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.
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.
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 Part 1, we argued that P(x)dx=dx/L (where L is an arbitrary length) applies to a particle at rest. If one views such a particle from a frame moving at constant -v, it is moving. We argued that in such a case, the relevant probability is not simply linked to the position x, but also to the momentum p at x because this delivers an impulse hit. In particular, probability is presumably to be used in a physical problem with interactions and probability and a dynamical particle requires a dynamical probability describing both x,t and E,p. This led to exp(-iEt+px) which actually predicts intervals in space hbar/p and time hbar/E. In other words, a probabilistic view point leads to these results. Here, we ask the question: What if one simply wishes to use deterministic physics to describe a moving particle? After all, Newtonian mechanics is deterministic and one does not really need to focus on P(x)=1/L or its modification for a moving particle. In such a case, a particle moving with constant v (starting at x=0, t=0) is described by x/t=v both in Newtonian mechanics and special relativity. There is no uncertainty in x and t here which begs the question: Why does it appear in the probabilistic formalism of Part 1? In particular, exp(-iEt+ipx) derived in Part 1 implies physical regions: dx = hbar/p and dt=hbar/E and these seem to be absent in x/t=v and this is a problem. We suggest that a probabilistic approach is buried in the Lagrangian formalism which leads to the result d/dt v = 0, i.e describes the particle’s motion in space. This motion is actually x/t=v, but this equation only presents a subset of the description of a particle that is moving with constant speed, even in a deterministic Lagrangian/Action approach. We have already pointed this out in (1). We briefly explain it again here, but the real question is why does one need to consider the Lagrangian in order to describe x/t=v? In other words, the probabilistic approach associated with v=x/t only emerges in the Lagrangian formalism using a relativistic action Lt = -Et+px, with x/t=v. Ignoring the action and Lagrangian leads to an absence of hbar/p and hbar/E and one is only concerned with x/t=v. We note that the Lagrangian is primarily linked to the result: p = dL/dv partial and so argue that one is not simply interested in motion (x/t=v), but also in interaction because p is associated with an impulse hit. Thus, the Lagrangian approach is more than simply a mechanism which describes x,t motion, it is associated with p and E (through Hamiltonian = E = pv-L.) Using the Lagrangian approach suggests that it does not suffice to consider x and t by themselves when p and E are present. One must consider all four variables x,t,E,p which is the same point made in Part 1. The physics of the problem dictates that it is not enough to simply think in terms of x,t,v. The Lagrangian approach, which is constructed to ultimately yield x/t=v, when written in terms of E, p, i.e. Lt = -Et+px demonstrates uncertainty in x, hbar/p and t, hbar/E as shown in (1). This uncertainty must then be written in terms of a probability which is Lorentz invariant and leads to exp(-iEt+ipx). The point is that one would not even consider a probability if one did not consider E,p and x,t together as a complete treatment of a moving particle. In other words, x/t=v is an incomplete description of the state of a particle moving with constant speed, even in a deterministic framework.. At first this might seem surprising, but it is well known that energy and p are linked to interactions. In special relativity, one does not consider x,t,v alone and then compute p and E from these, but considers four vectors and Lorentz invariant equations. It is possible to create a Lorentz invariant equation involving all four state variables x,t,E,p and it turns out that this is the relativistic Lagrangian which describes motion, so this approach is more general than simply stating x/t=v. Thus, delta x = hbar/p and delta t= hbar/E appear in both the probabilistic approach of Part 1 and the deterministic one of the Lagrangian used here.
