A Maxwell-Boltzmann gas with no external potential V(x) has a constant spatial density. The value of this spatial density alone does not yield the pressure because this also depends on T through the ideal gas law Pressure = density(x) RT, where T is temperature (associated with a conserved average energy), but also the average single particle pressure. Taking d/density(x) of Pressure yields the single particle average pressure value RT, yet at the same time single particle pressure is a stochastic value evaluated as the average of constant v p where p is momentum and v velocity. The quadratic form of p arises, we argue, because one has more than one particle, thus v distinguishes between the particles and relative speed becomes important. In the case of a single particle (as in a single free quantum particle) moving at constant p (speed as well) there is again a constant spatial density as in the MB gas with no V(x). Given the value of the constant density 1/L (L=length), one has no information about momentum p. In the case of a single particle, impulse and not pressure is the important quantity, we argue, because there is no need for a v as there is no relative motion. One only needs to know impulse which is proportional to p. If one speculates that a free quantum particle may follow the model of pressure in an MB gas, one would require a function F(p,x) such that d/d something F = p where p is the single particle impulse. In the MB case, “something” is density, but this is not appropriate for a single particle as density consists of many particles. Without changing T, one may change pressure in an MB gas by changing density i.e. changing the x value of particles. For a quantum particle, we suggest that the conjugate variable to p is x. One may change x of the single particle. Furthermore an association between d/dx and p exists in Fermat’s Least Time Principle. T Thus we suggest d/dx F(p,x) = p where F is unknown both in form and meaning. Pressure, however, is a statistical quantity so F should also be a statistical quantity i.e. one could have d/dx P(p,x) / P(p/x) = constant p where P(p,x) is a kind of unnormalized probability. This suggests, however, that P(p,x) is an eigenfunction of d/dx and represents a kind of unnormalized probability. This, we argue, leads to the notion of exp(ipx) because one does not want a monotonically growing or decreasing probability in x for a constant p. The “constant” is then i. Thus, we argue that if one mimics the pressure behaviour of a MB gas, and applies its properties to a single particle moving at constant speed, one obtains an unusual unnormalized kind of probability exp(ipx) i.e. the quantum free particle wavefunction. At the same time, a constant spatial probability or density 1/L exists. The two are linked in that exp(ipx) exp(-ipx) =1. The overall idea is that the stochastic force i.e. the entropic pressure in an MB gas has an analogy even for a single particle with constant momentum.
A Maxwell-Boltzmann gas with no external potential V(x) has a constant spatial density. The value of this spatial density alone does not yield the pressure because this also depends on T through the ideal gas law Pressure = density(x) RT, where T is temperature (associated with a conserved average energy), but also the average single particle pressure. Taking d/density(x) of Pressure yields the single particle average pressure value RT, yet at the same time single particle pressure is a stochastic value evaluated as the average of constant v p where p is momentum and v velocity. The quadratic form of p arises, we argue, because one has more than one particle, thus v distinguishes between the particles and relative speed becomes important. In the case of a single particle (as in a single free quantum particle) moving at constant p (speed as well) there is again a constant spatial density as in the MB gas with no V(x). Given the value of the constant density 1/L (L=length), one has no information about momentum p. In the case of a single particle, impulse and not pressure is the important quantity, we argue, because there is no need for a v as there is no relative motion. One only needs to know impulse which is proportional to p. If one speculates that a free quantum particle may follow the model of pressure in an MB gas, one would require a function F(p,x) such that d/d something F = p where p is the single particle impulse. In the MB case, “something” is density, but this is not appropriate for a single particle as density consists of many particles. Without changing T, one may change pressure in an MB gas by changing density i.e. changing the x value of particles. For a quantum particle, we suggest that the conjugate variable to p is x. One may change x of the single particle. Furthermore an association between d/dx and p exists in Fermat’s Least Time Principle. T Thus we suggest d/dx F(p,x) = p where F is unknown both in form and meaning. Pressure, however, is a statistical quantity so F should also be a statistical quantity i.e. one could have d/dx P(p,x) / P(p/x) = constant p where P(p,x) is a kind of unnormalized probability. This suggests, however, that P(p,x) is an eigenfunction of d/dx and represents a kind of unnormalized probability. This, we argue, leads to the notion of exp(ipx) because one does not want a monotonically growing or decreasing probability in x for a constant p. The “constant” is then i. Thus, we argue that if one mimics the pressure behaviour of a MB gas, and applies its properties to a single particle moving at constant speed, one obtains an unusual unnormalized kind of probability exp(ipx) i.e. the quantum free particle wavefunction. At the same time, a constant spatial probability or density 1/L exists. The two are linked in that exp(ipx) exp(-ipx) =1. The overall idea is that the stochastic force i.e. the entropic pressure in an MB gas has an analogy even for a single particle with constant momentum.
