Why is gravity a fictitious force?

[Demystifier]

Okay, then we're back to my original criticism: the first particle, not the second, is the one that's affected by the electric field. So any acceleration measured by the accelerometer should be assigned to the first particle, not the second, since it's the one that feels a force.

I don't see any problem with that. The second particle is a measuring "apparatus", which measures a property of the first particle. All measuring apparatuses are of this form. (Of course, the second particle also feels a force, the one due to the interaction with the first particle, but the second particle does not feel the force of the external electric field.)

I believe my toy model is a good model of an accelerometer, in the sense that it captures all essential properties of the real accelerometer, and yet does it in a very simple way.

 

[PeterDonis]

I don't see any problem with that.

But it's not what your description said. You said:

by observing the relative position $q_2-q_1$ one can determine the acceleration $\ddot{q_2}$. That's how the accelerometer measures the acceleration.

That says that the accelerometer is measuring the acceleration of particle 2. But now you appear to be agreeing with me that it's measuring the acceleration of particle 1. That means the math you should be showing should be for $\ddot{q}_1$.

 

[Demystifier]

But it's not what your description said. You said:


That says that the accelerometer is measuring the acceleration of particle 2. But now you appear to be agreeing with me that it's measuring the acceleration of particle 1. That means the math you should be showing should be for $\ddot{q}_1$.

You are right, I should have been more precise about that. In a direct sense it determines $\ddot{q}_2$. But indirectly it determines also $\ddot{q}_1$. How? Because I assume that the spring does not oscillate, i.e. that $q_1(t)$ and $q_2(t)$ are comoving, so $\ddot{q}_2=\ddot{q}_1$. That's what I tacitly assumed when I said that it is an accelerometer, because that is essentially how the real accelerometer works. To provide that there are no oscillations of the spring I could have added a dumping term which dumps the oscillations after a short time, which would make the model even more realistic, but I felt that this detail is not essential for my point. Now I see that maybe it is.

 

[Demystifier]

One additional note. My model of the accelerometer with only one apparatus degree of freedom $q_2$ is analogous to the von Neumann model of the measuring apparatus for quantum measurements, which also involves only one apparatus degree of freedom. The need for a dumping term to make the accelerometer more realistic is analogous to the need of decoherence to make a model of quantum measurement more realistic.

I could have made the analogy even more explicit by taking a different model of an accelerometer, by introducing the interaction Hamiltonian of the form
$$g(t) p_2 a_1$$
where $g(t)$ is a time-dependent coupling. Such an interaction establishes a direct correlation between the apparatus pointer variable $q_2$ (conjugated to the momentum $p_2$) and the acceleration $a_1$ of the measured object, in exactly the same way as in the von Neumann model. However, such a model would less resemble the working of the real accelerometer.