*Every real experiment is an interaction between a system and an observer, and the outcome depends on the physical properties of both. In particular, the result depends on the observer's mass.*

This may seem as an innocuous observation, until you realize that neither General Relativity (GR) nor Quantum Field Theory (QFT) make predictions that depend on the observer's mass. So clearly there must be a more general, observer-dependent, theory that reduces to GR and QFT in the appropriate limits. A little thought reveals that these limits are:

- In GR, the observer's heavy mass is assumed to be zero, so the observer does not disturb the fields.
- In QFT, the observer's inert mass is assumed to be infinite, so the observer knows where he is at all times. In particular, the observer's position and velocity at equal times commute.

The assumption about the small heavy mass is very intuitive: if the observer had a large heavy mass, he would immediately collapse into a black hole, which is not what happens in a typical experiment. However, the assumption about the infinite inert mass requires some more explanation. Here my philosophy is completely operational: in order to know something, you must measure it. In particular, in order to know where he is, the observer must measure his position, e.g. with a GPS receiver. In theory, that can be done with arbitrary accuracy. However, in order to know where he will be in the future, the observer must also be able to measure his velocity at the same time, but Heisenberg's uncertainty principle tells us that there is a limit to the precision with which the position and the momentum can be simultaneously known.

More precisely, let $\Delta x$ and $\Delta v$ denote the uncertainties in the observer's position and velocity, and assume that momentum and velocity are related by $p = Mv$, where $M$ is the observer's mass. Then

\[

\Delta x \cdot \Delta v \sim \hbar/M,

\]

where $\hbar$ is Planck's constant. Hence there are only two situations in which both the observer's position and velocity can be known arbitrarily well:

- If $\hbar = 0$, i.e. in classical physics including GR. In this limit we can assume $M = 0$.
- If $M = \infty$, which only makes sense if gravity is turned off. In this limit we have QFT in flat space.

An analogous statement can be made about other interactions than gravity: field theory assumes that the observer's charge is zero and his inert mass is infinite. However, this double limit does not pose a problem for non-gravitational interactions, because charge and mass are unrelated. So even if non-gravitational physics depends on the observer's charge in principle, this dependence is merely an experimental nuisance that can be eliminated. In gravity, where charge and mass are the same, this nuisance becomes a conceptual problem.

Rovelli makes the distinction between two types of observables: partial observables, which can be measured but not predicted, and complete observables, whose time evolution can be predicted by the theory and which are subject to quantum fluctuations. In Quantum Mechanics, there are two types of partial observables: $A$, the reading of the detector, and $t$, time measured by a clock, which combine into the complete observable $A(t)$. In QFT, there is a third type of partial observable: $x$, the position measured e.g. by a GPS receiver, and the complete observables are of the form $A(x,t)$. However, beneath this expression lies an unphysical assumption: that the pair $(x,t)$ is a partial observable that can only be measured but not predicted. A real observer obeys his own set of equations of motion, so his position at later times $x(t)$ can be predicted. Let us now change notation and call this observable $q(t)$ instead, because $x$ will be reused below.

In a physically correct treatment, we combine the three partial observables of QFT into two complete observables: $A(t)$ and $q(t)$, the readings of the detector and GPS receiver at a certain tick of the clock. However, we immediately notice that a lot of information is gone: we no longer sample data throughout spacetime but only along the observer's trajectory. Fortunately, this is not as disastrous at it might seem at first glance, because the available local data includes not only the values of the fields A but also their derivatives $d^m A/dx^m$ up to arbitrary order. We can assemble the local complete observables into a Taylor series,

\[

A(x,t) = \sum_m \frac{1}{m!} \frac{d^m A}{dx^m}(t) (x-q(t))^m

\]

This formula looks one-dimensional, but with multi-index notation it makes sense in higher dimensions as well. To the extent that we can identify an infinite Taylor series with the field itself, the original field has been recovered entirely, but with a twist: the Taylor series does not only depend on the field, but also on the expansion point $q(t)$, an operator which we identify with the observer's position.

In the next installment in this series this expression will be used to make the connection to locality and the multi-dimensional Virasoro algebra from the previous post.

Interesting argument but one may find difficult to understand the mass dependence. Suppose we have two labs with two different observers. In the first one, there is a human researcher with an obesity problem weighing around 100 kg,performing a standard Aharonov-Bohm type experiment while in the second lab, exactly the same experiment is performed with a set of micro-computers weighing no more than say, 10 kg. Both will record the same type of an electron phase shift How exactly will the mass difference get manifested between the two results.

SvaraRaderaAnother thing regarding the last statement on multi-index spaces, is that dimensionality may be somewhat artificial in such a case as any word combinatorics can always be assigned back to the integers via a recursive scheme on higher power symbolic alphabets.

Well, I have never thought much about non-local effects in QM, so I don't know. The AH effect tells us that there is some topological information about EM that is only encoded in the gauge potential, and not in the electric and magnetic fields themselves, right? There should be some extra fuzziness when a light observer (or an observer using a lighter probe) measures the fields, but I don't think that would change the AH effect.

RaderaPerhaps, one should also check the case of gauge based gravity then? I think there was an old attempt by some Chinese researchers but I have not seen anything important lately into this direction.

SvaraRadera