In massive free particles the square of Relativistic energy, E’2 = p2c2 + mo2c4/h2 divides into two: relativistic momentum p=m’vg and rest mass mo (where c is the speed of light and h is Planck’s constant}. The rest mass is a function of phase velocity vp; the momentum a function of group velocity vg. In each mass component, mo and (m’-mo) the velocity dependence is exclusive. This consequence of their formulations defines their functions in mass, in internal motion, in intrinsic spin, in superconductivity, in wavefunction collapse, in conservation laws such as momentum, spin, etc.
In massive free particles the square of Relativistic energy, E’2 = p2c2 + mo2c4/h2 divides into two: relativistic momentum p=m’vg and rest mass mo (where c is the speed of light and h is Planck’s constant}. The rest mass is a function of phase velocity vp; the momentum a function of group velocity vg. In each mass component, mo and (m’-mo) the velocity dependence is exclusive. This consequence of their formulations defines their functions in mass, in internal motion, in intrinsic spin, in superconductivity, in wavefunction collapse, in conservation laws such as momentum, spin, etc.
Dispersion dynamics applies wave-particle duality, together with Maxwell’s electromagnetism, and with quantization E = hν = ħω (symbol definitions in footnote) and p = h/λ = ħk, to special relativity E2 = p2c2 + m2c4. Calculations on a wave-packet, that is symmetric about the normal distribution, are partly conservative and partly responsive. The complex electron wave function is chiefly modelled on the real wave function of an electromagnetic photon; while the former concept of a “point particle” is downgraded to mathematical abstraction. The computations yield conclusions for phase and group velocities, vp⋅vg = c2 with vp ≥ c because vg ≤ c, as in relativity. The condition on the phase velocity is most noticeable when p≪mc. Further consequences in dispersion dynamics are: derivations for ν and λ that are consistently established by one hundred years of experience in electron microscopy and particle accelerators. Values for vp = νλ = ω/k are therefore systematically verified by the products of known multiplicands or divisions by known divisors, even if vp is not independently measured. These consequences are significant in reduction of the wave-packet by resonant response during interactions between photons and electrons, for example, or between particles and particles. Thus the logic of mathematical quantum mechanics is distinguished from experiential physics that is continuous in time, and consistent with uncertainty principles. [Footnote: symbol E = energy; h = Planck’s constant; ν = frequency; ω = angular momentum; p = momentum; λ = wavelength; k = wave vector; c = speed of light; m = particle rest mass; vp = phase velocity; vg = group velocity].
In Part I we argued that two flow equations for a free particle, namely: d/dx (partial) A(x,t) = p ((1a)) and d/dt (partial) A(x,t) = -E ((1a)) may be applied to both particles with rest mass and those without. In fact we argue that there are at least three scenarios associated with ((1a)) and ((1b)). The first is classical mechanics for a particle with rest mass mo and constant velocity. Using ((1a)) and ((1b)) with p=E(v)v this scenario yields special relativity. A second solution is to write ((1a)) and ((1b)) as eigenvalue equations: -id/dx (partial) exp(iA) = p exp(iA) and id/dt (partial) exp(iA)= E exp(iA) where A=-Et+px. This yields a quantum mechanical solution for a particle with rest mo if one assumes p=Ev. In such a case one may use the special relativistic result EE = pp + momo and establish the Klein-Gordon equation. A third solution pertains to objects with zero rest mass. As argued in Part I, these do not follow p=E(v)v, but rather the classical wave result: v=E/p. In this note we examine massless particles, namely waves on a string, photons and sound, in more detail. For the zero mass case, dA=0 so there is motion in such a way as to keep A constant. If one considers periodic motion and uses ((1a)) and ((1b)), a solution which is real and periodic is: d/dt d/dt cos(A) = vv d/dx d/dx cos(A) ((2)) A second order partial differential equation is needed as one insists the solution be real. This differs from the quantum mechanical solution of exp(-iEt+ipx) which is complex for a reason. The modulus must be linked to a probability of 1 for a free particle. The zero mass solution is one of a smeared-out physical wave. We note that ((1a)) and ((1b)) deal strictly with E energy and p momentum, but cos(A)=cos(-Et+px) is associated with a period proportional to 1/E and a wavelength proportional to 1/p. This raises two questions. First, what physical entity is periodic? Secondly, how does the energy appear in the problem as energy and not just 2*3.14 frequency of the physical entity? We try to consider this issue for a wave on a string, photon and sound wave.
