ЯДЕРНАЯ ФИЗИКА, 2011, том 74, № 12, с. 1813-1820
ЭЛЕМЕНТАРНЫЕ ЧАСТИЦЫ И ПОЛЯ
NEW REPRESENTATION FOR ENERGY-MOMENTUM AND ITS APPLICATIONS TO RELATIVISTIC DYNAMICS
©2011 R. M. Yamaleev*
Universidad Nacional de Mexico, Facultad de Estudios Superiores, Mexico;
Joint Institute for Nuclear Research, Dubna, Russia Received July 19,2010
In this paper we introduce the concept counterpart of rapidity and define energy and momentum of the relativistic particle as functions of the counterpart of rapidity. Formulae of the relativistic mechanics defined in such a way are regular near the zero-mass and speed of light state. This representation admits to attain a correct limit of the formulae of the relativistic mechanics, including the Dirac equation, at zero-mass point and explains violation of the parity at this state. On the other hand, the representation for energy—momentum can be realized as a mapping from the massless state onto the massive one which looks like a "q deformation". Hypothesis on quantization of the energy—momentum and the velocity near the light speed is suggested. The group of transformations using the counterpart of rapidity as a parameter of transformation is constructed.
1. INTRODUCTION
In the present paper we elaborate new expressions for the energy, momentum, and velocity of a relativistic particle regular at the point m = 0, v = = c. The formulae for the energy—momentum are presented as functions of some hyperbolic angle % dual to the rapidity which forms a counterpart of the rapidity.
In one-dimensional case the hyperbolic angles % and ^ are reciprocal quantities. However, in general, the rapidity and its counterpart physically and geometrically are quite different each from an other. The rapidity, is equal to zero at the rest state, v = 0, and goes to infinity when the velocity tends to v = c, whereas its counterpart, the hyperbolic angle %, is equal to zero at the point v = c and becomes infinity at the rest state, v = 0. Another important property of the counterpart of rapidity is its dependence of the proper mass: % = mc/n0. The quantity cn0 is interpreted as an energy of the massless state. Thus, the formulae for energy—momentum can be realized as a mapping from an energy of the massless state onto the energy of a particle with mass. These formulae look like as formulae of "q deformation". An analysis of this observation prompts to introduce a hypothesis on quantization of the velocity near the speed of light.
In a similar manner as translations of the rapidity form a part of the Lorentz group of transformations, translations of the counter-rapidity form some group of transformations.
Usefulness of the present theory we demonstrate exploring the Dirac equation at the limit m = 0. The representation for the energy—momentum via counter-rapidity allows to reach a correct limit at m = = 0 explicitly displaying violation of the parity in the Dirac equation at this limit.
2. REPRESENTATIONS OF ENERGY-MOMENTUM AS FUNCTIONS
OF RAPIDITY AND ITS COUNTERPART 2.1. Elements of Relativistic Dynamics of Charged Particle
Consider a motion of the relativistic particle with charge e in the external electromagnetic fields E and B. The relativistic equations of motion with respect to the proper time t are given by the Lorentz-force equations:
dP e ttv. , e r__.. T.1
dr
e ^ e r —Ep0 + —pxB,
mc m
dpo dr
e
mc
(E • p),
(2)
E-mail: iamaleev@servidor.unam.mx
dr p dt p0 dT m' dT mc' These equations imply the first integral of motion, the "mass-shell" equation:
p0 — (p • p)2 = M2c2. (3)
In the case of stationary potential field, i.e. when eE = = —W(r), the equations imply the other constant of motion, the energy of the relativistic particle
E = cpo + V (r). (4)
1813
1814
YAMALEEV
If the external electromagnetic field strengths are given in the covariant form, FßV, ¡i,v = 0,1,2,3, then Eqs. (1), (2) are written in the form of Minkowski force equations [1]:
d
dxß
= —Fßvvv, ui1 = 3-. (5)
dr
mc
dr
2 I 2
qi := cp0 — mc , q2 := cpo + mc
(6)
In the nonrelativistic limit the former is transformed into kinetic energy of the Newtonian particle
2 V2 cpo — mc —.
2m
(7)
Next, we shall restrict ourselves by considering only length of the momentum p = |p|. For that purpose let us use the project of the Lorentz-force equation (1) on the direction of motion, that is
dp dpo
#=P0' (8)
P p'
ÉÈ. = —E
dr mc '
E = ( n-E), n = -.
From the first two equations of (8) by taking into account (3), we find
p0 = mc cosh(^), p = mc sinh(^), (9)
where ^ = 0 corresponds to the rest state with p = 0, po = mc. Velocity with respect to coordinate time is defined by
v p
— = — = tanh(/0). c Po
(10)
The same expression is used for the mapping between rapidity ^ and velocity v in the Lorentz kinematics [4].
2.2. Hyperbolic Angle Dual to Rapidity
Conventionally formulae (9) are considered as formulae of parametrization of the mass-shell equation (3). In this context notice, however, that this is not unique form of parametrization. In fact, we can satisfy (3) by taking
p0 = mc coth (x), p =
mc
sinh (x) '
The parametrization as an objective does not give any interpretation of the parameter, however. Now,
let us consider the procedure of parametrization from another point of view.