In previous notes, we introduced a complex probability into Newtonian two-body elastic scattering to ensure that for a given (e1,e2) and (p1, p2) (momentum vectors), any (ei, ej) (pi, pj) have the same probability to occur. This ultimately led to the Lorentz invariant probability exp(-iEt+i p dot r). Here we begin with the delta functions delta( pinitial - pfinal) and delta( Einital - Final) and see that one may write these as Integral da C exp(-i pfinal + i p initial) and a similar expression for E. Here a is simply a math variable. (Note: This is only one math form for a delta function, but this particular form will prove useful from physical considerations later.) The delta function may be used in an expression Integral delta(pinitial - pfinal) dpfinal and only pfinal values which satisfy conservation of momentum yield a nonzero integral. Furthermore, this integral may be multiplied by various functions and still maintain conservation of momentum. At this point, the above is simply a calculation formalism. The question we ask is whether one may make the argument that a particle itself is responsible for conservation of momentum in an interaction. We suggest this may be the case based on Newton’s notion of action-reaction. It is not simply that one applies a force to a particle and it changes. The particle itself resists and this is the idea of inertia as well as action-reaction. The particle itself is strongly involved in a reaction and we suggest that a particle carries a property of ensuring conservation of momentum and energy itself. It is not something that is imposed by the interaction, otherwise there would be no need for the notion of action-reaction. The particle reacts and the manner in which it reacts leads to conservation of momentum and energy. We suggest that there seem to be grounds for considering a particle to actually be the driver of energy and momentum conservation and that this actually extends to further features, namely probabilistic results in interactions which do not appear in Newtonian mechanics. This property might then be taken as exp(i p a) and exp(i E b), where a and b are math variables used to establish the delta function. One, however, cannot really establish a property of a particle if math running variables a and b are present. The entire expression should consist of particle state variables. This seems to suggest the consideration of Lorentz invariance because E and p change when the particle is seen from a frame moving with -v. One may create a state based exp(-iEt+ipx) energy-momentum conserving property of a particle which is Lorentz invariant. We next note that p and E are linked to interactions, but ones that differ. In particular, p is linked with an impulse which has nothing to do with velocity , i.e. E. On the other hand, E = .5movv in the nonrelativistic limit and is dependent on v, i.e. a m1v1= m2v2 have the same impulse, but different kinetic energies. Thus, we suggest that exp(-iEt) and exp(ipx) be used separately depending on the type of interaction one considers. In a quantum single particle bound state, one is interested in impulse hits with the potential we argue and so does not need to consider exp(-iEt) as one cannot measure the particle moving through the potential in a classical manner. We argue that a constraint (momentum and energy conservation) may actually be due to properties of the particle and that this ultimately follows from Newton’s action-reaction. In (1), we argued that exp(-iEt+ipx) follows from Lorentz invariance which is based on the Newtonian idea of work increasing energy, i.e. 1/sqrt(1-vv/cc) and A= -Et+px have minus signs for this very Newtonian reason. We argued that dx = hbar/p and dt=hbar/E follow as fluctuations which leave A unchanged due to the presence of the minus sign.Here we again argue that a Newtonian idea is responsible for exp(-iEt+ipx), namely action-reaction. We show that the two approaches are identical.
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 argued that the Lorentz invariant -Et+px contains E and p which are physically different types of interaction variables. We argued that nature must somehow keep E and p independent in -Et+px (if nature also respects this Lorentz invariant). As a result, we suggested a math function exp(-iEt+ipx) which actually acts like a probability in terms of products and ensures conservation of E and p. Here, we wish to consider the situation from a different viewpoint. Like in Part 1, we do not postulate a priori any probability. Rather, we argue that there exist interactions in nature which conserve both E and p (elastic collisions) and that these two interaction variables are associated with different physical processes. For example, the notion of action-reaction is enough to see conservation of momentum, but conservation of energy is an independent process. In other words, E is like an enhanced inertia (mo and v), while p is like a moving inertia. An interaction which conserves both is complicated, yet we argue that the notion of the Lorentz invariant -Et+px should still hold for two interacting particles. For a single particle, the center-of-mass moves as x=vt and so one would expect that t and x in -Et+px are correlated. If one has two particles coming together to interact, then by conservation of E,p one has E1+E2 and p1+p2 (one dimension), but x and t for the two particles should be the same and it seems that one is not concerned about x=vt. Furthermore, a given outcome of the interaction E3+E4 and p3+p4 must also be linked to -Et+px and there must be conservation of total E and p. Thus: -(-E3-E4+E1+E2) t + (-p3-p4+p1+p2) x = 0. This suggests mathematically that t and x are independent variables in direct contrast to x=vt. Furthermore, the system must somehow ensure that x and t are independent for any Ei,Ej and pi,pk such that Ei+Ej=E1+E2 and pi+pj = p1+p2 as this is all the information that the Lorentz invariant -Et+px displays. Given that E and p are linked with different interactions, we suggest that there must exist an a priori property of a particle which allows for the above considerations. This would then explain how any Ei+Ej=E1+E2 and pi+pj = p1+p2 holds. In Part 1, we argued that this condition may occur if a particle is associated with fluctuating x and t values in terms of interactions given by exp(-iEt+ipx). The center-of-mass still moves as x=vt, but in an interaction it is the interaction process which is key, not the particle center-of-mass. One requires a description of such an interaction which can handle both E and p and is consistent with whatever information special relativity allows. We suggest that the exp(-iEt+ipx) is such a description, but that it carries the notion of periodicity in space and time and that this periodicity is then linked with E and p. This then becomes associated with Planck’s constant to link periodicity to E,p.
