and recently, the usefulness

of 3 + 1 decomposition techniques has shown that sometimes the

separation of space from time provides a better insight to what is

going on rather than the traditional fusion of space with time.

Imagine a universe in which nothing moves. In such a universe, the

notion of time disappears. So, space and motion in space are really

the primary concepts and time is a secondary concept. It should thus

be possible to construct a theory of Kinematic Relativity based on

the primary notions of space and motion instead of space and time.

1


2 Space, Time and Matter

In this monograph, we present a new formulation of both relati-

vistic kinematics and eld dynamics in an entirely 3-dimensional

space. The basic idea is to represent local physical observers as

non-singular 3-dimensional local vector elds and the dynamics

of physical elds by the ows of the vector elds, which are

1-parameter groups of local smooth transformations of the space.

The ow parameter thus plays the role of local time for each physi-

cal observer, and Lie-derivative replaces the time derivative.

In the kinematics part, we rst introduce the basic notions and

postulates of our formalism which take place in a 3-dimensional Rie-

mannian space (M 3, g). A physical observer is dened to be a non-

singular vector eld X on M3 with 0


Space and Time 3

In the dynamics part, we rst show how one can describe the time

evolution of certain classical elds by subjecting the 3-dimensional

elds to a transformation by the ow of a fundamental 3-dimensional

vector eld. We illustrate this with some simple 1-dimensional cases

such as the 1-dimensional Heat and Wave equations. We next apply

our basic procedure to the GaussEinstein equations, that is, the

Einstein eld equations in Gaussian normal coordinates, and obtain

a set of equations for the 3-metric g and a 3-dimensional vector eld

X. Every solution (g,X) of these equations determine uniquely a

space-time 4-metric solution of the Einstein eld equations.

Our rst example is the de Sitter solution. A generalization of the

de Sitter case leads us to a complex metric of Kasner type, which

in turn, provides us with a new (real) solution of the vacuum Ein-

stein eld equations, containing ve real parameters. We then give

another solution of our 3-dimensional equations which is mathemat-

ically interesting and non-trivial, but however, corresponds to the

physically trivial at space-time metric. Nevertheless, this solution

clearly illustrates the basic principles of our 3-dimensional approach.

Finally, we apply our formalism to the Maxwell equations and

obtain a corresponding 3-dimensional set of equations, which turn

out to be, not only concise, but also elegant.



1.1. The Hyperbolic Structure of the Spaceof Relative Velocities

We rst consider the geometry of the space of relative velocities in

conventional Special Relativity. This will suggest the formalism of a

theory of Kinematic Relativity on 3-manifolds which reproduces the

essential features of conventional relativity.

In the special theory of relativity, the velocity transformation for-

mula relating the relative velocities of three equivalent inertial sys-

tems u, v, w is given by (in units c = 1)

w2 = (u v)2 (u v)2

(1 u v)2 . (1.1)

Equation (1.1) has the following geometrical signicance. Let us

choose some reference inertial system and consider the velocities u =

(u1, u2, u3) of all other equivalent inertial systems relative to this

reference inertial system. Since for physical inertial systems u =[|(ui)2|]1/2 < 1, the space of relative velocities is topologically a 3-dimensional open disk

D3 = {u R3|u < 1}.

Note: For tachyons, the appropriate space to consider would be the

closure of the complement of D3, that is, R3\D3.Now, we introduce the standard (positive denite) hyperbolic met-

ric G on D3 by

ds2 =i,k

Gik(u)duiduk

=[1(ui)2][(dui)2] + [uidui]2

[1(ui)2]2 , (1.2)


Space and Time 5

that is,

Gik(u) =[1(ui)2]ik + uiuk

[1(ui)2]2 .Then, H3 =(D3, G) is the standard hyperbolic 3-space of con-

stant curvature 1. This is the so-called Poincare disk model ofH3 ([1]). Alternatively, one can introduce homogeneous coordinates

= (0, 1, 2, 3) by i = 0ui and dene for = (0, 1, 2, 3), =

(0, 1, 2, 3), the inner product (, ) = 00

i ii. Then,

ds2 =(d, )(, d) (, )(d, d)

(, )2. (1.3)

H3 is geodesically complete and has a distance function d given by

d(u, v) = cosh1

[1

iuivi

][1

i(u)2

]1/2 [1

i(v)2

]1/2 (1.4)

or, in homogeneous coordinates,

d(, ) = cosh1[ |(, )|(, )1/2(, )1/2

]. (1.5)

A comparison of Eq. (1.1) with Eq. (1.5) suggests that the dis-

tance d(u, v) is somehow related to the relative velocity w between

two inertial systems whose velocities relative to the reference inertial

system are u and v. To see the actual relationship, we should note

that the reference system is at rest relative to itself. Now, from (1.4)

d(0, v) = cosh1[

1(1 v2)1/2

]= tanh1 v.

That is, v = tanh d(0, v). We can, therefore, reinterpret Eq. (1.1) asa relation between the relative velocity w and the hyperbolic distance



4

3

2

10

1

2

3

4

y

4 2 2 4x

Fig. 1. The hypersurfaces M,M.

d(u, v) by

w = tanh[d(u, v)]. (1.6)

The isometry group of H3 is intimately connected with the Lorentz

group, as can be seen in what follows.

First, we consider an abstract R4 with the Lorentz metric gR4 given

by ds2 = (dx1)2+(dx2)2+(dx3)2(dx4)2. Consider now the hypersur-face M of R4: M = {x = (x1, x2, x3, x4) R4|(x1)2+(x2)2+(x3)2(x4)2 = 1}. M has two components: M = {x M|x4 1} andM = {x M|x4 1} (see Fig. 1; here (x, y) represents (x1, x4)).

Both M and M are dieomorphic to D3 and the dieomorphism

map is given by

(x1, x2, x3, x4) (u1, u2, u3),ui = xi/x4, (i = 1, 2, 3)

with its inverse

(u1, u2, u3) (x1, x2, x3, x4),

xi = uix4, x4 = [

1(1(ui)2)

]1/2.


Space and Time 7

Now, let gM be the induced Riemannian metric on M obtained

from gR4 on R4. In coordinates (ui), gM is given exactly by (1.2).

Thus, H3 = (D3, G) is isometric to (M , gM |M ) and (M , gM |M ).

The isometry group ofH3 was shown by Poincare to be PSL(2,C),

the projective special linear group in two complex dimensions and is,

thus, isomorphic to L+, the orthochronous Lorentz group. It is clear

from (1.5) that the isometry group of H3 leaves the inner prod-

uct (, ), in homogeneous coordinates, invariant. Let its action be

given by

(0, 1, 2, 3) (0, 1, 2, 3),

where

= L , L

L+ (, = 0, 1, 2, 3).

Since ui = i0, L+ acts on H

3 as a fractional linear transforma-

tion:

ui ui = (Li0 + Likuk)/(L00 + L0kuk).

Thus, the fundamental aspect of the Special Theory of Relativity,

namely, Lorentz invariance, is contained in the hyperbolic structure

of the space of relative velocities.



1.2. Relativistic Kinematics on 3-Manifolds

Let (M3, g) be a 3-dimensional space with a (positive-denite) Rie-

mannian metric g. By an observer (or a particle path), we usually

mean a smooth curve in M3, i.e., a smooth map c : I M3 from areal interval I into M3.

Since, to every such curve c, one can associate a local vector eld C

such that c is an integral curve of C ([2]). We shall regard the set of

all non-singular vector elds on M3 as the set of all observers in M3.

Non-singularity is necessary to avoid the possibility that an

observer comes to an absolute rest relative to M3. To each vector

eld X corresponds a local ow t , i.e., a local 1-parameter group

of local transformations of M3. Here, the parameter t measures the

ow of time as perceived by the observer represented by X. In this

sense, time is a local concept since X need not be globally dened

and is not necessarily complete unless M3 is compact ([3]).

Let us now consider only those non-singular vector elds X such

that g(X,X) < 1. Such observers will be called physical observers.

(Corresponding to the fact that no such observer can attain the veloc-

ity of light.)

From now on, we consider only physical observers and denote by

(M3) the set of all physical observers.

We dene a map V , called the relative velocity map, from a pair

of physical observers to a smooth function on M3 as follows:

V : (M3)(M3) C(M3),(X,Y ) V (X,Y ),


Space and Time 9

where

V (X,Y ) =[1 1

[F (X,Y )]2

]1/2,

F (X,Y ) =1 g(X,Y )

[1 g(X,X)]1/2[1 g(Y, Y )]1/2 .(1.7)

Here, C(M3) denotes the set of all smooth functions on M3.

Let us now dene an equivalence relation in (M3) by:

X Y if and only ifV (X,Y ) = constant function on M3.

(1.8)

In other words, the observers X and Y are said to be (inertially)

equivalent if the relative velocity function is constant on M3.

We shall now give theoretical denitions of space and time inter-

vals of a physical particle path relative to any physical observer X.

Let c : I M3, c() be a particle path and C any represen-tative corresponding vector eld associated with it such that c is an

integral curve of C. For a physical particle path, 0 < g(C,C) < 1

on c(I).

Definition. The time interval of c relative to an observer X, denoted

by tX , is given by the following integral over I:

tX =IF (X,C)d. (1.9)

Note that we can reparametrize c by tX , where dtX/d = F (X,C).



Definition. The space interval of c relative to the observer X,

denoted by dX , is given by

dX =IV (X,C)dtX =

IV (X,C)F (X,C)d. (1.10)

If (tY ,dY ) are time and space intervals of c relative to another

observer Y , then (tX ,dX) (tY ,dY ) provides a space-timetransformation from observer X to observer Y .


Space and Time 11

1.3. Class of Inertial Observers and GeneralizedLorentz Matrices

We now consider the problem of constructing a class of inertial

observers starting from a representative physical observer. In other

words, given a physical observer u, the problem of constructing an

observer w such that u and w are inertially equivalent, that is, the

relative velocity between u and w, V (u,w), as given by (1.7), is a

constant function on M 3.

