Biquaternion formulation of relativistic tensor dynamics

Transcription

1 Juli, 8, 2008 To be submitted... 1 Biquaternion formulation of relativistic tensor dynamics E.P.J. de Haas High school teacher of physics Nijmegen, The Netherlands epjhaas@telfort.nl In this paper we show how relativistic tensor dynamics and relativistic electrodynamics can be formulated in a biquaternion tensor language. The treatment is restricted to mathematical physics, known facts as the Lorentz Force Law and the Lagrange Equation are presented in a relatively new formalism. The goal is to fuse anti-symmetric tensor dynamics, as used for example in relativistic electrodynamics, and symmetric tensor dynamics, as used for example in introductions to general relativity, into one single formalism: a specific kind of biquaternion tensor calculus. Keywords: biquaternion, relativistic dynamics, Lorentz Force Law, Lagrange Equation, Lagrangian Introduction We start with a quote. One can say that space-time model and kinematics of the Quaternionic Relativity are nowadays studc 2008 E.P.J. de Haas

2 Juli, 8, 2008 To be submitted... 2 ied in enough details and can be used as an effective mathematical tool for calculation of many relativistic effects. But respective relativistic dynamic has not been yet formulated, there are no quaternionic field theory; Q-gravitation, electromagnetism, weak and strong interactions are still remote projects. However, there is a hope that it is only beginning of a long way, and the theory will mature. (Yefremov 2005) We hope that the content of this paper will contribute to the project described by Yegremov. Quaternions can be represented by the basis (1, I, J, K). This basis has the properties II = JJ = KK = 1; 11 = 1; 1K = K1 = K for I, J, K; IJ = JI = K; JK = KJ = I; KI = IK = J. A quaternion number in its summation representation is given by A = a a 1 I + a 2 J + a 3 K, in which the a µ are real numbers. Biquaternions or complex quaternions in their summation representation are given by C = A + ib = (a 0 + ib 0 )1 + (a 1 + ib 1 )I + (a 2 + ib 2 )J + (a 3 + ib 3 )K = a a 1 I + a 2 J + a 3 K + ib ib 1 I + ib 2 J + ib 3 K, (1) in which the c µ = a µ + ib µ are complex numbers and the a µ and b µ are real numbers. The complex conjugate of a biquaternion C is given by C = A ib. The quaternion conjugate of a biquaternion is given by C = A + ib = (a 0 + ib 0 )1 (a 1 + ib 1 )I (a 2 + ib 2 )J (a 3 + ib 3 )K. (2) In this paper we only use the complex conjugate of biquaternions.

3 Juli, 8, 2008 To be submitted... 3 Biquaternions or complex quaternions in their vector representation are given by or by C µ = c 0 1 c 1 I c 2 J c 3 K, (3) C µ = [c 0 1, c 1 I, c 2 J, c 3 K] (4) We apply this to the space-time four vector of relativistic biquaternion 4-space R µ as R µ = ict1 r 1 I r 2 J r 3 K = ir 0 1 r 1 I r 2 J r 3 K. (5) The space-time distance s can be defined as R µ R µ, or giving R µ R µ = [ ict1, r 1 I, r 2 J, r 3 K] [ict1, r 1 I, r 2 J, r 3 K], (6) R µ R µ = c 2 t 2 1 r 2 11 r 2 21 r 2 31 = (c 2 t 2 r 2 1 r 2 2 r 2 3)1. (7) So we get R µ R µ = s1 with the usual s = c 2 t 2 r 2 1 r 2 2 r 2 3 = r 2 0 r 2 1 r 2 2 r 2 3 (8) providing us with a (+,,, ) signature.

4 Juli, 8, 2008 To be submitted... 4 Adding the dynamic vectors The basic definitions we use are quite common in the usual formulations of relativistic dynamics. We start with an observer who has a given three vector velocity as v, a rest mass as m 0 and an inertial mass m i = γm 0 with γ = ( 1 v 2 /c 2 ) 1. We use the Latin suffixes as abbreviations for words, not for numbers. So m i stands for inertial mass and U p for potential energy. The Greek suffixes are used as indicating a summation over the numbers 0, 1, 2 and 3. So P µ stands for a momentum four-vector with components p 0 = 1U c i, p 1, p 2 and p 3. The momentum three-vector is written as p and has components p 1, p 2 and p 3. We define the coordinate velocity four vector as V µ = d dt R µ = ic1 v 1 I v 2 J v 3 K = iv 0 1 v 1 I v 2 J v 3 K. (9) The proper velocity four vector on the other hand will be defined using the proper time t 0, with dt = γt 0, as iγc1 U µ = d R µ = d 1 dt 0 dtr µ = γv µ = γv 1 I γv γ 2 J. (10) γv 3 K The momentum four vector will be P µ = m i V µ = m 0 U µ. (11) We further define the rest mass density as ρ 0 = dm 0 dv 0, (12)

