How to calibrate an accelerometer to measure absolute acceleration?

Question

I just watched a rather unreliable (or so I think) Physics YouTuber who asserted that absolute acceleration is impossible to measure, because it is impossible to ever calibrate an accelerometer accordingly, and I'm rather stumped by this argument.

My understanding for all these years has been that acceleration can be absolute, and it can be measured easily with an accelerometer, but the YouTuber asserts that all accelerometers that are calibrated are done so in a frame that may or may not be accelerating, and to "correct" for this acceleration means having knowledge of an "absolute" inertial frame, which is defined using the accelerometer in the first place, thereby making the logic circular and invalid.

Is he correct? Or is there a hole in his explanation? Or is this just an engineering problem, with a fairly easy fix in pure physics? What am I missing?

Answer 1

I think the key is you need to zero external forces on the accelerometer (not counting gravity). So if you put the accelerometer in deep space and calibrated it to read zero acceleration then it would be absolutely calibrated. Alternatively, you could drop it in free fall on Earth (ideally in a vacuum chamber so there is no air resistance) and calibrate it to be zero during the free fall.

Both techniques would absolutely calibrate the accelerometer. There would be systematic uncertainty in the calibration related to how well you are able to zero out external forces such as, e.g. air resistance or winds of particles.

Accelerometers calibrated while sitting on a table on Earth do have an absolute calibration problem. The accelerometer on the table is feeling an external normal force equal to its weight on the table. The problem is, $g$ varies across the surface of the Earth, so the accelerometer's calibration is spoiled if you move it somewhere else on Earth.

Answer 2

The video quotes a 1914 essay of Einstein's called "On the Relativity Problem" (Zum Relativitätsproblem), an English translation of which can be found here. On page 7 he says

One would try in vain to explain what it is that one should understand by the pure and simple acceleration of a body. One would succeed only in defining the relative accelerations of bodies with respect to each other. But on the other hand, we base our mechanics on the assumption that a force (cause) is necessary for creating an acceleration of a body, ignoring the fact that we are unable to explain what it is that we are to understand by "acceleration," precisely because only relative accelerations can be an object of perception.

(This is a different translation from the one used in the video, which I can't find online.)

When he wrote that, Einstein was enamored of what he called Mach's principle, which is roughly the idea that all sorts of motion, not just inertial motion, should be relative. He argued that just as special relativity satisfies a "special principle of relativity" applying only to inertial motion, general relativity satisfies a "general principle of relativity" that realizes Mach's principle.

But it just isn't true. Nongravitational acceleration is not relative in general relativity, nor, as far as we can tell, in the real world. There is no geometrical symmetry that relates different magnitudes of acceleration. You can calibrate an accelerometer without any outside reference. For example, the acceleration is zero if and only if light sent from one side of the accelerometer to the other is not Doppler shifted, and the Doppler shift can be measured with cesium clocks, or using the Mössbauer effect.

I think that Einstein backed off from this position later in his life. Whether he recanted it or not, the quoted text about the relativity of acceleration is wrong.

Answer 3

This isn't circular reasoning. If it were, then all calibration would be circular reasoning. How calibration works is that you have method1 of finding a value, you then use that to set a reference value, and then you use method2 to compare new values to that reference value. The argument "if you have method1, then you wouldn't need method2" fails because method2 is simpler/easier/cheaper/faster than method1.

For instance, a meter used to be defined as being the length of a physical rod. You could make a new meter stick by comparing it to the official reference rod (method1), and then you could use the new meter stick to measure distances (method2). Method2 was useful because it's not like the official meter rod is available every time someone wants to measure a distance. The official definition of a second is in terms of cesium atoms, but most watches don't have cesium atoms in them. They have some other measurement system that has been calibrated with reference to the official definition.

It's quite valid to have some method of defining a reference acceleration of "zero", then calibrating accelerometers to that, and then measuring how much acceleration differs from that. One could argue as to whether there is in fact some absolute "zero" acceleration, but circularity is not a counterargument against it.

