In previous notes, we presented a simple argument for obtaining the quantum free particle wavefunction exp(-iEt+ipx). We suggested that there are situations for which one cannot predict the outcome of an elastic collision and that given initial e1,e2 (energies) and p1,p2 (momentum vectors in 1-direction for simplicity), any ei,ej and pi,pj which conserve energy and momentum have the same product probability. Even in Newtonian elastic collisions of free particles, one sees calculations which allow any outcome which conserves momentum and energy. If the probability has the same real value for any free particle as no variable is being distributed and one wishes to impose Lorentz invariance, then exp(-iEt) and exp(ipx) represent the probabilities. This automatically gives rise to units in time and space, hbar/E and hbar/p (the hbar comes from considerations linked to a photon). One may ask, however: What is the real physics behind dx and dt? We argued in (1) that this physics is associated with force, in particular an intrinsic force. We considered a particle with rest mass mo and argued that this rest mass represents intrinsic force as it is proportional to Newton’s inertia and the force needed to set it in motion. If viewed from a moving frame, (0, mocc) maps to (E,p) with E containing the intrinsic inertial force as well as enhancements due to motion and p containing the intrinsic mo as well as the velocity vector. For a particle with rest mass, we suggest that a key idea is this notion of intrinsic force. We argued that one may consider the Lorentz invariant -Et+px = -mocc t1 and introduce dx=hbar/E, dt=hbar/p, dt1=hbar/(mocc) and dx1-> infinite as p->0. This leads to -E dt + p dx =0 or -E/dx + p /dt = 0. We then argued in (1) that p/dt is like an average force and from this we introduced exp(-iEt+ipx). Thus, dt and dx are linked with the notion of an average force. We suggest that this notion of intrinsic force is the reason behind linearizing -EE + p dot p = -momo cccc ((1)). -Edt+pdx=0 is too simple to show the geometry linked with rotational invariance etc in ((1)), yet at the same time ((1)) is not linear in the intrinsic force mo. This is the physical reason why ((1)) should be linearized via Dirac’s process, giving rise to spin ½. As a result, we postulate that a particle has an intrinsic force, in this case rest mass, which is directly linked with the probability exp(-iEt+ipx) and that one should be able to write an equation containing operators and the intrinsic force (in linear form) which also shows geometry. It is this equation which gives rise to spin. At this point, this is a postulate, i.e. speculation. It would be good if one could find an example in which these ideas arise naturally, i.e. a case in which one did not have to linearize an equation like ((1)) or make special arguments in order to obtain exp(-iEt+ipx). In that case, an equation linear in the intrinsic force would exist a priori and a direct link to exp(-iEt+ipx) would also naturally exist. We suggest that such an example does exist, namely for a photon. One may use the special relativistic equations developed for a particle with rest mass and set mo=0 to obtain E=[p] c for a photon. This is a fine equation, but unlike the rest mass case mo (proportional to the intrinsic force), there is no rest mass and no intrinsic force present for the photon. Certainly p (momentum) is linked with a force as both a particle with rest mass and a photon impart an impulse on a target, but momentum is not an intrinsic force, it is a consequential one. In the case of a rest mass, it arises because the rest mass (the intrinsic force) is seen to be moving. For a photon, one would then expect the existence of an intrinsic force which gives rise to momentum. We suggest that the electric field (together with the magnetic) represents this intrinsic force and that Maxwell’s equations allow one to directly write a linear equation in El (electric field) (no linearization required) which shows the geometry in the system. One may also explicitly show that El behaves as exp(-iEt+ipx) if one extends cos(-Et+ipx) to a more general complex number. In other words, the postulates made for the particle with rest mass mo seem to naturally arise in the case of the photon. This, we argue, is why the procedures for finding spin for a spin ½ particle with rest mass mo and a photon are so different even though both a photon and rest mass particle satisfy special relativity and are associated with exp(-iEt+ipx). It is the difference in the intrinsic force which is the key to finding spin, we speculate.