In Part I we suggested that free particle quantum mechanics seems to follow from special relativity. In this note, we focus on the idea of a rest frame defining a rest mass. In such a frame momentum is zero i.e. one has the 4-vector (0,mo)=(p,E). (x,t), however, is also a 4-vector which is usually assigned to a point in space time. In Part I, however, we considered (x,t)=(0,to) associating p=0 and x=0 and alternatively mo and to. Velocity is usually defined as dx/dt =v, but using the Lorentz transformation (1) (with v set to -v), (x,t)=(0,to) transforms to x’/t’ = v. The relationship between x,p and t,E is further manifested through the Lorentz invariant -Et+px in which E scales t and p, x. This we suggest breaks space into pieces proportional to 1/p and time (independently) into pieces proportional to 1/E We next consider particles “created” at different x points. In other words (0,mo) applies, but the particles are at different x positions. In order to keep the relationship p=0,x=0, one must have an x origin at all particle positions. Then spatial scaling of 1/p does not line up unless the particles are n/p (n=integer) apart. Otherwise the scaling in space is unaligned. For different p, scaling in space is unaligned already with (0,mo) and (0, to). Thus for two particles with the same p, but created at different x values at different times and arriving at x1, t1, if there is no alignment, special relativity suggests these two are distinguishable, while Newtonian mechanics does not distinguish. This distinguishability, however, depends on a physical mechanism i.e. a wavelength. Thus, even without considering different p (momentum) values, two identical momenta should be distinguishable if they are not aligned in terms of p scaling even for the same rest mass mo. This we argue is strictly a result of special relativity and is support for the exp(ipx) form of quantum free particle mechanics.
Momentum and velocity are two well-known physical observables in classical physics (and quantum physics as well because quantum particles have impulse and move through space in a certain amount of time based on an external clock and ruler). They are, however, measured in different ways. Velocity describes motion through x in some time t, while p is measured through impulses. Thus one does not need a clock to measure momentum. It may be measured in terms of the damage done to a foil, compression of a spring etc. Thus p=m1v1=m2v2 etc., i.e. a numerical number associated with momentum does not automatically yield the velocity unless one knows rest mass. In many situations rest mass is known (i.e. one deals with an electron etc) and so there seems to be a tendency to consider momentum and velocity to be almost the same thing when they are distinctly different. In fact, we argue that the notion of wavefunction W(x) and density P(x)=W*(x)W(x) are associated with momentum and velocity respectively. In fact, for a free particle these two measurables are linked with completely different spatial probabilities. Constant velocity impies P(x)=constant while exp(ipx) describes the two-dimensional x-probability associated with p which is based on a constant mo and v. In special relativity for a free particle (i.e. p constant, v constant), there are two well known Lorentz invariants: EE = pp + momo (c=1) and A = -Et + px. One may note that v does not appear explicitly even though one may write E and p in terms of v i.e. E = mo/sqrt(1-vv) and p=Ev. The first invariant describes E in terms of rest mass and p (measured through impulse). The second involves p,E and x,t with x and t appearing as independent even though ultimately x(t) and v=dx/dt=constant = x/t if x and t both start at 0. This independence is crucial because it means one may consider fluctuations which keep A constant separately. For example, if t→ t+ hbar/E and x→ x+hbar/p, then A remains constant. This leads to the idea of the quantum wavelength and a spatial probability independent of t, namely exp(ipx). One average, however, x=vt, but how does this appear in Lorentz invariants? There is no explicit v. We argue that the Lorentz invariant A = -Et + px must convert E from a function of p to one in v and that this occurs through a Legendre transformation. Thus the form of the invariant is itself a Legendre transformation with E being transformed into L, the Lagrangian. There is, however, a second important feature because p=dL/dv (partial) and given that p is constant, dp/dt=0 so d/dt dL/dv partial =0 which is in the form of an extremization. We argue that this is not an accident. dL/dv=p follows from E being constant, so, in other words, it follows from the px term. This same term leads to exp(ipx) which means that one does not have smooth x=vt motion within a wavelength. Thus, we argue that when one transforms to the second measurable v, the notion of this fluctuation behaviour must also be manifested. In other words, v=constant or dv/dt is not enough. The extremization behaviour of the Lagrangian, we argue, is thus not a mathematical convenience, but gives an indication that even though the extreme path x=vt is followed on average, other paths are sampled. We argue that this idea is incorporated into special relativity, in fact within a single Lorentz invariant A = -Et + px.
