Mach's principle says that it is impossible to tell if something is accelerating unless there is something else in the universe to compare that motion to, which seems reasonable. However, if you had one detector in the universe, you seem to be able to tell if it is accelerating because an accelerating detector would record radiation where a non-accelerating detector would not, due to the Unruh effect. So, my question is, does the Unruh effect provide a way to tell if something is accelerating, even if it is the only thing in otherwise "empty" space, thereby violating Mach's principle? (At least Mach's principle in its form stated above.)

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    $\begingroup$ A meta-comment: there are lots of much more practical and much less exotic effects that violate Mach's principle. For example, rotation is absolute in GR; you can detect whether your lab is rotating by using a gyroscope or the Sagnac effect. Many people seem to get the impression that Mach's principle holds in GR, which is just plain false. $\endgroup$
    – user4552
    Aug 12, 2011 at 13:16

3 Answers 3


Yes, the Unruh effect violates Mach's principle.

In effective quantum field theory, the Unruh effect arises because the ground state of the Hamiltonian $H_a$ naturally associated with an accelerating observer is different than the ground state of the Hamiltonian $H_s$ generating time translations in a "static" frame. The static and accelerating frames have different time coordinates $t$, and because the Hamiltonian generates infinitesimal changes of $t$, it is different. And different operators have different eigenstates, including the lowest-eigenvalue one (the ground state).

One vacuum may be viewed as a squeezed/coherent state built upon the other, which may be interpreted as particle production, and creation operators have to be accordingly mixed with the annihilation operators according to the other frame - by the so-called Bogoliubov transformation.

Although Mach's principle has never been properly defined or turned into a realistic theory, it seems pretty clear that according to all of its interpretations, it postulated that it was impossible to have two different notions of a "vacuum" depending on the acceleration of the observer. However, that's exactly the case because of the Unruh effect.

An acceleration-dependent definition of the ground state (the vacuum), as realized in the derivation of the Unruh effect, exactly means that the acceleration may be distinguished from no acceleration even in the vacuum - which is exactly what Mach's principle would like to prohibit.

The Unruh effect therefore violates Mach's principle. More precisely, it refutes it because Mach's principle is wrong and the Unruh effect is real. However, one doesn't really have to go to the quantum theory to see that Mach's principle has been showed incorrect by subsequent developments. The existence of a dynamical metric tensor - even in the classical, non-quantum theory - is enough. The gravitational waves are the simplest classical entities that show that the gravitational field is real even in the vacuum, and it allows the particles moving through it to distinguish free fall from accelerating motion.

  • $\begingroup$ IMO there are some oversimplifications in this answer. (1) Mach's principle in the absolute sense has been dead since ca. 1920. All interesting research on the issue since then has been on formulating and testing theories that are somewhat less Machian (like GR) or somewhat more Machian (like Brans-Dicke gravity). Therefore answering that Mach's principle fails to be absolutely true amounts to dodging the question. (2) The quantum-mechanical part of your answer seems to assume a flat background. Of course, if we know that all of spacetime is flat and we know that GR is true, then [...] $\endgroup$
    – user4552
    Aug 12, 2011 at 13:35
  • $\begingroup$ [continued] Mach's principle is trivially false. In an empty GR cosmology, I can determine the proper acceleration of my laboratory simply by hanging a weight from a spring scale. No need for anything exotic and impractical like Unruh radiation. But in a more Machian theory like B-D gravity, Minkowski space is a pathological example; generic spacetimes in B-D gravity (containing matter) always have Big Bang singularities, because otherwise the scalar field diverges. In such a context, does the Unruh radiation agree with the spring scale, or disagree? I don't think the answer is trivial. $\endgroup$
    – user4552
    Aug 12, 2011 at 13:44
  • $\begingroup$ "it seems pretty clear that according to all of its interpretations, it postulated that it was impossible to have two different notions of a "vacuum" depending on the acceleration of the observer" - wrong. What it says is that whether you are accelerating or not depends on other bodies. In empty space there can be no acceleration and no Unruh effect. Unruh effect arises due to influence of distant bodies. $\endgroup$
    – Anixx
    Sep 9, 2012 at 18:38

Luboš's Answer is perfectly fine, +1, but whether the Unruh effect violates Mach's Principle depends on whether one takes the vacuum state to be something or nothing. One can consider measurement of one's acceleration by taking advantage of the Unruh effect to be possible just because it is relative to the vacuum state.