5 so with Juli, 8, 2008 To be submitted... 5 dv = 1 γ dv 0 (13) and the inertial mass density as we get ρ i = dm i dv ρ i = dm i dv = dγm 0 1 dv γ 0 (14) = γ 2 ρ 0. (15) The momentum density four vector will be defined as i 1 u c i1 ig 0 1 G µ = g 1 I g 2 J = g 1 I g 2 J, (16) g 3 K g 3 K in which we used the inertial energy density u i = ρ i c 2. For this momentum density four vector we have the variations G µ = d dv P µ = dm i dv V µ = ρ i V µ = γ 2 ρ 0 V µ = γρ 0 U µ = γg proper µ. (17) The four vector partial derivative µ will be defined as i 1 c t1 µ = 1 I 2 J. (18) R µ 3 K

6 Juli, 8, 2008 To be submitted... 6 The electrodynamic potential four vector will be defined as i 1 φ1 ia c 0 1 A µ = A 1 I A 2 J = A 1 I A 2 J. (19) A 3 K A 3 K The electric four current density will be given by icρ e 1 ij 0 1 J µ = J 1 I J 2 J = J 1 I J 2 J = ρ ev µ, (20) J 3 K J 3 K with ρ e as the electric charge density. Adding the dynamic vector products, scalars The dynamic Lagrangian density L can be defined as L = Ṽ ν G ν = (u i v g)1 = u 0 1 (21) and the accompanying Lagrangian L as L = Ṽ ν P ν = (U i v p)1 = 1 γ U 01, (22) with u 0 as the rest system inertial energy density and U 0 as the rest system inertial energy. The latter is the usual Lagrangian of a particle moving freely in empty space. The Lagrangian density of a massless electric charge density current in an electrodynamic potential field can be defined as L = J ν A ν = (ρ e φ J A)1. (23)

7 Juli, 8, 2008 To be submitted... 7 On the basis of the Lagrangian density we can define a four force density as f µ L R µ = µ L = µ u 0. (24) In the special case of a static electric force field, and without the densities, the field energy is U 0 = qφ 0 and the relativistic force reduces to the Coulomb Force F = U 0 = q φ 0. (25) Using L = Ṽ ν G ν this four force density can be written as f µ = µ Ṽ ν G ν. (26) We can define the absolute time derivative d dt perfect fluid like, space/field quantity through of a continuous, V µ µ = Ṽ µ µ = v + t 1 = d 1. (27) dt Thus we can define the mechanic four force density as f µ d dt G µ = (V ν ν )G µ = V ν ( ν G µ ), (28) using the fact that biquaternion multiplication is associative. Adding the dynamic vector products, tensors

8 Juli, 8, 2008 To be submitted... 8 The stress energy density tensor T ν µ can be given as T ν µ = Ṽ ν G µ and gives T ν µ = [ iv 0 1, v 1 I, v 2 J, v 3 K] ig 0 1 g 1 I g 2 J g 3 K = v 0 g 0 1 iv 1 g 0 I iv 2 g 0 J iv 3 g 0 K iv 0 g 1 I v 1 g 1 1 v 2 g 1 K v 3 g 1 J iv 0 g 2 J v 1 g 2 K v 2 g 2 1 v 3 g 2 I iv 0 g 3 K v 1 g 3 J v 2 g 3 I v 3 g 3 1 (29) Its trace is T νν = Ṽ ν G ν = L. In relativistic dynamics we have a usual force density definition through the four derivative of the stress energy density tensor ν T ν µ = f µ (30) or ν V ν G µ = f µ (31) We want to find out if these equations still hold in our biquaternion version of the four vectors, tensors and their products.

