In previous notes, we introduced a complex probability into Newtonian two-body elastic scattering to ensure that for a given (e1,e2) and (p1, p2) (momentum vectors), any (ei, ej) (pi, pj) have the same probability to occur. This ultimately led to the Lorentz invariant probability exp(-iEt+i p dot r). Here we begin with the delta functions delta( pinitial - pfinal) and delta( Einital - Final) and see that one may write these as Integral da C exp(-i pfinal + i p initial) and a similar expression for E. Here a is simply a math variable. (Note: This is only one math form for a delta function, but this particular form will prove useful from physical considerations later.) The delta function may be used in an expression Integral delta(pinitial - pfinal) dpfinal and only pfinal values which satisfy conservation of momentum yield a nonzero integral. Furthermore, this integral may be multiplied by various functions and still maintain conservation of momentum. At this point, the above is simply a calculation formalism. The question we ask is whether one may make the argument that a particle itself is responsible for conservation of momentum in an interaction. We suggest this may be the case based on Newton’s notion of action-reaction. It is not simply that one applies a force to a particle and it changes. The particle itself resists and this is the idea of inertia as well as action-reaction. The particle itself is strongly involved in a reaction and we suggest that a particle carries a property of ensuring conservation of momentum and energy itself. It is not something that is imposed by the interaction, otherwise there would be no need for the notion of action-reaction. The particle reacts and the manner in which it reacts leads to conservation of momentum and energy. We suggest that there seem to be grounds for considering a particle to actually be the driver of energy and momentum conservation and that this actually extends to further features, namely probabilistic results in interactions which do not appear in Newtonian mechanics. This property might then be taken as exp(i p a) and exp(i E b), where a and b are math variables used to establish the delta function. One, however, cannot really establish a property of a particle if math running variables a and b are present. The entire expression should consist of particle state variables. This seems to suggest the consideration of Lorentz invariance because E and p change when the particle is seen from a frame moving with -v. One may create a state based exp(-iEt+ipx) energy-momentum conserving property of a particle which is Lorentz invariant. We next note that p and E are linked to interactions, but ones that differ. In particular, p is linked with an impulse which has nothing to do with velocity , i.e. E. On the other hand, E = .5movv in the nonrelativistic limit and is dependent on v, i.e. a m1v1= m2v2 have the same impulse, but different kinetic energies. Thus, we suggest that exp(-iEt) and exp(ipx) be used separately depending on the type of interaction one considers. In a quantum single particle bound state, one is interested in impulse hits with the potential we argue and so does not need to consider exp(-iEt) as one cannot measure the particle moving through the potential in a classical manner. We argue that a constraint (momentum and energy conservation) may actually be due to properties of the particle and that this ultimately follows from Newton’s action-reaction. In (1), we argued that exp(-iEt+ipx) follows from Lorentz invariance which is based on the Newtonian idea of work increasing energy, i.e. 1/sqrt(1-vv/cc) and A= -Et+px have minus signs for this very Newtonian reason. We argued that dx = hbar/p and dt=hbar/E follow as fluctuations which leave A unchanged due to the presence of the minus sign.Here we again argue that a Newtonian idea is responsible for exp(-iEt+ipx), namely action-reaction. We show that the two approaches are identical.
