We have argued in previous notes that C(T) exp(-ei/T) (the Maxwell-Boltzmann probability) and exp(-iEt) exp(ipx) (the free particle quantum probability called a wavefunction) have the same informational structure in that product probabilities for a given E1=Ei+Ej for the MB case and Ei,pi and Ej, pj for the quantum case lead to uniform distributions, i.e. ones with minimal information if energy and momentum are conserved. We suggest here that the root for this may be same. We have argued in previous notes that dp/dt = Force cannot have dt be infinitesimally small. This means that there exists a scale at which dt is not 0 or even tiny. We also suggested that the force (even though obeying action-reaction in collisions) may strike at some probabilistic time within dt. We then used special relativity to find that dt =hbar/E and dp=hbar/p. A probability which demonstrates these gradations is exp(-iEt+ipx) (and also Lorentz invariant). This is akin to the notion that p(e1)p(e2)=p(e3)p(e4) and p(p1)p(p2) = p(p3)p(p4) (one dimension for simplicity). In other words, the stochasticity introduced is one of a uniform distribution for product probabilities. One could obtain the same result by starting with this assumption and then suggesting Lorentz invariance. We suggest that this quantum uniform distribution product probability also holds for the Maxwell-Boltzmann case. There, the probability is not complex because p(ei) must be a real number linked to the number of particles with ei. Nevertheless, in an elastic collision of ei and ej, we suggest that the quantum result (and even dp/dt=F) suggest that there is stochasticity in x and t and that one cannot deterministically predict the outcome. Any outcome which conserves energy and momentum has the same likelihood. We suggest that it is the quantum mechanical assumption which is identical to the ideal gas case (which uses quantum mechanical gas molecules albeit at high temperatures). The high temperatures mean that one can only use the conservation laws and the associated stochasticity. We suggest that the stochasticity of interaction inherent in quantum mechanics carries over to collisions in the classical gas. It is not simply that one is approximating the situation in a gas by throwing away information. If one were to try to be as deterministic as possible at some scale (the quantum one) stochasticity arises and there is no full determinism in a collision. The best one can do is use conservation of momentum and sometimes conservation of energy in certain situations like high temperature. If this is the case, one may ask: Why is it possible to perform deterministic classical calculations such as particles bouncing off walls or mirrors or interacting with a potential V(x) in a completely deterministic manner. We suggest that these involve constrained interactions, i.e. the wall or V(x) is associated with a huge mass compared to the particle, unlike in a collision of two identical particles. Action-reaction in the former case is associated with the large mass and so always the same approximately 0 kinetic energy and hence no p(ei)p(ej) =p(ek)p(el), but in the latter, with a mass the same size and so the stochasticity may be seen clearly.
We have argued in previous notes that C(T) exp(-ei/T) (the Maxwell-Boltzmann probability) and exp(-iEt) exp(ipx) (the free particle quantum probability called a wavefunction) have the same informational structure in that product probabilities for a given E1=Ei+Ej for the MB case and Ei,pi and Ej, pj for the quantum case lead to uniform distributions, i.e. ones with minimal information if energy and momentum are conserved. We suggest here that the root for this may be same. We have argued in previous notes that dp/dt = Force cannot have dt be infinitesimally small. This means that there exists a scale at which dt is not 0 or even tiny. We also suggested that the force (even though obeying action-reaction in collisions) may strike at some probabilistic time within dt. We then used special relativity to find that dt =hbar/E and dp=hbar/p. A probability which demonstrates these gradations is exp(-iEt+ipx) (and also Lorentz invariant). This is akin to the notion that p(e1)p(e2)=p(e3)p(e4) and p(p1)p(p2) = p(p3)p(p4) (one dimension for simplicity). In other words, the stochasticity introduced is one of a uniform distribution for product probabilities. One could obtain the same result by starting with this assumption and then suggesting Lorentz invariance. We suggest that this quantum uniform distribution product probability also holds for the Maxwell-Boltzmann case. There, the probability is not complex because p(ei) must be a real number linked to the number of particles with ei. Nevertheless, in an elastic collision of ei and ej, we suggest that the quantum result (and even dp/dt=F) suggest that there is stochasticity in x and t and that one cannot deterministically predict the outcome. Any outcome which conserves energy and momentum has the same likelihood. We suggest that it is the quantum mechanical assumption which is identical to the ideal gas case (which uses quantum mechanical gas molecules albeit at high temperatures). The high temperatures mean that one can only use the conservation laws and the associated stochasticity. We suggest that the stochasticity of interaction inherent in quantum mechanics carries over to collisions in the classical gas. It is not simply that one is approximating the situation in a gas by throwing away information. If one were to try to be as deterministic as possible at some scale (the quantum one) stochasticity arises and there is no full determinism in a collision. The best one can do is use conservation of momentum and sometimes conservation of energy in certain situations like high temperature. If this is the case, one may ask: Why is it possible to perform deterministic classical calculations such as particles bouncing off walls or mirrors or interacting with a potential V(x) in a completely deterministic manner. We suggest that these involve constrained interactions, i.e. the wall or V(x) is associated with a huge mass compared to the particle, unlike in a collision of two identical particles. Action-reaction in the former case is associated with the large mass and so always the same approximately 0 kinetic energy and hence no p(ei)p(ej) =p(ek)p(el), but in the latter, with a mass the same size and so the stochasticity may be seen clearly.