Since M3 is 3-dimensional, it is parallelizable, that is, there exists a

global framing of M3 by means of three linearly independent vector

elds {e(i)}, i = 1, 2, 3, so that any vector eld u can be expressedas u = ui(x)e(i), and g(e(i), e(j)) = gij(x), where ui(x), gij(x) are

functions on M3. One can, of course, also consider everything locally

in some local coordinate system (xi), so that ui(x), gij(x) are func-

tions of the coordinate (xi). Since u is a physical observer, we have

g(u, u) = gikuiuk = uiui < 1, where ui = gikuk.

Let

u = [1 g(u, u)]1/2Au = (u 1)/g(u, u)

}(1.11)

and let us dene the 4 4 matrix functions L(u), , = 0, 1, 2, 3 asfollowsa:

L00(u) = u, L0k(u) = uuk

Lk0(u) = uuk, Lkj (u) =

kj + Auu

kuj.

(1.12)Note that in (1.8) uj dier from uk by the lowering of indices by

gjk and also that L are functions on M . These generalized Lorentz

aIn what follows Latin indices i, j, k, . . . = 1, 2, 3 and the Greek indices

, , , . . . = 0, 1, 2, 3.



matrices L(u) are easily seen to satisfy the following generalized

properties:

gikLi0L

k0 = (L

00)

2 1gmnL

m0 L

ni = L

00L

0i

gmnLmi L

nj = L

0i L

0j + gij

(1.13)and when (M3, g) = (R3, ), that is, when (M3, g) is the Euclidean

space R3 with the Euclidean metric gik = ik, L(u) reduces to

the usual Lorentz matrix L(u) with the pure boost given by u =

(u1, u2, u3) and satisfy the usual properties:

Li0Li0 = (L

00)

2 1Lm0 L

mi = L

00L

0i

Lmi Lmj = L

0i L

0j + ij .

(1.14)Let v = vi(x)e(i) be another vector eld where g(v, v) < 1.

We now dene a vector eld w = wi(x)e(i) by

wi =Li0(u) + L

ik(u)v

k

L00(u) + L0k(u)vk

. (1.15)

Then it follows that g(w,w) < 1, so that w also represents a physical

observer.

Using the generalized properties (1.9) one now shows that the

relative velocity function V (u,w) as given by (1.1) satises:

[V (u,w)]2 = g(v, v).

So, if v is chosen so that g(v, v) = const., the observers u and w

are inertially equivalent and, in this way, starting from u, we can

construct a class of inertially equivalent observers.


Space and Time 13

1.4. Classical Relativistic Field Dynamics

In classical eld theory, the dynamics of a eld is usually described

by a set of evolution equations (together possibly with some con-

straint equations) in some space-time coordinates. We shall now

demonstrate that in many cases the dynamics can also be described

in a purely spatial (i.e., 3-dimensional) setting without the neces-

sity of introducing explicitly a time-coordinate. Instead of the time-

coordinate and time-dependent quantities, the temporal evolution is

provided by the ow of a 3-dimensional vector eld representing a

physical observer.

Before considering the Einstein eld equations in some detail, we

shall rst illustrate our basic approach with some simple examples

from classical eld theory, such as the Heat and Wave equations. For

simplicity, we restrict our analysis to the 1-dimensional case only.

The generalization to three spatial dimensions is obvious.

1.4.1. One-dimensional Heat equation

Consider the 1-dimensional Heat equation with some initial data:

t = xx(x, 0) = (x).

}(1.16)

We look for solutions (x, t) of (1.16) which come from the initial

data (x) through a transformation by the ow of a vector eld

X depending on the spatial coordinate only, as follows. Let X be

given by

X = (x)d

dx. (1.17)



Consider its integral curves given by the dynamical system

dx

dt= (x). (1.18)

That is,dt =

dx

(x)= (x), say. So that x(t) = 1(t + c), c =

const., and x(0) = 1(c) or c = (x(0)). The ow t of X transforms

the initial point x(0) onto the point x(t), that is,

t : x t(x) = 1(t + (x)). (1.19)

Now letb

(x, t) = (t)(x)

= (t(x))

= (1(t + (x)))

(1.20)so that

(x, 0) = (x).

Then (x, t), which now also depends on the ow parameter t, is

the transformed initial data (x), which of course depends only on

the space variable x. Let us now substitute (x, t) in (1.16).

Put t + (x) = z, and (x) = y. Then, 1(y) = 1/(x) = (x);

1(y) = (x)(x). And

t(x, t) = (1(z)

)1(z)

t(x, 0) = (x)(x) = LX

}(1.21)

x(x, t) = (1(z)

)1(z) (x)

x(x, 0) = (x)1(y) (x) = (x)

}(1.22)

bGeometrical objects (e.g., scalar, vector, tensor fields and densities) can be

transformed by the flow of a vector field ([3]). We could have also transformed (x)

by t1 = t , by reversing the flow, that is, by letting (x, t) = (t

1(x)).This would have preserved the scalar character of under the flow.


Space and Time 15

xx(x, t) = (1(z)

)[1(z)

]2[ (x)

]2+(1(z)

)1(z)[ (x)]2

+(1(z))1(z) (x). (1.23)

xx(x, 0) = (x)[1(y)

]2[ (x)

]2 + (x)1(y)[ (x)]2+(x)1(y) (x)

= (x)[(x)

]2[(x)

]2+(x)(x)(x)

[(x)

]2(x)(x)(x)[(x)]2 = (x). (1.24)

If our (x, t) is to satisfy (1.16), we must have

(1(z)

)1(z) =

(1(z)

)[1(z)

]2[ (x)

]2+

(1(z)

)1(z)

[ (x)

]2+

(1(z)

)1(z) (x). (1.25)

Let us put 1(z) = u. So that again, 1(z) = 1(u) = (u),

1(z) = (u)(u). We can write (1.25) as

(u)(u) =[(u)

[(u)

]2 + (u)(u)(u)][ (x)]2+(u)(u) (x).

The variables x and u can thus be separated to give

1 (x)[ (x)]2

=(u)(u)

(u) + (u)

or

[(x)

]2 + (x) = (u)(u)

(u) + (u) = (const.) (1.26)



which must be satised for all x and u. In particular, for u = x, we

must have from (1.26)

(x)(x) = (x)

or LX = .

}(1.27)

Note that (1.27) also follows directly from (1.16) and the initial val-

ues in (1.21), (1.24), and that (1.27) can be obtained from (1.16)

by replacing (x, t) with (x) and the time-derivative t with the

Lie-derivative LX with respect to X. Any solution (,X) of (1.24)provides a solution of (1.16), because from (1.27), ([3]),

t =

t(t) = t(LX) = t() = (t) = . (1.28)

Since (1.26) or (1.27) are equations for both X and , the initial

data cannot be arbitrary and, therefore, our procedure provides only

special solutions of (1.16). We shall give two examples of such solu-

tions ([4]).

Example 1. (x) = ex, (x) = = const., X = ddx . Then (1.27)

is satised, and (x) = x , 1(y) = y, 1

(t + (x)

)= t + x. So

(x, t) = (1(t + (x))

)= (t + x) = e

2t+x. From (1.26), we

get 2 = . If we also allow to be negative, we also obtain complex

solutions (x, t) = e2tix with = i.

Example 2. (x) = x2, (x) = 1x , X =1x

ddx . Again, (1.27) is

satised and (x) = x2

2 , 1(y) = 2y, 1(t+(x)) = 2t + x2.

So (x, t) = (1(t + (x))

)= (2t + x2) = 2t + x2. This is a

well-known polynomial solution of (1.16).

In fact, it is easily possible to obtain the general solution of (1.26)

or (1.27).


Space and Time 17

1.4.2. One-dimensional Wave equation

Consider now the 1-dimensional Wave equation with some initial

data:

tt = xx(x, 0) = (x)

t(x, 0) = (x).

(1.29)Again let X = (x) ddx and its ow t be given by (1.16), where

(x) = 1(x)

, and (x, t) = (1(t + (x))

). Now since t(x, 0) =

LX, we must have

LX = . (1.30)

From (1.24)

tt(x, t) = (1(z)

)[1(z)

]2 + (1(z))1(z)tt(x, 0) = (x)

[(x)

]2 + (x)(x)(x).}

(1.31)

It follows from (1.24), (1.29), (1.31) that

(x)[(x)

]2 + (x)(x)(x) = (x)or LXLX = .

}(1.32)

Again (1.32) is obtained from (1.29) by replacing (x, t) by (x)

and the time-derivative t by the Lie-derivative LX . Any solution of(1.30), (1.32) provides a solution of (1.29). Note that the initial data

and now have to be related by the constraint equation (1.30)

and (,X) must satisfy (1.32). We give two special solutions of (1.30)

and (1.32), and hence of (1.29).

(i) (x) = +1, X = ddx , (x) = x, 1(t + (x)) = t + x, (x, t) =

(x + t) with (x) = (x),

(ii) (x) = 1, X = ddx , (x) = x, 1(t + (x)

)= t + x,

(x, t) = (x t) with (x) = (x).



It is also easily possible to obtain the general solution of (1.30)

and (1.29).