In previous notes, we introduced a complex probability into Newtonian two-body elastic scattering to ensure that for a given (e1,e2) and (p1, p2) (momentum vectors), any (ei, ej) (pi, pj) have the same probability to occur. This ultimately led to the Lorentz invariant probability exp(-iEt+i p dot r). Here we begin with the delta functions delta( pinitial - pfinal) and delta( Einital - Final) and see that one may write these as Integral da C exp(-i pfinal + i p initial) and a similar expression for E. Here a is simply a math variable. (Note: This is only one math form for a delta function, but this particular form will prove useful from physical considerations later.) The delta function may be used in an expression Integral delta(pinitial - pfinal) dpfinal and only pfinal values which satisfy conservation of momentum yield a nonzero integral. Furthermore, this integral may be multiplied by various functions and still maintain conservation of momentum. At this point, the above is simply a calculation formalism. The question we ask is whether one may make the argument that a particle itself is responsible for conservation of momentum in an interaction. We suggest this may be the case based on Newton’s notion of action-reaction. It is not simply that one applies a force to a particle and it changes. The particle itself resists and this is the idea of inertia as well as action-reaction. The particle itself is strongly involved in a reaction and we suggest that a particle carries a property of ensuring conservation of momentum and energy itself. It is not something that is imposed by the interaction, otherwise there would be no need for the notion of action-reaction. The particle reacts and the manner in which it reacts leads to conservation of momentum and energy. We suggest that there seem to be grounds for considering a particle to actually be the driver of energy and momentum conservation and that this actually extends to further features, namely probabilistic results in interactions which do not appear in Newtonian mechanics. This property might then be taken as exp(i p a) and exp(i E b), where a and b are math variables used to establish the delta function. One, however, cannot really establish a property of a particle if math running variables a and b are present. The entire expression should consist of particle state variables. This seems to suggest the consideration of Lorentz invariance because E and p change when the particle is seen from a frame moving with -v. One may create a state based exp(-iEt+ipx) energy-momentum conserving property of a particle which is Lorentz invariant. We next note that p and E are linked to interactions, but ones that differ. In particular, p is linked with an impulse which has nothing to do with velocity , i.e. E. On the other hand, E = .5movv in the nonrelativistic limit and is dependent on v, i.e. a m1v1= m2v2 have the same impulse, but different kinetic energies. Thus, we suggest that exp(-iEt) and exp(ipx) be used separately depending on the type of interaction one considers. In a quantum single particle bound state, one is interested in impulse hits with the potential we argue and so does not need to consider exp(-iEt) as one cannot measure the particle moving through the potential in a classical manner. We argue that a constraint (momentum and energy conservation) may actually be due to properties of the particle and that this ultimately follows from Newton’s action-reaction. In (1), we argued that exp(-iEt+ipx) follows from Lorentz invariance which is based on the Newtonian idea of work increasing energy, i.e. 1/sqrt(1-vv/cc) and A= -Et+px have minus signs for this very Newtonian reason. We argued that dx = hbar/p and dt=hbar/E follow as fluctuations which leave A unchanged due to the presence of the minus sign.Here we again argue that a Newtonian idea is responsible for exp(-iEt+ipx), namely action-reaction. We show that the two approaches are identical.
In Part 1, we considered Newtonian elastic scattering and proposed a probability exp(-iEt+ip dot r) to allow for equal product probability weights for any (ei, ej) (energy) and (pi, pj) (momentum vector) outcome pair. We insisted on the probability being a Lorentz scalar and argued that all free particles have the same real value weight, hence the modulus of 1. This notion is based on probability as it is applied to possible outcomes. This means that one does not have a fully deterministic situation because otherwise one would know the exact ei,ej, pi,pj outcome and would not require probabilities in the first place. In Part 2, we argued that a deterministic approach based on x=vt for a free particle contains no notion of probability and so this seems to contradict exp(-iEt+ipx) with its physical intervals: delta t = hbar/E and delta x = hbar/p. We then noted that these intervals and hence probability are found in the Action= Lagrangian * t = -Et+px with v=x/t. We concluded that x=vt, a solution of d/dt dL/dv partial - dL/dv = 0 loses some of the information present and that examining the action reveals more information because one is forced to deal with x,t, E and p at the same time. Here, we return to Newtonian elastic two-body scattering and assume that one has the following deterministic case: e1,e2, p1, p2 → e3,e4, p3,p4. One would then assume that no probability of any kind is present. We suggest that momentum is conserved due to action-reaction, something that Newton already proposed without any notion of probability required. Energy is conserved because it cannot be ultimately created or destroyed and we consider special elastic collisions. We then consider the Newtonian idea of a particle being acted upon by a force -dV(x)/dx, where V(x) is a potential. The interaction occurs for a dt, dx, but Newtonian physics assumes that dx→0, dt→0. Thus, Newtonian mechanics allows for a minimum dx->0 for conservation of momentum. By going into the center-of-mass frame of two particles which collide elastically, one has p and -p in this frame. When the two particles come together at x,t, for a tiny time one has a joint particle