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 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 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.
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.
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.
In previous notes, we presented a simple argument for obtaining the quantum free particle wavefunction exp(-iEt+ipx). We suggested that there are situations for which one cannot predict the outcome of an elastic collision and that given initial e1,e2 (energies) and p1,p2 (momentum vectors in 1-direction for simplicity), any ei,ej and pi,pj which conserve energy and momentum have the same product probability. Even in Newtonian elastic collisions of free particles, one sees calculations which allow any outcome which conserves momentum and energy. If the probability has the same real value for any free particle as no variable is being distributed and one wishes to impose Lorentz invariance, then exp(-iEt) and exp(ipx) represent the probabilities. This automatically gives rise to units in time and space, hbar/E and hbar/p (the hbar comes from considerations linked to a photon). One may ask, however: What is the real physics behind dx and dt? We argued in (1) that this physics is associated with force, in particular an intrinsic force. We considered a particle with rest mass mo and argued that this rest mass represents intrinsic force as it is proportional to Newton’s inertia and the force needed to set it in motion. If viewed from a moving frame, (0, mocc) maps to (E,p) with E containing the intrinsic inertial force as well as enhancements due to motion and p containing the intrinsic mo as well as the velocity vector. For a particle with rest mass, we suggest that a key idea is this notion of intrinsic force. We argued that one may consider the Lorentz invariant -Et+px = -mocc t1 and introduce dx=hbar/E, dt=hbar/p, dt1=hbar/(mocc) and dx1-> infinite as p->0. This leads to -E dt + p dx =0 or -E/dx + p /dt = 0. We then argued in (1) that p/dt is like an average force and from this we introduced exp(-iEt+ipx). Thus, dt and dx are linked with the notion of an average force. We suggest that this notion of intrinsic force is the reason behind linearizing -EE + p dot p = -momo cccc ((1)). -Edt+pdx=0 is too simple to show the geometry linked with rotational invariance etc in ((1)), yet at the same time ((1)) is not linear in the intrinsic force mo. This is the physical reason why ((1)) should be linearized via Dirac’s process, giving rise to spin ½. As a result, we postulate that a particle has an intrinsic force, in this case rest mass, which is directly linked with the probability exp(-iEt+ipx) and that one should be able to write an equation containing operators and the intrinsic force (in linear form) which also shows geometry. It is this equation which gives rise to spin. At this point, this is a postulate, i.e. speculation. It would be good if one could find an example in which these ideas arise naturally, i.e. a case in which one did not have to linearize an equation like ((1)) or make special arguments in order to obtain exp(-iEt+ipx). In that case, an equation linear in the intrinsic force would exist a priori and a direct link to exp(-iEt+ipx) would also naturally exist. We suggest that such an example does exist, namely for a photon. One may use the special relativistic equations developed for a particle with rest mass and set mo=0 to obtain E=[p] c for a photon. This is a fine equation, but unlike the rest mass case mo (proportional to the intrinsic force), there is no rest mass and no intrinsic force present for the photon. Certainly p (momentum) is linked with a force as both a particle with rest mass and a photon impart an impulse on a target, but momentum is not an intrinsic force, it is a consequential one. In the case of a rest mass, it arises because the rest mass (the intrinsic force) is seen to be moving. For a photon, one would then expect the existence of an intrinsic force which gives rise to momentum. We suggest that the electric field (together with the magnetic) represents this intrinsic force and that Maxwell’s equations allow one to directly write a linear equation in El (electric field) (no linearization required) which shows the geometry in the system. One may also explicitly show that El behaves as exp(-iEt+ipx) if one extends cos(-Et+ipx) to a more general complex number. In other words, the postulates made for the particle with rest mass mo seem to naturally arise in the case of the photon. This, we argue, is why the procedures for finding spin for a spin ½ particle with rest mass mo and a photon are so different even though both a photon and rest mass particle satisfy special relativity and are associated with exp(-iEt+ipx). It is the difference in the intrinsic force which is the key to finding spin, we speculate.