In previous notes, we presented a simple argument for obtaining the quantum free particle wavefunction exp(-iEt+ipx). We suggested that there are situations for which one cannot predict the outcome of an elastic collision and that given initial e1,e2 (energies) and p1,p2 (momentum vectors in 1-direction for simplicity), any ei,ej and pi,pj which conserve energy and momentum have the same product probability. Even in Newtonian elastic collisions of free particles, one sees calculations which allow any outcome which conserves momentum and energy. If the probability has the same real value for any free particle as no variable is being distributed and one wishes to impose Lorentz invariance, then exp(-iEt) and exp(ipx) represent the probabilities. This automatically gives rise to units in time and space, hbar/E and hbar/p (the hbar comes from considerations linked to a photon). One may ask, however: What is the real physics behind dx and dt? We argued in (1) that this physics is associated with force, in particular an intrinsic force. We considered a particle with rest mass mo and argued that this rest mass represents intrinsic force as it is proportional to Newton’s inertia and the force needed to set it in motion. If viewed from a moving frame, (0, mocc) maps to (E,p) with E containing the intrinsic inertial force as well as enhancements due to motion and p containing the intrinsic mo as well as the velocity vector. For a particle with rest mass, we suggest that a key idea is this notion of intrinsic force. We argued that one may consider the Lorentz invariant -Et+px = -mocc t1 and introduce dx=hbar/E, dt=hbar/p, dt1=hbar/(mocc) and dx1-> infinite as p->0. This leads to -E dt + p dx =0 or -E/dx + p /dt = 0. We then argued in (1) that p/dt is like an average force and from this we introduced exp(-iEt+ipx). Thus, dt and dx are linked with the notion of an average force. We suggest that this notion of intrinsic force is the reason behind linearizing -EE + p dot p = -momo cccc ((1)). -Edt+pdx=0 is too simple to show the geometry linked with rotational invariance etc in ((1)), yet at the same time ((1)) is not linear in the intrinsic force mo. This is the physical reason why ((1)) should be linearized via Dirac’s process, giving rise to spin ½. As a result, we postulate that a particle has an intrinsic force, in this case rest mass, which is directly linked with the probability exp(-iEt+ipx) and that one should be able to write an equation containing operators and the intrinsic force (in linear form) which also shows geometry. It is this equation which gives rise to spin. At this point, this is a postulate, i.e. speculation. It would be good if one could find an example in which these ideas arise naturally, i.e. a case in which one did not have to linearize an equation like ((1)) or make special arguments in order to obtain exp(-iEt+ipx). In that case, an equation linear in the intrinsic force would exist a priori and a direct link to exp(-iEt+ipx) would also naturally exist. We suggest that such an example does exist, namely for a photon. One may use the special relativistic equations developed for a particle with rest mass and set mo=0 to obtain E=[p] c for a photon. This is a fine equation, but unlike the rest mass case mo (proportional to the intrinsic force), there is no rest mass and no intrinsic force present for the photon. Certainly p (momentum) is linked with a force as both a particle with rest mass and a photon impart an impulse on a target, but momentum is not an intrinsic force, it is a consequential one. In the case of a rest mass, it arises because the rest mass (the intrinsic force) is seen to be moving. For a photon, one would then expect the existence of an intrinsic force which gives rise to momentum. We suggest that the electric field (together with the magnetic) represents this intrinsic force and that Maxwell’s equations allow one to directly write a linear equation in El (electric field) (no linearization required) which shows the geometry in the system. One may also explicitly show that El behaves as exp(-iEt+ipx) if one extends cos(-Et+ipx) to a more general complex number. In other words, the postulates made for the particle with rest mass mo seem to naturally arise in the case of the photon. This, we argue, is why the procedures for finding spin for a spin ½ particle with rest mass mo and a photon are so different even though both a photon and rest mass particle satisfy special relativity and are associated with exp(-iEt+ipx). It is the difference in the intrinsic force which is the key to finding spin, we speculate.