with no net momentum which is interacting and breaking apart. Then, in the 1D case, one particle moves out with p1 and the other with -p1 in the cm-frame. If one considers A= -Et+px for one particle, then Newton’s dt->0, dx->0 as well as x=0, t=0 yield A=0. It is known from special relativity, however, that there is a jump in x for a particle viewed from a moving frame. In particular, if the particle is at x=0 at t in the rest frame (cm frame), then in the moving frame: x’ = g(v) v t where g(v)=1/sqrt(1-vv/cc).This is linked to the person in the moving frame (-v) measuring the particle which seems to move to him/her. This measurement takes some minimum amount of time. Thus, it seems that one must reject dx→0 on the grounds of special relativity. The question then becomes: What is the smallest dx and dt one may have based on special relativity? We suggest that given the A=0 case, one may have delta x =hbar/p and delta t = hbar/E. Even in a deterministic elastic interaction, the notion that dt and dx cannot be 0 leads to minimal values for these which are probabilistic, i.e. they are based on special relativity and not on the actual interaction. If one has initial uncertainty in x and t, even if the particle then moves with x=vt, this uncertainty should also be propagated along as well, because the dx, dt uncertainty cannot simply vanish. We suggest that the notion of exp(-iEt+ipx) emerges even if one removes the initial probabilistic assumption of elastic scattering with unknown outcomes. Known outcomes still lead to uncertain dx=hbar/p and dt =hbar/E in keeping with the uncertainty contained in Lt= -Et+px, which is used for deterministic calculations. Thus, it is special relativity which places a constraint on how small dx and dt may be in a physical reaction and one cannot simply use the Newtonian dx→0, dt→0.
In Part 3, we started with the notion of a free particle probability linked to energy and momentum conservation in two body Newtonian elastic scattering. In particular, we argued that one needed a probability which shows an equal weight for any outcome (ei,ej) and (pi,pj) (momentum vector) set which conserves energy and momentum. This led to a first guess of exp(iC1 E) and exp(iC2 p), but in Part 3, we argued the momentum and energy are defined through a Lorentz transformation of mo at rest and such a transformation must also affect x,t → x’,t’. We concluded one cannot create any probability based on p and E without also considering x and t which transform according to the same transformation. This then allows one to create a Lorentz invariant probability. Here we take an opposite approach, i.e. one which does not involve p and E a priori. It is well known from classical physics that P(x)dx=dx/L, where L is an arbitrary length, describes a uniform distribution for particles at rest. One may, however, view these particles from two frames, one moving with constant v and the other with constant -v. This would seemingly introduce probabilities P(x1’,t1’,v) and P(x2’,t2’,-v) with no notion of energy or momentum present. It would seem that the product of the two for an x1’=x2’ and t1’=t2’ should equal the rest frame result. This suggests an exp(f(x,t,v)) solution, but there is no reason to have an ever increasing or decreasing exponential and so exp(i f(x,t,v)) seems like a solution. There is, however, a problem, because given P(x,t,v=0), and such a solution is then not P(x)= 1/L unless P(x,t,v)=1, which we reject as it assumes a moving particle is identical to one at rest in terms of probability.. In other words, there are two probabilities for the rest frame, P(x)=1/L and P(x,t,v=0) = exp(i f(x,t,v=0). Given that exp(i f(x,t,v) is based on Lorentz frames, one would also expect probability to be conserved and not be created or destroyed simply because a particle with rest mass is viewed from a rest frame or one moving at constant speed. This seems to force one to use relativistic variables linked to a Lorentz invariant, namely E,p as well as x,t. As a result, in Part 3 we argued that one cannot consider a probability of p,E without considering x,t as well and here we suggest that one cannot even consider a probability of x which is conserved in different constant moving frames without introducing t,p and E. Thus, P(x,t,v=0) which seemed like a “P(x)” type of probability in the rest frame is really a different probability, namely one in time exp(i f(x, v=0, t)), i.e. . for this to be Lorentz invariant, one has exp(-iEt+ipx) = exp(-iEt) for any x. In a sense, this is like P(x)=1/L in that each x carries the same weight, but there is still the time variable and the probability actually changes in time unlike P(x)=1/L. This seems to lead to a paradox because the given exp(-imocct) one may formally argue that exp(-px)=1 for p=0 which is the x part of the probability, yet above we argued that in order to obtain this value one should use P(x1’,t1’v)P(x2’,t2’,-v) at x1’=x2’ and t1’=t2’. The difference between P(x)=1/L is that the time measurement has infinite resolution, while exp(-imocc t) implies that it does not because otherwise information about x is the same. Thus, special relativistic considerations lead to the notion of uncertainty in time (hbar/E) and space hbar/p, but in the Newtonian scenario, these do not exist. We argue that the question then becomes: Why would one wish to change from a special relativistically related probability exp(-iEt+ipx) which shows physical uncertainty units of time and space, to one which shows infinite resolution for x and t, through exp(-iEt+ipx)exp(iEt+ipx)? We suggest that exp(-iEt+ipx) or exp(ipx) in time-independent problems is a dynamic probability needed to describe interactions. One may see that cos(px) and sin(px) may have positive and negative values, implying the removal of probability from one place and its addition somewhere else. One the interaction result has been obtained, one wishes to measure spatial results, i.e.count particles in bins in space. There is no need for the interaction information and it may be formally removed using exp(iEt-ipx) exp(-iEt+ipx).
