Mach's principle vs two counter-rotating buckets

[Roberto Pavani]

TL;DR

Two buckets spin in opposite directions at distance $R > c/\omega$. Both show curved water surfaces. From A's frame, B would need to orbit at $v > c$: impossible.
So A's rotation cannot be reduced to relative motion of B. Doesn't this rule out Mach's principle using only special relativity?

Watching a YouTube video about Newton's rotating bucket vs Mach's principle, a simple variant came to mind that I haven't seen discussed:

Take two buckets spinning in opposite directions, separated by a distance $R > c/\omega$. Both show curved water surfaces (centrifugal effect).
Now, from A's rest frame, B would have to "orbit" A at $v = \omega R > c$, which is impossible.

So A's rotation cannot be interpreted as "the rest of the universe rotating around me" , special relativity forbids it.

This seems to rule out Mach's principle without needing general relativity, distant stars, or any gravitational effect ($G$ plays no role here). Am I missing something, or is this argument already known?

 

[anuttarasammyak]

Now, from A's rest frame, B would have to "orbit" A at v=ωR>c, which is impossible.

We on planet Earth observe that the stars have rotation orbit of 24 hours period. $\omega R > c$ for R > 4 * 10^12 m　～　30 a.u. Is it OK in your point ?

 

[Roberto Pavani]

Good question. The key difference is:
When we observe stars rotating from Earth, we know it's a coordinate effect — we can detect Earth's rotation locally (Foucault pendulum, Sagnac effect). No physical object is actually moving at $v > c$.

But Mach's principle claims that "A rotates" and "everything orbits A" are physically equivalent, not merely a coordinate choice.
If they were truly equivalent, then from A's frame (where A is "not rotating"), B's water should be curved but A's should be flat.Yet both are curved.
The observed behaviour contradicts what Mach's equivalence would predict.

So in this scenario, Mach's principle cannot hold.

 

[renormalize]

So in this scenario, Mach's principle cannot hold.

I don't see the validity of your argument. Consider this quote about Mach's principle from S. Weinberg, Gravitation and Cosmology, pg. 17: "... if Mach is right, then the acceleration given a particle by a given force ought to depend not only on the presence of the fixed stars but also, very slightly, on the distribution of matter in the immediate vicinity of the particle." (Emphasis added.) The effect on the first rotating bucket, of the relatively small mass of the nearby second bucket, is sensibly negligible in comparison to the cumulative effect of the mass of the whole universe. In other words, the acceleration experienced by each bucket is due to each rotating individually with respect the distant stars, according to Mach's principle; the only difference is that one rotates, say, clockwise, while the other rotates counter-clockwise.

 

[Roberto Pavani]

Thank you, but I think the argument is being missed. The point is not about the gravitational effect of one bucket on the other; it's about the logical consistency of Mach's equivalence.
If rotation is purely relative (strong Mach), then "A rotates w.r.t. the stars" and "the stars rotate w.r.t. A" must be physically equivalent.
But in the second description, the distant stars orbit A at $v = \omega R_{\text{stars}} \gg c$, which is physically impossible.

Your answer: "each bucket rotates individually w.r.t. the distant stars", already concedes that rotation w.r.t. the stars is an absolute fact, not a relative one.
That's precisely the conclusion: rotation is absolute, Mach's equivalence fails.

Moreover, the argument holds even with $G = 0$: no gravitational interaction, no frame dragging, no distant-star influence, yet both buckets still show curved surfaces. The centrifugal effect does not depend on any coupling to external matter.

To be clear: this is not an argument against GR,
GR itself is not fully Machian (de Sitter space, Gödel's universe).
The point is simply that rotation is locally detectable and cannot be reduced to relative motion, even in principle.

 

[renormalize]

If rotation is purely relative (strong Mach), then "A rotates w.r.t. the stars" and "the stars rotate w.r.t. A" must be physically equivalent.
But in the second description, the distant stars orbit A at $v = \omega R_{\text{stars}} \gg c$, which is physically impossible.

Can you cite a physics reference that defines "strong Mach" in this way? I ask because, to my understanding, Mach's principle applies only to particle motion relative to the preferred frame in which the distant universe is at rest.

