Stupid question: given General Relativity, and in particular that all reference frames are equally valid , including rotating and accelerating ones, in what sense is heliocentrism more 'true' than geocentrism? E.g. you can pick a geocentric reference frame and all the math works out (albeit you'd need to define large fictitious forces, etc., but again, they are only 'fictitious' from the perspective of a different re…
Of all the answers offered here, the only ones that get to the heart of the matter are those that make the distinction between inertial (non-accelerating) frames of reference, and those that rotate (and therefore have at least centripetal acceleration). In Newtonian and post-Newtonian mechanics, only inertial frames are equivalent. A geocentric frame (with a rotating Earth) spins the whole universe once a year; fix t…
Addtionally, you can have locally inertial frames of reference that are inequivalent. Consider two space stations orbiting Earth at different altitudes. The occupants are in free-fall in both cases, but there is a relative acceleration between them.
Rotating frames of reference tend to distract users into confusing coordinate artifacts with physical ones. As a trivial example, suppose we vary the Minkowski metric by introducing a rotation into Cartesian coordinates: dS^2 = -\left(1 - \omega^2 ({x}^2 + {y}^2) \right){dt}^2 + 2 \omega (- y dx dt + x dy dt) + {dx}^2 + {dy}^2 + {dz}^2. Here a non-massless object far from the origin will have |d\vec{x}/dt| > 1, or in other words a coordinate velocity greater than c.
This is essentially the same as your last sentence.
However, when we calculate the four-velocity \eta^\mu of any non-massless object, even far from the origin, we will see that dS^2 We can certainly do Standard Model physics (or classical particle dynamics) in a rotating system of coordinates, using a number of approaches.
However, it is usually easier to use a different system of coordinates, and even better to abandon the idea of using any global system of coordinates in a spacetime that is not Riemann flat.