In Part 1, we argued that the Lorentz invariant -Et+px contains E and p which are physically different types of interaction variables. We argued that nature must somehow keep E and p independent in -Et+px (if nature also respects this Lorentz invariant). As a result, we suggested a math function exp(-iEt+ipx) which actually acts like a probability in terms of products and ensures conservation of E and p. Here, we wish to consider the situation from a different viewpoint. Like in Part 1, we do not postulate a priori any probability. Rather, we argue that there exist interactions in nature which conserve both E and p (elastic collisions) and that these two interaction variables are associated with different physical processes. For example, the notion of action-reaction is enough to see conservation of momentum, but conservation of energy is an independent process. In other words, E is like an enhanced inertia (mo and v), while p is like a moving inertia. An interaction which conserves both is complicated, yet we argue that the notion of the Lorentz invariant -Et+px should still hold for two interacting particles. For a single particle, the center-of-mass moves as x=vt and so one would expect that t and x in -Et+px are correlated. If one has two particles coming together to interact, then by conservation of E,p one has E1+E2 and p1+p2 (one dimension), but x and t for the two particles should be the same and it seems that one is not concerned about x=vt. Furthermore, a given outcome of the interaction E3+E4 and p3+p4 must also be linked to -Et+px and there must be conservation of total E and p. Thus: -(-E3-E4+E1+E2) t + (-p3-p4+p1+p2) x = 0. This suggests mathematically that t and x are independent variables in direct contrast to x=vt. Furthermore, the system must somehow ensure that x and t are independent for any Ei,Ej and pi,pk such that Ei+Ej=E1+E2 and pi+pj = p1+p2 as this is all the information that the Lorentz invariant -Et+px displays. Given that E and p are linked with different interactions, we suggest that there must exist an a priori property of a particle which allows for the above considerations. This would then explain how any Ei+Ej=E1+E2 and pi+pj = p1+p2 holds. In Part 1, we argued that this condition may occur if a particle is associated with fluctuating x and t values in terms of interactions given by exp(-iEt+ipx). The center-of-mass still moves as x=vt, but in an interaction it is the interaction process which is key, not the particle center-of-mass. One requires a description of such an interaction which can handle both E and p and is consistent with whatever information special relativity allows. We suggest that the exp(-iEt+ipx) is such a description, but that it carries the notion of periodicity in space and time and that this periodicity is then linked with E and p. This then becomes associated with Planck’s constant to link periodicity to E,p.
