Newtonian mechanics deals with two types of force delivery schemes, one linked to the variable x and the other to t. They may be written as: Integral Fdt (impulse) and Integral Fdx (work) and appear to be symmetric in the two variables. The problem is that expressed in terms of dp one sees an immediate difference linked to discrete versus continuous change. In particular: dp= Fdt ((1a)) and vdp = Fdx ((1a)). ((1a)) allows for a discrete change in p in a tiny interval of time. “X” does not come into play at all in the equation and one might think of the impulse occurring at a point x. Due to the discrete nature of integral dp changes, one may think in terms of probabilities because a given pA may have been obtained from discrete impulses: p1→p2→p3 or p4+p5. etc. The final p state is the same and so not only does there exist probability but also the notion of product probabilities such that exp( pi b) exp( pj b) = exp( (pi+pj) b) where b is a constant. Given that exp(pb) increases exponentially, one might rather use the periodic probability exp(i p b). One may note how impulse type hits and their discrete momentum results lead to stochasticity in a Maxwell-Boltzmann gas. The form vdp = Fdx is very different in that it does not allow discrete jumps of p because v=dx/dt. If one assumes a very large F at x1 and dx–0 such that the product is not 0, one might consider dp changing discretely, but what is then used for v? Instead one considers a range of x with a corresponding range in t such that x(t) in order to compute v. Thus this is a deterministic situation because x and t cannot be decoupled. It leads to the notion of a kinetic energy pp/2m associated with p. This kinetic energy is calculated by considering p change from 0 to p. An interesting feature arises because impulse = Integral Fdt may be thought of as discrete allowing p1 to jump to p2. Nevertheless, the deterministic approach states there is a kinetic energy associated with p1 and p2, namely p1p1/2m, p2p2/2m, even though no deterministic path in time and space was taken. The impulse was said to occur at x in a very tiny time range. Thus there seems to be a contradiction. pp/2m requires a range of x and t with x(t). The impulse only occurs in time, with x decoupled but there must be an associated x range which is then not deterministic i.e. a decoupled x range. This suggests a characteristic length associated with a momentum p in which one does not follow time deterministically. If one does not follow x in t in this special length, then one has a probability distribution. A similar argument occurs for time. Consider Fdx with time absent. If dx becomes very tiny and F(x) very large , one may think that work is done at an instant in time, but there must be a time range for pp/2m. Thus there is a special time period not associated with x in which one has probabilistic behaviour i.e. not x(t). One may note from the notion of impulse that a given p may be obtained through different series of smaller impulses, e.g. {p1,p2,p3} and {p4,p5}. These are associated with kinetic energy series {p1p1/2m, p2p2/2m,p3p3/2m} and {p4p4/2m, p5p5/2m}. The final result is the same. Thus the discrete nature of changes leads to a probability scheme such that P(p1)P(p2) = P(p1+p2) and P(e1)P(e2)=P(e1+e2). This we argue leads to: P(p) = exp(i p b1) and P(e) = exp(i e b2) where b1 and b2 at first seem to be constant. We noted, however, that an impulse implies a probabilistic range in x and Fdx→finite when dx→0 a probabilistic range in t. Thus b1 may be replaced with x and b2 with t. At any given point in time or space, x and t are constants i.e. b1 and b2. Next, we consider the situation of special relativity i.e. a Lorentz boost. We consider an initial state of energy= mo, x=0, t=to and view it from a frame moving at a constant velocity v. Although one may use infinitesimal velocities, a main point we wish to make is that special relativity allows for the use of finite velocities which give the sense of discrete changes like in the impulse case.This should again be associated with probabilities. Special relativity yields the Lorentz invariant A = -Et+px where v=x/t. Nevertheless, x and t are written in the invariant as if they were independent. It may be noted that dA=0 if t changes on its own as hbar/E and x on its own as hbar/p. Thus one may obtain specific x and t probability ranges.This is again consistent with the probabilistic picture or exp(-iEt) and exp(-ipx). It may be noted that even with the probabilistic picture, a particle with constant p moves determinisitcally on average as x= p/m t. If x = hbar/p, then 1/t is not hbar/E and if t=hbar/E then x is not hbar/p. Thus the values hbar/E and hbar/p seem not to be linked with the current deterministic motion, but rather with the creation of p in the first place. E=pp/2m and this is obtained by considering p changing from 0 to p. Thus the “frequency” E seems to capture the information of the creation of p and we argue that the wavelength captures the creation of x as a probability distribution in x which one would usually associate with force/acceleration.
