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.

 

[PeterDonis]

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

I think you mean damping here, and yes, that is important, for the reason you give (and it's the reason why real accelerometers have damped springs).

 

[Demystifier]

I think you mean damping here, and yes, that is important, for the reason you give (and it's the reason why real accelerometers have damped springs).

Yes, sorry for the misspelling!

 

[Demystifier]

Where did Einstein say this? Since many things were measured prior to the development of the corresponding theory, that seems like a pretty difficult claim to justify. Even if Einstein did say it.

A quote from the book A. Becker, What is Real?

Safely ensconced in his apartment, Einstein finally asked Heisenberg what he really wanted to know.
"You assume the existence of electrons inside the atom, and you are probably quite right to do so. But you
refuse to consider their orbits.… I should very much like to hear more about your reasons for making such
strange assumptions."
"We cannot observe electron orbits inside the atom," replied Heisenberg. He pointed out that only the
spectrum of light from an atom is really observable and concluded with a rather Machian statement.
"Since a good theory must be based on directly observable magnitudes, I thought it more fitting to restrict
myself to these."
In Heisenberg's later retelling of this encounter, Einstein was shocked at this. "But you don't seriously
believe that none but observable magnitudes must go into a physical theory?”
"Isn't that precisely what you have done with relativity?" replied Heisenberg.
"Possibly I did use this kind of reasoning, but it is nonsense all the same," said Einstein. "In principle, it is quite wrong to try founding a theory on observable magnitudes alone. In reality the very opposite happens. It is the theory which decides what we can observe."

 

[PeterDonis]

"Possibly I did use this kind of reasoning, but it is nonsense all the same," said Einstein. "In principle, it is quite wrong to try founding a theory on observable magnitudes alone. In reality the very opposite happens. It is the theory which decides what we can observe."

Sounds like typical Einstein, like when someone asked him what he would have done if experiments had not confirmed the predictions of relativity, and he said he'd feel sorry for God, but the theory is correct.

https://skeptics.stackexchange.com/...ld-feel-sorry-for-the-good-lord-the-theory-is

 

[PeterDonis]

"In principle, it is quite wrong to try founding a theory on observable magnitudes alone. In reality the very opposite happens. It is the theory which decides what we can observe."

One could interpret this as saying that the theory might have to make use of unobservable quantities, but it will predict what we will and will not be able to observe. But we still have to test those predictions.

 

[Dale]

A quote from the book A. Becker, What is Real?

Which is not a professional scientific publication.

It is the theory which decides what we can observe

And this remains a false statement. Many seminal observations were made prior to the existence of the relevant theory.

 

[Demystifier]

And this remains a false statement. Many seminal observations were made prior to the existence of the relevant theory.

But then the opposite statement, that the theory does not decide what we can observe, is also false, because many seminal observations would not be possible without a relevant theory. Gravitational waves, for instance. Which poses an interesting version of the question of this thread: If gravity is a fictitious force, then what force causes stretching of the LIGO arms in the gravitational wave detection?

 

[PAllen]

But then the opposite statement, that the theory does not decide what we can observe, is also false, because many seminal observations would not be possible without a relevant theory. Gravitational waves, for instance. Which poses an interesting version of the question of this thread: If gravity is a fictitious force, then what force causes stretching of the LIGO arms in the gravitational wave detection?

I think earlier it is mentioned that tidal gravity, the actual manifestation of curvature, is not fictitious. GW would similarly not be fictitious.

 

[PeterDonis]

what force causes stretching of the LIGO arms in the gravitational wave detection?

The test masses inside LIGO are in free fall, at least in the horizontal direction (of course they are suspended vertically and have nonzero proper acceleration in that direction, but that's orthogonal to the direction of the "stretching" of the arms). So no force is required to "stretch" the arms. The spacetime curvature changes as a result of the GW passing, but that is not describable as a force.

 

[Dale]

But then the opposite statement, that the theory does not decide what we can observe, is also false

I won’t argue against that. In either case, the local gravitational force, like fictitious forces, is not measurable either way. It can only be inferred through motion.

 

[Demystifier]

I have no problems with accepting that gravity is a fictitious force according to GR. But physicists are used to think that the theories they have are only effective approximate theories, which one day may be superseded by better theories. So suppose that one day GR is replaced by a better theory, in which the equivalence principle and diffeomorphism invariance are not fundamental principles. (There are indications that string theory is such a theory.) In such theory, at the most fundamental level, gravity might be a true force. The effective action for such a theory could be written schematically as "GR+correction". Now my question is the following. What is the true force in such a theory, is only the "correction" a true force, or is the full "GR+correction" a true force? In such a context, would it be correct to say that the GR part is also a true force, due to the existence of the correction?