In previous notes, we argued that one may find the source of exp(-iEt+ip dot r) in Newtonian mechanics by considering two body elastic scattering. Even though Newtonian mechanics is deterministic, probability is routinely introduced in such elastic scattering events because one cannot deterministically predict the outcome in most cases. Rather, one relies on conservation of energy and momentum and any possible conforming solution has equal probability. We argued that a free particle does not have a real valued probability weight because each is independent of the other (unlike in an ideal gas). As a result, we suggested the probabilities exp(iE C1) and exp(ip C2), with the constants C1 and C2 accounting for units. We then went on to make arguments linked to time reversal symmetry for exp(ipC2) and overall Lorentz invariance to finally arrive at exp(-iEt+ip dot r). Here, we wish to examine these considerations in more detail as we think there is a more general principle present, namely that of a state probability. We first note that even though exp(iEC1) and exp(ipC2) are consistent with conservation of energy and momentum, i.e. exp(iC1 E1) exp(iC1 E2) = exp(iC1 E3) exp(iC1 E4) and a similar expression for p, one only requires such conservation if all particles are classically at the same x and t, i.e. the collision point. As a result, simply given exp(iC1 E1) and exp(iC1 E2), the particles with E1 and E2 may be far apart in space. As a result, it is not clear why one would create a product probability with them unless they were actually interacting at the same x,t classically, but mathematically there is nothing preventing this and claiming conservation even though there is no reaction. It is this idea which suggests that exp(iC1 E) and exp(iC2 p) should be linked with x and t.Then, conservation only occurs in products if x and t are the same for different particles, otherwise it does not follow. This might suggest: exp(i E * (f(x)+g(t)) ) and exp(i p * (f(x)+g(t)). Here p is taken as a one-dimensional number along the direction of momentum. It may be positive or negative.f and g are unknown functions. We next note that x is measured from an origin. Thus, x1 and x2 are distances measured from an origin as is x1+x2. If one writes exp(i E * (f(x)+g(t)), this seems to imply that x and t are both measured from zero points. If one multiplies by exp(i E * (f(x2)+g(x2)), it seems one would wish this result to be equivalent to exp(i E (f(x+x2)+g(t+t2)). This leads to f and g being linear functions in x and t. Now one may have noticed earlier that including E,p,x and t in the “conservation” probability for a free particle means that one has completely defined the state of the particle. This probability is a state probability and so even though it is complex, one may argue that this is a property which should not change through a Lorentz transformation, just like particle number. This then leads to the notion of exp(-iEt+ip dot r). Thus, one initially attempts to describe the probability to have a particular p and E such that conservation of momentum and energy holds. In such a case, there is no need to introduce x and t and a Lorentz transformation changes the probabilities, but not their conservation properties as is shown. Thus, it is the desire to enforce conservation of energy and momentum only when a classical reaction occurs with all particles at the same x and t which introduces x and t into the probability along with E and p making it a state probability which should be Lorentz invariant. We suggest that there is a further important consequence of introducing a state probability. Given that it does not change under a Lorentz boost, it is like a quantity and may be used in OR situations a priori, even though it does not represent an obvious physically conserved quantity. This allows one to immediately write: Aexp(ipx)+Bexp(-ipx)=Cexp(ip2x) at x=0 for 1-D reflection-refraction from an n1-n2 index of refraction junction. (One may write a second equation involving p as exp(ipx) represents the probability to have p at x.) This then leads to a quantum mechanical manner of dealing with an interacting free particle.
In physics (including classical physics), conservation of momentum is always considered, but one must be more careful with conservation of energy as it may include forms like sound etc. As a result, there is a tendency to focus on conservation of momentum all of the time, and conservation of particle energy only in elastic collisions. In addition to this, one may think of a particle striking a target, in which case two equal momentum (but different energies due to different v speed and mo rest mass) cause the same impact. In the case of an interaction with a potential, however, it is strictly energy considerations which are important classically, i.e one uses p dot p /2m + V(x) = E. Knowing momentum without rest mass is not sufficient. One may consider it unusual that one may consider two different scenarios, one which focuses only on momentum (impulse) or conservation of momentum in collisions while only energy is the focus in classical bound states or motion in V(x). Given the classical focus on momentum, imagine that one wishes to describe all free particles with different momenta as a uniform probability distribution such that a product of two probabilities for p1,p2 is the same as for p3,p4 if p1+p2=p3+p4. (We use 1-dimension for simplicity.). One cannot use P= 1/n because p is absent and n->infinite. In one sense, all particles must have the same probability, but at the same time must