1.4.3. GaussEinstein equations

In conventional General Relativity the space-time metric g satises

the Einstein eld equations, which are basically evolution equations

with certain constraints, and it is well known ([5]) that the vacuum

Einstein eld equations: R = 0 equivalent to

G0 = 0 (constraints equations)

Rik = 0 (evolution equations)

}(1.33)

provided that g00 = 0; for example, on a non-null hypersurface. HereG R 12gR is the Einstein tensor. A slight modicationis necessary if one wishes to include a non-zero cosmological con-

stant in the eld equations. The vacuum Einstein eld equations

with a cosmological constant: R = g also split up into a set of

constraint and a set of evolution equations and they take a particu-

larly simple form if one uses a Gaussian normal coordinate system in

which g00 = g00 = 1 and g0i = g0i = 0. Then the eld equations:

R0i = 0

R00 = Rik = gik

(1.34)are equivalent to:

G0i = 0

G00 =

}(constraint equations)

Rik = gik (evolution equations)

(1.35)


Space and Time 19

because (1.34) implies (1.35), since G0i = g00Ri0 + g0kRik = 0 and

G00 =12g

00R00 12gikRik = 12(g00g00 gikgik) = .Conversely, (1.35) implies (1.34), since if G0i = g

00Ri0+g0kRik = 0,

then from (1.35) Ri0 = 0; and if G00 =12g

00R00 12gikRik = , thenfrom (1.35), 12g

00R00 32 = , that is, R00 = .The corresponding matter eld equations with an energy momen-

tum tensor T : G + g + T = 0 or, equivalently, R =12Tg T + g take the following form (in the case of dustlikematter with density , pressure p = 0, and also in a Gaussian normal

coordinate system):

R0i = 0

R00 = 12

Rik =(1

2 +

)gik.

(1.36)

Equation (1.36) is then equivalent to

G0i = 0

G00 =

}(constraint equations)

Rik =(1

2 +

)gik (evolution equations).

(1.37)

Gaussian normal coordinates can be introduced in general always,

and with x = (x0 = t, x1, x2, x3), the metric takes the form: ds2 =

dt2 + gik(x)dxidxk. So that g00 = g00 = 1, g0i = g0i = 0. The4-metric g thus determines a 3-metric gik, with its inverse gik,

which is positive denite on the hypersurfaces t = const. The eld

equations (1.34) are then explicitly, in a Gaussian normal coordinate



system ([6]):

G0i 12(gi,0); 12(g

kgk,0),i = 0, (1.38)

G00 12R +

18gmgm,0g

ikgik,0

18gigkmgik,0gm,0 = , (1.39)

Rik Rik 12gik,00 14gmgm,0gik,0 +

12gmgi,0gkm,0

=(1

2 +

)gik (1.40)

where

gik,0 gik/x0 = gik/t,gik,00 2gik/t2,Rik Ricci-tensor of the 3-metric gik,R Ricci-scalar of the 3-metric gik,

A; covariant derivative of A.We shall refer to Eqs. (1.38)(1.40) as the GaussEinstein eld equa-

tions.c

cSolutions of these equations have been considered by Marsden and Fischer

([7]) in an infinite dimensional setting. Our approach to the problem is, however,

entirely finite dimensional.


Space and Time 21

1.5. Relativistic Field Dynamics on 3-Manifolds

1.5.1. Three-dimensional field equations andrelationship with Einstein equations

As we have seen that in classical eld theory, the dynamics of a eld

is usually described by a set of evolution equations (together pos-

sibly with some constraint equations) in some space-time manifold.

We shall now demonstrate that in many cases the dynamics can also

be described in a purely 3-dimensional setting without the neces-

sity of explicitly introducing a time-coordinate. Instead of the time-

coordinate and time-dependent quantities, the temporal evolution is

provided by the ow of a 3-dimensional vector eld representing a

fundamental observer.

The starting point of our dynamical theory would be a set of dif-

ferential equations for a 3-metric g and a fundamental 3-vector eld

X on a M3. Let X = Xi(x)/xi be a vector eld on M3 (in some

local coordinate system (xi)), and h = LXg, the Lie-derivative of gwith respect to X.

So that

h = hikdxi dxk where hik = (LXg)ik .

Let

f,i, f;i partial and covariant derivative of f , respectively,Rik Ricci-tensor of the 3-metric gik,R Ricci-scalar of the 3-metric gik.

Then our basic equations are:

hi; (gkhk),i = 0, (1.41)



12R +

18(gmhm)(gikhik)

18(gihik)(gkmhm) = 0, (1.42)

Rik 12(LXh)ik 14(gmhm)hik

+12(gmhi)hkm = 0. (1.43)

We regard Eqs. (1.41)(1.43) as a set of dierential equations for

both the 3-metric g and the 3-dimensional vector eld X on M3.

Every solution (g,X) of (1.41)(1.43) determines uniquely a solution

of the vacuum Einstein eld equations as follows:

The integral curves of X determine a ow , which are local

1-parameter group of local dieomorphisms ([5]) of M3. Thus, for

each value of the ow parameter , denes a point transformation:

x x = (x) with 0(x) = x. The ow transformed metric g = (g) is given by gik(, x) = x

xixm

xk gm(x), and is thus -dependent.

Theorem 1. Let (gik;Xi) be a solution of (1.9)(1.11) andgik(, x) the -dependent metric transformed by the ow of X.

Then, the 4-dimensional metric,

ds2 = d2 + gik(, x)dxidxk (1.44)

is a solution of the vacuum Einstein equations in a Gaussian normal

coordinate system (x0 = , xi).

Proof. Let

K any tensor eld, (K) the ow transformed tensor eld K,LXK the Lie-derivative of K with respect to X.


Space and Time 23

Then one has ([5])

(LXK) =

( (K)). (1.45)

Therefore, by applying to Eqs. (1.41)(1.43), one can replace the

Lie-derivatives of gik in (1.41)(1.43) by the derivatives of gik(, xi)

with respect to , and obtain

(gkgik,0); (gkgk,0),i = 0, (1.46)

12R +

18gmgm,0g

ikgik,0 18 gigkmgik,0gm,0 = 0, (1.47)

Rik 12 gik,00 14gmgm,0gik,0 +

12gmgi,0gkm,0 = 0. (1.48)

(Here, Rik, R and the covariant derivatives are all relative to the

3-metric gik(, xi), and gik,0, gik,00 the rst and second partial deriva-

tives of gik with respect to .) These are precisely the vacuum

Einstein eld equations for the metric (1.44) in the Gaussian nor-

mal coordinate system (x0 = , xi) ([6]).

Note that the signature of the 4-metric (1.44) would depend on the

signature of the 3-metric gik. Thus, may or may not be the physical

time-coordinate.



1.6. Flat-Space Solutions

We now consider some special solutions of (1.41)(1.43). First, note

that if, either X = 0 or X is a Killing vector eld of g, then h =

LXg = 0. So that (1.41) is identically satised. Furthermore, if = = 0, then (1.43) implies that Rik = 0 which in turn implies that

Rijk = 0 (since M3 is 3-dimensional), that is, M3 is at with the

metric g = (i.e., gik = ik in some coordinate system). So, {X = 0,g = } and {X is Killing, g = } are both solutions of (1.41)(1.43)with = = 0. These are trivial at-space solutions.

For non-trivial at-space solutions, we put gik = gik = ik, so that

Rik = 0, R = 0, and

hik = X i,k + Xk,i,

(LXh)ik = hik,X + hiX,k + hkX,i,

hi; = hi,; (gkhk),i =

(

h

),i

, (1.49)

18(gihik)(gkmgm) =

18(h211 + h

222 + h

233)

+14(h212 + h

213 + h

223).

Therefore, (1.41)(1.43) becomes (together with (1.49)):

12(hi1,1 + hi2,2 + hi3,3)

12(h11,i + h22,i + h33,i) = 0, (1.50)

14(h11h22 + h11h33 + h22h33)

14(h212 + h

213 + h

223) = , (1.51)


Space and Time 25

12(hik,X + hiX,k + hkX

,i) +

12(hi1hk1 + hi2hk2 + hi3hk3)

14(h11 + h22 + h33)hik =

(1

2 +

)ik. (1.52)

In terms of Xi and its derivatives, Eqs. (1.50)(1.52) are

12(Xi, + X

,i), (X,),i = 0, (1.53)

12(Xi,i)

2 18(Xi,m + X

m,i )(X

m,i +X

i,m) = , (1.54)

12(Xi,k + X

k,i),X

12(Xi, + X

,i)X

,k

12(X,k + X

k,)X

,i

12X,(X

i,k + X

k,i) +

12(Xi,m + X

m,i )(X

m,k +X

k,m)

=(1

2 +

)ik. (1.55)

In spite of the complexity of the above equations, it is possible to

nd some non-trivial solutions as we shall see next.



1.7. The de Sitter Solution

As a non-trivial solution of (1.53)(1.55) or (1.49)(1.52), let = 0

and Xi = xi, where is a constant to be determined. Then by(1.49) (LXg)ik = hik = 2ik and

12(LXh)ik = 12[hix

,k hkx,i] = 22ik,

14(gmhm)hik = 14(

mhm)hik = 32ik,12(gmhi)hkm =

12(mhi)hkm = 22ik.

So (1.50) is identically satised and (1.51)(1.52) or (1.42)(1.43),

become

32 = , (1.56)

22ik + 22ik 32ik = ik. (1.57)

These equations are satised by taking =

3 . So that {g = ,

X =

3 x

i xi }, = 0 is a solution of (1.49)(1.52), that is, a

at-space solution of (1.41)(1.43).

The ow generated by the vector eld X = xi xi is simplyxi xi = etxi, so that xixk = etik and the transformed time-dependent 3-metric is therefore gik(x, t) = e2tik. The corresponding

4-metric (in coordinates (t, xi)):

ds2 = dt2 + gikdxidxk = dt2 + e2tikdxidxk

with =

3 is the well-known de Sitter solution of the Gauss

Einstein equations (with = 0).


Space and Time 27

1.8. A New Solution of the Vacuum EinsteinField Equations

As our next non-trivial solution of (1.49)(1.52), we make the ansatz:

= = 0 and X1 = 1x1, X2 = 2x2, X3 = 3x3, i.e., Xi =ixi (no summation), where is are constants to be determined.Then hik = (i + k)ik (no summation), i.e., h11 = 21, h22 =22, h33 = 23, hik = 0 if i = k.

