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Two particles[or micro-observers] A and B are in relative uniform motion wrt each other in the x-x’ direction. The “observer” A decides to deduce[or interpret] the Lorentz Transformations wrt to B. Accurate knowledge of the position of B makes its[that of B] momentum highly uncertain.[$\Delta x \Delta p\ge h/2\pi$]

How does the observer A go about his job? Rather how do we interpret the Lorentz Transformation in the micro-world?

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The simultaneous existence of position coordinate and velocity at the said point is a mental picture of the classical physicist. In the process of measurement there is some interaction involving accelerations, derivatives of acceleration etc. The frames themselves may not have a uniform relative velocity between themselves, violating the basic conditions required for the Lorentz Transformation.But the "mental picture" of the classical physicist is perhaps not an incorrect one. The process of interaction may produce a complicated result but in a consistent manner –  Anamitra Palit Aug 3 '12 at 6:47
    
Photons used for viewing may produce upsetting accelerations etc between the frames themselves.We simply do not have the Lorentz transformation now. The picture is quite consistent. –  Anamitra Palit Aug 3 '12 at 6:51
    
The Posting in the following link[especially the comments] is relevant to the issue at hand:physics.stackexchange.com/questions/32385/…: The System and the Measuring Gadget –  Anamitra Palit Aug 3 '12 at 7:03
    
If one particle is to "view" another one it has to receive signals from the other one[which may emit or reflect signals]Receiving and emitting of signals is likely to cause relative accelerations or even no zero derivatives of the relative acceleration. –  Anamitra Palit Aug 3 '12 at 7:26
    
Interestingly in a micro system comprising only two particles, the signal received by one is emitted by the other.The emitting particle "knows" the type of impact its signal will have on the other one. It possibly does not have to wait for any measurement on the second one even in in the highest type of non-local situation[provided we have only two particles and their own signals] –  Anamitra Palit Aug 3 '12 at 7:32

2 Answers 2

Lorentz transformation is used to recalculate some characteristics of a system from one reference frame to another. The reference frame itself is a coordinate system used to specify positions of events (including time).

Usual way to select a reference frame is to say that some specific particle rests at the origin in terms of that frame. Such reference frame is said to be bound to the particle.

In quantum mechanics one can not say that some particle rests somewhere precisely but one can say that it is true at the average. In this case the reference frame bound to the particle is the frame where the average velocity and coordinates of the particle are zero.

The reference frame must always have determined coordinates and velocity. In quantum mechanics this is achieved by using average values.

The observer A in the question uses the reference frame where the average coordinates of the particle A are constant and equal to zero. The reference frame bound to B has the velocity equal to the average velocity of the particle B w.r.t. the average position of A.

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Accurate answer involves Dirac theory of electron which has been discovered when he tried to make Schroedinger equation Lorenz invariant. I recommend you to read corresponding paragraph in any [good] textbook. If you try to merge non-relativistic quantum mechanics and relativity, you may discover a lot of paradoxes just due to the fact that "classical" QM is not Lorenz invariant.

However, it seems in your question you just mix up uncertainty of the value with value itself. In principle, $\Delta p$ should not transform in accordance with Lorenz. And why would it?

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The Klein Gordon and the Dirac equations are Lorentz covariant.But the Schrdinger eqn is not Lorentz covariant.The non-relativistic form of the uncertainty relation goes with the Schrodinger equation..Would you like to get an up updated/upgraded version of the uncertainty principle to keep up with the Klein Gordon or the Dirac equations?Is that your suggestion to remove the difficulties in deriving the Lorentz transformations—the difficulties indicated in the questions? –  Anamitra Palit Feb 24 '12 at 18:37
    
"However, it seems in your question you just mix up uncertainty of the value with value itself. In principle, Δp should not transform in accordance with Lorenz. And why would it?"-----If the position coordinate is accurately measured the value of $p$ and hence $v$ can be any thing. Given that ,how would you derive the Lorentz Transformations?You could think of transforming $p$ or even $\Delta p$. Why not?But again the uncertainty relation will cause problems to you.You have to upgrade the thing or sacrifice it. –  Anamitra Palit Feb 24 '12 at 18:43
    
Yes, you totally miss the point. Momentum is connected with energy. Energy is involved in uncetrainty relations with time. Measurement of momentum assumes that first you synchronize your clock. Hoops... To follow these dependencies accurately you should do something better than handwaving. –  Misha Feb 24 '12 at 20:10
    
Lorentz Transformations require simultaneous knowledge of the position coordinate and velocity.If an upgraded version of the uncertainty principle is considered it should include concepts like simultaneous measurement of position coordinate and velocity so that the Lorentz transformations become consistent with the upgraded uncertainty principle.Such concepts are obviously incorrect! –  Anamitra Palit Feb 25 '12 at 10:58
    
You answer suggests an up-gradation or a sacrifice of the uncertainty principle-----at least I have made this interpretation [of your answer]and this interpretation is present in my previous comments.Unfortunately such upgradation[or sacrifice] will not work as indicated in my last comment.QM is a successful theory with the uncertainty principle [as we know it] as a basic principle –  Anamitra Palit Feb 25 '12 at 11:11

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