Newtonian mechanics deals with two types of force delivery schemes, one linked to the variable x and the other to t. They may be written as: Integral Fdt (impulse) and Integral Fdx (work) and appear to be symmetric in the two variables. The problem is that expressed in terms of dp one sees an immediate difference linked to discrete versus continuous change. In particular: dp= Fdt ((1a)) and vdp = Fdx ((1a)). ((1a)) allows for a discrete change in p in a tiny interval of time. “X” does not come into play at all in the equation and one might think of the impulse occurring at a point x. Due to the discrete nature of integral dp changes, one may think in terms of probabilities because a given pA may have been obtained from discrete impulses: p1→p2→p3 or p4+p5. etc. The final p state is the same and so not only does there exist probability but also the notion of product probabilities such that exp( pi b) exp( pj b) = exp( (pi+pj) b) where b is a constant. Given that exp(pb) increases exponentially, one might rather use the periodic probability exp(i p b). One may note how impulse type hits and their discrete momentum results lead to stochasticity in a Maxwell-Boltzmann gas. The form vdp = Fdx is very different in that it does not allow discrete jumps of p because v=dx/dt. If one assumes a very large F at x1 and dx–0 such that the product is not 0, one might consider dp changing discretely, but what is then used for v? Instead one considers a range of x with a corresponding range in t such that x(t) in order to compute v. Thus this is a deterministic situation because x and t cannot be decoupled. It leads to the notion of a kinetic energy pp/2m associated with p. This kinetic energy is calculated by considering p change from 0 to p. An interesting feature arises because impulse = Integral Fdt may be thought of as discrete allowing p1 to jump to p2. Nevertheless, the deterministic approach states there is a kinetic energy associated with p1 and p2, namely p1p1/2m, p2p2/2m, even though no deterministic path in time and space was taken. The impulse was said to occur at x in a very tiny time range. Thus there seems to be a contradiction. pp/2m requires a range of x and t with x(t). The impulse only occurs in time, with x decoupled but there must be an associated x range which is then not deterministic i.e. a decoupled x range. This suggests a characteristic length associated with a momentum p in which one does not follow time deterministically. If one does not follow x in t in this special length, then one has a probability distribution. A similar argument occurs for time. Consider Fdx with time absent. If dx becomes very tiny and F(x) very large , one may think that work is done at an instant in time, but there must be a time range for pp/2m. Thus there is a special time period not associated with x in which one has probabilistic behaviour i.e. not x(t). One may note from the notion of impulse that a given p may be obtained through different series of smaller impulses, e.g. {p1,p2,p3} and {p4,p5}. These are associated with kinetic energy series {p1p1/2m, p2p2/2m,p3p3/2m} and {p4p4/2m, p5p5/2m}. The final result is the same. Thus the discrete nature of changes leads to a probability scheme such that P(p1)P(p2) = P(p1+p2) and P(e1)P(e2)=P(e1+e2). This we argue leads to: P(p) = exp(i p b1) and P(e) = exp(i e b2) where b1 and b2 at first seem to be constant. We noted, however, that an impulse implies a probabilistic range in x and Fdx→finite when dx→0 a probabilistic range in t. Thus b1 may be replaced with x and b2 with t. At any given point in time or space, x and t are constants i.e. b1 and b2. Next, we consider the situation of special relativity i.e. a Lorentz boost. We consider an initial state of energy= mo, x=0, t=to and view it from a frame moving at a constant velocity v. Although one may use infinitesimal velocities, a main point we wish to make is that special relativity allows for the use of finite velocities which give the sense of discrete changes like in the impulse case.This should again be associated with probabilities. Special relativity yields the Lorentz invariant A = -Et+px where v=x/t. Nevertheless, x and t are written in the invariant as if they were independent. It may be noted that dA=0 if t changes on its own as hbar/E and x on its own as hbar/p. Thus one may obtain specific x and t probability ranges.This is again consistent with the probabilistic picture or exp(-iEt) and exp(-ipx). It may be noted that even with the probabilistic picture, a particle with constant p moves determinisitcally on average as x= p/m t. If x = hbar/p, then 1/t is not hbar/E and if t=hbar/E then x is not hbar/p. Thus the values hbar/E and hbar/p seem not to be linked with the current deterministic motion, but rather with the creation of p in the first place. E=pp/2m and this is obtained by considering p changing from 0 to p. Thus the “frequency” E seems to capture the information of the creation of p and we argue that the wavelength captures the creation of x as a probability distribution in x which one would usually associate with force/acceleration.