Newtonian force is based on a change in momentum in time, but momentum is a vector and so may change in magnitude and direction. In many cases, the change is of one form or the other. For example, for a bound particle the change is in kinetic energy i.e. in magnitude, while for a particle moving at constant speed in a circle or pressure in a Maxwell-Boltzmann gas, the change is in direction. For an MB gas, the probability for p and -p are the same so the average momentum at any point x is 0 and thus chunks of the gas do not move, even though a single particle in the gas moves. The pressure, on the other hand, is the average of 2pv (2p for impulse, v for flux) and so pressure for p and -p do not cancel. A quantum free particle is based on exp(ipx) which does not contain time and is a dynamical probability, This is both an eigenfunction of the momentum operator -id/dx and the kinetic energy operator -1/2m d/dx d/dx, yet as pointed out in Part II, there are two pieces (cos(px),sin(px)) with a spatial density and hence a kind of pressure/force. These, however, combine to give the sense of constant motion in one direction. We argue this is due to a dynamical probability, unlike a static balance of pressures/forces found in an MB gas or Newtonian statics. There is a density with pressure distribution cos(px), but this is continuously changing because the particle moves in one direction with constant p. Thus the two pressure distributions “balance” so to speak to create a constant p. Interference of exp(ipx) and say exp(-ipx) may “bring out” a “static” spatial density i.e. 2cos(px), but behind the scene the dynamical probabilities still exist. It is these which “hold” the cos(px) together which appears strange from a static point of view. For example, a quantum bound state has a static W*(x)W(x) spatial density at the same time that the single particle has an average classical kinetic energy KE(x)=En-V(x). Interference also gives rise to the notion of an impulse-pressure in quantum as one ignores the isin(px) and i sin(-px) terms. Creating a pressure scenario with p and -p leads to two considerations. First, there is an impulse-pressure (defined without flux) at each x point and secondly, this impulse-pressure changes with x. The time-independent Schrodinger equation is formulated in terms of energy (i.e. a scalar): KE(x)+V(x)=En where KE(x)= {-1/2m d/dx dW/dx} \ W and W=Sum over p a(p)exp(ipx). Thus, as in Newtonian mechanics, one thinks of -dV/dx solely accelerating a particle. The quantum particle however, is described by an ensemble W(x) which has pressure. This pressure is due to a combination of the form exp(ipx) and V(x), so V(x) does not simply change root mean square speed. exp(ipx), by itself, creates pressure when interference occurs as in a two slit scenario. We try to examine these ideas in this note.
For a Maxwell-Boltzmann gas (MB) it is known that T (temperature) is a statistical parameter related to the average energy. It appears in f(ei) = C exp(-ei/T) so T has the same units as ei, a single particle energy. If T increases, ei increases identically if ei/T is treated as constant. Thinking in terms of ei=pp/2m, this means the momentum spread (i.e. variance of the Gaussian increases) and one expects entropy to increase In (1) we argued that exp(-ei/T) for an MB gas and exp(ipx) for a quantum free particle (constant p), follow from the same statistical rule P(a1)P(a2)=P(a1+a2), but ei is a scalar and p a vector. Furthermore exp(ipx) is associated with spatial invariance and the generator of translations d/dx. In this note, we try to argue that a constant p may act as a statistical parameter just as T, although in this case hbar/p has the behaviour of a length i.e. the wavelength. If p increases (just like T is increased), the wavelength decreases meaning there are more peaks in a fixed length or more entropy, we argue. Thus there seems to be a statistical mechanical similarity to increasing T for an MB gas and to increasing p for a single free quantum particle. The association of x and p in exp(ipx) has implications, not only for the resolution of x space, but the increased entropy suggested by more peaks of cos(px). In addition, this means more peak places for impulses to occur. Thus a large p not only represents a large impulse, but is associated with more peak points for this impulse to be delivered and more entropy. In (2) we argued that the humps of cos(px) encode information about the impulse/force used to create p and which may be delivered if p hits an object. Thus there is a somewhat unusual relationship between p and x that one does not find in Newtonian mechanics, but has its roots in special relativity i.e. in the Lorentz invariant -Et+px. We suggest that just like exp(-V(x)/T) in an MB gas means that for a large force=-dV/dx, exp(-V/T) changes rapidly in x, a large p, associated with a large force should also imply that the humps in exp(ipx) rise (and fall) sharply, thus leading to sharp peaks. Thus sharp peaks are indicative of high internal force and increased entropy for a free particle exp(ipx), we argue.