Answer 4

Summary in the front:

1. All observed accelerations are relative acceleration.
2. Force-free state is conventional.
3. Universality of 'free'fall gives a natural choice to force-free convention.
4. Relativity provides the same as 3.

We shall all agree on that measurements of position and velocity are always relative. Therefore as a direct consquence of mathematical definition of acceleration $\vec{a}:=d\vec{v}/dt$, measurements of acceleration are relative too.

However, we think there's a special type of acceleration that is measured relative to inertial frames: objects that are force-free. Indeed, we have an intuition of mechanics that forces are real which means we could in principle tell if there is any force at all -- except the poor fictitious forces :-(.

With this intuition you may think that we thought we could make an object to be force-free by putting it into a vacuum box. IMO this means our intuitive notion of force is strongly related to contact force and we believe we could tell if there is any thing contacting the object under consideration -- what do you mean by isolating an object? Well, if you are smart enough you could realize that it is our best theory of physics that tells us an object is isolated -- it's not a bare truth! (Just like Jagerber48 mentioned. )

Even realizing this is not enough. How exactly does our best theory produce a value of a force?

Kinematically, we could multiply object's acceleration relative to force-free(!) object by its inertial mass or using change in momentum if you prefer:

$$\vec{f}=m\vec{a},\,\vec{a}\,relative\,to\,inertial\,frame.$$

Dynamically, we could use the law of the specific force acting on the object such as:

$$\vec{f}=q(\vec{E}+\vec{v}\times\vec{B}).$$

But where does the law come from in the first place? -- By kinematical research relative to force-free(!) object. See the epistemological circular?

Acccumulation mentioned that the ZERO of acceleration is just like zero of many other quantities, a convention. My point here is that the zero of force is also a convention!

Seriously, you must define a state of objects in your lab to be state with zero force, i.e. a reference state, then you can meaningfully talk about value of force by measuring the deviation from the pre-defined reference state.

A problem naturally arises: how to unify different definition of reference states in different labs? Or ask in this way: under which definition we could restore the old good Newtonian mechanics?

The natural choice stems from universality of freefall. Now the (dynamical) prefix 'free' in freefall haven't got a concrete meaning. What actually matters is the kinematical feature, universality (not necessarily be 'falling'), which means any object regardless of its intrinsic structure (well, we still need the 'small enough' condition) would have exactly the same motion $\vec{x}(t)$ providing same initial condition (position and velocity that have been independently defined) .

If (by metaphysical? aesthetical? consideration) we assume a force must have different effects for at least some different objects, then since all objects experience same effect (if there is any effect) we should define freefall motion as force-free motion. Now the prefix 'free' in freefall has a concrete and fondamental meaning in physics.

You may notice that this formulation leads to the famous (or infamous) statement: Gravity is not a force. A price to take for escaping logical circular: no pre-general-relativity theory of gravity. (Well, you may only take those universal motions that are along surface of planets or remote in space to be reference state to avoid this. )

You see that I attempt to totally discard the notion of a convention-independent absolute acceleration while there's a natural choice of convention for zero of force -- freefall.

IMO, this formulation can be seemlessly integrated into relativity. Indeed, in relativity we could define a seemingly absolute $4-acceleration$ which encodes the deviation of object's worldline from geodesics. But the reason is we inherited the convention of force in Newtonian mechanics. You could say the geodesics are special from mathematical POV, but to attach it a physically special meaning we need more.

The same logic as above applys here: we need to pick up a kind of worldlines as reference worldline first, only then to research and formulate the law of four-dimensional forces. As Heisenberg said:

What we observe is not nature itself, but nature exposed to our method of questioning.

It turns out, geodesic convention is the same as freefall convention. In the end, you could still ask: Why are there universal motions: freefall? That is a story for another day (another century? ).

