Why is gravity a fictitious force?

[Dale]

As Einstein said, it is theory which determines what is measurable.

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.

According to the Newtonian theory of gravity, the gravitational force can be measured.

How?

 

[Demystifier]

How?

E.g. by weighing scale.

 

[Dale]

E.g. by weighing scale.

That doesn’t measure the gravitational force, even in Newtonian physics. In the ISS the Newtonian gravitational force is almost the same as it is on the ground, but a weighing scale reads 0.

In Newtonian theory there is no way to measure the gravitational force. All you can do is infer it from the motion. Just like a fictitious force.

 

[Herman Trivilino]

I can't understand why we compare elevator in the space and elevator on the surface of the Earth

The elevator in space is being accelerated by an external force. For example a rope is tied to the roof of the elevator and pulls on the elevator with a constant force.

As long as the elevator is small enough there is no way to distinguish between being in the elevator in space and the elevator on Earth.

 

[haushofer]

My 2 cents: a fictitious force is due to your (accelerating) frame of reference; there is no interaction involved (if you don't subscribe to Mach's principle, that is). Gravity, although having the mentioned properties of fictitious forces (proportional to mass and frame-dependent), does involve an interaction we call "spacetime curvature". That's why we usually don't call gravity a fictitious force.

 

[A.T.]

Gravity, although having the mentioned properties of fictitious forces (proportional to mass and frame-dependent), does involvevan interaction we call "spacetime curvature".

Spacetime curvature is not an interaction, but a property of spacetime. And it's related to tidal gravity, not gravity itself, which can also arise in flat spacetime in an accelerating frame.

 

[Demystifier]

That doesn’t measure the gravitational force, even in Newtonian physics. In the ISS the Newtonian gravitational force is almost the same as it is on the ground, but a weighing scale reads 0.

In Newtonian theory there is no way to measure the gravitational force. All you can do is infer it from the motion. Just like a fictitious force.

I think we disagree on what does it mean to measure something. In my view, to measure something means to infer it from observing something else, which is correlated with the thing one wants to measure. So inferring the force from the motion, or from the position of a static object, is measuring the force.

 

[Dale]

I think we disagree on what does it mean to measure something. In my view, to measure something means to infer it from observing something else, which is correlated with the thing one wants to measure. So inferring the force from the motion, or from the position of a static object, is measuring the force.

That is reasonable. However, any quantity that is inferred from motion is relative to the reference frame that defines the motion. Its value can be changed at a whim by changing the reference frame, before or after the experiment. For such quantities it is required to explicitly state the frame when describing the quantity.

So we don’t say that “we measured the velocity” but instead we say “we measured the velocity relative to the ground”. Or we don’t say “the duration of the process” but rather “the duration of the process in Bob’s frame”.

Similarly with gravity under your approach. You cannot say that some experiment “measures gravity”, but that it “measures gravity relative to some frame”. The statement is incomplete without the specification of the frame.

Other forces do not require this specification.

 

[Demystifier]

Other forces do not require this specification.

How would you measure the electric force, without a reference to a frame? Note that the Coulomb theory of electric force is almost identical to the Newton theory of gravitational force.

 

[A.T.]

Note that the Coulomb theory of electric force is almost identical to the Newton theory of gravitational force.

Except that the electric force cannot be always locally transformed away by a frame change, because it's not proportional to mass like Newton's force gravity.

 

[Dale]

How would you measure the electric force, without a reference to a frame?

The electromagnetic force on an otherwise isolated charge can be measured with an accelerometer. This measurement is frame-independent.

Note that the Coulomb theory of electric force is almost identical to the Newton theory of gravitational force.

Which is one reason that the Coulomb force is not actually a force, but just a component of a force. The actual force is the Lorentz force.

 

[Demystifier]

The electromagnetic force on an otherwise isolated charge can be measured with an accelerometer.

That works for a macroscopic body, but not for a microscopic particle.

 

[Dale]

That works for a macroscopic body, but not for a microscopic particle.

Why not? It just means that you have to redesign your accelerometer.

In any case, it doesn’t change the fact that the gravitational force is relative to a specified reference frame (like fictitious forces) and real forces are not.

 

[Demystifier]

In any case, it doesn’t change the fact that the gravitational force is relative to a specified reference frame (like fictitious forces) and real forces are not.

I still disagree to some level. If we define force as mass times acceleration, the acceleration is always relative to a specified reference frame, so even the electric force is relative in this sense.

Of course, you are right if by acceleration you mean proper acceleration, which is not relative. The proper acceleration along a geodesic is zero, which is a more precise way to say that the gravitational force is not a proper force. But this point of view is only valid in the context of Einstein relativity theory. I don't see how to make this argument precise in a wider context, including the context of the 19th century physics.

