# Lorentz contraction in the Quantum realm

1. Feb 27, 2013

### fermi

Special Relativity states that a ruler flying by at near the speed of light will appear to be shortened as observed by a stationary observer in the lab frame. This is, of course, exactly what happens to macroscopic objects. The question is how far can we push this into microscopic realm without contradicting with Quantum effects? So let us replace the ruler by a single gold atom (or better yet by a single proton). I carefully avoid elementary particles here which manage to act point like. I am considering microscopic objects which have finite size and which have structure underneath. So a gold atom or a proton both fit the bill.

The classical proof that a ruler appears shrunk requires that we can distinguish the right edge of the ruler (say in the direction of motion) from the left edge of the ruler. The rest is easy by the application of Lorentz transformations to show that it appears shrunk. But I suspect that there is trouble here when I replace the ruler by a proton. At rest, the proton is an extended object that is spherically symmetric (apart from its spin.) Classical physics predicts that a proton appears very much like a ‘pancake’ in the Lab frame due to Lorentz contraction. However, there is one big difference between a proton and a macroscopic ruler: the right and the left edges of a proton cannot be distinguished. Classically, every particle can be distinguished from every other. By extension, the left and right edges of a sphere can also be distinguished even if the sphere is very small. This is not true in the Nature (in Quantum mechanics.) There is no such thing as the right edge of a proton. In order to see the right edge of a proton you need a gamma ray of which wavelength is a fraction of the proton radius, say 10^(-14) cm, or less. This photon will have an energy larger than 10 GeV, more than an order of magnitude larger than the proton rest mass. If the photon interacts with the proton, it will create a shower of particles after which the meaning of the right edge of the proton will have no meaning at all.

So does proton appear shrunk even though we cannot distinguish the right edge of the ruler (proton) from the left edge? Can I still trust the classical derivation of Lorentz contraction?

2. Mar 1, 2013

3. Mar 3, 2013

### tom.stoer

The next logical step is relativistic quantum mechanics, i.e. Dirac- and Klein-Gordon-equation. A solution, i.e. a wave function ψ in frame S is related to a solution ψ' in frame S'. For the Dirac eq. the relation is difficult due to spin, but for a scalar particle the relation is due to Lorentz-transformation only, i.e. ψ'(x') = ψ(Lx) so all what one has to do is to Lorentz-transform the coordinates x to x' (of course x means x,y,z,t).

That means that via a Lorentz-transformation in a certain direction the solution ψ' in frame S' is Lorentz-contracted w.r.t. solution ψ in frame S.