In Part 1, we tried to understand why spin arises. The basic conclusion was that it is linked with an intrinsic force which in turn is associated with E,p (energy momentum) and geometry (rotational symmetry and other symmetries). In practical terms, this means finding an equation linear in intrinsic force (as force appears linearly in nature), but still tied to E,p and geometry. Interestingly, this leads to two very different results for particles with rest mass (spin ½) and photons, although both satisfy the same special relativistic equations (albeit the photon with rest mass mo=0) and both have exp(-iEt+ipx) free wavefunctions. In the photon case, the Electric field El is the intrinsic force (together with the magnetic B) which gives rise to both energy and momentum density. Specifically, energy density= .5eo El dot El + .5/uo B dot B and momentum density = 1/uo El x B. In the photon case, however, there is no rest frame and so the classical vector El is directly associated with E (energy) and p (momentum) without one needing to consider special relativity. Instead one directly uses Maxwell’s electromagnetic equations in order to find a linear equation describing the classical vector El. This equation involves d/dt partial and grad partial which are linked to the quantum operators for E and p (id/dt, -igrad) and El itself is proportional on exp(-iEt+ipx) (actually cos(-wt+kx)), also from Maxwell’s equations. Given this scenario, one has a classical vector El which is perpendicular to B and both lie in a plane perpendicular to motion. For circular polarization, El moves in a circle and given that it gives rise to energy, this energy circulates like classical angular momentum. For the 3-vector for El, one says that one has spin 1, but for circular polarized light, this is classical angular momentum. As a result, spin in the photon case follows from the behaviour of the intrinsic force (a linear equation in this force as shown in Part 1), but one must also have a clear link between the intrinsic force and the consequential energy and force, because it is the energy which actually has to rotate. It is not enough to have El rotate. In the case of a particle with rest mass, we argued in Part 1 that mo (Newton’s inertia) is the intrinsic force which together with velocity gives rise to the consequential energy E and momentum p. The situation is that one actually has two physical frames, the rest frame with the intrinsic force mo and the moving frame in which one has E,p. In order to link the intrinsic force mo to E, p, one must use a Lorentz invariant. We suggested in Part 1 that -Et+px = -mocc t1 is associated with -E dt + p dx = -mocc dt1 + (p=0) * (dx1->inifite) =0. Then dx=hbar/p, dt=hbar/E and E/dx and p/dt are average forces as is mocc/dt1. Tthe idea of force is very much the cornerstone of exp(-iEt+ipx) we argue. In order to see spin arise, one needs an equation linear in mo (the intrinsic force), but linked to E,p and showing geometrical invariances, according to the recipe of Part 1. -EE + cc p dot p = -momocc ((1)) links E,p, mo and shows invariances, but is not linear in mo. One must then linearize it. This means that one compares to Lorentz frames, the rest frame containing mo and the moving containing E,p. Both mo and p are linked to forces, but mo is the intrinsic force and p, the consequential. As a result, they must somehow be separated as they are in ((1)), in particular they are perpendicular if one considers: EE = pz pz + momo (c=1, p->px). In order to linearize without using unit vectors, one cannot use numbers, but must use matrices (as Dirac showed). The operator linked with pz must anticommute with that linked with mo. The operators associated with the consequential and intrinsic forces differ and in fact anticommute in the rest mass case. One may show that one requires at least a 2x2 matrix (the full Dirac treatment uses a 4-vector made of 2 spinors), which means that there is a two vector state. This is presumably associated with motion, if exp(ipz z) is linked to motion as well. In order to analyze this 2 vector state (spinor), we first compare the situation with the photon case. There, one had a 3-vector El which means that one knows an exact position in the plane. The second piece of information must be the direction in which this vector is moving in a circle (for circularly polarized light). There are two a priori given pieces of information which together with the link between El and energy density yields classical angular motion. In the rest mass case, one again expects rotational motion in a plane perpendicular to the z axis (because pz describes all motion along z). (1,0) and (0,1), however, describe classical rotational direction, clockwise and counterclockwise, but this spin ½ case cannot be classical rotation because the physical vector in the plane is not given (as it is by El the intrinsic force). Thus, one has only the direction of rotation. We argue that if exp(ipx) means that one has p hits at different x within hbar/p with probability exp(ipx), one may imagine different x,y positions of energy in the plane (with probabilities) as long as the rotation for (1,0) for each is in the clockwise direction (for example). This means that a physical vector may appear at some x,y in the plane and then at an x1,t1 which would be considered a classical time in the past as long as the motion is clockwise. In this way, one may have an angle of 720 degrees (instead of 360 degrees) describe the motion. In terms of interaction with a B field, one has the clear two directions (spin up clockwise and spin down counterclockwise as examples). Thus, spin must exist in the rest mass case, because one must have mo perpendicular to p if special relativity is to hold. If one does not want to have spin, then one must disregard the equivalence of frames moving at constant speeds with respect to each other and that seems unphysical. (Note: There are spin 0 objects (Higgs particle), but we do not consider that here.