In Newtonian mechanics, force and energy are clearly separated. For example, pp/2m = Integral F dx = Integral dp/dt dx = m/2 vv = pp/2m. The equation pp/2m + V(x) = E is considered to be strictly an energy equation. In special relativity, mocc is considered to be strictly an energy. We have suggested in previous notes that mo is proportional to inertia which is a force. Thus, mo and mocc are thought of in two very different manners simultaneouly, i.e. as an energy and as a force. We propose that it might be beneficial to think of both mo and V(x) in terms of energy and force distribution simultaneously. In particular, we consider that V(x) is not only potential energy at x, but is also linked to force or interaction at x through -d/dx V(x). The reason this duality of energy-force may be important is that it is directly linked to quantum mechanics, we argue. In general, it is argued that classical mechanics is a high energy level limit of quantum mechanics, We argue, however, that for a single particle bound state, the energy linked with a momentum p (at any energy level) is pp/2m (nonrelativistic) which is deterministic (and also follows from special relativity) and has nothing to do with quantum mechanics. The difference between classical mechanics and quantum, is that there exists an interaction (impulse hit) probability of exp(ipx) which means that instead of a hit occurring at the center-of-mass point, it may occur within an hbar/p range. Thus, one may have a particle interact and move through either of two slits if they are separated by about hbar/p. If p, however, is changed in an interaction, p1->p2, then the change in energy p2p2/2m - p1p1/2m (one dimension) is classical-deterministic. It is, however, the duality of V(x) with force that force one to consider not only energy pp/2m (deterministic), but also its associated force distribution given by exp(ipx) / W(x), where W(x) = Sum over p a(p)exp(ipx). As a result, one creates the average kinetic energy term: {Sum over p a(p) pp/2m exp(ipx) } / W(x). One may call exp(ipx)/W(x) a probability and so formally one has KEave(x), but physically exp(ipx) represents a force distribution. The balance to V(x) is a combination of a deterministic energy pp/2m with a force distribution exp(ipx)/W(x). This, we argue, is a physical statement and helps to describe why V(x) may be expanded as a series of virtual particles (e.g. photons etc). It is not just energy, but force that must be considered together. It is only in classical physics, pp/2m + V(x) = E that one solely considers the energy aspects, i.e. pp/2m and V(x) only as energy, but the more general picture, we argue, is to consider not only energy, but force distribution together. This is the quantum picture and it incorporates not only probability, but deterministic Newtonian mechanics (or special relativity) through the use of pp/2m.