 

[haushofer]

Spacetime curvature is not an interaction, but a property of spacetime. And it's related to tidal gravity, not gravity itself, which can also arise in flat spacetime in an accelerating frame.

I don't see that clear distinction between "interactions" and "spacetime curvature"; Fierz-Pauli is an interactive description of GR, and Newton-Cartan theory describes Newtonian gravity as spacetime curvature.

 

[Mike_bb]

I don't see that clear distinction between "interactions" and "spacetime curvature"; Fierz-Pauli is an interactive description of GR, and Newton-Cartan theory describes Newtonian gravity as spacetime curvature.

Spacetime curvature is geometric property.

 

[Dale]

you are right if by acceleration you mean proper acceleration, which is not relative.

Yes, that is what I mean.

this point of view is only valid in the context of Einstein relativity theory. I don't see how to make this argument precise in a wider context, including the context of the 19th century physics.

I will point out that we are having this conversation in the Special and General Relativity subforum, so relativity is clearly assumed. However, this point of view is valid in any theory which uses a manifold for spacetime. All extant physics can be cast in that framework, including Newtonian gravity (which is the Newton-Cartan gravity that @haushofer mentioned).

 

[PeterDonis]

If we define force as mass times acceleration, the acceleration is always relative to a specified reference frame

Not if you define it as invariant mass times proper acceleration--which is how relativity defines it. Both of those are invariants.

 

[PeterDonis]

this point of view is only valid in the context of Einstein relativity theory.

I don't agree. What an accelerometer measures is not theory dependent. Newtonian gravity agrees that an accelerometer attached to a body moving solely under the influence of gravity (for example, in a free-fall orbit around a planet) will read zero. So even Newtonian gravity recognizes that gravity is in a different category from forces like electromagnetism.

The difference between Newtonian gravity and GR is that the former draws a distinction between gravity. considered as a force, and "fictitious" forces like Centrifugal and Coriolis, whereas GR draws no such distinction; all such "forces" are artifacts of using a non-inertial frame.

 

[Demystifier]

@Dale and @PeterDonis I'm still somewhat confused, so I will ask a slightly more general question. Consider a Newton-like theory of gravity, but with inertial mass different from the gravitational mass. In such a theory, is the "gravitational" force a true force or a fictitious force? Can such a theory be written in the Newton-Cartan form?

 

[Dale]

Consider a Newton-like theory of gravity, but with inertial mass different from the gravitational mass. In such a theory, is the "gravitational" force a true force or a fictitious force?

I haven’t studied such theories, so I am not able to answer this. I am sure there is some scientific literature on the topic, but I don’t know it.

 

[PAllen]

@Dale and @PeterDonis I'm still somewhat confused, so I will ask a slightly more general question. Consider a Newton-like theory of gravity, but with inertial mass different from the gravitational mass. In such a theory, is the "gravitational" force a true force or a fictitious force? Can such a theory be written in the Newton-Cartan form?

Metric theories of gravity all obey all but the strongest form of equivalence principle. So, no Newton-Cartan theory could not accommodate this. However, conventional Newton’s gravity clearly could, in which case gravity would not be a fictitious force. It could be measured comparing free fall bodies of the same inertial mass but different gravitational mass. (In conventional Newton’s gravity, the m in the law of gravitation would be a different m than the one in F=ma.)

 

[Demystifier]

In order to better explain all this to myself (and hopefully to others as well), I will present a simple toy model. Consider two point particles with positions $q_1(t)$ and $q_2(t)$ moving in one dimension. Suppose that the Lagrangian has the form
$$L = \frac{m_1\dot{q}_1^2}{2} + \frac{m_2\dot{q}_2^2}{2} - V_1^{\rm ext}(q_1) - V_2^{\rm ext}(q_2) - \frac{k(q_1-q_2)^2}{2} $$
The external potentials $V_1^{\rm ext}$ and $V_2^{\rm ext}$ describe external forces on the particles, e.g., the Newton gravitational force, the Coulomb electric force, etc. The last term describes the internal force between the particles, which models the working of an accelerometer made of a spring attached to a moving object. The equations of motion are
$$m_1 \ddot{q_1} = F_1^{\rm ext}(q_1) -k (q_1-q_2)$$
$$m_2 \ddot{q_2} = F_2^{\rm ext}(q_2) -k (q_2-q_1)$$

Now consider a special case in which $F_2^{\rm ext}(q_2) \equiv 0$, meaning that the external force acts only on the first particle. An example is the electric force, for the case in which the first particle is charged while the second particle is neutral. In this case the second equation of motion can be written as
$$\ddot{q_2} = -\frac{k}{m_2} (q_2-q_1)$$
so 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.