Consider the quantities q2, qi as solutions of a quadratic equation
(11)
X2 — 2po X + p2 = 0,
Correspondence with nonrelativistic equations gives an interpretation of the constant of motion M2 as a squared mass of the particle, so that M2 = m2. For the massless particle M2 = 0 [2]. It is important to emphasize that the relativistic dynamics of charged particle is formed by the pair of energies [3]:
where
2po = qi + q2, p2 = qiq2, 2mc = q2 - qi. (12)
Notice, this quadratic equation is another form of the mass-shell equation (3). In fact, translation of X by X = mc + p0 leads to (3). According to Hamilton—Cayley theorem general solution of Eq. (11) can be represented by the following matrix
E :=
0 — p 2 1 2po
(13)
eigenvalues of which are roots of polynomial equation (11). This matrix generates some evolution along parameter 0 [3, 5, 6]. Write the Euler formula for exponential function of the matrix E0
exp(E0) = go(0; po,p2) + Egi(0; po,p2). (14)
In terms of the eigenvalues this equation is decoupled into a pair of equations
exp(qi0) = go(0; po,p2) + qigi(0; po,p2), (15) exp(q20) = go(0; po,p2)+ q2gi(0; po,p2).
Theorem [7].
The following equation holds true
— = exp(2mc0o), qi
where 2mc = q2 — qi.
Proof:
exp(2mc0) = exp(q2^) giq2 + go
(16)
(17)
q2 — U
exp(qi0) giqi + go qi — U'
where
U = ——. gi
Notice, simultaneous translation of the eigenvalues by qi = qi + U, i = 1,2 remains unchanged the difference between them 2mc = q2 — qi, or, in other words, it remains the mass unchanged. Hence, this translation induces a corresponding translation of the hyperbolic argument:
exp(2 mc(0 + 5{U))) = — = exp(2mc0o). (18)
qi
Inversely, translation of 0o by 0 = 0o + 5 has to remain 2m unchanged because m does not depend
on 0, whereas qi,q2 will undergo some translation simultaneously by q2 = q2 + A, q1 = q1 + A.
End of proof.
Thus, we got some interrelation between the fraction and the hyperbolic exponential function, or between simultaneous translation of two quantities of the fraction and the hyperbolic rotation. Write (18) as follows
Po + mc p0 — mc
= exp(2mc^),
(19)
where the mass m is a constant of the evolution with respect to parameter 0, whereas p0,p depend of 0 and this dependence is given by hyperbolic trigonometry
Po = mccothimcd)), p = —r^-—-r. (20)
sinh(mc0)
It is interesting to apply the present theorem to the following fraction
Po + P = c + v Po — p c — V '
where p and p0 are variables of the evolution. Let p > 0, then (21) we have to re-write as follows
i 20 I i
PO+P _ p + 1
P0 — P
Po _ I '
On making use of the theorem we come to the following equality
£0+1 P
PO _ I p
= exp(2^).
(22)
From (22) by taking into account (3) we get
p0 = mc cosh(^), (23)
v
p = mc sinh(ip), - = tanhf^),
c
which is nothing else but Eqs. (9), (10).
We possess now with two different representations for the energy and momentum via hyperbolic trigonometry:
(I) p0 = mc cosh(^), p = mc sinh(0);
(II) po = mccoth(mcé), p = —-—-r.
smh(mc^)
An essential feature of the latter is its regularity at the point m = 0, v = c. At this point the hyperbolic argument is equal to the energy—momentum of a massless particle:
p(m = 0)=po(m = 0) = ^=7To. (24)
Thus, the value cn0 is the energy of the relativistic system at the point m = 0, v = c. We should emphasize some difference between two hyperbolic
angles. At the rest, ф = 0, but ф = то, and vice versa, when v = c, ф = 0 but ф = то. The particle with m > > 0 is not able to attain the state of speed of light, conversely, the particle possessing n0 > 0 cannot fall to the rest state. In the rest state the energy is equal to the proper inertial mass (in energy units) and, in the same manner, in the state of the light speed the energy is equal to n0. Thus, the kinetic energy of the relativistic particle is governed, besides the inertial mass m, with some internal energy we denoted by n0. The massive particle moving with velocity less than light velocity possesses both parameters, m and n0. The parameter n0 is, in some sense, counterpart of the inertial mass which determines the value of the kinetic energy of the motion. This quantity corresponds to the energy of the particle in its massless state [8].
Let v be the velocity of a particle with respect to coordinate time. This velocity is essentially less than the light velocity, v < c. Besides v let us introduce some complementary velocity V obeying the equation
v2 + V2 = c2. (25)
Now let us express v via the parameter (x = тсф). We get
p
v' = c'-v' = c'[l-1-)=c' tanh (%). (26) V po/
Substitute (23) and (26) into (25), this gives
c2 tanh2(^) + c2 tanh2({) = c2. (27)
Notice that v is expressed via hyperb
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