The vacuum state is a somewhat awkward object for 19th Century Physics, in that it does not allow us to measure velocity relative to it, but that it does allow measurement of acceleration relative to it can be said to make it something. The something that it is has a Poincaré invariant description, not a Galilean invariant description. This can be regarded as just what one has to do if one wants to save Mach's Principle. One doesn't have to save Mach's Principle, one can just let it go, but one can save it.

The vacuum state can easily be taken to be something rather than nothing if one takes the view that there are quantum fluctuations, which are distinct from thermal fluctuations because of their different symmetry properties. It is possible to do a lot of Physics while maintaining that there are no quantum fluctuations, essentially considering the vacuum to be nothing, but it is not essential that we take this point of view, we can use either or both points of view (we can choose to look at a table with one eye or with both, right?).

When one moves to General Relativity, which requires a manifestly generally covariant description, we do not have an adequate theory of the vacuum state, so we have yet to see the fate of Mach's Principle in that conceptual background.

  • $\begingroup$ * ... whether the Unruh effect violates Mach's Principle depends on whether one takes the vacuum state to be something or nothing ... * my thoughts exactly. +1 $\endgroup$
    – user346
    May 1, 2011 at 11:58
  • $\begingroup$ there is certain beauty in this argument $\endgroup$
    – lurscher
    May 1, 2011 at 13:08
  • $\begingroup$ Dear Peter, you're surely right that you may circumvent the contradiction of Mach's principle with the Unruh effect - or with anything - if you declare that what looks like the vacuum isn't really empty and has objects in it. But in that case, Mach's principle becomes totally vacuous. You might also say that there is always absolute space inside the vacuum, and it's another object, so the acceleration may be measured with respect to this object. Clearly, Mach et al. didn't want to be vacuous. What they meant is that what looks like a vacuum has to be empty - and it's not true. $\endgroup$ May 2, 2011 at 7:12
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    $\begingroup$ @Luboš: "What they meant is that what looks like a vacuum has to be empty - and it's not true." I don't think this is tenable. Mach's principle was extremely poorly defined in the 19th century, and one of the reasons was that Mach never provided clear answers to obvious questions like what qualifies as "looking like" a vacuum. E.g., it's not clear whether an empty Newtonian universe permeated with a uniform electric field would qualify. (There is no matter, and no preferred frame, unless you really prefer a frame with zero magnetic field.) [...] $\endgroup$
    – user4552
    Aug 12, 2011 at 13:53
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    $\begingroup$ [continued] The whole discussion of Mach's principle becomes interesting if and only if you look at it in a modern context (post-Brans-Dicke). $\endgroup$
    – user4552
    Aug 12, 2011 at 13:53

The Unruh effect does not violate Mach's principle, although it seems to superficially. Nothing violates Mach's principle, in my opinion, because Mach's principle is just true. Appropriately stated, it is one classical interpretation of the holographic principle.

I should qualify this by saying that Minkowski space, considered as a solution of GR, or any other asymptotically flat solution, violates Mach's principle if it is stated naively. This is unavoidable, because the scale-invariance of classical GR implies that you can scale up a small region of any global solution to be flat, and then even if the rotation of objects is relative to far-away matter in the original solution, teensy objects in the scaled up solution are rotating relative to matter infinitely far away, so they end up rotating relative to the boundary conditions on the metric at infinity.

To make Mach's principle reasonable, one must be clear as to what constitutes "matter" which one can be rotating relative to. "Matter" does not mean that the vacuum equations are not satisfied, because then black holes wouldn't be matter. If you extend that to extremal horizons, and you believe string theory is correct so that electrons, quarks, are little weak-dual quantum versions of extremal black-holes, then Mach's principle is stating that all motion is relative to a distant horizons.

