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arXiv:0904.3180 · Old and New Concepts of Physics

Packet Spreading and Einstein Retardation

M. I. Shirokov

Published Oct 1, 2009• 1 citation
arXivHTMLPDF

Abstract

According to the classical special theory of relativity any nonstationary system moving with velocity $v$ must evolve (e.g., decay) $1/γ$ times slower than the system at rest, $γ=(1-v^2)^{-1/2}$ (the Einstein retardation ER). Quantum mechanics allows one to calculate the evolution of both systems separately and to compare them thus verifying ER. It is shown here that ER is not valid for a simple system: the spreading packet of the free spinless particle. Earlier it was shown that ER does not hold for some other systems. So one may state that ER is not a universal kinematic law in quantum mechanics.

DOI
10.2478/v10005-009-0012-3

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A Lorentz boost which then invalidates exp(ip), would only apply to the total rest mass of the bound particle system which has no overall momentum. It seems that one does apply the Lorentz transformation to the interaction of exp(ipx)’s with V(x). One is only interested in the overall bound system as a single entity for special relativity. As a result, one may only impose momentum considerations as V(x) is a spatial function and one must link it with exp(ipx). In fact, one may write V(x) = Sum over k Vk exp(ikx) to associate it with momentum hits linked with Vk energies, which already moves away from the idea of special relativity as Vk and k are not linked as a 4-vector. The idea of exp(ip1x)exp(i k1x) = exp(ip2x)exp(ik2x), however, seems to still hold (as it does for free particles), linking these two values (i.e. Vk1 and Vk2) with (p1+k)(p1+k) exp(i (p1+k)x). Lorentz invariance for a particle with rest mass seems to be linked with the idea that one ultimately has a rest frame for the particle. One then establishes a relation between the rest mass in one frame and its new E and p as seen by another frame. One does not, however, need to impose Lorentz invariance conditions on the interaction which creates the rest mass it seems. For example, exp(-iEt+ipx) is the Lorentz form used to describe a free particle, but exp(ipx) is the only portion used to describe its interaction with V(x) in the rest frame of the large mass creating V(x). Furthermore, exp(-iEt+ipx) ensures conservation of energy and p for a free particle, but in an interaction case, that does not happen as the potential may absorb energy.