A classical bound system may be said to have a spatial density proportional to the amount of time dt a particle spends in a constant dx region i.e. C/v(x) (1) We stress that in this scenario there is a well defined and constant dx so that “dt” may be linked to velocity v(x). We also note that there is no spatial representation of action-reaction here even though this exists in Newton’s theory. In a classical Maxwell-Boltzmann gas there are many particles which each act dynamically according to .5mvv+V(x) = E (where v and E are particular to the particle when it is not colliding). At the same time there is a spatial density which “represents” macroscopic static chunks of matter which are balanced through pressure differences (P=spatial density(x) RT) and -dV/dx density where V(x) is the potential. From the pressure equation, T is proportional to energy which in turn is proportional to RMS speed squared. RMS speed may then be linked to a dt if a fixed dx is understood. In the classical statistical case, however, spatial density is directly linked to the number of particles in a region. Time enters in the pressure expression through a term proportional to the average kinetic energy i.e. pv where p is like an impulse and v a flux. What happens in a single particle quantum bound state? In a sense it is like a classical bound state as the equation: prms(x)prms(x)/2m + V(x) = En holds for any energy level. Can one then take vrms(x), choose a fixed dx and argue that spatial density should be equal to dt or C/vrms(x)? We argue that one cannot due to the lack of defined “dx” quantity. Why is dx problematic to define? We argue that it is because quantum mechanics is based on action-reaction x-probaiblity distributions linked to impulse. For example, exp(ipx) for a free particle captures only p or impulse (not velocity) and cos(px) and sin(px) are its action-reaction regions.These lead ultimately to a density is given by the well-known W*(x)W(x) where W(x)=Sum over p a(p)exp(ipx), although kinetic energy with pp/2m = .5pv is used to find W(x) as well. . W*(x)W(x) has crests and troughs similar to cos(px) in exp(ipx), except that these now have varying heights and widths. We try to argue that even though each x point maps to a different KEave(x), the hump physically represents action-reaction in the system (just as the humps in cos(px) do except for a single p) while KE(x) represents an average of p’s. This action-reaction is linked with impulses. In general, if one takes d/dx of the time-independent Schrodinger equation one finds an average impulse term, -dV/dx and KE(x) dW/dx / W where dW/dx / W is the average impulse. At peaks, dW/dx =0 and so an average force through time*impulse term balances -dV/dx=Newtonian force. These peaks are also associated with W*(x at peak)W(x at peak) which is the spatial density. We interpret this as representing the average time the single particle spends near x at peak. This value, however, follows from ideas of impulse and not from speed as there is no well defined dx even though there is a well defined vrms(x). Thus the ratio of W*(x)W(x) at two peaks is not the inverse of the ratio of vrms(x) except apparently when the energy level n becomes very high. We investigate these ideas.
Classical Newtonian mechanics separates a particle (effect) from a force (cause). A particle has properties at a given x (and time t), but if it interacts with a force, this interaction occurs over dt and dx. Given that many forces do not change in time, we argue that a notion of a force is inseparable from the idea of dx. If one gives the value of a force at x, this value acts within a dx, which in Newtonian mechanics tends to 0. We have argued before that a particle with momentum represents a force or impulse itself, but at the same time may receive impulse hits from other sources (e.g. a Newtonian force -dV/dx). If a particle represents a force(impulse) it should be inseparable from the notion of dx, even if it has a constant p (momentum) and is traveling through space. Thus the idea of a momentum defining an internal ruler gradation through a probability distribution in x which we have presented in previous notes is compatible with the idea of describing a particle as a force. If one only considers a particle as an effect, there would be no need to worry about dx until a force(x) appears. An internal dx is not defined by an external ruler and does not need to tend to 0 as in Newtonian mechanics. Rather it is statistical in nature as p is a fixed impulse, but the probability to be at different x points within dx (called a wavelength) differ. In (1) we compared a classical Maxwell-Boltzmann gas to a quantum free particle. We argued that d/d density Pressure = RT = average pressure of a particle. We suggested this is analogous to the quantum free particle situation: (-id/dx probability(p,x) ) / probability(p,x) = p where probability(p,x) is unnormalized. In other words a change in a probability yields a deterministic value p. The notion of a deterministic value p associated with a probability distribution in what would be another deterministic variable x in Newtonian mechanics (except now one has a dx distribution because the particle is a force/impulse) has consequences. If one wishes to have a particle sit at x=0, say in an oscillator system, one needs to use a wavelength which is fairly small and centered at x=0. The wavelength, however, is associated with momentum and even being small spreads out from x=0 meaning there is potential energy .5kxx as well. Thus one may use an ensemble of forces/impulses Sum over p a(p)exp(ipx) to create a “dx” about x=0, but at the same time this ensemble represents an average particle (defined at x with prms(x)) which must satisfy KE(x) + V(x) = En i.e. the classical conservation expression. In other words, one sees again that average statistical quantities may behave as deterministic ones.