 

[Dale]

In such a context, would it be correct to say that the GR part is also a true force, due to the existence of the correction?

We are already in that situation with GR alone. So why should we hypothesize about unknown possible future theories when we can already answer the question with current theories?

In GR the gravitational tidal force is a true force which is the part of gravity modeled by spacetime curvature. Only the local part of GR is fictitious. And when we speak precisely, that is precisely what we say. You can see above in this discussion how this already works.

 

[Demystifier]

Only the local part of GR is fictitious.

Excellent! I like the idea that, at some small distance scale, the space is discrete rather than continuous. If true, then the fundamental theory does not have a local part at all, suggesting that there is no fictitious force at all. I know that you don't like wild speculations, but I couldn't resist.

 

[Mike_bb]

the space is discrete rather than continuous.

Why are you so sure?

 

[Dale]

I like the idea that, at some small distance scale, the space is discrete rather than continuous.

And once there is some actual evidence supporting your preference then it will be an acceptable topic for PF. But even then it would be off topic for this thread.

 

[Demystifier]

Why are you so sure?

Why do you think I'm sure? But as Dale said, that's off topic.

 

[Sagittarius A-Star]

In GR the gravitational tidal force is a true force

I don't think so. If for example an asteroid breaks in tidal gravity, the related forces are electrical forces between parts of it.

 

[PAllen]

I don't think so. If for example an asteroid breaks in tidal gravity, the related forces are electrical forces between parts of it.

You can’t transform away tidal gravity. In every frame there some non EM force that the EM force resists until it can’t. Even more simply, just consider two bodies at mutual rest in free fall. With tidal gravity, they will either get closer together or move apart, depending on orientation. This must be accounted for by a force.

 

[Sagittarius A-Star]

With tidal gravity, they will either get closer together or move apart, depending on orientation. This must be accounted for by a force.

Why? I think a true force creates a deviation from the geodesic.

 

[PAllen]

Why? I think a true force creates a deviation from the geodesic.

You may adopt that definition. But the definition accepted by most in this thread is whether it can be measured. Tidal effects and energy transferred by GW can be measured directly.

Another example of direct measurement of tidal gravity as a force: 2 balls connected by a spring. Move it from low to high tidal gravity, and the spring stretches or compresses.

 

[Sagittarius A-Star]

You may adopt that definition.

Is it a matter of adoption of a definition, what a true force is?

But the definition accepted by most in this thread is whether it can be measured.

I think a true force can be measured with an accelerometer.

 

[PAllen]

Is it a matter of adoption of a definition, what a true force is?

Definitely. If you define a force as that which is mediated by particle exchange, and treat gravity as a theory of gravitons, then all aspects would be defined as true force, even when not measurable.

I think a real force can be measured with an accelerometer.

The spring device I described, I think, counts as an accelerometer. It just has to be big enough or sensitive enough.

 

[Sagittarius A-Star]

The spring device I described, I think, counts as an accelerometer. It just has to be big enough or sensitive enough.

I think in GR tidal gravity, it just has to be small enough and insensitive enough.

 

[PeterDonis]

the gravitational tidal force is a true force which is the part of gravity modeled by spacetime curvature

Tidal gravity can be measured--it's not "fictitious"--but in GR it is not a force. A force is something that causes nonzero proper acceleration. Tidal gravity does not. Tidal gravity is geodesic deviation--the behavior of objects in free fall.

What is informally called "tidal force" is internal, non-gravitational forces inside objects that cause the worldlines of parts of those objects to not be geodesics, when that happens because of spacetime curvature. But that is not modeled by spacetime curvature; it's modeled by some kind of material model of the object and its internal forces.

 

[PeterDonis]

If you define a force as that which is mediated by particle exchange,

Then you are doing quantum field theory, not classical GR. This is the classical GR forum, so here, we use the GR definition of force.

 

[Dale]

Tidal gravity can be measured--it's not "fictitious"--but in GR it is not a force.

Sure, that is a matter of definitions and categories. Nevertheless, the tidal force is the physical part of gravity, the part that does not share the characteristics of fictitious forces.

What is more important than the label is how it behaves. It is not proportional to the mass of the object on which it acts. It cannot be transformed away. And (most importantly in my opinion) it produces physical strain.

The “tidal force” is not a “force”. That’s fine, I don’t object. Language is inherently untidy. And recognizing that, I also don’t feel very mistaken for calling the “tidal force” a “force” anyway.