0
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ More on Mach's principle. $\endgroup$
    – Qmechanic
    Aug 19 at 13:17
  • $\begingroup$ What about angular momentum conservation? If your universe was not rotating initially you cannot spin it up as a whole (with a single direction of rotation) without providing some mechanism of obtaining angular momentum. $\endgroup$
    – A.V.S.
    Aug 19 at 15:09

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Brill&Cohen really does not correspond to the scenario described by the OP (even aside from “the Universe” part), since their paper deals with a stationary solution while OP was interested in transition from static, nonrotating to stationary. Consider EM analogue: charged shell when it is spinned up would produce magnetic field inside, this magnetic would induce currents in superconductor inside the shell because the $∂B/∂t$ term would induce electric fields. So, I would expect that nonstationary effects would be essential in these scenarios also in GR just like in EM. $\endgroup$
    – A.V.S.
    Aug 20 at 10:53
  • $\begingroup$ @A.V.S. I did/do not interpret the question as being about the transition period, but rather about the steady state - essentially referring to Mach’s principle. $\endgroup$
    – J. Murray
    Aug 20 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.