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.
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.
In (1), the notion of a stochastic classical action is introduced and a probability is linked to exp( constant * action). The objective of (1) is not to derive quantum mechanics, but to derive stochastic dynamics based on maximum entropy in action space. We have argued in a number of previous notes (2) that one may use the classical action to create a quantum free particle probability. This classical action, however, is not chosen a priori, but rather arises due to other considerations. We suggest that stochasticity is present in Newtonian mechanics at some time scale because dt is finite in dp/dt= Force (Newton’s second law) and so postulate that force delivery at some x,t is probabilistic within some dt and dx. A priori, dt and dx are not known, but if there is stochasticity, there should be a free particle probability, i.e. a probability(x,t) where x,t represent an x, t interaction point. Furthermore, the notion of dx and dt should arise naturally. We note that the form exp(i function(x,t,p,E) may be introduced because free particles have no distributed variable (like energy in the Maxwell-Boltzmann case). exp(i function(x,t)) immediately introduces periodicity, but we have argued that stochasticity can only exist within tiny, fixed dx, dt linked to dp/dt=F. In particular, one classical Newtonian scales this stochasticity is not seen. The periodicity should be directly in x and t space, suggesting exp( i A1 t + i A2 x). This expression, however, should be Lorentz invariant, and A1 and A2 should be constants, at least for a fixed mo and v. The reason that the one must have periodicity is that the dx and dt ranges of stochasticity in dp/dt=Fr must appear naturally. At this point, one is led to exp(-iEt+ipx) with -Et+px being a Lorentz invariant which satisfies the above criteria. This Lorentz invariant happens to be the classical action in both the relativistic and nonrelativistic cases, hence giving rise to action as being a stochastic variable (of (1)) because it is a function of the two scholastic variables x and t. Alternatively, one may choose the exp(iA1x+iA2t) due to its conservation properties. One may then argue that A1 and A2 could also be conserved for x,t held constant, i.e. A1 and A2 could be linked to Newtonian energy and momentum in elastic collisions. This allows one to solve for the special relativistic g(v) =1/sqrt(1-vv/cc). Given the idea of a conserved quantity form for a probability, one may argue that exp(i Action) satisfies a maximization of Shannon’s entropy subject to a conserved constraint (which is again a link with (1).) The question then becomes: Why should the Lorentz invariant A= -Et+px be the classical action.One may note that x,t usually refer to trajectory values and so for a free particle, one has x=vt. This means that one may factor out t, i.e. A= t( -E+pv), but this is automatically the form of the Lagrangian with E being the Hamiltonian. This holds relativistically and non-relativistically with the form: dL/dv partial = p. The remaining question is: Why should the classical action be Lorentz invariant? We consider E(v)-p(v) v = -L such that dE/dv - v dp/dv -p = -dL/dv partial. This means that dE/dv - v dpd/v = 0 ((1)). A Lorentz transformation, however, changes mo in a rest frame into p=mov/sqrt(1-vv/cc) and E=mocc/sqrt(1-vv/cc). One may see that this solution may be obtained from ((1)) showing a direct link between the classical action approach and special relativity. The connection may be carried further as Lt = -Et+px is a Lorentz invariant, but also the classical action (in both the relativistic and nonrelativistic cases). We argue that the use of the action in (1) as a stochastic variable is directly related to special relativity. In summary, from finite dt in dp/dt=F, t and x are stochastic variables within some naturally appearing dx and dt. One may write a linear combination of x and t and to create a stochastic Lorent invariant variable which should appear in exp(i Lorentz invariant) and create the dx and dt units. It may then be shown that this Lorentz invariant in E,p,x,t should be equivalent to the classical action.
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.
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.
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 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.