9 Juli, 8, 2008 To be submitted... 9 We calculate the left hand side and get for ν T ν µ = ν Ṽ ν G µ : [ i c t1, 1 I, 2 J, 3 K = ] v 0 g 0 1 iv 1 g 0 I iv 2 g 0 J iv 3 g 0 K iv 0 g 1 I v 1 g 1 1 v 2 g 1 K v 3 g 1 J iv 0 g 2 J v 1 g 2 K v 2 g 2 1 v 3 g 2 I iv 0 g 3 K v 1 g 3 J v 2 g 3 I v 3 g 3 1 i 1 c tv 0 g 0 1 i 1 v 1 g 0 1 i 2 v 2 g 0 1 i 3 v 3 g c tv 0 g 1 I 1 v 1 g 1 I 2 v 2 g 1 I 3 v 3 g 1 I 1 c tv 0 g 2 J 1 v 1 g 2 J 2 v 2 g 2 J 3 v 3 g 2 J 1 c tv 0 g 3 K 1 v 1 g 3 K 2 v 2 g 3 K 3 v 3 g 3 K = i( 1 c tv 0 g v 1 g v 2 g v 3 g 0 )1 ( 1 c tv 0 g v 1 g v 2 g v 3 g 1 )I ( 1 c tv 0 g v 1 g v 2 g v 3 g 2 )J ( 1 c tv 0 g v 1 g v 2 g v 3 g 3 )K (32)

10 Juli, 8, 2008 To be submitted Using the chain rule this leads to ν T ν µ = i( 1 c v 0 t g 0 + v 1 1 g 0 + v 2 2 g 0 + v 3 3 g 0 )1 ( 1 c v 0 t g 1 + v 1 1 g 1 + v 2 2 g 1 + v 3 3 g 1 )I ( 1 c v 0 t g 2 + v 1 1 g 2 + v 2 2 g 2 + v 3 3 g 2 )J ( 1 c v 0 t g 3 + v 1 1 g 3 + v 2 2 g 3 + v 3 3 g 3 )K i( 1( c tv 0 )g 0 + ( 1 v 1 )g 0 + ( 2 v 2 )g 0 + ( 3 v 3 )g 0 )1 ( 1( c tv 0 g 1 ) + ( 1 v 1 )g 1 + ( 2 v 2 )g 1 + ( 3 v 3 )g 1 )I ( 1( c tv 0 g 2 ) + ( 1 v 1 )g 2 + ( 2 v 2 )g 2 + ( 3 v 3 )g 2 )J ( 1( c tv 0 g 3 ) + ( 1 v 1 )g 3 + ( 2 v 2 )g 3 + ( 3 v 3 )g 3 )K This can be abbreviated to So = ( 1 c v 0 t + v v v 3 3 ) ( 1 c tv v v v 3 ) ig 0 1 g 1 I g 2 J g 3 K ig 0 1 g 1 I g 2 J g 3 K = (Ṽ ν ν )G µ + ( ν Ṽ ν )G µ (33) ν (Ṽ ν G µ ) = (Ṽ ν ν )G µ + ( ν Ṽ ν )G µ (34) ν T ν µ = (Ṽ ν ν )G µ + ( ν Ṽ ν )G µ. (35) We have Ṽ ν ν = d and if we assume the bare particle velocity dt continuity equation ν Ṽ ν = 0, then we get ν T ν µ = d dt G µ = f µ. (36)

11 Juli, 8, 2008 To be submitted Electrodynamic vector products If we apply this to the case in which we have a purely electromagnetic four momentum density G µ = ρ e A µ then we have and L = Ṽ ν G ν = Ṽ ν ρ e A ν = J ν A ν, (37) The relativistic force equation T ν µ = Ṽ ν G µ = J ν A µ. (38) ν T ν µ = (Ṽ ν ν )G µ + ( ν Ṽ ν )G µ. (39) can be given its electrodynamic expression as ν T ν µ = ( J ν ν )A µ + ( ν J ν )A µ. (40) If the charge density current continuity equation ν J ν = 0 can be applied, then this reduces to ν T ν µ = ( J ν ν )A µ = (J ν ν )A µ = J ν ( ν A µ ). (41) The electrodynamic force field tensor B ν ν is given by In detail this reads B ν µ = B ν µ = ν A µ. (42) [ i 1 c t1, 1 I, 2 J, 3 K ] i 1 c φ1 A 1 I A 2 J A 3 K = 1 c 2 t φ1 i 1 c 1φI i 1 c 2φJ i 1 c 3φK i 1 c ta 1 I 1 A A 1 K 3 A 1 J i 1 c ta 2 J 1 A 2 K 2 A A 2 I i 1 c ta 3 K 1 A 3 J 2 A 3 I 3 A 3 1 (43)

