Yes and no. On the one hand, relativistic time dilation is path independent, but I do not believe it is necessarily quantum spin independent, because quantum spin / entanglement itself is at once the fastest process (not a velocity) in the universe, and also the basis of time. The twin paradox works as well for a long ride on the rim of a relativistic merry-go-round (or more accurately, a centrifuge), as it does with orbiting spacecraft. A relativistic merry-go-round is much larger than atomic structure, but it is true that even radioactive decay of unstable atoms would proceed more slowly on the rim of that relativistic merry-go-round than they would at its center. It could be argued that average v=0 on the relativistic merry-go-round as well, or for a plane with an atomic clock on board, like the experiment my former mentor Carroll O. Alley performed. His experiment was an early confirmation of relativity, well before we had GPS satellites using it every day. I would argue the case for v=0 differently for atoms only because the relativistic merry-go-round is usually not treated as a wave function. Bound energy that is moving <c or at rest persists in time because of entanglement. Unbound energy propagating at c persists in time because of entanglement also. Energy is conserved because both forms of energy persist in time and are related by E=mc^2 and entanglement. What is really needed here is an equivalent of the Schrödinger wave equation for bound energy. Hint: it won't be propagating at c, and this is the REAL reason that relativity does not apply to atoms the same way it does to a relativistic merry-go-round. After a radius of appropriate length, it appears to an observer on the rim a good deal longer for the merry-go-round to complete one revolution than it does to an observer at its center. Even the stationary estimation of the constant pi is different than it appears to an observer moving with the rim. The meter sticks closest to the observer appear unaffected, of course, but looking at meter sticks on the far side of the outer rim, half of them are Lorentz compressed because a component of v is in the opposite direction of relative motion. This is the case with the Milky Way galaxy as a relativistic merry-go-round as well. The stars we view 180,000 light years away at the most distant rim have already completed a greater amount of rotation than we are observing now, and also they are experiencing a much greater amount of time dilation than the stars, nebulae and black holes nearer the center. The stars on the opposite side of the Milky way also will appear to be much closer together (more densely packed) on the far rim than they actually are. Atomic structure may not be a relativistic merry-go-round, and I don't think it would be a "structure" in any sense of the word without entanglement. Entanglement inteacts most strongly with the basis of time and direction of propagation here, or else atomic structure itself would not be possible. It's not all about electric charge, because electrons, if they are moving, still exist in free space. What force or mechanism holds them together? This was a very, very good follow-up question exchemist.