What if $F_2^{\rm ext}(q_2)$ is not zero? In that case there is no simple relation between the particle positions and the acceleration, so the accelerometer cannot work in a way we are used to. In principle the acceleration still can be determined, but it's much more complicated and depends on details of the external force.

The above suggests that perhaps, in some complicated way, the accelerometer could measure even the gravitational force. To see if this is true, let us explore that case in more detail. For the Newton gravitational force we have
$$ F_i^{\rm ext}(q_i) = m_i g(q_i)$$
where $g(x)=-\partial_x\phi(x)$ and $\phi(x)$ is the external gravitational potential. The equations of motion are then
$$m_1 \ddot{q_1} = m_1g(q_1) -k (q_1-q_2)$$
$$m_2 \ddot{q_2} = m_2g(q_2) -k (q_2-q_1)$$
or equivalently
$$m_1 (a_1 - g(q_1))= -k (q_1-q_2)$$
$$m_2 (a_2 - g(q_2)) = -k (q_2-q_1)$$
where $a_i \equiv \ddot{q_i}$. Summing the equations we get
$$m_1 (a_1 - g(q_1)) + m_2 (a_2 -g(q_2)) =0$$
But $m_1$ and $m_2$ are both positive, and we assume that the two particles are not too far from each other so that we can use the homogeneous field approximation $g(q_1)= g(q_2)$. Hence, assuming also the equilibrium $a_1=a_2$, the equation above implies
$$a_1 - g(q_1)=a_2 -g(q_2)=0$$
so the equations of motion imply
$$q_1-q_2=0$$
Thus the accelerometer always shows a zero reading, independently on the acceleration $a_1=a_2$. That is the dynamical reason why the gravitational force cannot be determined by the accelerometer.

 

[Dale]

What if F2ext(q2) is not zero?

That depends. If the force is proportional to the mass, $F_2 = g m_2$, and if $F_1=g m_1 + …$ then you have a good accelerometer design. Otherwise your accelerometer design is poor and doesn’t function as an accelerometer should.

That is the dynamical reason why the gravitational force cannot be determined by the accelerometer.

Yes.

 

[PeterDonis]

That's how the accelerometer measures the acceleration.

I think you have it backwards. In your example, particle 2 is the one "moving only under the influence of gravity", so its proper acceleration is zero. Particle 1, which has an elecromagnetic force on it, will have a nonzero proper acceleration. So the accelerometer in your toy example would be assigning the nonzero reading it gets to particle 1.

 

[Demystifier]

In your example, particle 2 is the one "moving only under the influence of gravity",

No it isn't. There is also the interaction with particle 1 due to the last term in the Lagrangian. That is exactly the reason why I constructed an explicit toy model, to see such things explicitly.

 

[PeterDonis]

There is also the interaction with particle 1 due to the last term in the Lagrangian.

Hm. I see what you mean, but now I question the physical relevance of this toy model, because now you're saying particle 2 isn't affected by any external force at all, including gravity (since you're modeling gravity as an external force). But the whole point of the force of gravity, even in Newtonian physics, is that it affects everything. There is no such thing as a particle that isn't affected by it. So I don't see how your toy model is relevant to the workings of an actual accelerometer, which has to use actual particles, not imaginary ones that aren't affected by gravity.

 

[Demystifier]

Hm. I see what you mean, but now I question the physical relevance of this toy model, because now you're saying particle 2 isn't affected by any external force at all, including gravity (since you're modeling gravity as an external force). But the whole point of the force of gravity, even in Newtonian physics, is that it affects everything. There is no such thing as a particle that isn't affected by it. So I don't see how your toy model is relevant to the workings of an actual accelerometer, which has to use actual particles, not imaginary ones that aren't affected by gravity.

No, you misunderstood. I didn't say that the second particle is not affected by gravity. Instead, I considered several cases. In the first case only the electric field, not the gravitational one, is present. (It can be a good approximation in space far from Earth and Sun.) In the last case the gravitational field is present and both particles are affected by it.

 

[PeterDonis]

In the first case only the electric field, not the gravitational one, is present.

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.

 

[haushofer]

@Dale and @PeterDonis I'm still somewhat confused, so I will ask a slightly more general question. Consider a Newton-like theory of gravity, but with inertial mass different from the gravitational mass. In such a theory, is the "gravitational" force a true force or a fictitious force? Can such a theory be written in the Newton-Cartan form?

I don't think so, but I have to think about this a bit more. My gut reaction would be that this is similar to the question whether electrostatics with the Coulomb force can be recasted as spacetime curvature (since electric charge doesn't equal inertial mass), and I'm pretty sure it can't. Unless one is thinking about some embedding like Kaluza-Klein, which makes it contrived.