cesiumfrog said:
No, if you spatially translate your observation by one (light-)year, you will certainly not see the explosion.
I agree about that, but I regard this fact as mere technical annoyance rather than something fundamental.
cesiumfrog said:
but all of my existing experience supports the presumption that changing my velocity does not change what else is real.
I also agree that nobody has seen directly any dynamical effect of boosts. So, apparently these effects are rather weak (at accessible velocities). However, there is an argument, which, in my opinion, demonstrates conclusively that these effects should be real.
In relativistic quantum theory we must construct an unitary representation of the Poincare group U_g in the Hilbert space of observed system. This representation tells us how to connect operators of observables in two different reference frames
F' = U_g F U_g^{-1}...(1)
In particular, when g is time translation
U_g = \exp(\frac{i}{\hbar} Ht)
where
H = H_0 +V .....(2)
is the interacting Hamiltonian. Then eq. (1) tells us how observable F evolves in time. The fact that the Hamiltonian contains interaction V explains non-trivial interacting dynamics, such as reactions, decays of apples, etc. Without this interaction term the dynamics of particles would be trivial and boring (all particles move with uniform velocities along straight lines independent on the presence of other particles)
When g is boost with rapidity \vec{\theta}, then
U_g = \exp(\frac{i}{\hbar} \mathbf{K} \vec{\theta})
and eq. (1) tells us how observable F is seen from different moving frames of reference. In particular, eq. (1) should describe for us Lorentz transformations (for time-position, energy-momentum, etc.). It is not difficult to demonstrate that if the operator of boost is non-interacting
\mathbf{K} = \mathbf{K}_0... (3)
then eq. (1) would lead to usual Lorentz transformation formulas. These formulas would be linear and universal (i.e., independent on the type of observed physical system and interactions acting there).
However, if the boost operator contains interaction term
\mathbf{K} = \mathbf{K}_0 + \mathbf{W}...(4)
then boost transformations of observables computed by formula (1) become non-trivial and interaction-dependent.
The punchline is this: It is known from the theory of Poincare group representations (You can find this discussion, for example, in Weinberg's "The quantum theory of fields", vol. 1)that if the Hamiltonian contains interaction dependence (2), then the boost operator cannot remain interaction-independent as in (3). It must have interaction terms, like in (4). So, necessarily, boost transformations of observables must depend on interactions, i.e., they must be "dynamical".
Eugene.