•Dec 8, 2025•0 citations
According to the classical special theory of relativity any nonstationary system moving with velocity vvv must evolve (e.g., decay) 1/γ1/γ1/γ times slower than the system at rest, γ=(1−v2)−1/2γ=(1-v^2)^{-1/2}γ=(1−v2)−1/2 (the Einstein retardation ER). Quantum mechanics allows one to calculate the evolution of both systems separately and to compare them thus verifying ER. It is shown here that ER is not valid for a simple system: the spreading packet of the free spinless particle. Earlier it was shown that ER does not hold for some other systems. So one may state that ER is not a universal kinematic law in quantum mechanics.
According to the classical special theory of relativity any nonstationary system moving with velocity v must evolve (e.g., decay) 1/γ times slower than the system at rest, γ = (1 − v 2) −1/2 (the Einstein retardation ER). Quantum mechanics allows one to calculate the evolution of both systems separately and to compare them thus verifying ER. It is shown here that ER is not valid for a simple system: the spreading packet of the free spinless particle. Earlier it was shown that ER does not hold for some other systems. So one may state Experiments show that moving unstable particles (e.g., µ-mesons, π-mesons) decay slower than the particles at rest. More explicitly, let N0(t) = N exp(−t/τ0) be the decay law of the particles at rest, τ0 being life-time. Then the life time of particles moving with velocity v is τv = τ0γ, γ = (1 −v 2) −1/2, and the decay
Relativistic quantum theory shows that the known Einstein time dilation (ED) approximately holds for the decay law of the unstable particle having definite momentum p (DP). I use a different definition of the moving particle as the state with definite velocity v (DV). It is shown that in this case the decay law is not dilated. On the contrary, it is contracted as compared with the decay law of the particle at rest. It is demonstrated that ED fails in both DP and DV cases for time evolution of the simple unstable system of the kind of oscillating neutrino. Experiments are known which show that ED holds for mesons. The used theory may explain the fact by supposing that the measured mesons are in DP state.
We study the decay law for a moving unstable particle. The usual time-dilatation formula states that the decay width for an unstable state moving with a momentum ppp and mass MMM is Γ~p=ΓM/p2+M2\tildeΓ_{p}=ΓM/\sqrt{p^{2}+M^{2}}Γ~p​=ΓM/p2+M2​ with ΓΓΓ being the decay width in the rest frame. In agreement with previous studies, we show that in the context of QM as well as QFT this equation is \textit{not} correct provided that the quantum measurement is performed in a reference frame in which the unstable particle has momentum ppp (note, a momentum eigenstate is \textit{not} a velocity eigenstate in QM). We then give, to our knowledge for the first time, an analytic expression of an improved formula and we show that the deviation from Γ~p\tildeΓ_{p}Γ~p​ has a maximum for p/M=2/3,p/M=\sqrt{2/3},p/M=2/3​, but is typically \textit{very} small. Then, the result can be easily generalized to a momentum wave packet and also to an arbitrary initial state. Here, we give a very general expression of the non-decay probability. As a next step, we show that care is needed when one makes a boost of an unstable state with zero momentum/velocity: namely, the resulting state has zero overlap with the elements of the basis of unstable states (it is already decayed!). However, when considering a spread in velocity, one finds again that Γ~p\tilde{Γ}_{p}Γ~p​ is typically a very good approximation. The study of a velocity wave-packet represents an interesting subject which constitutes one of the main outcomes of the present manuscript.\textbf{ }In the end, it should be stressed that there is no whatsoever breaking of special relativity, but as usual in QM, one should specify which kind of measurement on which kind of state is performed.
We discuss the relation between the quantum-mechanical survival probability of an unstable system in motion and that of the system at rest. The usual definition of the survival probability which takes into account only the time evolution of an unstable system leads to a relation between the survival probability of the system in motion and that of the system at rest which is different from the standard relation based on relativistic time dilation. This approach led other authors to claim non-standard quantum-mechanical effects which are in clear contradiction with Special Relativity. We show that an appropriate relativistic definition of the survival probability which takes into account also the space evolution of an unstable system leads to the standard relation between the survival probability of the system in motion and that of the system at rest, in agreement with Special Relativity. We present a rigorous derivation of this result based on a wave packet treatment.
Special relativity seems to be associated with linking physical relations as seen in one frame with those as seen in a frame moving at a speed relative to such a frame. A constant speed is often used for the moving frame. This suggests that classically there is a special frame for a particle with rest mass called a rest frame. In such a case, one simply introduces a parameter for rest mass and does not worry about the complicated motion and interactions which actually create the rest mass. It seems that even quantum mechanics, which considers a particle to be moving as one does not know its location, one uses the idea of a potential V(x) which in principle is associated with a very large rest mass, presumably at rest as no variables of such a rest mass appear in the Klein-Gordon or Dirac equations. It seems that one uses a preferred frame when solving relativistic quantum mechanical equations. Furthermore, given V(x) and a bound state with exp(-i E bound t), E applies to the overall bound system, not to a possible exp(ipx) constituent. The special relativistic invariant equation seems to exist at the system level, like with the rest mass and one may apparently calculate E in the special rest frame of the large mass and hence find a rest mass of the bound system, i..e large rest mass*cc + mo cc + E, mo is the rest mass of the single bound particle. As a result of the use of the special rest mass frame for the V(x) producing large rest mass, it seems that one does not have to worry about special relativistic constraints on the interaction with V(x), as only one frame is involved. In other words, a bound calculation is about finding a new rest mass for an overall new system. Then this system may be used in Lorentz boosts. One simplifies by working in the rest frame of the mass creating V(x). Perhaps surprisingly then, we note that exp(-iEt+ipx) for a free particle is Lorentz invariant and in fact must be Lorentz invariant for the following reason. Imagine that one wishes to create a probability for free particles which is linked with conservation of momentum. Any p must have the same real value weight because for free particles there is no bias for one p or the other. There is, however, a bias for a p1 and p2. It should have the same probability as a p3 and p4, if p1+p2 = p3+p4. (We use one dimension for simplicity.) As a result, exp(ip) seems to be a probability which respects these conditions. The problem is that one may have p=m1v1=m2v2 hence (E1,p) and (E2,p). With respect to momentum, it is irrelevant if one particle has E1 and the other E2, but not if one does a Lorentz boost. Then, in one case: p→ g(v) p + g(v) v E1 and in the second, p→ g(v) p + g(v)vE2. Here g(v) =1/sqrt(1-vv/cc). In other words, p is the same for each particle in one frame and not in the other which violates conservation of momentum because in frame 1 both particles can complete the same conservation of momentum equation, but in the second frame this is not the case, because each has a different momentum. Thus, exp(ip) cannot hold. One requires a linear form in p which is Lorentz invariant, i.e. exp(-iEt+ipx), we argue (and have done so in previous notes). We see now that momentum probability is linked with space in exp(ipx). The idea seems to be that for free colliding particles there is no preferred frame and the probabilities used must reflect this Lorentz invariance. As a result, we argue that a free particle probability in E,p arises because of Lorentz invariance, but then apply it to a V(x) problem and no longer worry about Lorentz invariance at the interaction level. V(x) with no associated rest mass variables assumes a rest frame for the V(x) producing mass. A Lorentz boost which then invalidates exp(ip), would only apply to the total rest mass of the bound particle system which has no overall momentum. It seems that one does apply the Lorentz transformation to the interaction of exp(ipx)’s with V(x). One is only interested in the overall bound system as a single entity for special relativity. As a result, one may only impose momentum considerations as V(x) is a spatial function and one must link it with exp(ipx). In fact, one may write V(x) = Sum over k Vk exp(ikx) to associate it with momentum hits linked with Vk energies, which already moves away from the idea of special relativity as Vk and k are not linked as a 4-vector. The idea of exp(ip1x)exp(i k1x) = exp(ip2x)exp(ik2x), however, seems to still hold (as it does for free particles), linking these two values (i.e. Vk1 and Vk2) with (p1+k)(p1+k) exp(i (p1+k)x). Lorentz invariance for a particle with rest mass seems to be linked with the idea that one ultimately has a rest frame for the particle. One then establishes a relation between the rest mass in one frame and its new E and p as seen by another frame. One does not, however, need to impose Lorentz invariance conditions on the interaction which creates the rest mass it seems. For example, exp(-iEt+ipx) is the Lorentz form used to describe a free particle, but exp(ipx) is the only portion used to describe its interaction with V(x) in the rest frame of the large mass creating V(x). Furthermore, exp(-iEt+ipx) ensures conservation of energy and p for a free particle, but in an interaction case, that does not happen as the potential may absorb energy.