12 Juli, 8, 2008 To be submitted To see that this tensor leads to the usual EM force field biquaternion, we have to rearrange the tensor terms according to their biquaternionic affiliation, so arrange them according to the basis (1, I, J, K). This results in ( 1 c 2 t φ 1 A 1 2 A 2 3 A 3 )1 ( 2 A 3 3 A 2 + i 1 c 1φ + i 1 c ta 1 )I ( 3 A 1 1 A 3 + i 1 c 2φ + i 1 c ta 2 )J ( 1 A 2 2 A 1 + i 1 c 3φ + i 1 c ta 3 )K (44) This equals F µ = ν A ν 1 (B 1 i 1 c E 1)I (B 2 i 1 c E 1)J (B 3 i 1 c E 1)K = F 0 1 F 1 I F 2 J F 3 K. (45) For this biquaternion to be the exact match with the EM force field, one has to add the Lorenz gauge condition F 0 = ν A ν = 0. The operation of rearranging the tensor terms according to their biquaternion affiliation is external to the mathematical physics of this paper. We try to develop a biquaternion version of relativistic tensordynamics. The above operation destroys the tensor arrangement of the terms involved. It is alien to the system we try to develop in this context. It may be a useful operation in others areas though, for example in quantum physics. The electrodynamic force field tensor Bµ ν can also be given by B ν µ = µ A ν. (46) This leads to the same EM force field biquaternion.

13 Juli, 8, 2008 To be submitted The combination of Eq.(41) and Eq.(42) leads to ν T ν µ = J ν B ν µ, (47) which is valid if charge is conserved so if ν J ν = 0. We can write Eq.(41) also as ν T ν µ = ( J ν ν )A µ = ρ e (Ṽ ν ν )A µ = ρ d dt A µ. (48) The two EM force expression we gave in this and the previous sections based on f µ = ν T ν µ and f µ = µ L do not result in the well known Lorentz Force. But we can establish a relationship between these force expressions and the Lorentz Force. The Lorentz Force Law The Lorentz Force Law in its density form is given by the expression f µ = J ν ( ν A µ ) ( µ Ã ν )J ν, (49) or So if charge is conserved we also have f µ = J ν B ν µ ( µ Ã ν )J ν. (50) f µ = ν T ν µ ( µ Ã ν )J ν (51) as an equivalent. If we examen this last part ( µ Ã ν )J ν in more detail, an interesting relation arises. We begin with the equation µ L = µ J ν A ν. (52)

14 Juli, 8, 2008 To be submitted Now clearly we have J ν A ν = Ãν J ν = u 0 1 as a Lorentz invariant scalar. Together with the chain rule this leads to µ J ν A ν = ( µ J ν )A ν + ( µ Ã ν )J ν. (53) This equation is crucial for what is to come next, the connection of a Lagrange Equation to the Lorentz Force Law. So we have to prove it in detail, provide an exact proof, specially because biquaternion multiplication in general is non-commutative. We start the proof with µ Ã ν : µ Ã ν = i 1 c t1 1 I 2 J 3 K [ ia 01, A 1 I, A 2 J, A 3 K] = 1 c ta 0 1 i 1 c ta 1 I i 1 c ta 2 J i 1 c ta 3 K i 1 A 0 I 1 A A 2 K 1 A 3 J i 2 A 0 J 2 A 1 K 2 A A 3 I i 3 A 0 K 3 A 1 J 3 A 2 I 3 A 3 1 (54) In the next step we calculate ( µ Ã ν )J ν and use the fact that