 

[gmax137]

If rotation is purely relative (strong Mach), then "A rotates w.r.t. the stars" and "the stars rotate w.r.t. A" must be physically equivalent.

I don't think I've seen that before, where does it come from?

 

[Roberto Pavani]

I don't think I've seen that before, where does it come from?

It comes directly from the meaning of "relative." If rotation is purely relative (as Mach claims), then "A rotates w.r.t. the universe" and "the universe rotates w.r.t. A" must be physically equivalent, just as "A moves at velocity $v$ w.r.t. B" and "B moves at $-v$ w.r.t. A" are equivalent in special relativity for inertial motion.

This is not an exotic formulation. It's discussed in Bondi & Samuel, Phys. Lett. A 228 (1997) 121, and in Barbour & Pfister, Mach's Principle: From Newton's Bucket to Quantum Gravity (Birkhäuser, 1995), which catalogues multiple inequivalent versions.

If you deny this equivalence; i.e., if you say "A rotates w.r.t. the stars" is physically different from "the stars rotate w.r.t. A", then you are already conceding that rotation is absolute, which is my conclusion (Newton's statement).

 

[renormalize]

If you deny this equivalence; i.e., if you say "A rotates w.r.t. the stars" is physically different from "the stars rotate w.r.t. A", then you are already conceding that rotation is absolute, which is my conclusion (Newton's statement).

Nope. Your Bondi & Samuel reference enumerates 10 versions of Mach's principle (some of which are contradictory!) and discusses their applicability to the theories of Newtonian absolute space (N) and Einstein in asymptotically-flat (EA) and cosmological (EC) universes. Their "Mach3" is most relevant to the rotating "bucket experiment":

So based on Mach3, your statement that "rotation is absolute, which is my conclusion (Newton's statement)" is simply wrong.

 

[Roberto Pavani]

Mach3 states: "The local inertial frame is completely determined by the matter distribution of the universe."

This is precisely what my argument challenges. With $G \to 0$, there is no mechanism by which the matter distribution can influence the local inertial frame; yet the centrifugal effect persists unchanged. The water is on the walls regardless of $G$ (using closed barrels instead of buckets).

If the inertial frame were "completely determined" by the matter distribution (Mach3), then removing the coupling ($G \to 0$) should eliminate inertial effects. It doesn't. Therefore Mach3 is falsified by this thought experiment.

As for "rotation is absolute" being "simply wrong", could you clarify what determines the local inertial frame in the $G \to 0$ limit, if not the fact that rotation can always be detected locally?

 

[renormalize]

If the inertial frame were "completely determined" by the matter distribution (Mach3), then removing the coupling ($G \to 0$) should eliminate inertial effects. It doesn't. Therefore Mach3 is falsified by this thought experiment.

Water climbing the walls of a bucket rotating relative to the fixed stars is an empirical result observed in our universe where $G\neq 0$. How can you claim to know by mere thought the result of the same experiment conducted in some hypothetical alternative universe where $G=0\,$? Is this a personal theory?

 

[Roberto Pavani]

No personal theory needed, just Newton's second law.
The centrifugal effect in a rotating frame is $F = m\omega^2 r$.
This expression does not contain $G$. The water climbs the walls because of $\omega$, not because of the gravitational coupling to distant stars.
Or better when using closed barrels the water stays close to the barrel walls if any $\omega \neq 0$ .
Only the pressure will change.

This is not a hypothetical alternative universe, it's a statement about which parameters enter the formula.
If $G$ does not appear in the expression for the centrifugal force, then the centrifugal force cannot depend on $G$, regardless of its value.

 

[renormalize]

The centrifugal effect in a rotating frame is $F = m\omega^2 r$.
This expression does not contain $G$.
...
This is not a hypothetical alternative universe, it's a statement about which parameters enter the formula.
If $G$ does not appear in the expression for the centrifugal force, then the centrifugal force cannot depend on $G$, regardless of its value.

That argument is not persuasive. How do you know that the inertial mass $m$ doesn't depend on the gravitational constant $G$ such that $m\rightarrow 0$ as $G\rightarrow 0\,$, so that the force $F$ vanishes in that limit?