Newtonian mechanics describes forward and backwards motion by x=vt and x= -vt. In order to understand this motion, one needs to create dx and dt units such that a particle with v moves from x,t to x+dx, t+dt. X and t are necessarily coupled and dx and dt are not zero (even in the dx->0, dt->0 limit). They are just very small. For a particle with v=0 (at rest), x and t are decoupled. Given that the particle may be at rest at any x in L, it is natural to define a probability P(x)dx = dx/L, even in Newtonian mechanics. For x=vt and x=-vt, however, x and t must be decoupled, but given different initial points, one can have different sets of x,t measurements. The key point we stress is that motion is linked with dx and dt and these do not have a fixed numerical value in Newtonian mechanics, but nevertheless the concept exists. In principle, Newtonian mechanics is probabilistic because one does not really know what occurs within dx and dt. It is this essential probability, we argue, that is manifested in quantum mechanics by defining actual dx and dt values, although in essence it is already present in Newtonian mechanics. Given that probability P(x) is well-defined and applies to v=0, one may postulate a dynamic probaiblity P1(v) as we have done in previous notes. If one wishes probability to be Lorentz invariant, one may write P1(-Et+px). At first, based on Newton’s x=vt and x=-vt, one would expect that x and t must be coupled through x=vt. The problem with this is that one may form an AND (product) scenario in probability theory. Furthermore, P1(-Et+px) does not contain initial point information.If one writes P1(v)*P1(-v), one should obtain P(x)=1/L because there is no motion. P(x) decouples x,t and so P1 must do the same. In other words, P1(-Et+px) contains decoupled x,t. This differs from Newton’s coupled x(t). If P1 does not increase or decrease indefinitely in x,t, then a solution is P1(-Et+px) = exp(-iEt+ipx). The question we ask is: How does exp(-iEt+ipx) contain more information than x=vt, x=-vt from the Newtonian scenario? In the Newtonian picture, in order to have a v and -v particle at the same x,t, they must be at different x,t values before and after. This suggests a disconnect in dx (and dt) for x and t. This disconnect is present in exp(ipx) and exp(-ipx), at least in the sin(px) term. The only new information contained in exp(-ipx-iEt) is that dx=hbar/p and dt=hbar/E. The uncertainty in x and t (due to dx and dt) is already present in Newtonian mechanics, but one ignores it by considering dx->0 and dt->0. The notion that dx and dt may be finite does not arise experimentally in classical calculations, but becomes apparent on atomic and nuclear scales. As a result, we argue that the probabilistic nature of a moving particle exp(-iEt+ipx) is already present in Newtonian mechanics, but one does not know how to fix dx and dt. As a result, it is not really the case that Newtonian mechanics is so greatly different from quantum free particle mechanics. The dx and dt lengths may be associated with uncertainty even in Newtonian mechanics and if this is considered, then Newtonian mechanics would not be dissociated with 2-slit interference and 1-D reflection-refraction at an n1-n2 index of refraction interface.