In this note we argue there exist two force delivery schemes which are already well-known in classical mechanics namely: F(x)dx and F(t)dt. The former is associated with work and kinetic energy, is deterministic (yields x(t)) and requires information about velocity (or rest mass). It deals with F(x) where F=Force. The second is associated with an impulse which changes momentum. In this case, one does not need to know the rest mass or velocity. For example, a particle with momentum p (this allows for many mo,v pairs with the same p) strikes a surface and is reflected back with -p. It is also associated with probability as such a scenario exists for light reflecting and refracting from a surface. One does not know a priori the probabilities for these two phenomena so the problem is probabilistic. Classical mechanics primarily deals with the F(x) force delivery scheme, e.g. solving a differential equation(s) which arise directly from dp/dt = F(x) or using Lagrange’s or Hamilton’s formalism. Ultimately one obtains the deterministic result x(t). Although the notion of an impulse is well-known in Newtonian mechanics, it does not seem to be used much in calculations because an explicit form of F(t) is usually not known i.e. the force acts in a very tiny dx region in a very short period of time. As noted above, both schemes i.e. both deterministic and probabilistic situations appear in the macroscopic classical world, yet for some reason it is suggested that Newtonian mechanics is only deterministic. Newton considered a photon to be a particle with momentum p. A single photon reaching a surface with a different index of refraction has a certain probability to reflect and another to refract. This is a probabilistic, not deterministic, scenario governed by an impulse delivery. The probabilistic problem of light reflecting and refracting in a macroscopic situation may be solved by using continuity of the periodic electric field and its derivative from Maxwell’s solution for a photon from his electromagnetic equations. We have suggested in previous notes, however, that no knowledge of Maxwell’s result is needed. A steady state reflection/refraction problem removes time (unlike the x(t) deterministic picture), but seems to require some kind of balance on the right and left hand sides of the medium. Furthermore two equations are needed to solve for the two unknowns (i.e. reflection and refraction probabilities). Intensity/flux does not balance because the incident and reflected rays have positive flux. For n1<n2, it would be good to have a balance function which allows partial cancellation of the incident and reflected fluxes. The solution exp(ipx) allows exactly this and this solution only requires knowledge of p, momentum, not of rest mass (i.e. velocity). Thus there seem to be two force delivery mechanisms in the macroscopic classical world with very different solutions. The first is the F(x) delivery yielding the deterministic x(t) and requiring knowledge of mo (rest mass). The second is the impulse delivery which is probabilistic, makes use of exp(ipx), and does not require rest mass. If rest mass is known, however, exp(ipx) is linked with the average equation x = p/m t. At first the two delivery schemes seem mutually exclusive, but in the impulse situation, we try to show that a series of impulses may create the semblance of an F(x). Thus one has the exp(ipx) from the probabilistic force delivery scheme, but at the same time an F(x) emerges and so an x(t) type solution should also follow. How can this be? exp(ipx) is probabilistic and one may create averages with probabilities. Such an average or root mean square average may exhibit deterministic behavioiur. Thus exp(ipx) may be used together with a potential V(x) to create a solution W(x)= Sum over p a(p)exp(ipx) such that the prms(x) behaves like a Force(x) delivery result, yet at the same time one has the features of the impulse delivery as exp(ipx)s are used. Thus the exp(ipx) impulse approach may be used in strict impulse delivery problems and in F(x) delivery problems. This is exactly what occurs in quantum mechanics.
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.