In Part 1, we argued that Newtonian mechanics essentially contains the notion of probability found in the free particle wavefunction exp(-iEt+ipx). The difference is that the wavefunction/probability quantifies dx=hbar/p and dt=hbar/E. The standard Newtonian approach is to have dx->0 dt->0 in order not to deal with the lengths of these intervals. This limiting feature may work in the classical world (but not for photons reflecting and refracting through windows), as the actual sizes of dx and dt may in general be very small in the classical world, but they are not small in the case of atomic or nuclear lengths. The question is: In Newtonian mechanics, without any reference to exp(-iEt+ipx), how would one obtain finite values for dx and dt? Furthermore, why should one wish to obtain dx,dt values associated with x,t in x=vt. We suggest that because one deals with a moving object, and not a stationary one, any x,t on a x=vt trajectory should be associated with a dx,dt to verify its existence. Unlike the stationary case, for which one may strike the particle and make it move, dx,dt for the moving particle should be dynamical and associated with Lorentz invariance. We suggest that it is meaningless to speak of x,t for a moving particle without a corresponding dx,dt. If one can obtain dx,dt values, then one can argue, as we did in Part 1, that Newtonian mechanics contains the notion of an uncertainty dx about x and dt about t for a free particle. This is essentially the uncertainty which allows the 2-slit interference to occur and solves the probabilistic reflection-refraction problem at an n1-n2 index of refraction interface. Of course, if probability must exist within dx and dt, one requires a math expression for such a probability (i.e. exp(-iEt+ipx). One cannot simply use Newtonian mechanics because it does not make use of such a probability, but at least it predicts that such a probability should exist, we argue. Here we show that one may directly use special relativity to show that dx=hbar/p and dt=hbar/E in Newtonian mechanics, such that one has a dx and dt associated with the same value of the classical action as x,t for which x=vt. The classical action describes, for various x,t values describes the x(t)) trajectory in spacetime and so having built-in dx,dt values in this same action ( which is the Lorentz invariant -Et+px both nonrelativistically and relativistically) shows that Newtonian mechanics must be associated with finite dx,dt. The reason that we consider a dx and dt in the first place is that v=dx/dt. The Newtonian approach is to take the limit dx->0 and dt->0. This, we argue, is associated with a coarse-grained equation x=vt which holds both relativistically and nonrelativistically. This is associated with a coarse-grained classical action.Each x,t on the trajectory x=vt yields a different action value for the action (x/v*.5mvv or -x/v mo sqrt(1-vv/cc) or -Et+px where x/t=v. One may consider x=vt and x+dx1 and t+dt1 as the next points in the trajectory. According to Newtonian mechanics, one may make these values arbitrarily small. Furthermore, x=vt holds both nonrelativsitically and nonrelativisitically. How then can one propose a finite valued dx and dt? We argue that the manner in which this may be achieved is to have a dx,dt associated with x,t which does not change the action value for x,t, i.e. does not change -Et+px. This set dx,dt is finite valued even though the next x,t considered may have a dx1, dt1 value such that dx1<<dx and dt1<<dt. If, however, dx and dt are to be taken seriously, the dx1 and dt1 less than dx,dt are simply math values, not physical ones. In other words, one needs a dx,dt to justify the existence of x,t and if x=vt allows any x+dx1 (dx1<<hbar/p), then this is a coarse-grained equation. The coarse-grained action, however, allows for the defining dx=hbar/p and dt=hbar/E even if the action itself is coarse-grained.