Classical statistical pressure is calculated by averaging 2pv where 2p is the momentum change related to an impulse hit Integral F dt (i.e. the particle bounces back) and v is related to flux which in turn is linked to time or the number of particles with momentum p in a certain volume which strike a surface in a given time interval. The time units in the flux term v cancel dt time units in the impulse. We have argued in previous notes that within a quantum wavelength hbar/p (from exp(ipx)) there is no internal time (although there is an external one linked with x=p/m t.). The two dimensions of exp(ipx) account for the direction of motion and so time is not needed internally. [Given that exp(ipx) contains p which in turn is linked to v, there is an external time in the picture, but not an internal one, we argue.] The quantum bound state problem is statistical, but there is only one particle and no time. Thus instead of considering an average of pv, one may consider an average of the impulse i.e. average momentum given by -2idW/dx / W where W is the wavefunction.(A bound state wavefunction is real so the i factor indicates that the momentum is not associated with the entire system translating i.e. it is a bound state.) Impulse, however, is Integral F dt in classical physics, so an average 2p treated as an average impulse contains some kind of average time term. We argue that this seems to be hbar/E in a quantum system. In order for the quantum system to mimic a Maxwell-Boltzmann gas, we argue that the force -dV/dx at x should match hbar/E * average impulse at x. There is no spatial density in W(x) (W*(x)W(x) is spatial density) so one does not need to consider a little box, as in the classical statistical mechanical case. Interestingly, there is a quantum case in which this situation arises, namely the oscillator ground state.
Historically Newtonian mechanics appeared before the statistical mechanical treatment of a Maxwell-Boltzmann (MB) gas. In this note, however, we initially investigate a MB gas in terms of information which we define as -ln(density(x)). In equilibrium there are no macroscopic changes in time so there is only one variable namely x in one dimension. The density may be changed so it is variable while a potential V(x) is not. Thus we argue that there should be a fixed piece of information based on x in the system i.e. V(x). We then argue (as in (1)) that equilibrium involves minimizing the information so we suggest: -ln(density(x)) = C V(x). This ultimately leads to the form d(x) = C1 exp(-V(x)/T). Next we focus on a Newtonian system, but unlike the MB gas two variables x and t are needed to describe the macroscopic changes for the particle even though x suffices to describe V(x). {In particular one may send in a particle with an initial speed at xo and it will move through x,t in a certain manner. Sending another particle with the same initial speed at xo at a later time leads to the same motion. Thus we argue there is a given V(x) function in the system. } We also suggest there should exist a density d(x) and an information given by -ln(d(x)). One may see from (2) that a Newtonian system may be described by a density(x) even though it is usually only described by x(t), velocity, acceleration etc. The density is linked to a variable quantity of “information” in the system. Given the presence of V(x) we divide the x axis into equally spaced dx units. Thus d/dx the translation generator applies to V(x). For the moving object, however, one has a corresponding dt(x) for each equal dx and the dt’s are usually not the same. In other words the ratio dx/dt = function(x), but it is dt which changes. We place dt in the denominator so that it mimics the time translational generator d/dt. One cannot simply equate function(x) and C V(x) because function(x) does not change as d/dx, rather it changes because dt values change. Thus we suggest using d/dt to express its changes. As a result we propose: d/dt (dx/dt) = C dV/dx ((1)) as both sides of the equation must be vectors. Now d/dt= dx/dt d/dx so both sides of ((1)) may be expressed as functions of x. One then obtains function(x) d/dx function(x) = C dV/dx where function(x) = dx/dt. We call this v(x). v(x) is adjustable so we equate it with information i.e. -ln(d(x)) where d(x) is density as there is no reason to have two different functions of x. Thus d(x)=C/v(x). In other words density(x) is proportional to the “information” dt at each x. In (2) a density is ascribed to a Newtonian system using dt as the definition of density. Here we wish to link density to information= -ln(d(x)). As a result one finds that ((1)) involves information squared unlike the MB gas which involves information -ln(d(x)) to the first power. As a result: d/dx ( v(x)v(x)) = C dV/dx which may be integrated to yield Newton’s conservation of energy: .5v(x)v(x)+ C V(x) = constant. In other words we suggest it may be possible to obtain Newton’s second law using ideas of information. There is further loss of information as the constant applies to all x points so the the information from .5v(x)v(x) combines with CV(x) to eliminate x information. (x information, however, is present in each term i.e. .5v(x)v(x) and CV(x)).