There is no such thing as continuous space time and therefore continuous motion in a trajectory. Those are classic concepts not realized in the quantum world. Essentially consider this; The notion of a particle that exists out there traveling in a well defined trajectory x=f(t), with nobody there to observe it but truly generating an objective reality in that idealization of a path that is a well known as function of time, doesnt exist.
To exist you must be observed/interacted with.
The uncertainty principle secures that you cant know the location and speed of a particle (momentum) with infinite precision which is exactly what is required by claiming you know x=f(t) (momentum is related to the perfectly known derivative etc) .
Big systems are collection of small particles that are quantum systems and for which such trajectories dont exist but only as classical limits. The big systems are quantum systems as well but they are examined from a macroscopic perspective that ignores the fine structure eg of their edges.
To me there is absolutely no problem at all. There is nothing there to worry about. As you split distances and observation segments you will at some point arrive at the inconsistency of the classical picture.
For example you can imagine both particles as wavepackets with some different speeds (here v1 and v2 separated by a distace of d units) starting as gaussian wavepackets.
Again i suggest seeing the link on wavepackets that involves such a solution for an original gaussian wavepacket at t=0 that enjoys the ideal uncertainty relation (ie the equality of the general inequality Δx*Δpx >=ħ/2)
http://en.wikipedia.org/wiki/Wave_packet
Such wavepacket is the solution of the Schrodinger equation and the square of its amplitude, when you solve the equation in time with the initial condition that it starts as a Gaussian wavepacket corresponding to the initial positions (eg at 0 with speed v) , is the probability density to find the particle in some location x,t.
It comes out as (in some simple units selected) ;
(which is not properly normalized here yet, also they use k0 for the speed v )
Notice that its still more likely in that picture to find the particle in the classic trajectory x=v*t. But it is not a given that you will find it exactly there. Its experimental position is a random variable obeying a normal distribution now.
An electron for example that has speed 0.1c has momentum uncertainty if you demand to localize it within one micron that corresponds to a speed uncertainty of 50m/s. You cant know both with as much accuracy as you like.
If you consider 2 wavepackets now, 2 ahead of 1 by d and with different speeds v2, v1 the probability densities of their locations as function of time will be; (it remains a Normal curve but the width and average evolve with time and the width in particular spreads with time now)
f1(x,t)=(2/Pi)^(1/2)*1/(1+4t^2)^(1/2)*Exp(-2*(x-v1*t)^2/(1+4*t^2))
f2(x,t)=(2/Pi)^(1/2)*1/(1+4t^2)^(1/2)*Exp(-2*(x-d-v2*t)^2/(1+4t^2))
For the above time evolved wavepacket assumption.
One can always ask what is the chance 1 (the faster) is ahead of 2 if we measured them and that will be given by a combined probability density P(x1>x2) same as P(x1-x2>0). Since both f1,f2 are normal and while spreading traveling remain normal with widening widths (standard deviations) the difference is also normal with the average the difference of averages ie the value d-(v1-v2)*t (their classical distance in time) and a new sd since both have common sd just 2^(1/2)*the sd of each above which was 1/2*(1+4t^2)^(1/2) for a general time t.
As they approach it will start becoming obvious as the 2 densities start to overlap that well in advance of the classical trajectories v1*t and v2*t+d meeting that the chance that 1 is ahead of 2 is nonzero and significant now (which was close to 0 initially in that picture of wave-packets for all practical purposes).
As a result you never have to get to any infinitesimal small distance between the 2 with the conviction than 2 is still ahead. Well in advanced of some extraordinarily small divisions of distances you will have arrived in finite time at situations where which one is ahead is a cloudy statement already answered by the new normal;
f21(x=x2-x1,t)=1/(Pi)^(1/2)*1/(1+4*t^2)^(1/2)*Exp(-(x-(d-(v1-v2)*t))^2/(1+4*t^2))
The following charts show the 2 wavepackets at t=0 and the distribution of their difference of locations by any experiment (clearly a bell curve around 10 as expected for an initial distance of 10 units).
And now the chart when 1 sec has passed and classically object 2 is still ahead (at 11 and the fast at 10 where it was before ) but any experiment now can go either way with significant probability to claim 1 is ahead already.
Notice the area on the left of x=0 below the curve is the chance now that 2 is already ahead in measurements even if classically 1 is still ahead but they are close enough now for the ambiguity of the quantum nature of the particles to come in play.
As expected at 1.11111... =d/(v1-v2)=10/(10-1) (the classical meeting point in time) the distribution is 50-50 for who is ahead and the 2 wavepackets are perfectly identical.
(In this example i assumed both to be traveling in 2 parallel lines to avoid any interaction between the 2 particles that will happen in real life if collinear)
Last edited by masque de Z; 05-10-2015 at 10:16 PM.