Again, (1.50) is identically satised, and (1.51) implies

12 + 13 + 23 = 0. (1.58)

Now 12(LXh)ik = 12(hiX,k + hkX,i). If i = k, for example,i = 1, k = 2, 12(LXh)12 = 0, 12 (h11h21 + h12h22 + h13h23) = 0 and14(h11 + h22 + h33)h12 = 0. So that (1.52) is identically satisedif i = k. If i = k, for example, i = 1, k = 1, 12(LXh)11 = 221,12(h11h11 + h12h12 + h13h13) = 2

21 and 14(h11 + h22 + h33)h11 =

1(1 + 2 + 3). So that (1.52) implies1(1 + 2 + 3) = 02(1 + 2 + 3) = 03(1 + 2 + 3) = 0.

So, we would have a solution of the form X i = ixi (no summation)provided

1 + 2 + 3 = 0

and

12 + 13 + 23 = 0



which is equivalent to:

1 + 2 + 3 = 0

21 + 22 +

23 = 0

(1.59)which is possible only if, either all i are zero or some of the iare complex. Let us entertain the possibility that some of the i are

complex.

The ow generated by the vector eld X = (1x1 x1 +2x2 x2 +3x

3 x3

)is now xi xi = eitxi (no summation), so that xixk =

eitik (no summation) and the transformed time-dependent 3-metric

is gik(x, t) = e2itik (no summation). The corresponding 4-metric (in

coordinates (t, xi), after removing the tilde)

ds2 = dt2 + e21t(dx1)2 + e22t(dx2)2 + e23t(dx3)2 (1.60)

with (1, 2, 3) satisfying (1.59), is therefore a solution of the

vacuum GaussEinstein equations (with =0), that is, R =0.

This can also easily be checked by direct computation. That this,

the solution is non-trivial can be seen by computing, for exam-

ple, the 1212-component of the Riemann curvature tensor. We get

R1212 = 12e2(1+2)t = 0 (if 1, 2 = 0).The complex solution (1.53) is related to the so-called (real) Kas-

ner ([8]) solutions in the following manner. Recall that the Kasner

solutions for the vacuum Einstein equations R = 0 are given by:

ds2 = t21(dx1)2 + t22(dx2)2 + t23(dx3)2 t20dt2 (1.61)

where the satisfy the two equations

1 + 2 + 3 = 0 + 1

21 + 22 +

23 = (0 + 1)

2.

}(1.62)


Space and Time 29

If one takes 0 = 0, one gets the standard Kasner solution

ds2 = dt2 + t21(dx1)2 + t22(dx2)2 + t23(dx3)2 (1.63)

with

1 + 2 + 3 = 1

21 + 22 +

23 = 1.

}(1.64)

A special solution of (1.64) is 1 = 23 , 2 =23 , 3 = 13 . The other

obvious special solution 1 = 1, 2 = 3 = 0, i.e.,

ds2 = dt2 + t2(dx1)2 + (dx2)2 + (dx3)2 (1.65)

is actually a at metric as can be seen by the following transformation

of coordinates: t = t coshx1, x = t sinhx1, x2 = x2, x3 = x3, which

transforms (1.65) into

ds2 = dt 2 + (dx1)2 + (dx2)2 + (dx3)2.

If we replace t by t = et in (1.61), we get

ds2 = e21 t(dx1)2 + e22 t(dx2)2 + e23 t(dx3)2 e2(0+1)t(dt )2. (1.66)

If we now take 0 = 1 in (1.62) and (1.66), we get precisely(1.60), with t instead of t,

ds2 = e21 t(dx1)2 + e22 t(dx2)2 + e23 t(dx3)2 (dt )2

with

1 + 2 + 3 = 0

21 + 22 +

23 = 0.

}(1.59)

Consider now the complex solutions of (1.59). They are

1 =12(1 + i

3)3

2 =12(1 i

3)3

(1.67)



where 3 can be real or complex. We shall choose 3 = r, a real

number. Then 1 = p + iq, and its complex conjugate 2 = p iq,with p = r2 , q = r

3

2 .

So that the non-zero (complex) metric tensor components for

(1.60) are

g11 = e2(p+iq)t

g22 = e2(piq)t

g33 = e2rt

g00 = 1

with p = r2, q = r

32

.

(1.68)

We now make a complex coordinate transformation: x x toobtain a real metric as follows:

x1 = (a + ib)12 x1 + (c + id)

12 x2

x2 = (a ib) 12 x1 + (c id) 12 x2

x3 = x3

t = t.

(1.69)

Here a, b, c, d are any real numbers, such that det(x

x ) = 0. Thetransformation matrix for (1.69) and its determinant are

(x

x

)=

(a + ib)

12 (c + id)

12 0 0

(a ib) 12 (c id) 12 0 00 0 1 0

0 0 0 1

, (1.70)


Space and Time 31

det(x

x

)=[(a + ib)(c id)] 12 [(a ib)(c + id)] 12

= i2D

where

D = Im[(a + ib)12 (c id) 12 ]. (1.71)

So the condition on a, b, c, d is that D = 0.The transformed non-zero (real) metric tensor components are

then

g11 =xi

x1xk

x1gik

=(

x1

x1

)2g11 +

(x2

x1

)2g22

= (a + ib)e2(p+iq)t + (a ib)e2(piq)t

= 2e2pt(a cos 2qt b sin 2qt).

Similarly, g22 = 2e2pt(c cos 2qt d sin 2qt) and

g12 =xi

x1xk

x2gik =

x1

x1x1

x2g11 +

x2

x1x2

x2g22

= (a + ib)12 (c + id)

12 e2(p+iq)t

+(a ib) 12 (c id) 12 e2(piq)t

= 2e2pt(A cos 2qtB sin 2qt)

where

A = Re[(a + ib)12 (c + id)

12 ] (1.72)

B = Im[(a + ib)12 (c + id)

12 ] (1.73)

and

g33 = g33 = e2rt, g00 = g00 = 1.



We thus now have a real 4-metric g containing ve real param-

eters a, b, c, d, r (in coordinates (t, xi), after removing the bar):

g11 = 2ert(a cos(r

3t) b sin(r3t))

g22 = 2ert(c cos(r

3t) d sin(r3t))

g12 = 2ert(A cos(r

3t)B sin(r3t))

g33 = e2rt

g00 = 1.

(1.74)

This is a solution of the vacuum Einstein equations R = 0, as

can also be checked by direct computation.

The corresponding real vector eld X is then given by X =

Xi(x) xi , where X

i(x) = xi

xk Xk(x). From (1.70), we have

xi

xk=

(c id)1/2

i2D(c + id)

1/2

i2D0

(a ib)1/2

i2D(a + ib)1/2

i2D0

0 0 1

,

so that,

X1(x) =

(c id)1/2i2D

(1x1) (c + id)1/2

i2D(2x2)

= (p iq)(c id)1/2

i2D[(a + ib)1/2x1 + (c + id)1/2x2

]+

(p + iq)(c + id)1/2

i2D[(a ib)1/2x1 + (c id)1/2x2]

=Im[(p iq)(c + id)1/2(a ib)1/2]

Im[(a + ib)1/2(c id)1/2] x1 q(c

2 + d2)1/2

Dx2

= qx1 q(c2 + d2)1/2

Dx2. (1.75)


Space and Time 33

Similarly, X2(x) = q(a2+b2)1/2

D x1 + qx2 and X3(x) = rx3. After

removing the bar again, the real vector eld X is given by

X =

(qx1 q(c

2 + d2)1/2

Dx2

)

x1

+

(q(a2 + b2)1/2

Dx1 + qx2

)

x2 rx3

x3. (1.76)

We, thus, have a real solution (g,X) of (1.41)(1.43) with g =

and X given by (1.76).

A special case of (1.74) is given by a = 1, b = c = 0, d = 1 with

A = B = 1/2,D = 1/2,g11 = 2er cos(r

3)

g22 = 2er sin(r3)

g12 =2er

(cos(r

3) sin(r3))

g33 = e2r

g00 = 1.

(1.77)

This solution can also be checked by direct calculation.d

Notes on signature: The signature of the metric (1.77) is non-

Lorentzian and the vector eld X is real. A Lorentzian metric is

obtained by changing the sign of g33, i.e., g33 = e2r . This can beachieved by taking x3 = ix3 in (1.69). However, then, the correspond-

ing vector eld has to be complex. In both cases, the corresponding

3-metrics must have indenite signatures.

dThis was done on MAPLE, the symbolic computation language. See

Appendix C.



1.9. A Solution of Mathematical Interest

We present now another solution of our equations which is mathe-

matically interesting but turns out to be physically trivial. However,

this solution illustrates the basic principle of our approach.

Again, suppose = = 0 and now we assume

hik = ikf (1.78)

where i are constants and f is a function of (xi) to be determined.

Then (1.51) is identically satised, whereas (1.50) implies

12(i1f,1 + i2f,2 + i3f,3) 12(

21 +

22 +

23)f,i = 0. (1.79)

We can satisfy (1.71) by assuming

if,k = kf,i (1.80)

which implies that

f(x) = F (u), where u = ixi (1.81)

and F is a function of the single variable u. Now, consider Eq. (1.52)

in view of (1.78) and (1.80). It becomes

12(LXh)ik = 14ikf

2j

2j

or

ikf,X + ifX,k + kfX

,i =

12ikf

2j

2j . (1.82)

Let us put

Xi =12iG(u). (1.83)


Space and Time 35

Then

hik = Xi,k + Xk,i = ikG

,

so that

F = G (1.84)

and (1.82) becomes

(F G + 2FG)j

j = F 2j

2j ,

or[GG+ 2(G)2

]j

j = (G)2j

2j .(1.85)

We can satisfy (1.85) by taking, say,j

j =j

2j = 1 (1.86)

and by making sure that G satises the following equation

GG + (G)2 = 0, (1.87)

the general solution of which is: G(u) = (au + b) 12 , where a, b arearbitrary constants.