We have argued in previous notes that special relativity is often developed from the point of view of a particle with rest mass, leading to the Lorentz invariant -EE + cc p dot p = -momo cccc ((1)). Then, as a second step, one may obtain the equation E= [p] c by setting mo=0. This appears to be a straightforward math step to obtain a photon relevant result, but we have suggested in (1) that mo represents an inherent force, called inertia (Newton). Thus, special relativity for a particle with rest mass is really linked to this inherent force. In fact, we have suggested in (1), that spin ½ is obtained by finding an equation linear in a particle’s inherent force (rest mass) which is linked to E and p from special relativity and geometry. This, we suggest, is why Dirac linearized ((1)) to find spin ½. The reason for linearization is to have an equation linear in the inherent force, because force appears in F=dp/dt in a linear fashion. The rest mass case seems to be largely based on mo, inertia, which we call the inherent force. We further argued in previous notes that free particle quantum probability arises from -Et+px = Lorentz invariant ((2)) because x,t and x+hbar/p, t+hbar/E give the same result and we consider -Et+px to act somewhat like a free particle potential. This suggests that a photon should also have a probability exp(-iEt+ipx), which is verified as both a photon and particle with rest mass, with the same p, give rise to the same 2-slit interference. The problem in the photon case is that mo=0 and so no inherent force appears in special relativity. We argue that physically, there must be an inherent force as interactions (as in the 2-slit case) depend on it. In (1) we argued that the inherent force of a photon is the electric field, but this does not appear in ((1)) or ((2)), but emerges from Maxwell’s electromagnetic equations. This is why the photon has spin 1 and not spin ½ even though it has the probability exp(-iEt+ipx). (See (1) for details on the specific equation which gives rise to spin 1.) Here, we argue that for a particle with spin ½ (i.e. rest mass), exp(-iEt+ipx), which is a probability which conserves energy and momentum, the momentum and energy are proportional to rest mass (inertia or inherent force) and so exp(ipx) is really a probability distribution for inherent force (augmented to E,p due to the Lorentz boost). In the case of the photon, one still has exp(ipx), but the force is due to the electric field. This begs the question: What is the spatial distribution of the electric field? Is it free to be any function? We argue that given that exp(-iEt+ipx) is the energy/momentum conserving probability from special relativity, the inherent force cannot contradict this distribution. The electric field, which is a real valued function, must be cos(-Et+px). We note that one may formally obtain a classical wave equation from Maxwell’s EM equations to find El proportional to cos(-Et+px), but this does not explain the complex probability exp(-iEt+ipx) which holds for both a photon and particle with rest mass. Our objective is to argue that given exp(-iEt+ipx) for a photon, the electric field must behave as cos(-Et+px). Interestingly, the photon momentum density 1/uo Electric field x Magnetic field is proportional to cos(-Et+px) * cos(-Et+px). This does not follow Real(exp(-iEt+ipx), but it is the electric field and not momentum in the photon case which is linked to force. We argue that ultimately it is force which governs interactions, even though momentum (not momentum density) is conserved.