Newtonian mechanics describes forward and backwards motion by x=vt and x= -vt. In order to understand this motion, one needs to create dx and dt units such that a particle with v moves from x,t to x+dx, t+dt. X and t are necessarily coupled and dx and dt are not zero (even in the dx->0, dt->0 limit). They are just very small. For a particle with v=0 (at rest), x and t are decoupled. Given that the particle may be at rest at any x in L, it is natural to define a probability P(x)dx = dx/L, even in Newtonian mechanics. For x=vt and x=-vt, however, x and t must be decoupled, but given different initial points, one can have different sets of x,t measurements. The key point we stress is that motion is linked with dx and dt and these do not have a fixed numerical value in Newtonian mechanics, but nevertheless the concept exists. In principle, Newtonian mechanics is probabilistic because one does not really know what occurs within dx and dt. It is this essential probability, we argue, that is manifested in quantum mechanics by defining actual dx and dt values, although in essence it is already present in Newtonian mechanics. Given that probability P(x) is well-defined and applies to v=0, one may postulate a dynamic probaiblity P1(v) as we have done in previous notes. If one wishes probability to be Lorentz invariant, one may write P1(-Et+px). At first, based on Newton’s x=vt and x=-vt, one would expect that x and t must be coupled through x=vt. The problem with this is that one may form an AND (product) scenario in probability theory. Furthermore, P1(-Et+px) does not contain initial point information.If one writes P1(v)*P1(-v), one should obtain P(x)=1/L because there is no motion. P(x) decouples x,t and so P1 must do the same. In other words, P1(-Et+px) contains decoupled x,t. This differs from Newton’s coupled x(t). If P1 does not increase or decrease indefinitely in x,t, then a solution is P1(-Et+px) = exp(-iEt+ipx). The question we ask is: How does exp(-iEt+ipx) contain more information than x=vt, x=-vt from the Newtonian scenario? In the Newtonian picture, in order to have a v and -v particle at the same x,t, they must be at different x,t values before and after. This suggests a disconnect in dx (and dt) for x and t. This disconnect is present in exp(ipx) and exp(-ipx), at least in the sin(px) term. The only new information contained in exp(-ipx-iEt) is that dx=hbar/p and dt=hbar/E. The uncertainty in x and t (due to dx and dt) is already present in Newtonian mechanics, but one ignores it by considering dx->0 and dt->0. The notion that dx and dt may be finite does not arise experimentally in classical calculations, but becomes apparent on atomic and nuclear scales. As a result, we argue that the probabilistic nature of a moving particle exp(-iEt+ipx) is already present in Newtonian mechanics, but one does not know how to fix dx and dt. As a result, it is not really the case that Newtonian mechanics is so greatly different from quantum free particle mechanics. The dx and dt lengths may be associated with uncertainty even in Newtonian mechanics and if this is considered, then Newtonian mechanics would not be dissociated with 2-slit interference and 1-D reflection-refraction at an n1-n2 index of refraction interface.
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.
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.
We try to consider free particle and single bound state quantum mechanics from the viewpoint of classical physics. We first assume that a free particle moves as x=vt. Secondly, we assume that dx and dt cannot be 0 as in Newtonian mechanics, but must be finite. This begs the question: How does one measure dx and dt? If one can measure a particle at a given x and t, then dx→0, dt→0. Thus, finite dx, dt means that a given p=constant and pp/2m=kinetic energy (nonrelativistic) interact with a force over a finite dx. In other words, one does not know where the interaction occurs within dx. As long as the free particle moves without interacting, then x=vt and there is no interest in any dx or interactions. For example, one may measure very large x and t ranges. Even if there are dx and dt effects at the start and end points, these are small compared to the large x and t values. As a result, one has a F(x,p) probability within dx, which repeats itself for the next dx (i.e. is periodic), where F is a probability for a p impulse hit at x within dx. At the same time, the particle moves with x=vt it seems within dx. In general, one might then think that F(x,p) and x=vt are decoupled. In previous notes, we have argued that one way to couple F(x,p) and x=vt is through special relativity, i.e. A = -Et+px. X and t values used would presumably satisfy x=vt, but for any x,t one has x+hbar/p and t+hbar/E as solutions yielding the same A value and so might consider dx= hbar/p and dt=hbar/E. The question we ask here is whether one may arrive at this conclusion without using special relativity. The answer seems to be that one can if one suggests that F(x,p) must contain information making it consistent with x=vt, i.e. with every x and t point having the same weight. Clearly, however, each x,t do not have the same weight within dx and dt, but are rather governed by F(x,p). This means, however, that F(x,-p) also represents a probability distribution and so if one has F(x,p)F(x,-p) there is no momentum at a given x point. In such a case, the distribution within dx must disappear and one must have a uniform distribution. This holds even if F(x,p) has the same dx value for any p. Thus, F(x,p)F(x,-p) = constant and must be periodic in x. This implies that F(x,p) = 1/F(x,-p) for constant=1. Given that F(x,p) =0 at the initial and final points of dx, then F(x,-p) = infinite at these points and that is not allowable. Thus, one is forced to use complex variables and try exp(i g(p) x). If g(p) is a function of p (i.e. not a constant), then conservation of momentum implies that exp(i g(p1) x) exp(i g(p2) x) = exp(i g(p1+p2) x), suggesting exp(ipx). What, however, if g(p) = constant? In such a case, one has issues creating an average kinetic energy: ( Sum over p a(p) exp(i constant x) pp/2m) / (Sum over p a(p) exp(i constant x) ((1)) This leads to a kinetic energy which does not depend on x and cannot be used with V(x) in a conservation of energy equation. Thus, we choose exp(ipx). Note: there is no reason to choose a(p,x) because the physics of finite dx and dt is restricted to exp(i constant x). Finally, we try to explain why the same dx=hbar/p and dt=hbar/E together with x=vt follow from special relativity.