A solution of (1.49)(1.52) is thus provided by (g,X) where g =

and X = Xi(x)/xi with Xi(x) = 12iu1/2, where i satisfy (1.86)

and u = ixi. Next, we shall derive the corresponding 4-metric solu-

tion of the vacuum eld equations (1.38)(1.40) by transforming the

3-metric g = by the ow of X. The ow of X is given by the

solution of the dynamical system:

dxi

dt=

12iu

12 = i(1x1 + 2x2 + 3x3)

12

xi(0) = xi0

(1.88)



together with (1.86). To solve (1.88), note that, according to (1.86)

and (1.88) dudt = idxi

dt =12(i

2i )u

12 = 12u

12 and u(0) = u0 = ixi0.

This implies that u12 = t4 + u

120 , and so

x2(t) =

122

(t

4+ u

120

)dt = x20 +

122u

120 t +

116

2t2,

x3(t) =

123

(t

4+ u

120

)dt = x30 +

123u

120 t +

116

3t2,

x1(t) =u

1 2

1x2(t) 3

1x3(t)

=11

(t

4+ u

120

)2 2

1

(x20 +

122u

120 t +

116

2t2

) 3

1

(x30 +

123u

120 t +

116

3t2

)= x10 +

121u

120 +

116

1t2.

Under the ow of X, the point (xi0) is mapped into the

point (xi(t)). The ow generated by X is thus t : xi xi = xi+

12iu

12 t + 116it

2. The inverse transformation is: xi xi = xi 12i(u

1/2 t4)t 116it2, where u = ixi. In view of (1.86), the Jacobianmatrix of the transformation turns out to be

xi

xk= ik ik; = t/4(ixi) 12

with det(

xi

xk

)= 1 w.

(1.89)

Under the ow of X, the at 3-metric gik(x) = ik is transformed

into gik(x, t), according to:

gik(x, t) =x

xixm

xkm.


Space and Time 37

Thus,

gik(x, t) = ik + ik2 2ik. (1.90)

And, we have a 4-metric solution g of the vacuum eld equations

R = 0 (in coordinates (t, xi), after removing the tilde in (1.90))

g =

(ik + ik2 2ik 0

0 1

)

= t/4(ixi)12 ;

i

i =

i

2i = 1

(1.91)

with det(g) = (1 )2. However, not only R = 0, but alsoR = 0. The solution thus represents a at space-time!



1.10. The Schwarzschild Solution

As the rst example of a non-trivial at 3-space solution of (1.41)

(1.43), consider a general spherically symmetric 3-metric g and a

spherically symmetric vector eld X in coordinates (x1 = , x2 =

, x3 = ):

g = gikdxidxk = f()2d2 + 2(d2 + sin2 d2),

X = Xi

xi= a()

.

(1.92)

Substituting (1.92) in (1.41)(1.43), we obtain the following

ordinary dierential equations for the unknown functions f()

and a()

4

(d

df()

)a()

f()= 0, (1.93)

42(

d

df()

) + (f())3 f() + 2

(d

df()

)(a())2(f())2

2(f())3

+42 (f())3

(d

da()

)a() + (a())2(f())3

2(f())3= 0, (1.94)

(2

d

df() + (f())2

(d2

d2f()

)(a())2

+3(

d

df()

)a() (f())2

(d

da()

)

) (f())1

a()(f())3

(d2

d2a()

) + (f())3

(d

da()

)2

f()


Space and Time 39

2(

d

df()

)(a())2(f())2 + 2(f())3

(d

da()

)a()

f()= 0,

(1.95)

(d

df()

) + (f())3 f() + (a())2(f())3

(f())3

2 (f())3

(d

da()

)a() +

(d

df()

)(a())2(f())2

(f())3= 0.

(1.96)

[Eq. (1.96)] (sin())2. (1.97)

Equation (1.93) implies that, for non-trivial a(), f() = const.

(which we set equal to one). Equation (1.94) then becomes

2(

d

da()

)a()

+a()2

2= 0 (1.98)

whose solution is a() = k1/2 , k = const. The remaining Eqs. (1.95)

(1.97) are then identically satised.

Thus the at 3-metric

g = d2 + 2(d2 + sin2 d2)

and the vector eld X = k12

(1.99)is the only solution of Eqs. (1.93)(1.97).

1.10.1. The Schwarzschild metric

We now wish to derive the space-time 4-metric which corresponds

to our solution (1.99), according to Theorem 1, by transforming the

3-metric g by the ow of the vector eld X. In order to calculate



the ow of X, it is convenient to transform X rst to a simpler form

by a simple change of coordinates. For example, the transformation

(, , ) (R, , ), where = (32kR)23 transforms the metric as

well as the vector eld in (1.99), into

g = k2(32kR

) 23

dR2 +(32kR

) 43

(d2 + sin2 d2)

X =

R.

(1.100)

It is clear from the invariant form of Eqs. (1.41)(1.43), that a

change of coordinates provides another solution of the metric and

the vector eld. It can also be checked directly that (1.100) is indeed

a solution of (1.41)(1.43). And, of course, the 3-metric in (1.100) is

still at.

The ow of X is given by: : R R = R + , = , = . Therefore, according to Theorem 1, (1.25) corresponds tothe space-time 4-metric (in Gaussian normal coordinates (, R, , )):

ds2 = d2 + k2[32k(R )

] 23

dR2

+[32k(R )

] 43

(d2 + sin2 d2). (1.101)

That (1.101) is a solution of the vacuum Einstein eld equations

can also be checked directly. In fact, (1.101) is the Schwarzschild

solution in the so-called Lemaitre coordinates ([911]) if we take

k = (2m)12 .

This can be seen by considering the following transformation:

(, R, , ) (t, r, , ),


Space and Time 41

where

= 2( r2m

) 12 + 2m log

r 2mr +

2m

t = (r, t)R =

23r32 (2m)

12 + = R(r, t)

= , =

(1.102)

with the inverse transformation:

(t, r, , ) (, R, , ),

where

r = (2m)13

[32(R )

] 23

= r(R, )

t = 2( r2m

) 12 + 2m log

r 2mr +

2m

= t(R, ) = , = .

(1.103)

(1.102) or (1.103) transforms (1.101) into the Schwarzschild metric:

ds2 = (1 2m

r

)dt2 +

(1 2m

r

)1dr2 + r2(d2 + sin2 d2).

(1.104)

Incidentally, we have thus proved a version of Birkos Theorem in

our formalism, namely, that the only spherically symmetric solution

of (1.41)(1.43) leads to the Schwarzschild space-time.

One might think that if (X, g) is a solution of (1.41)(1.43), then

adding a Killing vector eld of g to X would again give us another

solution. In fact, one has the following proposition.

Proposition 1. Let (X, g) be a solution and X0 a Killing vector

eld of g. Then, (X +X0, g) is also a solution provided [X,X0] = 0.



Proof. Let X = X + X0. Then, LXg = LXg and LX(LXg) =LX(LXg)+LX0(LXg). Now, [LX0,LX ] = L[X0,X]. Hence, [X0,X] = 0implies that LX0(LXg) = LX(LX0g) = 0, and therefore, we have alsoLX(LXg) = LX(LXg).

However, this does not generate a new space-time 4-metric in view

of the following Proposition.

Proposition 2. (X, g) and (X +X0, g), where [X,X0] = 0, gener-

ate equivalent space-time 4-metrics.

Proof. Since X commutes with X0, the ow of X + X0 is a com-

position (as maps) of the ow of X and the ow of X0. Since X0 is

Killing, the ow of X0 has no eect on the metric g. Therefore, the

eect of the ow of X +X0 on g is the same as that of X.

For example, consider the solution (1.99) together with the two

Killing vectors X0 and X1 of the 3-metric:

g: ds2 = d2 + 2(d2 + sin2 d2),

X = (2m/)12

,

X0 = sin

+ cot cos

,

X1 = sin cos

+ (cos cos/)

(cosec sin/)

.

Here, the Killing vector eld X0 corresponds to rotational isometry,

whereas X1 corresponds to translational isometry. [X,X0] = 0, but

[X,X1] = 0. (X + X0, g) is again a solution, but (X + X1, g) is not.The 4-metrics corresponding to (X, g) and (X+X0, g) are equivalent.


Space and Time 43

Our denition of a physical observer was that of a vector eld X on

(M3, g) such that 0 < g(X,X) < 1. Note that the vector eld X =

[2m ]12

in (1.99), therefore, ceases to be physical when 2m or

when R 43m, even though X has a mathematical singularity only at = 0. = 2m is, of course, the Schwarzschild horizon corresponding

to r = 2m in Schwarzschild coordinates (t, r, , ).



1.11. From Space-Time 4-Metric to 3-Metricand 3-Vector Field

There exists a converse to Theorem 1.

Theorem 2. Suppose a space-time 4-metric solution of the vacuum

Einstein eld equations has the form

ds2 = d2 + gik(, xi)dxidxk

in the Gaussian normal coordinates (, xi), where gik(, xi) is the

transformed -dependent metric by the ow of some 3-vector eld

X acting on a 3-metric gik(xi). Then (gik,X) is a solution of our

Eqs. (1.41)(1.43).Proof. By assumption, g = (g) satises Eqs. (1.46)(1.48).

Again, from (1.45), one has

(LXK)|=0 =

( (K))=0

.

Since 0 = Identity, (1.46)(1.48) implies (1.41)(1.43).

Corollary. In particular, if a space-time 4-metric solution of the

vacuum Einstein equations has the form:

ds2 = d2 + gik(x1 , x2, x3)dxidxk (1.105)

in the Gaussian normal coordinates (, xi), then (gik = gik(x1, x2,

x3),X = 1 x1 ) is a solution of (1.41)(1.43).