In Part 3, we started with the notion of a free particle probability linked to energy and momentum conservation in two body Newtonian elastic scattering. In particular, we argued that one needed a probability which shows an equal weight for any outcome (ei,ej) and (pi,pj) (momentum vector) set which conserves energy and momentum. This led to a first guess of exp(iC1 E) and exp(iC2 p), but in Part 3, we argued the momentum and energy are defined through a Lorentz transformation of mo at rest and such a transformation must also affect x,t → x’,t’. We concluded one cannot create any probability based on p and E without also considering x and t which transform according to the same transformation. This then allows one to create a Lorentz invariant probability. Here we take an opposite approach, i.e. one which does not involve p and E a priori. It is well known from classical physics that P(x)dx=dx/L, where L is an arbitrary length, describes a uniform distribution for particles at rest. One may, however, view these particles from two frames, one moving with constant v and the other with constant -v. This would seemingly introduce probabilities P(x1’,t1’,v) and P(x2’,t2’,-v) with no notion of energy or momentum present. It would seem that the product of the two for an x1’=x2’ and t1’=t2’ should equal the rest frame result. This suggests an exp(f(x,t,v)) solution, but there is no reason to have an ever increasing or decreasing exponential and so exp(i f(x,t,v)) seems like a solution. There is, however, a problem, because given P(x,t,v=0), and such a solution is then not P(x)= 1/L unless P(x,t,v)=1, which we reject as it assumes a moving particle is identical to one at rest in terms of probability.. In other words, there are two probabilities for the rest frame, P(x)=1/L and P(x,t,v=0) = exp(i f(x,t,v=0). Given that exp(i f(x,t,v) is based on Lorentz frames, one would also expect probability to be conserved and not be created or destroyed simply because a particle with rest mass is viewed from a rest frame or one moving at constant speed. This seems to force one to use relativistic variables linked to a Lorentz invariant, namely E,p as well as x,t. As a result, in Part 3 we argued that one cannot consider a probability of p,E without considering x,t as well and here we suggest that one cannot even consider a probability of x which is conserved in different constant moving frames without introducing t,p and E. Thus, P(x,t,v=0) which seemed like a “P(x)” type of probability in the rest frame is really a different probability, namely one in time exp(i f(x, v=0, t)), i.e. . for this to be Lorentz invariant, one has exp(-iEt+ipx) = exp(-iEt) for any x. In a sense, this is like P(x)=1/L in that each x carries the same weight, but there is still the time variable and the probability actually changes in time unlike P(x)=1/L. This seems to lead to a paradox because the given exp(-imocct) one may formally argue that exp(-px)=1 for p=0 which is the x part of the probability, yet above we argued that in order to obtain this value one should use P(x1’,t1’v)P(x2’,t2’,-v) at x1’=x2’ and t1’=t2’. The difference between P(x)=1/L is that the time measurement has infinite resolution, while exp(-imocc t) implies that it does not because otherwise information about x is the same. Thus, special relativistic considerations lead to the notion of uncertainty in time (hbar/E) and space hbar/p, but in the Newtonian scenario, these do not exist. We argue that the question then becomes: Why would one wish to change from a special relativistically related probability exp(-iEt+ipx) which shows physical uncertainty units of time and space, to one which shows infinite resolution for x and t, through exp(-iEt+ipx)exp(iEt+ipx)? We suggest that exp(-iEt+ipx) or exp(ipx) in time-independent problems is a dynamic probability needed to describe interactions. One may see that cos(px) and sin(px) may have positive and negative values, implying the removal of probability from one place and its addition somewhere else. One the interaction result has been obtained, one wishes to measure spatial results, i.e.count particles in bins in space. There is no need for the interaction information and it may be formally removed using exp(iEt-ipx) exp(-iEt+ipx).
