Does the equivalence principle hold for rotational motion?

Question

This question made me think about linear vs. rotational motion. The general view expressed in the answers is that rotational motion is not the same as linear motion.

It is claimed that in a rotational motion absolute accelerations are present, but if we make the connection with a linearly accelerated motion, can't we say then that also rotational motion is relative?

Let me explain. I accelerate in empty space, and experience a force pushing me to the floor. If I have no knowledge of the rockets pushing me, I can say I am at rest in a uniform gravitational field. All stars and galaxies fall freely in my frame. I can also say I'm accelerated in empty space. In both cases there is force, and thus acceleration.

For a sphere, can't we apply the same argument? I can say I am at rest in my restframe and the stars and galaxies move in a gravitational around me. But I could claim just as well I rotate and the stars are at rest. In this case no rocket thrusters are needed to accelerate me, as in the linear case. But I'm still accelerated (without increase in kinetic energy, because thruster absence.

So, can we say rotational motion is equivalent to accelerated linear motion? According to the equivalence principle, a linearly accelerated motion is equivalent to a uniform gravitational field. Is a rotational motion equivalent to a gravitational field just the same?

Edit In other words. In the linear acceleration case I can say there is an infinite massive plane causing gravity and I'm stationary in the field. The stars fall freely.
In the same vain, I can think there is a spherical shell of mass causing a gravitational field.

Answer 1

The centripetally accelerated frame is similar to the linearly accelerated frame in terms of spacetime geometry but your idea about gravitationally rotating stars is wrong - or at least so pointlessly difficult to work with as to have no practical application. That is: in terms of clock synchronization, it doesn't matter what direction you accelerate, but the only model of the universe in which the stars can free-fall in a circle at $v\gg c$ is one in which the entirety of spacetime is rotating in a circle except for the observer. That's just another way of saying that the observer is the one that is rotating with infinitely harder math.

A linearly accelerated observer can invent an improbably large, distant, and ancient mass with respect to which her rocket is keeping her stationary, and apply the laws of physics consistently everywhere assuming the existence of that mass. But a centripetally accelerated observer must either admit that she is rotating or define laws of physics that change based on where and when you are in spacetime.

Answer 2

A rotational motion can also simulate an artificial gravity. But one difference with respect to linear accelerated motion is the presence of the Coriolis fictitious acceleration.

In a spaceship in the form of a rotating cylindrical shell with radius R, any object of mass $m$ inside is under a fictitious centrifugal force $F = m\omega^2R$. If $\omega$ and $R$ are such that $\omega^2R = g$, it could simulate the Earth gravitational acceleration without the need of burning fuel.

Suppose a person let an object fall from the hands. For an external inertial frame, this person's feet move in an arc: $l_f = \omega Rt$. The object travels in a straigth line for the inertial frame, and its length should be (expanding the $sin(\omega t)$ in Taylor series)$$l_o' = Rsin(\omega t) \implies l_o' \approx \omega Rt(1 - \frac{1}{6}\omega^2 t^2)$$ to fall at the new position of the person's feet.

But the object initial (and constant) velocity for the inertial frame for a height $h$ is $$l_0 = \omega(R-h)t = \omega Rt(1 - \frac{h}{R})$$

For a perfect simulation of gravity, $t$ should be such that: $$h = \frac{1}{2}gt^2 = \frac{1}{2}\omega^2Rt^2 \implies l_o = \omega Rt(1 - \frac{1}{2}\omega^2t^2)$$

So, $l_0 < l_0'$, what means that for the expected time of the parabolic motion the object has not touched the ground yet. And when there is contact, it is a little bit behind the person' feet. But the difference can be made very small if the artificial gravity is simulated with small $\omega$ and big $R$.

About the celestial bodies rotating around, I think that the notion of gravity loses its meaning. All of them have the same angular velocity, no matter its size or distance. The situation is like in the old geocentric model, modelled as a kind of clock, a mechanism.

Answer 3

In this question, it does depend on exactly what you consider the equivalence principle to say. There are many statements of the equivalence principle, as nicely summarized here: https://en.m.wikipedia.org/wiki/Equivalence_principle

Most of those statements don’t really apply to your question. In fact, there is only one that I can see which has any bearing on it:

The local effects of motion in a curved spacetime (gravitation) are indistinguishable from those of an accelerated observer in flat spacetime, without exception.

If you interpret “accelerated observer” as including rotating observers then it would seem that this formulation does include rotational motion. But if you interpret “accelerated observer” as not including rotating observers then it would not.

I tend to think that the choice of interpretation is not terribly important in this case. I think your more interesting question is this one

It is claimed that in a rotational motion absolute accelerations are present, but if we make the connection with a linearly accelerated motion, can't we say then that also rotational motion is relative?

The equivalence principle has nothing to do with this. Both linear acceleration and rotational motion are “absolute” (meaning invariant) completely independently of the equivalence principle. This is because both can be detected directly and locally using common experimental devices like accelerometers and gyroscopes. If you are in a small closed laboratory you can determine your invariant acceleration with an accelerometer and your invariant rotation with a gyroscope, but you cannot determine your velocity with any device.

They are thus not relative like velocity in a direct experimental sense, regardless of the equivalence principle.