This is because the ow of X = 1 x1 is simply : x1 x1 =

x1 + , x2 x2 = x2, x3 x3 = x3.As an example of Theorem 2, we shall now illustrate how to obtain

the equivalent 3-metric g and 3-vector eld X from a space-time 4-

metric which satises the vacuum Einstein eld equations and which


Space and Time 45

has the specic form given by (1.105). The rst example will be the

Schwarzschild black hole metric and the procedure below will be help-

ful when we consider next the Kerr black hole in our 3-dimensional

formalism.

Consider the Schwarzschild metric gx in coordinates x = [x1, x2,

x3, x4]

gx =

x12

x12 2mx1 0 0 00 x12 0 0

0 0 x12 sin2 x2 0

0 0 0 1 + 2mx1

(1.106)

and, in anticipation of (1.103), make a coordinate transformation of

the form

[x1, x2, x3, x4] [y1, y2, y3, y4]x1 = F (y1 y4), x2 = y2,

x3 = y3, x4 = H(F (y1 y4)) y4,(1.107)

where the functions F and H would be determined by imposing suit-

able conditions in order to bring the metric into the appropriate form.

Then, the transformed 4-metric components gy (, = 1, 2, 3, 4)

are (see Appendix D)

gy11 = (D(F )(y1 y4))2[(F (y1 y4))2

+(D(H)(F (y1 y4)))2(F (y1 y4))2

4(D(H)(F (y1 y4)))2mF (y1 y4)

+ 4(D(H)(F (y1 y4)))2m2]/F (y1 y4)(F (y1 y4) 2m),(1.108)



gy12 = 0, (1.109)

gy13 = 0, (1.110)

gy22 = F (y1 y4)2, (1.111)gy23 = 0, (1.112)

gy33 = (F (y1 y4))2(sin(y2))2, (1.113)gy14 = D(F )(y1 y4)[D(F )(y1 y4)(F (y1 y4))2

+(D(H)(F (y1 y4)))2D(F )(y1 y4)(F (y1 y4))2

4 (D(H)(F (y1 y4 )))2D(F )(y1 y4)mF (y1 y4)+ 4 (D(H)(F (y1 y4)))2D(F )(y1 y4)m2

+D(H)(F (y1 y4))(F (y1 y4))2

4D(H)(F (y1 y4))mF (y1 y4)+ 4D(H)(F (y1 y4))m2]/F (y1 y4)(F (y1 y4) 2m),

(1.114)

gy24 = 0, (1.115)

gy34 = 0, (1.116)

gy44 = [(D(F )(y1 y4))2(F (y1 y4))2

+(D(H)(F (y1 y4)))2(D(F )(y1 y4))2(F (y1 y4))2

4(D(H)(F (y1 y4)))2(D(F )(y1 y4))2mF (y1 y4)+ 4(D(H)(F (y1 y4)))2(D(F )(y1 y4))2m2

+2D(H)(F (y1 y4))D(F )(y1 y4)(F (y1 y4))2

8D(H)(F (y1 y4))D(F )(y1 y4)mF (y1 y4)+ 8D(H)(F (y1 y4))D(F )(y1 y4)m2 + (F (y1 y4))2

4mF (y1 y4) + 4m2]/F (y1 y4)(F (y1 y4) 2m),(1.117)

where D is the derivative operator.


Space and Time 47

To reduce it to the appropriate Gaussian normal form we need to

impose the conditions

gy14 = 0, (1.118)

gy44 = 1. (1.119)

One can solve (1.118)(1.119) for the derivatives D(F ), D(H) in

terms of F (y1 y4) to obtain

D(F )(y1 y4) =2

m

F (y1 y4) , (1.120)

D(H)(F (y1 y4)) =2

m

F (y1 y4)(1 2 m

F (y1 y4))1

.

(1.121)

Substituting (1.120)(1.121) back into (1.108)(1.117) we obtain the

desired Gaussian normal form for the Schwarzschild metric (provided

F and H satisfy (1.120)(1.121))

gy14 = gy24 = gy34 = 0, gy44 = 1gy11 = 2

m

F (y1 y4)gy12 = gy13 = gy23 = 0

gy22 = (F (y1 y4))2

gy33 = (F (y1 y4))2(sin(y2))2.

(1.122)

According to the Corollary of Theorem 2 the above 4-metric is

generated by the 3-metric gz in coordinates z = [z1, z2, z3] (one

simply replaces y1 y4 by z1, y2 by z2 and y3 by z3)

gz =

2m/F (z1) 0 0

0 F (z1)2 0

0 0 F (z1)2(sin(z2))2

(1.123)



and the 3-vector eld

Xz = 1

z1. (1.124)

One can now easily solve (1.120) to nd that F (z1) = ((3/2)

(2m)1/2z1)2/3 (apart from a constant), and thus recover essentially

(1.100).

However, we do not need to solve (1.120) explicitly for F (z1) in

order to obtain (1.99). We simply make a sort of inverse transforma-

tion of the coordinates [z1, z2, z3] back to the original coordinates.

In other words, a transformation

[z1, z2, z3] [w1, w2, w3]z1 = F1(w1), z2 = w2, z3 = w3

}(1.125)

where z1 = F1(w1) is the inverse function of w1 = F (z1), so

that F (F1(w1)) = w1 and dF1

dw1 = 1/(dFdz1 ) = 1/(2m/w1)

1/2 from

(1.120). Under (1.125), (1.123)(1.124) is, therefore, transformed into

gw11 = 2m(

dF1

dw1

)2 /F (F1(w1)) = 1

gw12 = gw13 = gw23 = 0

gw22 = F (F1(w1))2 = w12

gw33 = F (F1(w1))2(sin(w2))2 = w12(sin(w2))2,

(1.126)

Xw =(1/(

dF1

dw1

))

w1= (2m/w1)1/2

w1(1.127)

which is essentially (1.99).


Space and Time 49

1.12. The Kerr Solution

As for the second example of Theorem 2, we shall now demonstrate

that the axially symmetric stationary Kerr solution [12] of the vac-

uum Einstein equations can also be brought into the appropriate

Gaussian normal form by a procedure similar to the one outlined in

the previous section, and thus can also be formulated in terms of

a 3-metric and a 3-vector eld on a 3-manifold. Consider the Kerr

metric in BoyerLindquist coordinates [x1, x2, x3, x4]

gx11 =x12 + a2(cos(x2))2

x12 2mx1 + a2 ,

gx12 = gx13 = gx23 = gx14 = gx24 = 0,

gx22 = x12 + a2(cos(x2))2,

gx33 =((x12 + a2)2 (x12 2mx1 + a2)a2(sin(x2))2)(sin(x2))2

x12 + a2(cos(x2))2,

gx34 = 2 amx1 (sin(x2))2

x12 + a2(cos(x2))2,

gx44 = 1 + 2 mx1x12 + a2(cos(x2))2

,

(1.128)

and make a coordinate transformation of the form:

[x1, x2, x3, x4] [y1, y2, y3, y4],x1 = F (y1 y4, y2), x2 = y2, x3 = K(y1 y4, y2) + y3,x4 = H(F (y1 y4, y2)) y4.

(1.129)

We choose the unknown function F,K,H in such a way that the

following conditions are satised for the transformed metric compo-

nents. (The expressions for the transformed metric components are



too long to reproduce here. See Appendix D.)

gy14 = gy24 = gy34 = 0, gy44 = 1. (1.130)

These equations can now be solved for the following derivatives of

F,K,H, in terms of F (y1 y4) to give

D[1](F )(y1 y4, y2)

=2m

F (y1 y4, y2)a2 + (F (y1 y4, y2))3(F (y1 y4, y2))2 + a2(cos(y2))2 , (1.131)

D[1](K)(y1 y4, y2)

= 2amF (y1 y4, y2)/[((F (y1 y4, y2))2 + a2(cos(y2))2)

((F (y1 y4, y2))2 2mF (y1 y4, y2) + a2)], (1.132)

D(H)(F (y1 y4, y2))

=2m

F (y1 y4, y2)a2 + (F (y1 y4, y2))3(F (y1 y4, y2))2 2mF (y1 y4, y2) + a2 . (1.133)

Here D[1](F ), D[1](K) are the rst partial derivatives of F,K with

respect to the rst argument y1 y4, respectively. Note that thereare no conditions on the rst partial derivatives of F,K with respect

to the second argument y2.

The above conditions (1.131)(1.133) are not only necessary but

also sucient for (1.130) to hold. In principle, therefore, one can solve

(1.131) for F (y1 y4, y2) and, then, (1.132)(1.133) to determineK(y1 y4, y2), H(F (y1 y4, y2)). Unfortunately, (1.131) cannot beintegrated explicitly as it involves elliptic integrals.

Let us assume the above conditions (1.131)(1.133). The compo-

nents of gy then have the appropriate Gaussian normal form given by

(1.105). They depend on functions of (y1y4, y2) and y2 only. There-fore, according to the Corollary of Theorem 2, gy is generated by the


Space and Time 51

3-metric gz in coordinates [z1, z2, z3] (where one replaces y1 y4 byz1, y2 by z2 and y3 by z3 in gyik (i, k = 1, 2, 3) to get gzik) and the

3-vector eld

Xz = 1

z1(1.134)

gzik (i, k = 1, 2, 3) contain only F (z1, z2), K(z1, z2) and their rst

partial derivatives.

To obtain an explicit form of the 3-metric we make an inverse

transformation of the coordinates [z1, z2, z3] (analogous to the

Schwarzschild case) back to the original coordinates as follows:

[z1, z2, z3] [w1, w2, w3]z1 = F1(w1, w2), z2 = w2,

z3 = w3K(F1(w1, w2), w2)(1.135)

where F1 is the inverse function of w1 = F (z1, z2) with respect to

the rst variable, that is, F (F1(w1, w2), w2) = w1.