Answer 5

Acceleration is not absolute, it is measured in some reference frame and may be different in a second reference frame if this second frame is accelerating in the first.

Example: A passenger on a train has zero acceleration in the train's frame of reference, but may be accelerating in the train station's frame (together with the train). In the accelerating frame, however, you may feel a pseudo-force (pressing the passenger into her seat, for example) - which may give you a way of identifying accelerating systems. As long as you can distinguish the force from a gravitational one, cf. the equivalence principle. Perhaps this is what the youtuber was referring to?

Answer 6

Arguably$^{§}$ the (magnitude of) "absolute acceleration" of some specific participant $A$, at some specific event $\varepsilon_{A \Xi}$ (which is identified by $A$ having met and passed another identifiable participant $\Xi$ at this particular event), $| \mathbf a_A[ \, \Xi \, ] |$, is proportional to the inverse of the radius of curvature of the (timelike) worldline $\mathcal W_A$ traced by participant $A$ at this event,

$$| \mathbf a_A[ \, \Xi \, ] | \sim 1 \, / \, \rho[ \, \mathcal W_A, \varepsilon_{A \Xi} \, ].$$

Referring to a flat region (whose events have their geometric relations between each other expressable as spacetime interval values) this radius of curvature can be expressed in terms of spacetime interval values:

$$\normalsize \rho[ \, \mathcal W_A, \varepsilon_{A \Xi} \, ] \! := \! \! \left( \! \! \! \overset{{\huge \text{lim}}}{\small \overset{}{\begin{matrix}\varepsilon_{A \Psi} \in \mathcal P_A[ \, \Xi \, ], \cr \varepsilon_{A \Phi} \in \mathcal F_A[ \, \Xi \, ] : \cr s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] \rightarrow 0 \end{matrix}}} \! \! \! \left[ \, \frac{\begin{matrix} \, \cr -{\, \, \, \overset{}{| \, s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Xi} \, ] \, s^2[ \, \varepsilon_{A \Xi}, \varepsilon_{A \Phi} \, ] \, s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] \, | \, \, \, \, }} \cr \, \cr \, \end{matrix}}{ \begin{vmatrix} 0 & 1 & 1 & 1 \cr 1 & 0 & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Xi} \, ] & s^2[ \, \varepsilon_{A \Xi}, \varepsilon_{A \Phi} \, ] \cr 1 & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Xi} \, ] & 0 & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] \cr 1 & s^2[ \, \varepsilon_{A \Xi}, \varepsilon_{A \Phi} \, ] & s^2[ \, \varepsilon_{A \Psi}, \varepsilon_{A \Phi} \, ] & 0 \end{vmatrix}} \, \right] \right)^{\! \! \! \left(\frac{1}{2}\right)}$$

where $\mathcal P_A[ \, \Xi \, ] := \{ \, \varepsilon_{A \Psi} \in \mathcal W_A : \varepsilon_{A \Psi} \ll \varepsilon_{A \Xi} \, \}$ denotes the subset of $A$'s world line consisting of events which chronologically precede event $\varepsilon_{A \Xi}$;
and $\mathcal F_A[ \, \Xi \, ] := \{ \, \varepsilon_{A \Phi} \in \mathcal W_A : \varepsilon_{A \Xi} \ll \varepsilon_{A \Phi} \, \}$ the subset of $A$'s world line consisting of events which chronologically follow event $\varepsilon_{A \Xi}$.

Accordingly, $\rho[ \, \mathcal W_A, \varepsilon_{A \Xi} \, ]$ is manifestly invariant (wrt. evaluation in reference to any particular reference frame); and any value of (the magnitude of) "absolute acceleration" reported by some "accelerometer" should be compared to the true value $c \, / \, \rho[ \, \mathcal W_A, \varepsilon_{A \Xi} \, ].$

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$§$: Whether or not that's actually and strictly true may well be worth another explicit question.