Netwonian mechanics and quantum mechanics seem to be very different theories, even though Newtonian concepts like momentum, energy and their conservatioon carry into quantum mechanics. The difference is not even due so much to the case that Newtonian mechanics is deterministic and quantum mechanics, statistical, we argue. We suggest that quantum mechanics utilizes a turly dynamic probability which treats forwards and backwards motion differently thorugh exp(ipx) and exp(-ipx), while Newtonian mechanics treats them in essentially the same manner, with the exception of the direction. We thus suggest that Newtonian mechanics is really based on the notion of stationary objects. In particular, x(t) means that a particle is at x in dx at t, as if it were stationary with the exception that it is translating with a fixed velocity. This stationary viewpoint might describe why the P(x)=1/L probability for stationary objects is sometimes also applied to objects moving with constant speed. The x(t) approach really means at t, the particle is frozen at x for dt before appearing frozen at x+dx at t+dt with a new velocity. There is of course the notion of acceleration, but for a particle to be at x at t, one must measure it and this means that it is as if it is stationary for an instant. The velocity is then considered to be constant at x (i.e. a translation) with a new translation value at x+dx. As a result, the notion of quantum mechanics does not really follow directly from this Newtonian stationary viewpoint. One has to think in terms of a very different idea, namely that exp(ipx and exp(-ipx) are not merely translations in different directions. A truly dynamic theory is needed. As we have argued in previous notes, P(x)=1/L represents a stationary object and so if one had an AND condition for p and -p (i.e not speed), then exp(ipx)/sqrt(L) exp(-ipx)/sqrt(L) = 1/L. Freezing out speed using quantum dynamic probability exp(ipx) (exp(-ipx)) leads to a stationary picture, but the actual physics of exp(ipx) is very different from anything in Newtonian mechanics. This leads to the question: What happens in the high energy (i.e. classical limit) if Newtonian mechanics and quantum mechanics are so fundamentally different? Quantum mechanics differs from classical physics in that there are fixed uncertainties of dt = hbar/E and dx=hbar/p for a particle with E,p, while Newtonian mechanics uses infinite resolution. One may note that exp(ipx) and exp(-ipx) are very different, but that is only if one can resolve x regions on the order of hbar/p. If one has worse resolution, one would not notice that exp(ipx) and exp(-ipx) differ. They are both possibly present in a dx which is the resolution limit. Thus, it is actually not the infinite resolution of Newtonian mechanics that is manifest, but rather the approximate finite resolution which means that on classical scales, a tiny dx is actually larger than hbar/prms where prms is the root means square momentum of a quantum bound solution. The quantum bound state solution is based on finding energy levels and a wavefunction W(x) = Sum over p a(p)exp(ipx). It includes exp(ipx) and exp(-ipx), but physically in the classical limit, the particle moves to the right for some time and then later to the left. Quantum mechanics cannot change this measured situation. As noted, exp(ipx)*exp(-ipx) =1 and this is linked with a stationary object even though exp(ipx) and exp(-ipx) both represent dynamic probabilities. The dynamic probabilities allow one to correctly solve problems at scales of hbar/p. W*(x)W(x) is an attempt to create a stationary distribution which is consistent with the kind of measurements made in the classical world. It is W(x), however, that is actually used in solving problems. As the energy level increases, the peaks of W*(x)W(x) become increasingly narrow and so even this stationary-quantum approach is replaced by the simpler, approximate exp(i prms x) * exp(-i prms x), leading to a stationary free particle distribution. This is why we suggest that Newtonian mechanics is really a stationary free particle solution with the particle then translating from one x (within dx) to the next in dt. In other words, x(t) means that the particle is at rest in t (for an instant as it must be measured). The v(x) = dx(t)/dt is called the constant speed in dx (as it must be measured) even though there is acceleration occurring. It seems that Newtonian mechanics is a static kind of viewpoint which is very different from the dynamic exp(ipx) and exp(-ipx).