Under (1.135) the vector eld Xz is transformed into

Xw =[1/

F1

w1(w1, w2)

]

w1

+ [D[1](K)(F1(w1, w2), w2)]

w3. (1.136)

The transformed components gwik contain F (F1(w1, w2), w2)

and only the following derivatives:

F1

w1(w1, w2),

F1

w2(w1, w2),

D[1](K)(F1(w1, w2), w2),

D[1](F )(F1(w1, w2), w2),

D[2](F )(F1(w1, w2), w2).



These can be expressed explicitly in terms of w1, w2 as follows. From

(1.131) and (1.132), we have

F (F1(w1, w2), w2) = w1, (1.137)

D[1](K)(F1(w1, w2), w2)

=K

z1= 2

amw1(w12 + a2(cos(w2))2)(w12 + a2 2mw1) ,

(1.138)

F1

w1(w1, w2)

= 1/

F

z1= 1/2

(w12 + a2(cos(w2))2)2

m

w1(w12 + a2). (1.139)

Now, from (1.131), by integrating once

w1z1

=2m

w1(w12 + a2)w12 + a2(cos(w2))2

,

we have

z1 =1

(2m)1/2

w12 + a2(cos(w2))2

w1(w12 + a2)dw1,

apart from a function of w2, which we set equal to zero. Therefore,

z1w2

= 2a2 sin(w2) cos(w2)

(2m)1/2

dw1

w1(w12 + a2)

or

F1

w2(w1, w2) = 2a

2 sin(w2) cos(w2)(2m)1/2

EL(w1), (1.140)

whereEL(w1) is a standard elliptic integral. And, again from (1.131),

D[1](F )(F1(w1, w2), w2)

=F

z1=2m

w1(w12 + a2)w12 + a2(cos(w2))2

. (1.141)


Space and Time 53

To calculate D[2](F )(F1(w1, w2), w2) = Fz2 , we dierentiate both

sides of (1.137) with respect to w2, to get

0 =w1w2

=F

z1F1

w2(w1, w2) +

F

z2z2w2

=

[2m

w1(w12 + a2)w12 + a2(cos(w2))2

]

[a

2 sin(w2) cos(w2)EL(w1)2

m

]+

F

z2 1

so that

D[2](F )(F1(w1, w2), w2)

= 2a2 sin(w2) cos(w2)EL(w1)

w1(w12 + a2)

w12 + a2(cos(w2))2. (1.142)

Finally, the explicit expressions for the 3-metric gw and the

3-vector eld Xw, which generate the Kerr solution, are:

gw11 =w12 + a2(cos(w2))2

w12 + a2 2mw1

2mw1(w12 + a2)(w12 + a2(cos(w2))2 2mw1)

(w12 + a2(cos(w2))2)(w12 + a2 2mw1)2(1.143)

gw12 = gw23 = 0, (1.144)

gw13 = 2 a(sin(w2))22m3/2w1

w13 + a2w1

(w12 + a2(cos(w2))2)(w12 + a2 2mw1) , (1.145)

gw22 = w12 + a2(cos(w2))2, (1.146)

gw33 =(w12 + a2 2mw1 + 2 mw1(w1

2 + a2)w12 + a2(cos(w2))2

)(sin(w2))2,

(1.147)



Xw =

[2mw13 + a2w1

w12 + a2(cos(w2))2

]

w1

+[2

amw1(w12 + a2(cos(w2))2)(w12 + a2 2mw1)

]

w3.

(1.148)

One can now verify (Appendix E) that the above (g,X) is indeed

a solution of our Eqs. (1.41)(1.43).

Note that, (1.143)(1.145) reduce to the Schwarzschild case

(1.126)(1.127) when a 0. When m 0, we obtain

gw =

w12 + a2(cos(w2))2

w12 + a20 0

0 w12 + a2 0

(cos(w2))20 0 (w12 + a2)

(sin(w2))2

,

(1.149)

Xw = 0. (1.150)

The above 3-metric (1.149) is at, and therefore (1.149)(1.150) is

equivalent to the at space-time.

If a = 0, the 3-metric (1.143)(1.145) is not at, in contrast to(1.126), the 3-metric in the Schwarzschild case. Its scalar curvature

is given by

Rw =2a2mw1(3(cos(w2))2 1)

w12 + a2(cos(w2))2(1.151)

and the 3-vector eld Xw in (1.148) has the length

g(X,X) = gw(Xw,Xw) =2mw1

w12 + a2(cos(w2))2. (1.152)


Space and Time 55

Thus X ceases to be physical when g(X,X) = 1, i.e.,

2mw1w12 + a2(cos(w2))2

= 1 (1.153)

which corresponds to the so-called stationary limit of the Kerr

black hole.



1.13. The Maxwell Equations

Our basic approach to evolution in a 3-dimensional manifold can also

be applied to the Maxwell equations.

We rst write the Maxwell equations on a space-time

curlE = Bt

curlH = 4j+Dt

divD = 4

divB = 0

(1.154)

for the electromagnetic variables E,H,B,D, j, (the usual elec-

tric, magnetic, elds and inductions, current and charges) in terms

of 3-dimensional time-dependent dierential forms on a 3-manifold

(M3, g) as follows. Dene the following (in some local coordinates

(xi) of (M3, g)):

1-forms, , 1(M3):

=

i

idxi, E = (1, 2, 3),

=

i

idxi, H = (1, 2, 3).

2-forms, , , 2(M3):

=i


Space and Time 57

3-form: 3(M3):

= dx1 dx2 dx3.

Note that these 3-dimensional forms are time-dependent with the

time t as a parameter. Using the exterior derivative operator d the

Maxwell equations (1.154) can then be written as:

d = t

d = 4 +

t

d = 4

d = 0.

(1.155)

As we did in the case of GaussEinstein equations, let us now

replace the time-derivative t in (1.155) by the Lie-derivative LXwith respect to some vector eld X on M3. Then (1.155) becomes

d = LX, (1.156)d = 4 + LX, (1.157)d = 4, (1.158)

d = 0. (1.159)

Here, , , , , , are to be regarded now as (time-independent)

purely 3-dimensional dierential forms on M3. Equations (1.156)

(1.159) are thus analogous to (1.38)(1.40), corresponding to the

GaussEinstein equations, and are to be considered as a set of equa-

tions for the forms , , , , , and the vector eld X. Every solu-

tion of (1.156)(1.159) determines uniquely a solution of (1.155) and,

thus, of (1.154).e One simply has to transform the time-independent

eAgain, the converse need not be true.



dierential form , for example, by the ow t of X to obtain a time-

dependent form: = t(), etc. The vector eld X thus provides the

temporal evolution of the electromagnetic elds.

The above equations take a surprisingly simple (and elegant) form

if one uses the relationship between the Lie derivative LX and theexterior derivative d and the contraction operator iX relative to X:

LX = iX d + d iX (1.160)

to obtain from (1.156) and (1.159)

d = (iX d + d iX) = d(iX)

or

d = 0 where = + iX. (1.161)

Similarly, from (1.87) and (1.88)

d = 4 + (iX d + d iX) = 4( + iX) + d(iX)

or

d = 4 where = iX, = + iX. (1.162)

Thus, Eqs. (1.156)(1.159) become

d = 0

d = 4

}(1.163)

where

= + iX

= iX = + iX.

(1.164)


Space and Time 59

In vacuum, = = = 0; thus, the corresponding vacuum

equations are

d = 0

d = 0.

}(1.165)

The continuity equation follows by taking the exterior derivative

of the second equation in (1.163), i.e., d = 0. From (1.164)

0 = d( + iX) = d + (LX iX d).

Since is a 3-form on a 3-manifold, d = 0. Thus

LX + d = 0 (1.166)

which is the continuity equation relative to the vector eld (observer)

X if we keep in mind that the Lie-derivative LX corresponds to thetime-derivative t and d is the divergence of the current j.



References

[1] G. D. Mostow, Strong Rigidity of Locally Symmetric Spaces, Ann.

Math. Series (Princeton Univ. Press, Princeton, 1973).

[2] S. Helgason, Dierential Geometry and Symmetric Spaces (Academic,

New York, 1972).

[3] S. Kobayashi and K. Nomizu, Foundations of Dierential Geometry,

Vol. 1 (Interscience, New York, 1983).

[4] D. V. Widder, The Heat Equation (Academic Press, New York, 1975).

[5] A. Lichnerowicz, Theories relativistes de la gravitation et de

lelectromagnetisme (Masson, Paris, 1955).

[6] J. S. Synge, Relativity, the General Theory (North-Holland, Amster-

dam, 1960).

[7] A. E. Fisher and J. E. Marsden, J. Math. Phys. 13 (1972) 546568.

[8] E. Kasner, American J. Math. 43 (1921) 217.

[9] G. Lemaitre, Lunivers en expansion, Ann. Soc. Sci. Bruxelles I A 53

(1933) 51.

[10] A. Z. Petrov, Einstein Spaces (Pergamon Press, Oxford, 1969).

[11] D. K. Sen, Class. Quant. Gravity 12 (1995) 553577.

[12] D. K. Sen, J. Math. Phys. 41 (2000) 75567572.


2. MATTER

2.0. Introduction

The basic constituents of all matter, namely, the elementary particles,

have the following salient features.

The massive ones, that is, those with non-zero rest mass, can be

classied into two basic categories: the lighter leptons and the heav-

ier baryons.. In each category, there is a stable particle, the lepton

electron and the baryon proton. The unstable leptons and baryons all

spontaneously decay eventually into the stable ones.

Then, there are the so-called zero rest mass particles, the photon

and the neutrino, which are essentially carriers of energy and vehicles

of interaction between the particles.

The so-called standard model has had considerable success as a

unied theory of all elementary particles. Together with the Higgs

boson and Higgs mechanism, we are able to explain the existence

and masses of several new bosons.