This principle is true and it is predictive. If you have just one cosmological horizon and no other horizon (no other matter), say a completely empty deSitter space, then you cannot set it rotating. There is nothing it can rotate relative to. If you have a black hole inside a de-Sitter horizon (a dS-Schwartschild solution), you can set the black hole rotating, but only at the cost of making the de-Sitter horizon rotate in the opposite direction. The Gibbons/Page/Pope solution (hep-th/0404008) has the property--- the number of rotation parameters is equal to the number of rotation parameters of a single object.

It is easy to object "but the cosmological horizon isn't matter!", but this is specious. If you make the deSitter black hole area bigger and bigger, it becomes symmetric with the cosmological horizon exactly when they have equal areas (nothing but the usual coordinates go bad at this point), and afterwards switches places with the cosmological horizon (this process violates entropy if you try to actually do it, but imagine two little black holes at opposite ends Einstein static universe racing one against the other to become the cosmological horizon. Whichever one becomes biggest, wins. Did one of them stop being "matter" once it won?)

So Mach's principle, properly stated, tells you that rotation/acceleration is relative to a distant horizon. There is an Olbers paradox objection to Mach's principle--- there is not enough glowing matter to determine the local metric, otherwise every line of sight would end on matter. But this objection is specious. Every line of sight in our universe which doesn't end on matter ends on the cosmological horizon, so it also ends on "horizon matter". The holographic principle tells you that all motion should be thought of as projected on the holographic screen, and this is just the statement that any angular motion in the interior of the universe has a visible counterpart in the motion of the constituent holographic counterpart on the screen.

In Unruh's radiation, when you accelerate, there is a gigantic acceleration horizon at a finite distance from you--- a black wall stretching from one end of the universe to the other. This horizon is what you are moving relative to. If you slowly stop moving, this horizon recedes to infinity, but the asymptotically flat solutions all violate Mach's principle anyway. If you are accelerating in de-Sitter, what happens is just that the accelerating observer has a different point of view regarding what the de-Sitter horizon is. If we had a good holographic description of deSitter space, then two observers which are in communication should be able to map their holographic descriptions back and forth, much as the infalling observer and the outside observer in a black hole can do.

While this does require some extrapolation of new physics, I believe it is safe to say that Mach's principle is just a not-quite-right forshadowing of the holographic principle, and is therefore correct.

  • $\begingroup$ "This principle is true and it is predictive. If you have just one cosmological horizon, say a deSitter space, then you cannot set it rotating. There is nothing it can rotate relative to." How do you deal with the example of a perturbed FRW solution that rotates, as in this paper? adsabs.harvard.edu/full/1985MNRAS.213..917B $\endgroup$
    – user4552
    Aug 12, 2011 at 13:59
  • $\begingroup$ FRW can rotate if the cosmological horizon rotates the other way. This is also true in deSitter space--- if you have a black hole in deSitter, it can rotate. It's just empty de-Sitter that can't rotate. To be clear about what "the other way" means (it isn't obvious in curved space), I mean that if you take a Nariai limit of dS-Schwartzschild (BH as large as possible), then the space-time is a cylinder (dS2 cross S2) and there is a meaningful symmetry between the two S2 horizons at the edge of the dS2. If you rotate one horizon slightly, the other rotates badkwards the same amount. $\endgroup$
    – Ron Maimon
    Aug 12, 2011 at 14:33
  • $\begingroup$ I edited the answer to make it clear the deSitter space is empty. $\endgroup$
    – Ron Maimon
    Aug 12, 2011 at 14:42
  • $\begingroup$ This would imply that the higher apparent temperature of the cosmological horizon, the weaker inertia in the space should be. $\endgroup$
    – Anixx
    Sep 9, 2012 at 13:09
  • $\begingroup$ @Anixx: I don't understand why it implies this--- higher temperature horizons are closer. It is just saying that classical GR already has holography, and we call it 'Mach'. $\endgroup$
    – Ron Maimon
    Sep 9, 2012 at 15:44

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