Classically, probability is considered in terms of the number of possible outcomes. For example, a die toss has six possible outcomes and a coin toss, two. Probability calculations make use of probabilities and hence of the total number of possible outcomes. Thus, they are not deterministic calculations which follow “what happens” as in Newtonian mechanics. We suggest that if free particle quantum mechanics is probabilistic, it must also involve the notion of possible outcomes in an experiment and not represent a deterministic approach. We suggest, however, that something interesting applies to quantum mechanics because we have suggested that free particle quantum mechanics arises from the Lorentz invariant A = -Et+px. We note that there exist x,t trajectory points which follow x=vt. We also note, however, that one may introduce the idea of intervals dx, dt (because Newton’s dx→0, dt→0 is an idealization) and that A is unchanged for x,t and x+hbar/p , t+hbar/E. In other words, we suggest that there are intervals with probability about each x,t trajectory point. A trajectory point is associated with deterministic physics, but probability about this point is not. In other words, one has a kind of hybrid probability scheme. Thus, exp(ipx) and exp(-iEt) really represent fluctuation probabilities about a trajectory point x and t. These ranges of possible x and t fluctuations about a trajectory point x,t allow for further probability to occur in certain interactions. We consider two examples. The first is 2-slit interference which involves a deterministic particle moving along a straight line towards a 2-slit mechanism perpendicular to its path. At the 2-slit apparatus, one considers a probabilistic calculation as the uncertainty in x, i.e. dx=hbar/p allows for possible interaction with both slits. This involves considering all possible outcomes (i.e. motion through one slit or the other which is a classical situation), but exp(i p dot r1) and exp(i p dot r2) “sit on top” of these trajectories at the 2-slit apparatus and one has interference of the probabilities linked with each classical possible outcome. Thus, there are two sets of probabilities which are intertwined, the probability range hbar/p for each path and the two possible paths which have equal probability. One must consider the possible outcomes for both sets together. We suggest that a similar analysis applies to the case of 1-D reflection-refraction for x=0 at an n1-n2 index of refraction junction. Thus, we suggest that a deterministic free particle with x=vt may encounter probabilistic interaction situations, such as a two-slit apparatus or an n1-n2 index of refraction junction. These are probabilistic only because there is a probability fluctuation scheme in x associated with each deterministic x point. It is this probability which allows for a second probability to arise due to interaction with the slits or at the 1-D n1-n2 junction. In such a case, one must consider all possible outcomes which means that one does not only consider the exp(ipx) type probabilities, but also those linked with the physical interaction, i.e. two possible paths, one from each slit, or an incident, reflected and refracted photon at x=0 (n1-n2 junction). It is a combination of all possible paths which must be used in calculations. We thus suggest that even though there is a sense of determinism, i..e free particles move as x=vt and these exist in the 2-slit case and the 1-D reflection-refraction case, in a probabilistic calculation one must consider all probabilistic outcomes consistent with these. Thus, quantum mechanical calculations are true probabilistic calculations in the sense that they involve all possible outcomes in an analysis.
