Today I will repeat the argument from yesterday using a slightly different formulation.
Every experiment is an interaction between a system and an observer, and the outcome of the experiment depends on the physical properties of both. In particular, it depends on the observer's mass. Neither general relativity nor Quantum Field Theory (QFT) make predictions that depend on the observer's mass, so some hidden assumptions must have been made. It is clear what those assumptions are:
- In general relativity, the observer's heavy mass is zero, so the observer does not disturb the fields. This observer is essentially what Einstein calls a test particle.
- In QFT, the observer's inert mass is infinite, so the observer knows where he is at all times. In particular, the observer's position and velocity at equal times can be measured simultaneously to arbitrary precision.
Hence quantum gravity requires that the observer's mass is finite and nonzero. Then we can not ignore the interaction between the observer and the gravitational field, and we can not ignore horizontal fuzziness.
Let us spell out the second assumption in more detail. Let \(q\) be the observer's position and \(p\) his momentum. The accuracy with which they can be simultaneously measured is limited by the uncertainty principle:
\[
\Delta q \cdot \Delta p \geq {\hbar\over2},
\]
where \(\hbar\) is Planck's constant divided by \(2\pi\). The observer's velocity \(v\) is given by \(p = Mv\), where \(M\) is the observer's mass. Hence
\[
\Delta q \cdot \Delta v \geq {\hbar\over2M}.
\]
In order to know the observer's position at all times, we must simultaneously know his position and velocity at equal times without any uncertainty. Hence the right-hand side of the equation above must vanish, i.e. \(\hbar/M = 0\). There are two ways to achieve this:
- \(\hbar = 0\). This is classical field theory including classical gravity.
- \(M = \infty\). This is quantum field theory without gravity.
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