The Lorentz Transformations and the Uncertainty Principleby Anamitra Tags: lorentz, principle, transformations, uncertainty 

#1
Mar1311, 09:01 AM

P: 621

Two observers A and B are in relative motion with a constant velocity[for example, along the xx' direction].If A knows the the position of B accurately , the motion of B gets enormously uncertain[and vice verse] in his calculations/considerations.How is he going to derive the Lorentz transformations[or what is he supposed to understand of the Lorentz Transformations] in view of the uncertainty principle, I mean in the statistical sense?
[The observers may be assumed to be particles or microsystems] 



#2
Mar1811, 07:31 AM

P: 232

Since your dealing with just particles which are either just points or strings I am not sure how much the Lorentz transformation would apply. Don't get me wrong you would still have time dilation and mass increase. But if the Lorentz transformation did apply then if you knew about it, and could measure it, then that would mean that you would become very uncertain about the particles momentum.




#3
Mar1811, 12:15 PM

P: 205

To do the Lorentz transformation you take each individual purefrequency Fourier wave and do the transformation on the [w, k] four vector and on the [x, t] fourvector. You also need to do another transformation because the wavefunction is a spinor rather than a scalar, but it will be easier for you if we oversimplify. The difficulty that you feel is due to the mystical treatment of the Uncertainty Principle. It should be treated as nothing more than a Fourier phenomenon involving the spread of a function in frequency space behaving roughly inversely to its spread in position space...but unfortunately it is still treated in the crude mystical form from the primitive days of Quantum Physics. 



#4
Mar2011, 01:00 AM

P: 621

The Lorentz Transformations and the Uncertainty Principle
A pair of inertial frames S and S’ are considered. They are in motion along the xx’ direction with a relative speed v. The particle is at rest at the origin of the primed frame.The wave function is denoted by:
[tex]{\delta}{(}{x}^{'}{)}[/tex] The position of the particle does not change in the primed frame. [We shall denote px by p in the subsequent calculations] We carry out a double Fourier Transformation on the above function [tex]{F}{(}{\delta}{x}^{'}{)}{=}{\int }{\int}{\delta}{(}{x}^{'}{)}{exp}{(}{}{i}{(}{p}^{'}{x}^{'}{}{E}^{'}{t}^{'}{)}{)}{dx}^{'}{dt}^{'}[/tex] Or, [tex]{F}{(}{\delta}{(}{x}^{'}{)}{)}{=}{\delta}{(}{E }^{'}{)}[/tex] [tex]{\psi}^{'}{(}{p}^{'}{,}{E}^{'}{)}{=}{\delta}{(}{E }^{'}{)}[/tex]  (1) The primed psi on the left is the transformed psi function in terms of momentum and energy. We have delta functions clustered around E'=0 and stretching all along the paxis. On transforming to the unprimed frame, the RHS of eqation(1) works out to [tex]{\delta}{\gamma}{(}{E }{}{pv}{)}{=}{1}{/}{\gamma}{\delta}{(}{E}{}{pv}{)}[/tex] The psifunction in the unprimed frame is given by: [tex]{\psi}{(}{x}{,}{t}{)}{=}{\frac{1}{{2}{\pi}}{\int}{\int}{\delta}{(}{\gam ma}{(}{E}{}{pv}{)}{)}{exp}{i}{(}{px}{}{Et}{)}{dp}{dE}[/tex] [tex]{=}{\frac{1}{{2}{\pi}{\gamma}}{\int}{exp}{(}{px}{}{pvt}{)}{dp}[/tex] [tex]{=}{\frac{1}{{2}{\pi}{\gamma}}{\delta}{(}{x}{}{vt}{)}[/tex]  (2) The above result is a consistent one , since x=vt describes the motion of the particle [at the origin of the primed frame] wrt the origin of the unprimed frame. But if v is taken as some function of p the problem will stay.. Incidentally “p” can take on any value as indicated in the second statement after eqn(1). But E should be localized as indicated by eqn(1). One may think of different values of p subject to this condition[for high speed particles,the kinetic energy E=pc and for low speed ones E=(p^2)/2m] That should impose some restriction on the uncertainty principle itselfthe maximum extent to which we may consider the spreading of p.We cannot use arbitrary large values for delta_p. If one uses the uncertainty principle in an unrestricted form there could be some trouble.[only if v is constant we may take gamma outside the integral in eqn(2) The second point to mention is that I have not taken into account the time evolution of the initial delta function. One may express this as a linear combination of the energy eigenfuntions[eigenfunctions of the Hamiltonian of the free particle] and then consider the time evolution of each energy eigen function. If the delta function changes to the extent of getting delocalized the “particle” will not represent a particle at the future instants. 


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