Nevertheless, it cannot be considered as a complete and fully sat-

isfactory theory of all elementary particles. It cannot, for example,

explain intrinsically why the proton is about 1836 times heavier than

the electron.

61



In the standard model, the phenomenon of neutrino oscillation

([1]) requires that neutrinos have non-zero mass.

In 1957, Heisenberg ([2, 3]) tried to formulate (without much

success) a unied theory of all elementary particles starting from a

nonlinear 4-component spinor equation with a built-in fundamental

constant.

Here, we ([4, 5]) suggest that massless 2-component Weyl neutri-

nos, instead of 4-component spinors, are probably more fundamental

than previously thought. We consider a composite system consist-

ing of a massless positively oriented 2-component Weyl neutrino and

a massless negatively oriented 2-component Weyl neutrino with a

certain specic symmetry-breaking interaction between the two.

We assume that the observable physical particles manifest as

energy states of the resulting 4-component system. A simple quan-

tum mechanical treatment shows that such a model should exhibit

2-fold branching and energy defects, which could then be interpreted

as formation of particles of non-zero rest mass.

Such a model can also provide a qualitative, alternative non-

standard explanation of the dierent avors of amassless 4-component

neutrino and thus, of neutrino oscillation without assuming a neutrino

mass.


Matter 63

2.1. The Photon and the Weyl Neutrinos

A long time ago, de Broglie ([6]) and Jordan ([7]) have tried to

construct a neutrino theory of light based on the formers idea of

fusion. Both these authors tried to explain the photon as a com-

bination of two 4-component neutrinos of essentially the same kind.

Pryce ([8]) and Barbour et al. ([9]) however showed that such a pho-

ton would be longitudinally polarized. For a historical review of the

neutrino theory of light, see [10].

We rst show that the photon is intimately connected not to the

4-component Dirac neutrino, but to the 2-component positively and

negatively oriented Weyl neutrinos.

The Maxwell equations in vacuum can be written as ([11, 12])

df = 0, d f = 0 (2.1)

where f is the electromagnetic dierential 2-form, d is the exterior

derivative operator and f is the Hodge dual of f . We shall use space-time coordinates x(= x, y, z, ict), = 1, 2, 3, 4, so that the metric

tensor in a Minkowski space-time is and consider only proper

Lorentz transformations. Then, in components, the Hodge dual of f

is given by

f = 12f (2.2)

where is the completely anti-symmetric LeviCivita symbol.

In general, for a p-form w on an n-dimensional pseudo-Riemannian

manifold, one has

w = (1)p(np)w (2.3)



so that f = f for the electromagnetic 2-form f on the 4-dimensionalMinkowski space-time. It is thus possible to decompose f in an invari-

ant manner into a self-dual part f s and an anti-self-dual part fa, as

follows

f = f s + fa, f s =12(f + f), fa = 1

2(f f) (2.4)

where f s = f s, fa = fa. The Maxwell equations (2.1) are thenequivalent to

f = f s + fa, df s = 0, dfa = 0. (2.5)

For an anti-symmetric tensor, the self-duality condition (in compo-

nents) f s = f s implies that

f s12 = fs34, f

s23 = f

s14, f

s31 = f

s24 (2.6)

so that it has only three independent components. Now, a self-dual

(or an anti-self-dual) anti-symmetric tensor cannot by itself describe

a physical electromagnetic eld, because in that case there should

exists a 4-vector A = (A, i), such that f s = AA. No such4-vector with three real components and one imaginary component

can exist if f s were to be self-dual, because (2.6) would then imply

that

yAx xAy = (1/ic)tAz iz. (2.7)

Consider, however, a 2-component quantity =(

12

)and a 4-vector

s =

1

i1

2i2

(2.8)


Matter 65

and set f s = s s. Then, the self-duality condition (2.6)

gives basically two equations

f s12 = 21 i11 = 42 i32 = f s34, (2.9)

f s23 = i31 + 22 = 41 i12 = f s14. (2.10)

The third condition in (2.6) gives again (2.10). Multiplying (2.9) by

i and (2.10) by i, we get two equations for (1, 2):

11 + i21 32 + i42 = 012 i22 + 31 + i41 = 0

(2.11)which by introducing the Pauli spin matrices

1 =

(0 1

1 0

), 2 =

(0 ii 0

), 3 =

(1 0

0 1

)(2.12)

can be written as a single equation (k = 1, 2, 3)

(kk + i4) = 0 (2.13)

which is nothing but the Weyl equation for the 2-component left-

handed neutrino L. It is easily seen that if s is to be a 4-vector,

must transform as a 2-component spinor ([4]).

Following a similar procedure for the anti-self-dual case, let

a =

2i1

1

i2

(2.14)

where =(

12

)is 2-component quantity and set fa =

aa.



Then, the anti-self-duality condition for fa implies that

fa24 = fa13, f

a34 = f

a21, f

a14 = f

a32 (2.15)

which in turn implies two equations for (1, 2).

11 i22 + 32 + i41 = 012 + i21 + 31 + i42 = 0.

(2.16)Introducing the matrices

1 =

(1 0

0 1

), 2 =

(0 ii 0

), 3 =

(0 1

1 0

), (2.17)

(2.16) can be written as

(kk + i4) = 0. (2.18)

Equation (2.18) diers from Eq. (2.13) only in the respect that the

set k diers from the Pauli matrices k only in the interchange of

the indices 1 and 3. As a result, while the Pauli matrices satisfy

12 = i3, 23 = i1, 31 = i2, (2.19)

the set k satisfy

12 = i3, 23 = i1, 31 = i2. (2.20)

Therefore, Eq. (2.18) describes a 2-component right-handed

neutrino R.

Conversely, if and satisfy (2.13) and (2.18), respectively, then

f = f s + fa with f

s =

s s and fa = a a,

where s and a are given by (2.8) and (2.14), would automatically

satisfy the Maxwell equations in vacuum ([8]). In other words, the

photon can be regarded in some sense as a fusion of a left-handed

and a right-handed neutrino L and R.


Matter 67

This suggests that the left-handed and right-handed neutrinos are

perhaps the fundamental constituents of all elementary particles and

that one should consider composite LR models with some fun-damental interaction between the two. Ideally, such a model should

be treated in the framework of quantum eld theory. However, as a

rst step, in this work, we shall consider a simple, heuristic quantum

mechanical model. This will give us an idea of what we can expect

in a rigorous quantum eld theoretical model.



2.2. A Neutrino Theory of Matter

2.2.1. Composite LR system withoutinteraction

Consider rst a composite LR system without interaction. TheHamiltonian of a left-handed (massless) neutrino L described by

(2.13) is given by HL = ic( ) with = (1, 2, 3) and itseigenfunctions EL satisfy

HLEL = ELEL

or ( )EL = (iEL/c)EL .

(2.21)From now on, we shall adopt the conventional units in which c =

= 1.

The solutions of (2.21) are well-known:

EL(x,p) = a(p)eixp

where

p2 p21 + p22 + p23 = E2L

and

a(p) =

(p1 ip2EL p3

). (2.22)

They describe a left-handed neutrino L with energy EL.

The spectrum of HL is not discrete and hence, the eigenfunctions

have to be normalized by the delta-function.

EL(x,p)|EL(x,p) = a(p)a(p)(p p). (2.23)


Matter 69

This may give rise to questions of rigor and technical diculties in

what follows. One can always enclose the neutrino in a cubical box

of length L, in which case the eigenfunctions would be

EL(x,n) = a(n)ei(2/L)xn,

where n = (n1, n2, n3), n21 + n22 + n

23 = E

2L, with integer ni, and

therefore, discrete energy values. We shall not follow this procedure

and proceed as if we are dealing with a discrete problem.

The eigenfunctions are also -fold degenerate, since p can takeany value on the energy shell. We shall remove this degeneracy by

integrating EL(x,p) over the energy shell S2EL : p21 + p

22 + p

23 = E

2L

which is a 2-sphere of radius EL. We get (with a slight abuse of

notation)

EL(x) =

S2EL

a(p)eixpdSp (2.24)

as a surface integral over S2EL (see later). Furthermore, we shall sup-

pose that EL(x) are normalized.

Similarly, the corresponding eigenfunctions for the right-handed

neutrino R with Hamiltonian HR = i( ) with energy ER aregiven by

ER(y,q) = b(q)eiyq

where

q21 + q22 + q

23 = E

2R

and

b(q) =

(q3 iq2ER q1

).

(2.25)



We use y for the coordinate of R and note that b(q) diers from

a(p) by an interchange of the indices 1 and 3. A similar integration

over the energy shell S2ER : q21 + q

22 + q

23 = E

2R gives

ER(y) =

S2ER

b(q)eiyqdSq. (2.26)

Let HL and HR be the respective Hilbert spaces for L and R.Consider now a composite LR system without interaction givenby the tensor product H = HL HR with the Hamiltonian Ho =HL + HR (or more precisely, HL I + I HR) ([13]). Since Ho =i( L) i( R), where L,R act on x,y, respectively, andthus does not depend on x and y explicitly, not only

1(x,y) = EL(x) ER(y) (2.27)

but also

2(x,y) = ER(x) EL(y) (2.28)

are both eigenfunctions of Ho with energy EL +ER. Thus, the com-

posite system LR behaves like a system of identical particles eventhough l is not identical to R. We remark that if the universe were

spatially non-orientable, L would be indistinguishable from R. But

in an orientable universe, they would be distinct particles. We thus

have a 2-fold degeneracy similar to the He-atom.

2.2.2. Composite LR system with interaction

Consider now an interaction V (x,y) between L and R of the form

V (x,y) = F (|x|)H(|x| |y|) + F (|y|)H(|y| |x|) (2.29)


Matter 71

where F is a function to be specied and H is the Heaviside function.

Thus

space, time and matter

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