In previous notes, we argued that one could describe elastic two particle collisions, even in Newtonian mechanics, in a classical statistical manner. In particular, we argued that one could introduce a probability which depends on E,p, but at the same time has the same weight (in a sense) for any free particle. Product probabilities would be linked with a uniform distribution in the sense that exp(-iE1)exp(-iE2)= exp(-iE3)exp(-iE4) if E1+E2=E3+E4 and a similar expression for momentum. If one respects the special relativistic invariant -Et+px, then a possible solution is exp(-iEt+ipx), but even though this yields the free particle wavefunction (probability), this does not explain quantum mechanics. These are simply ideas of classical probability theory with the enforced special relativistic invariant -Et+px. Here, we argue that the real feature leading to quantum mechanics (the idea of hbar/p and hbar/E) arises from the notion of inertia (which we have argued in previous notes also gives rise to spin). We argue that a single force (inertia) in a rest frame becomes two independent interaction notions, E and p as well as creating x’ as a running variable. Thus, mo,t and E,t’ and p,x’ are intertwined. We suggest that there must be a physical mechanism which keeps these two concepts E,p independent and so one cannot simply rely on x=vt. Given the Lorentz invariant and the classical notion of probability exp(-iEt+ipx), one must go one step further and postulate that there are independent fluctuations in t’ and x’ in such a calculation such that hbar/E and hbar/p emerge. We argue that this is a key feature of quantum mechanics and that it does not immediately arise from exp(-iEt+ipx) if one only considers x=vt. One is forced to allow x,t to fluctuate independently which leads to a fluctuation in mocc in the rest frame as well. We argue that this feature is actually associated with special relativity as it allows a single force inertia to manifest itself as two independent interaction quantities E,p. As a result, we suggest that one may obtain exp(-iEt+ipx) without any notion of probability, but only by postulating the existence of a physical mechanism (fluctuation of x’,t’ or t in the rest frame) which keeps E,p independent because -Et+px does not ensure their independence if one only considers x=vt. One may then interpret exp(-iEt+ipx) as having the form of a classical probability which respects special relativity.
We have argued in Parts 1 and 2, that the free particle quantum form exp(-iEt+ipx), which we call a complex probability, arises from special relativity in order to allow for conservation of energy and momentum (in an elastic collision). One may see from single particle bound states how exp(-iEt) plays the part of a binding energy. We have noted in Parts 1 and 2, that exp(-iEt+ipx) follows from the Lorentz invariant -Et+px, but that one does not consider the deterministic x=vt equation which applies to the center-of-mass, but rather treats t and x as independent variables which force the independence of E and p (two different interaction variables) and also force separate E and p conservation. The question becomes: How is it possible to have x,t play two such different roles in special relativity? Here we try to answer this question. We argue that special relativity is based on determinism, i.e. x=vt. One may create a Lorentz transformation for a particle with rest mass mo at x=0 with t, a running variable. As seen from a frame moving with -v, one has x’=vt’. In other words, a separate x’,t’ is associated with each v. In Newtonian mechanics, it is possible to have mov1 and mov2 in the same frame (i.e. a frame with a ruler and clock), but in special relativity, each mo in the rest frame maps to a different x1’,t’ and x2’,t2’. This, however, seems to lead to a contradiction. If one considers two particles p1,E1 and p2,E2 (one dimensional motion) which elastically collide, then before the collision, they may be thought of as a composite particle (for more than a single x,t point) with energy E1+E2 and momentum p1+p2. Given the validity of special relativity, we argue that one may associate a Lorentz invariant -(E1+E2)t + (p1+p2) x with this composite particle. Furthermore, any outcome Ei+Ej = E1+E2 and pi+pj = p1+p2 would satisfy this same invariant. At this point there is no contradiction. The contradiction arises if one considers a single t and x for a composite particle, while internally there exist t1’,x1’ and t2’, x2’. The composite particle, in terms of special relativity, is not compatible with each -E’t’+p’x. The energies and momenta, however, add for each particle (in conservation equations) and so we argue that one should have a relativistically compatible approach which allows one to link each particle to a composite particle and see conservation of E and p. We note that it is precisely x=vt for each particle which leads to x1’,t1’ and x2’,t2’ and so argue that if one gives up this notion of determinism for x and t and introduces independent x, t values (fluctuations), then one can indeed retain the -Et+px form for each particle and use it to create composite particles. This seems to give rise to an AND situation which does not exist for both p,x and E,t in the deterministic relativistic approach. In that approach, one can only think of p’s and E’s adding separately, without any linking of x’s and t’s. In the fluctuation approach, it is possible to have the same x and t fluctuation points for each particle. We argue that it is this kind of thinking which leads to the probabilistic form exp(-iEt+ipx) which still respects the Lorentz invariant form and at the same time allows for conservation of E,p with the same t,x fluctuation points allowed. x=vt does not allow one to combine x1’,t1’ and x2’,t2’ for a composite particle because these may coincide for only one x,t point and one wishes to describe a composite particle with an x,t which exists over more than one x,t. This is why we suggest the physical fluctuation scenario which exists together with x=vt. x=vt describes the deterministic motion of the center-of-mass, while exp(-iEt+ipx) describes probabilistic hits in x and energy in t. It is special relativity which forces one to associate time fluctuations with E and spatial ones with p. As a result, we argue that special relativity gives rise to both probabilistic and deterministic versions, both of which respect the form -Et+px. The deterministic version is linked with x=vt center-of-mass motion, while exp(-iEt+ipx) represents probabilistic appearances of E in t and p in x, i.e. there is a spatial probability for impulse hit delivery.
