Since Einsteins GR tells us all the frames of reference are equal, is there anything invalid about treating the Earth as unmoving and the universe itself rotating?

Other than the fact that the mathematical model is much more complex, is there anything that is wrong with it?

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    $\begingroup$ GR doesn't state that all frames are equivalent, but only that they're equivalent locally, at a point. For example, the acceleration at a point of a rotating circle is equivalent to gravity from outside. However, the acceleration at a different point of the circle is equivalent to gravity from a different direction. There is no such a source of gravity, even hypothetical, that would pull the circle apart in all directions, because gravity inside a heavy cylinder is zero (same as why no field in a charged conductive sphere). Thus the answer is no, rotation is not relative in GR, it is absolute. $\endgroup$ – safesphere Jan 2 '18 at 9:21

It's an oversimplification to say that GR treats all frames if reference as equal. In particular if we take any frame (strictly speaking any coordinate system) we can calculate the proper acceleration of an observer at rest in the frame and the result will be zero or non-zero depending on the frame. If the proper acceleration is zero then the frame we have chosen is locally equivalent to an inertial frame, while if the proper acceleration is non-zero the frame is locally equivalent to a non-inertial frame.

So for example if we take the frame of an observer at rest on the surface of the Earth this frame is locally non-inertial. That means freely moving objects won't move in straight lines i.e. if you throw a stone it will move in a curve (approximately a parabola) and a pendulum will rotate its plane of swing by $2\pi$ every $24$ hours. From a Newtonian perspective there will be fictitious forces acting.

Even in Newtonian mechanics there's nothing wrong with using non-inertial frames. They are just (as you say) more complicated to do calculations in. Likewise in GR there's absolutely no problem in choosing a frame corotating with the Earth, but it will make any attempts to do calculations more complicated than they need be. For example in this frame the flat spacetime metric:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$


$$ ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2 $$

  • $\begingroup$ To be clear: Whether using a non-inertial frame makes computations harder is highly dependent on the calculation. Some aspect of the calculation clearly will become harder, others might become easier if you choose the "right" non-inertial frame. For example, if you want to work out the relativistic corrections to a GPS localization, this is a lot easier if you use a frame where the GPS satellites, ground stations, and receiver are all standing still. $\endgroup$ – mmeent Jan 4 '18 at 15:12

Yes! Every coordinate system in GR is "treated as equal". But the quantities you measure, by means of "clocks and rods" (of course with much more complicated devices), are independent on your choice of coordinates. Like the metric itself; indeed the metric $g_{\mu\nu}$ is NOT measurable, since is an object which depends on coordinates (as in your case the coordinates by which the earth is not rotating). But the $\textit{space-time interval}$ $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ is instead measurable since it is a scalar. Now, this is an over-semplification, since what I just wrote holds only locally, but there are several spacetimes in which "local" means "global" since I can cover the entire space-time manifold with just one chart (system of coordinates).


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