15 Juli, 8, 2008 To be submitted ( 3 A 1 )J 2 = J 2 ( 3 A 1 ): ( µ Ã ν )J ν = 1 c ta 0 1 i 1 c ta 1 I i 1 c ta 2 J i 1 c ta 3 K i 1 A 0 I 1 A A 2 K 1 A 3 J i 2 A 0 J 2 A 1 K 2 A A 3 I i 3 A 0 K 3 A 1 J 3 A 2 I 3 A 3 1 [ij 0 1, J 1 I, J 2 J, J 3 K] = ( i 1J c 0 t A 0 + i 1J c 1 t A 1 + i 1J c 2 t A 2 + i 1J c 3 t A 3 )1 (J 0 1 A 0 J 1 1 A 1 J 2 1 A 2 J 3 1 A 3 )I (J 0 2 A 0 J 1 2 A 1 J 2 2 A 2 J 3 2 A 3 )J (55) (J 0 3 A 0 J 1 3 A 1 J 2 3 A 2 J 3 3 A 3 )K Now clearly ( µ Ã ν )J ν and ( µ J ν )A ν behave identical, only J and A have changed places, so ( µ J ν )A ν = ( i 1 c A 0 t J 0 + i 1 c A 1 t J 1 + i 1 c A 2 t J 2 + i 1 c A 3 t J 3 )1 (A 0 1 J 0 A 1 1 J 1 A 2 1 J 2 A 3 1 J 3 )I (A 0 2 J 0 A 1 2 J 1 A 2 2 J 2 A 3 2 J 3 )J (A 0 3 J 0 A 1 3 J 1 A 2 3 J 2 A 3 3 J 3 )K (56)

16 Juli, 8, 2008 To be submitted If we add them and use the inverse of the chain rule we get ( µ Ã ν )J ν + ( µ J ν )A ν = i 1( c tj 0 A 0 t J 1 A 1 t J 2 A 2 t J 3 A 3 )1 ( 1 J 0 A 0 1 J 1 A 1 1 J 2 A 2 1 J 3 A 3 )I ( 2 J 0 A 0 2 J 1 A 1 2 J 2 A 2 2 J 3 A 3 )J ( 3 J 0 A 0 3 J 1 A 1 3 J 2 A 2 3 J 3 A 3 )K i 1 c 1 1 I 2 J 3 K = (J 0A 0 J 1 A 1 J 2 A 2 J 3 A 3 ) = µ ( J ν A ν ) (57) Thus we have given the exact proof of the statement So we get µ J ν A ν = ( µ J ν )A ν + ( µ Ã ν )J ν. (58) µ L = µ J ν A ν = ( µ J ν )A ν + ( µ Ã ν )J ν. (59) We now have two force equations, fµ L = µ L = µ u 0 and fµ T = ν T ν µ = d G dt µ. We combine them into a force equation that representing the difference of these two forces: f µ = f T µ + f L µ = ν T ν µ + µ L. (60) For the purely electromagnetic case this can be written as and leads to f µ = ν J ν A µ µ J ν A ν (61) f µ = ( J ν ν )A µ + ( ν J ν )A µ ( µ J ν )A ν ( µ Ã ν )J ν. (62)

17 Juli, 8, 2008 To be submitted If we have ν J ν = 01 and µ J ν = 0 then this general force equation reduces to the Lorentz Force Law f µ = ( J ν ν )A µ ( µ Ã ν )J ν. (63) This of course also happens if ν J ν = µ J ν, so if the RHS of this equation has zero non-diagonal terms. The Lagrangian Equation If the difference between fµ T and fµ L is zero, we get the interesting equation ν T ν µ = µ L. (64) For the situation where µ Ṽ ν statement so we get which equals We will prove that = 0 we already proven the ν T ν µ = d dt G µ, (65) d dt G µ = µ L, (66) d dt G µ = L. (67) R µ G µ = Ṽ ν G ν = L, (68) see Appendix A for the proof and its limitations. Combined with the forgoing equation, this leads us to d dt ( L ) = L R µ (69)

18 as equivalent to Juli, 8, 2008 To be submitted ν T ν µ = µ L. (70) A canonical Lagrangian density If we choose a canonical Lagrangian density as L = Ṽ ν G ν + J ν A ν = v g J A u i + ρ e φ, (71) and an accompanying stress energy density tensor T ν µ = Ṽ ν G µ J ν A µ, (72) then our force equation f T µ = f L µ can be split in an inertial LHS and an EM RHS ( f T µ + f L µ ) inertial = ( f T µ + f L µ ) EM. (73) For situations were (f L µ ) inertial = µ u 0 = 0 this results in f inertial µ = f Lorentz µ. (74) as d dt G µ = J ν ( ν A µ ) ( µ Ã ν )J ν. (75) Maxwell s inhomogeneous equations We end with the formulation of the two inhomogeneous equations of the set of four Maxwell Equations, as they can be expressed in our terminology. They read ν ν A µ µ ν A ν = µ 0 J µ. (76)