contain extra information related to p for the conservation of p part of the problem. The notion of a uniform distribution is independent of conservation of momentum, hence of p momentum. Conservation of momentum requires an explicit presence of p. As a result, there seems to be a paradox. One may, however, resolve this paradox by arguing that a probability is a mathematical way of presenting a subset of the information present. If one argues for a uniform probability, then it does not need to contain p, and one would have 1 as an unnormalized probability for any free particle p. A different probability (representing a different subset of information) which respects all free particles as having the same weight, but includes p can also be constructed, namely exp(i C p). C is a constant introduced for unit issues. In this case, the modulus gives all free particle probabilities with p the same weight, but the presence of p allows one to have conservation of momentum. Given that momentum is the key variable that is conserved in classical physics, one should be satisfied with exp(i C p). One might argue that there is no need to consider energy E. We note, however, that if one has p, E1 and -p, E2 in a frame, then in a frame moving with -v, one has gp+gvE1= p1’ and p2’ = -gp + gvE2. Adding yields gv(E1+E2) and so energy automatically appears and cannot be discarded. Furthermore for p, E1 and -p, E2, exp(i C 0) is exp(i C gv(E1+E2)) as seen in the moving frame and so one is forced to consider conservation of energy together with momentum and include both p and E in the more detailed probability. In other words, special relativity automatically links p and E in a 4-vector, leading to the idea that both are important. We argue that one cannot simply consider exp(i C p) due to special relativity. Both E and p must be considered together and this is a key idea, we argue, because if one wishes to have a Lorentz invariant probability, it must be in E and p, i.e. -Et+px. Instead of having a probability for an interaction with p given by exp(i C p) , one has exp(-iEt+ipx) and we argue that this must have the same value in all frames for a particular interaction. One may note that E and p are, on the one hand, separated allowing for impulse hit versus E related interactions, but that overall the probability for interaction includes both and it is the full -Et+ipx which must be invariant. This further suggests that if one does not artificially force an interaction to occur at the center-of-mass, then x,t and x+hbar/p, t+hbar/E are equally good points for the interaction. We argue that there is no need to force x=vt because exp(-iEt+ipx) * exp(iEt+ipx) already shows that one has uniformity, not only for any p,E, but also for any x, t, as there are now four variables. Thus, there is no reason to enforce x=vt in exp(-iEt+ipx), when considering an interaction, we argue. This means that one has two possible interactions at different x,t values, i.e. x,t and x+hbar/p, t+hbar/E. If one is unable to follow the interaction in time, but may follow it in space as in the case of 2-slits about hbar/p apart or for a potential V(x), then one may use the exp(ipx) part of the interaction probability piece by itself. Thus, we argue that special relativity does not allow one to treat p independently of E. For example, for p values that are the same in one frame, but have E1, E2, will be different in another frame. One must consider E and p together and so instead of creating a probability for p alone, one must create one for E,p and then impose Lorentz invariance which leads to different x,t pairs yielding the same probability value for an interaction, i.e. free particle quantum mechanics.
In a number of previous notes, we argued that free particle quantum mechanics, i.e. exp(-iEt+ipx), follows from the Lorentz invariant A = -Et+px. For t=0,x=0, A=0, but we argued that there is another solution, dx = hbar/p and dt=hbar/E (we use intervals because x=0, t=0 is a trajectory point). We then noted that this solution is linked with the probability exp(-iEt+ipx) which is Lorentz invariant and creates the above intervals in t and x. Finally, we remarked on the fact that exp(-iEt) conserves energy and exp(ipx), momentum, when one uses these for 2-body collisions, i.e. AND situations. Here we point out that there are numerous dt= f(p,E) and dx=g(p,E) (f, g are arbitrary functions) which yield A=0. For example, dt=p and dx=E is an example. Only dx=hbar/p and dt =hbar/E, however, lead to a probability form which conserves momentum and energy and this is something which we did not stress clearly in previous notes. Thus, there seem to be a number of issues involved in obtaining a free particle probability exp(-iEt+ipx) and these are independent requirements, which then lead to a particular solution. We argue that these requirements are: ((1a)) One wishes to preserve Lorentz invariance ((1b)) One desires discrete dx, dt intervals instead of Newton’s intervals which tend to 0. ((1c)) These intervals need to be linked with a particle related probability which defines them. This probability is critical physically because if the theory is to define photons and particles equally well ( as special relativity ((1a)) applies to both), one must account for the probabilistic effects of 2-slit interference and 1-D reflection-refraction at an n1-n2 index of refraction junction. ((1d)) One must have conservation of momentum and energy. We argue that it is the combination of these requirements that ultimately leads to the free particle probability exp(-iEt+ipx).