Special relativity seems to be associated with linking physical relations as seen in one frame with those as seen in a frame moving at a speed relative to such a frame. A constant speed is often used for the moving frame. This suggests that classically there is a special frame for a particle with rest mass called a rest frame. In such a case, one simply introduces a parameter for rest mass and does not worry about the complicated motion and interactions which actually create the rest mass. It seems that even quantum mechanics, which considers a particle to be moving as one does not know its location, one uses the idea of a potential V(x) which in principle is associated with a very large rest mass, presumably at rest as no variables of such a rest mass appear in the Klein-Gordon or Dirac equations. It seems that one uses a preferred frame when solving relativistic quantum mechanical equations. Furthermore, given V(x) and a bound state with exp(-i E bound t), E applies to the overall bound system, not to a possible exp(ipx) constituent. The special relativistic invariant equation seems to exist at the system level, like with the rest mass and one may apparently calculate E in the special rest frame of the large mass and hence find a rest mass of the bound system, i..e large rest mass*cc + mo cc + E, mo is the rest mass of the single bound particle. As a result of the use of the special rest mass frame for the V(x) producing large rest mass, it seems that one does not have to worry about special relativistic constraints on the interaction with V(x), as only one frame is involved. In other words, a bound calculation is about finding a new rest mass for an overall new system. Then this system may be used in Lorentz boosts. One simplifies by working in the rest frame of the mass creating V(x). Perhaps surprisingly then, we note that exp(-iEt+ipx) for a free particle is Lorentz invariant and in fact must be Lorentz invariant for the following reason. Imagine that one wishes to create a probability for free particles which is linked with conservation of momentum. Any p must have the same real value weight because for free particles there is no bias for one p or the other. There is, however, a bias for a p1 and p2. It should have the same probability as a p3 and p4, if p1+p2 = p3+p4. (We use one dimension for simplicity.) As a result, exp(ip) seems to be a probability which respects these conditions. The problem is that one may have p=m1v1=m2v2 hence (E1,p) and (E2,p). With respect to momentum, it is irrelevant if one particle has E1 and the other E2, but not if one does a Lorentz boost. Then, in one case: p→ g(v) p + g(v) v E1 and in the second, p→ g(v) p + g(v)vE2. Here g(v) =1/sqrt(1-vv/cc). In other words, p is the same for each particle in one frame and not in the other which violates conservation of momentum because in frame 1 both particles can complete the same conservation of momentum equation, but in the second frame this is not the case, because each has a different momentum. Thus, exp(ip) cannot hold. One requires a linear form in p which is Lorentz invariant, i.e. exp(-iEt+ipx), we argue (and have done so in previous notes). We see now that momentum probability is linked with space in exp(ipx). The idea seems to be that for free colliding particles there is no preferred frame and the probabilities used must reflect this Lorentz invariance. As a result, we argue that a free particle probability in E,p arises because of Lorentz invariance, but then apply it to a V(x) problem and no longer worry about Lorentz invariance at the interaction level. V(x) with no associated rest mass variables assumes a rest frame for the V(x) producing mass. A Lorentz boost which then invalidates exp(ip), would only apply to the total rest mass of the bound particle system which has no overall momentum. It seems that one does apply the Lorentz transformation to the interaction of exp(ipx)’s with V(x). One is only interested in the overall bound system as a single entity for special relativity. As a result, one may only impose momentum considerations as V(x) is a spatial function and one must link it with exp(ipx). In fact, one may write V(x) = Sum over k Vk exp(ikx) to associate it with momentum hits linked with Vk energies, which already moves away from the idea of special relativity as Vk and k are not linked as a 4-vector. The idea of exp(ip1x)exp(i k1x) = exp(ip2x)exp(ik2x), however, seems to still hold (as it does for free particles), linking these two values (i.e. Vk1 and Vk2) with (p1+k)(p1+k) exp(i (p1+k)x). Lorentz invariance for a particle with rest mass seems to be linked with the idea that one ultimately has a rest frame for the particle. One then establishes a relation between the rest mass in one frame and its new E and p as seen by another frame. One does not, however, need to impose Lorentz invariance conditions on the interaction which creates the rest mass it seems. For example, exp(-iEt+ipx) is the Lorentz form used to describe a free particle, but exp(ipx) is the only portion used to describe its interaction with V(x) in the rest frame of the large mass creating V(x). Furthermore, exp(-iEt+ipx) ensures conservation of energy and p for a free particle, but in an interaction case, that does not happen as the potential may absorb energy.