Newton’s law F= dp/dt is considered to be associated with deterministic kinematics. The problem, we argue, is that dt is not really zero (only very small on Newtonian scales). If, for the sake of argument, one considers a finite dt, one must then consider the workings of force in a very tiny time and there is no reason to believe this is a deterministic situation. It is more likely that there is some stochasticity at the level of dt and dx and so t and x within dt and dx are stochastic and associated with probability. On Newtonian scales, this would not be noticed, but on other scales (quantum ones), it could be. A key point seems to be that even if one has x=vt ( a deterministic equation), one does not really know the x and t values exactly, there is a stochastic uncertainty of dt, dx (because the particle was put in motion by an interaction). This means that even a free particle should have this inherent stochasticity in x and t, even if p and E are considered exact. Now expressions for p and E formally follow from special relativity and so if there is some dx,dt stochasticity and it has a certain range (i.e. a maximum value for dx and dt in a sense), this range should appear in special relativity as one must know dx,dt and dx’, dt’. We have suggested in previous notes that possible dx,dt ranges in fact appear in the Lorentz invariant A = -Et+px because for x,t (center of mass description) yielding a certain. Values, off center-of-mass values, x1=x+hbar/p and t1=t+hbar/E also yield the same A. We suggest that these are the “maximum” dx and dt values with uncertainty because the momentum or energy interaction can occur with probability within dx and dt and need not occur at the center-of-mass. It is dx and dt which differ in different frames, e.g. hbar/p’ and hbar/E’, but there is uncertainty in these ranges, i.e. a probability distribution. This uncertainty is associated with a free particle and should manifest itself in an interaction (e.g. 2-slit interference), but it also means that two body elastic scattering cannot ever be considered fully deterministic. There is still the stochastic, uncertain dx, dt. If one wishes to consider conservation of energy and momentum, however, one uses the notion of a free particle x=vt. Even though actual interaction hits may occur at different places, if one has two particles P(p1,x1) P(p2,x2) in an AND situation, one must have x1=x2 in order to show that momentum is conserved at the same x. This leads to the notion of exp(ipx) as a probability, or more fully exp(-iEt+ipx) which is consistent with the special relativistic condition on the range of dx and dt. If one were to consider special relativity, then two particles together have E1+E2 and p1+p2 and one uses the same x and t and so dx= hbar/(p1+p2) and dt= hbar/(E1+E2). This matches the P(p1,x)P(p2,x) scheme for ranges of dx and dt. Alternatively, one may consider that one has a uniform distribution for any outcomes of E1,E2, p1,p2 elastic scattering which conserve energy and momentum. This leads to exp(iC1E) and exp(iC2p), but these must display the dt, dx ranges and so one has exp(-iEt+ipx) in order to have a Lorentz invariant probability. The above arguments have been based on Newton’s second law dp/dt=F and special relativity which means that the two must exist together, which in fact they do. In particular, special relativity is not simply about viewing particles from different frames, but must also be linked with interactions which create the equivalent motion. As shown in (1), one may start with number and vector concepts E,p without knowing their explicit forms and then use special relativity to calculate them. In (1), however, this involved the notion of dp = Fdt and dke= Fdx even for the form of p unknown. Thus, dp/dt =F and special relativity are directly intertwined and the notion of a finite stochastic dt and hence dx in dp/dt=F should have effects on special relativity. One, however, has to directly impose this stochasticity in dt (and hence dx) on dp/dt=F by arguing that dt has a finite value with probabilistic t value within and so one must also look for the stochastic dx and dt range in special relativity, i.e. by considering A= -Et+px as described above. The stochasticity is an assumption, but the finite dt and t+hbar/E, x+hbar/p are not.