19 Juli, 8, 2008 To be submitted It can be written as ( c 2 d 2 dt 2 )A µ µ ( t φ A) = µ 0 J µ, (77) so as the difference between a wave like part and the divergence of the Lorenz gauge part. Conclusions We have presented a specific kind of biquaternion relativistic tensor dynamics. We formulated the general equation ν T ν µ + µ L = 0. (78) The stress energy density tensor of a massive moving charged particle in a potential field was formulated as T ν µ = Ṽ ν G ν + J ν A ν with an accompanying Lagrangian density L as its trace L = T νν. Under curtain continuity conditions for the four current and the four velocity, this leads to the Lorentz Force Law and to the usual equations of relativistic tensor dynamics. One advantage of our specific kind of biquaternion formalism is that it is very akin to the usual relativistic space-time language. Acknowledgments I would like to thank xxx for valuable comments. Appendix A Proof We want to proof that, under curtain conditions, we have L = Ṽ ν G ν = G ν. (79)

20 Juli, 8, 2008 To be submitted The chain rule as we have used and shown before gives a first hunch. The chain rule leads us to Ṽ ν G ν = ( Ṽ ν )G ν + ( V Gν )V ν. (80) µ As before, we cannot assume this, because it uses commutativity, so we have to prove it. We start the proof with Ṽ ν : Ṽ ν = v 0 v 0 1 i v 1 v 0 I i v 2 v 0 J i v 3 v 0 K i v 0 1 v 1 I v 2 J v 3 K i v 0 v 1 I v 1 v 1 1 v 2 v 1 K v 3 v 1 J [ iv 01, v 1 I, v 2 J, v 3 K] = i v 0 v 2 J v 1 v 2 K v 2 v 2 1 v 3 v 2 I i v 0 v 3 K v 1 v 3 J v 2 v 3 I (81) v 3 v 3 1 Now we use the fact that we have an orthogonal basis, so v µ v ν = δ µν : Ṽ ν = 11 0I 0J 0K 0I 11 0K 0J 0J 0K 11 0I 0K 0J 0I 11 (82)

21 Juli, 8, 2008 To be submitted Then we multiply G ν with the result, giving 11 0I 0J 0K 0I 11 0K 0J 0J 0K 11 0I 0K 0J 0I 11 The result of this part is For the second part, ( Ṽ ν )G ν = [ig 01, g 1 I, g 2 J, g 3 K] = ig 0 1 g 1 I g 2 J g 3 K = G µ. (83) ( Ṽ ν )G ν = G µ. (84) ( Gν )V ν, (85) we have two options. The first is the easiest, assuming particle velocity and particle momentum to be independent properties, which makes this part zero and gives us the end result L = Ṽ ν G ν = ( Ṽ ν )G ν = G µ. (86) In the case that L = J ν A ν this assumption is allowed.

22 Juli, 8, 2008 To be submitted The second option is that particle velocity and particle momentum are mutually dependent through the relation G ν = ρ i V ν, with ρ i as the inertial mass density. In that case we have to go back to the original equation. If we assume a velocity independent mass density this gives L = Ṽ ν G ν = ρ (Ṽ ν V ν ) = ρ (v 2 0 v 2 1 v 2 2 v 2 3) = 2G µ. (87) The last situation is assumed in relativistic gravity, where the stress energy tensor is given by ρ i Ũ ν U ν. In that situation could be tempted to choose the Lagrangian as L = 1 2 ρ iũ ν U ν in order to preserve the outcome L = G ν. (88) This is done for example by Synge in his book on relativity (Synge, 1965, page 394). But that is outside our scope. So we have to restrict the use of L = Ṽ ν G ν = G ν (89) to the situations in which V µ and G µ are independent of each other. References [1] A.P. Yefremov, Quaternions and biquaternions: algebra, geometry, and physical theories. arxiv: mathph/ , 2005.

23 Juli, 8, 2008 To be submitted [2] J.L. Synge, Relativity: The Special Theory. North-Holland Pub. Co, Amsterdam, 2nd ed, 1965.