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Newton's bucket experiment is a way to tell if you are in a rotating frame. If you have a bucket of water which is spinning the water surface will form a concave shape. Likewise, if the bucket is not rotating the surface will be flat. I was wondering how this experiment is viewed in modern physics, i.e. with General Relativity.

So my question is: what happens if I have a stationary bucket and I make the universe rotate around it? For this setup start with a stationary bucket and then make the universe rotate around it by strapping a rocket on every massive object in the observable universe and having it fire such that the masses start revolving around the bucket. Then finally we have to wait until this information has had time to reach the bucket.

How will the bucket look like? Will it have flat surface or concave?

Qmechanic
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2 Answers2

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To make the problem a bit more physically reasonable, you might consider a toy model of a hollow spherical shell of mass $M$ and radius $R$ which is spinning about its polar axis with angular velocity $\omega$ - a problem which was solved to first order in $\omega$ by Brill and Cohen in 1966.

They treat the problem as a linear perturbation about the ordinary Schwarzschild solution $$\mathrm ds^2 = - V^2 \mathrm dt^2 + \psi^4(\mathrm dr^2 + r^2[\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2])$$ which is expressed in isotropic coordinates rather than the usual Schwarzschild coordinates. We will use units in which $G=c=1$. In the interior region, $V$ and $\psi$ take the form $$V = (1+\alpha/R) \qquad \psi =\frac{R-\alpha}{R+\alpha}$$ where $\alpha \equiv \frac{M}{2}$ is the Schwarzschild radius, and where we assume that $R>\alpha$ (otherwise we'd have a black hole). If we take the sphere to be rotating with angular velocity $\omega$, then the perturbed metric becomes

$$\mathrm ds^2 = -V^2 \mathrm dt^2 + \psi^4(\mathrm dr^2 + r^2[\mathrm d\theta^2 + \sin^2(\theta) [\mathrm d\phi - \color{red}{\Omega(r) \mathrm dt}]^2])$$

On the interior region, the Einstein equations yield that $\Omega(r)=\Omega_0$ is a constant given by

$$\Omega_0 \equiv \frac{\omega}{1+\left[3\frac{R-\alpha}{4M(1+\beta)}\right]} \qquad \beta \equiv \frac{\alpha}{2(R-\alpha)}$$

Because $\Omega(r)=\Omega_0$, the metric on the interior is related to the unperturbed metric by simple change of coordinates $\phi \mapsto \phi-\Omega t$.


The physical meaning of this is that for an observer sitting inside the shell, non-inertial effects (such as the water creeping up the sides of the bucket) are absent when the bucket is rotating at a rate $\Omega_0\neq 0$. The fact that $\Omega_0 \neq \omega$ is because the asymptotic spacetime (the $r\rightarrow \infty$ limit in the exterior region) is non-rotating, so there is a kind of "competition" between the asymptotically static spacetime and the dragging influence of the rotating shell. However, as the mass of the shell increases and $\alpha \rightarrow R$, we see that $\Omega_0 \rightarrow \omega$ and the "rotation" of the interior region becomes totally dependent on the rotation of the shell.

In other words, on the interior of a sufficiently massive spherical shell, the (locally) inertial frames are the ones rotating along with the mass. A stationary observer who sees the shell rotating will also observe non-inertial effects like the water climbing the walls of their bucket.

Albatross
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Inertial frames are aligned with the "fixed stars" (or better, the homogeneous and isotropic mass that models the universe). This has been shown by Weinberg as I described in this post that as of now has not received answers, unfortunately.

The alignment is direct, without time delay and without regard of (black hole or cosmological) event horizons, as per the Einstein Field Equations. In other words, the connection between inertial frames and the universe is through static gravity not dynamic gravity such as when you attached and started moving the fixed stars with your rockets. Moreover, action is minus reaction, so to move a star in one direction there will be an equal momentum in the other direction, and the net effect for the energy-momentum tensor might be zero. You would never be able to send rockets to stars beyond the cosmological event horizon that also affect the inertial frames here.

In other words, I don't believe that you can generate the Machian effect by using rockets as you described. That said, the inertial frames are linked to the universe. If the bucket rotates relative to the universe, regardless of how this happens, the level in the bucket will be concave.

HenkSB
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