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 Part 1, we suggested that exp(ipx) represents a spatial probability for the rendering of momentum hits. A single particle has a center-of-mass which undergoes x=vt, but momentum hits are not rendered from this point, but are rather delivered in x according to the complex probability exp(ipx) which is linked with the resolution delta x = hbar/p. This accounts for the fact that a particle can probabilistically interact with each slit of a 2-slit apparatus if they are about hbar/apart. If one is only interested in energy, pp/2m, the classical result holds for both exp(-ipx) and exp(ipx). In a classical bound state, one considers only energy p(x)p(x)/2m + V(x)=E because each p(x) exists at x in dx->0. One cannot distinguish any probabilistic hit distribution. exp(ipx) and exp(-ipx) differ even though they are linked with the same kinetic energy pp/2m (nonrelativistic case). We argued in Part 1, that the Newtonian conservation of energy equation actually includes a force profile linked with V(x). The catch is that for V(x) one has a force at each x, dx->0. Such is not the case for exp(ipx). We argued in Part 1 that one must consider not only pp/2m, but the associated spatial probability distribution for hits (i.e. energy together with force). Here, we note that this idea gives rise to an equilibrium calculation. This is somewhat startling because classically, and even quantum mechanically for higher energy levels, a particle moves to the right part of the time and then to the left for other times. As a result, a calculation which respects time would not average over exp(ipx) and exp(-ipx). Only an equilibrium calculation in time would do so. We suggest that one must consider the Newtonian conservation of energy equation as applying to both p(x) and -p(x) and so from an energy point of view one may consider it to be an average. The catch is that each term p(x) and -p(x) yields the same kinetic energy profile in the Newtonian case and the same V(x). For a quantum particle the kinetic energy with probabilistic force distribution is not the same for p and -p. As a result, to use V(x) which applies to forward and backwards motion, one is forced to consider the average and this leads to an equilibrium scenario. We show that the average kinetic with probabilistic force hit profile in space is not the same for +p and -p. Measurements, however, take time and may be averages of much data and so the notion of an average or equilibrium seems to make sense. We note, however, that an average is very different from the Newtonian forwards-backwards motion picture. Only in the high energy limit does the energy pp/2m with probabilistic force profile become the same for both p and -p. Thus, the main point of this note is to stress the notion of a single particle bound state as an average and to point out the consequences of this.
Newton’s second law dp/dt = F(x)=-dV/dx may be written in terms of x alone i.e. dp/dx v(x) = -dV/dx or 1/(2m) d/dx pp = -dV/dx leading to the idea of kinetic energy pp/2m and potential energy V(x). In (1) we argued that there is velocity, rest mass mo, and momentum in quantum mechanics. A free particle, however, does not undergo smooth acceleration as in classical mechanics, and we argued there exists a momentum space wave exp(ipx) accompanying the moving particle. In such a scenario, force is considered to be probabilistic momentum hits of the form Vk exp(ikx) leading to a free particle momentum distribution W(x)= Sum over p a(p) exp(ipx). Both of these expressions involve interference due to momentum waves which do not track a particle in time. A mass flux mov or energy flux p=Ev relativistically seems to be the main consideration. Nevertheless, summing Vkexp(ikx) over k to obtain V(x) yields a function varying in x alone. This is a kind of average based on interference and Sum k Vk exp(ikx) is an average linked to momentum hits. We speculate that average momentum hits i.e. -d/dx V(x) are linked to momentum flux changes. In other words, even though an impulse is based on p and does not distinguish v i.e. p=m1v1=m2v2, momentum flux is velocity dependent and appears for an average momentum hit i.e. -dV/dx. In such a case (pave+dpave)(vave + dv ave) = 2 pave dv(ave) . This expression, however, contains both forward and backward motion which is part of quantum mechanics. Moving to an average picture seems to introduce time and direction i.e one uses pave d v(ave). This is already a change from a quantum to a classical scenario albeit for averages (which have not been specifically defined). Thus we consider pave dv(ave). If this is the case, then: d/dx pave 1/m space = -dV/dx or 1/(2m) pave pave = -d/dx V(x). Instead of treating pave pave as two separate averages one considers pave=prms or pavepave=<pp>. Thus the quantum bound picture (at any energy level) becomes a classical type of scenario which includes time and velocity. Note that for a two slit interference pattern, one does not need velocity and time because the interaction with the potentials from both slits is based on exp(ipx). One does not even need the form of the potential. Interference is the same for p=m1v1=m2v2 (even though the process takes place during different time intervals, the pattern is not affected). In other words, we argue that Newton’s second law might follow from a picture of momentum hits in which a flux i.e. p=mov is treated as a quantity which in an average scenario leads to a second flux i.e. an average momentum flux which reintroduces direction, time and velocity.. This is somewhat like a hydrodynamics scenario.