In Part 1, we considered Newtonian elastic scattering and proposed a probability exp(-iEt+ip dot r) to allow for equal product probability weights for any (ei, ej) (energy) and (pi, pj) (momentum vector) outcome pair. We insisted on the probability being a Lorentz scalar and argued that all free particles have the same real value weight, hence the modulus of 1. This notion is based on probability as it is applied to possible outcomes. This means that one does not have a fully deterministic situation because otherwise one would know the exact ei,ej, pi,pj outcome and would not require probabilities in the first place. In Part 2, we argued that a deterministic approach based on x=vt for a free particle contains no notion of probability and so this seems to contradict exp(-iEt+ipx) with its physical intervals: delta t = hbar/E and delta x = hbar/p. We then noted that these intervals and hence probability are found in the Action= Lagrangian * t = -Et+px with v=x/t. We concluded that x=vt, a solution of d/dt dL/dv partial - dL/dv = 0 loses some of the information present and that examining the action reveals more information because one is forced to deal with x,t, E and p at the same time. Here, we return to Newtonian elastic two-body scattering and assume that one has the following deterministic case: e1,e2, p1, p2 → e3,e4, p3,p4. One would then assume that no probability of any kind is present. We suggest that momentum is conserved due to action-reaction, something that Newton already proposed without any notion of probability required. Energy is conserved because it cannot be ultimately created or destroyed and we consider special elastic collisions. We then consider the Newtonian idea of a particle being acted upon by a force -dV(x)/dx, where V(x) is a potential. The interaction occurs for a dt, dx, but Newtonian physics assumes that dx→0, dt→0. Thus, Newtonian mechanics allows for a minimum dx->0 for conservation of momentum. By going into the center-of-mass frame of two particles which collide elastically, one has p and -p in this frame. When the two particles come together at x,t, for a tiny time one has a joint particle with no net momentum which is interacting and breaking apart. Then, in the 1D case, one particle moves out with p1 and the other with -p1 in the cm-frame. If one considers A= -Et+px for one particle, then Newton’s dt->0, dx->0 as well as x=0, t=0 yield A=0. It is known from special relativity, however, that there is a jump in x for a particle viewed from a moving frame. In particular, if the particle is at x=0 at t in the rest frame (cm frame), then in the moving frame: x’ = g(v) v t where g(v)=1/sqrt(1-vv/cc).This is linked to the person in the moving frame (-v) measuring the particle which seems to move to him/her. This measurement takes some minimum amount of time. Thus, it seems that one must reject dx→0 on the grounds of special relativity. The question then becomes: What is the smallest dx and dt one may have based on special relativity? We suggest that given the A=0 case, one may have delta x =hbar/p and delta t = hbar/E. Even in a deterministic elastic interaction, the notion that dt and dx cannot be 0 leads to minimal values for these which are probabilistic, i.e. they are based on special relativity and not on the actual interaction. If one has initial uncertainty in x and t, even if the particle then moves with x=vt, this uncertainty should also be propagated along as well, because the dx, dt uncertainty cannot simply vanish. We suggest that the notion of exp(-iEt+ipx) emerges even if one removes the initial probabilistic assumption of elastic scattering with unknown outcomes. Known outcomes still lead to uncertain dx=hbar/p and dt =hbar/E in keeping with the uncertainty contained in Lt= -Et+px, which is used for deterministic calculations. Thus, it is special relativity which places a constraint on how small dx and dt may be in a physical reaction and one cannot simply use the Newtonian dx→0, dt→0.