# B Is Gravity a Force?

#### kfielder

Summary
Are gravitons consistent with general relativity?
Hi all, another newbie question for you. I often hear gravity described as a force, including speculation that gravitons are its messenger particle, but is that consistent with general relativity? I thought relativity implies it's more like rolling down a hill than being pulled by a force.

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#### jbriggs444

Homework Helper
Summary: Are gravitons consistent with general relativity?

Hi all, another newbie question for you. I often hear gravity described as a force, including speculation that gravitons are its messenger particle, but is that consistent with general relativity? I thought relativity implies it's more like rolling down a hill than being pulled by a force.
The models of Newton (gravity as a force) and of Einstein (gravity as space-time curvature) are consistent in the sense that they make approximately the same predictions for situations we encounter in day to day life. Apples separate from trees and make contact with the ground in both theories.

The question of whether gravity is really a force or really space-time curvature does not arise. They are two different ways of describing the same behavior.

Of course, as it turns out, the "force" description does not result in agreement with precise experimental results in some scenarios (such as the gravitational deflection of light or the precession of the orbit of Mercury) so we adopt the description provided by general relativity when the difference matters.

#### hyunxu

What do we call force? In simple words force is a pull or push on object.

we are standing on the surface of the earth. Because of a force exerted by it. So that force pulls us down. Hence we stand on the surface of the earth.Obviously this force is nothing but gravity.

so I conclude that gravity is a force.

#### PeterDonis

Mentor
we are standing on the surface of the earth. Because of a force exerted by it. So that force pulls us down.
Not according to GR. According to GR, the only force you are experiencing when you stand on the surface of the earth is the force of the earth's surface pushing up on you. Were it not for that force, you would be in free fall, feeling no force at all, and the thing that determined your trajectory, which is what the word "gravity" refers to, would be the geometry of spacetime.

In other words, GR's definition of "force" is different from the Newtonian one; in GR, only things you actually feel as a force are forces. You don't feel the "force" of gravity, even in Newtonian mechanics; if gravity is the only "force" acting on you (if, for example, you are an astronaut in orbit in the ISS), you are weightless, in free fall, feeling no force at all. GR explains this by simply saying you feel no force because there is no force; "gravity" is not a force, it's the geometry of spacetime.

#### hyunxu

In other words, GR's definition of "force" is different from the Newtonian one; in GR, only things you actually feel as a force are forces. You don't feel the "force" of gravity, even in Newtonian mechanics; if gravity is the only "force" acting on you (if, for example, you are an astronaut in orbit in the ISS), you are weightless, in free fall, feeling no force at all. GR explains this by simply saying you feel no force because there is no force; "gravity" is not a force, it's the geometry of spacetime.
Thank you for correcting my mistake.I have learnt much from as a normal higher secondary school student!

#### Martin Scholtz

Gold Member
I am not sure how your question is related to gravitons (see your Summary).

Gravity is the force in a classical sense. By force we mean the acceleration, i.e. the change of velocity. Suppose you hold a book in your hand 2 meters above the ground. It doesn't change the position with respect to you, it has zero velocity. As soon as you release it, the book starts to fall down to the ground. Now it changes the position with respect to you, so it has non-zero velocity. Moreover, this velocity increases in time. Hence, we observe that the book changed its velocity from zero to a non-zero value and it even grows in time. This is kinematics: we observe the change of velocity and call it acceleration.

Now we would like to explain why this happens and we say there is some interaction between the Earth and the book, in this case called gravity. This is dynamics: we wish to formulate a quantitative law which relates parameters of both bodies to the acceleration. The reason why bodies accelerate is called force.

Hence, gravitational force is a true force, it accelerates objects.

Another thing is that Einstein provided geometrical description of the origin of this force, general relativity (GR). He employed the notion of spacetime introduced by Minkowski. Einstein in special relativity (SR) shown that spatial distances and time intervals are relative. For example, in Newtonian physics the velocity is defined by
$\mathbf{v} = \dfrac{d\mathbf{r}}{dt}$
where $\mathbf{r}$ is the position vector, $t$ is the time and #d# stands for the derivative.

Similarly, the acceleration is defined by
$\mathbf{a} = \dfrac{d\mathbf{v}}{dt}=\dfrac{d^2\mathbf{r}}{dt^2}$
but here both numerator and denominator should transform according to reltivistic laws. Hence, the acceleration obviously depends on the state of motion of the observer. Hence, description in terms of space and time separately is relative, observer dependent.

Minkowski (1908) has shown that if you instead introduce a spacetime interval
$ds^2 - c^2\,dt^2 - dx^2-dy^2-dz^2$
and apply Einstein's postulates of SR, you obtain an invariant, observer-indeendent quantity, although all coordinates in the expression are relative. It can be easily shown that for an observer moving with subluminal velocity, quantity dsds has the meaing of the proper time, i.e. the time measured by the clocks which are at rest w.r.t. the object we observe. If we observe the propagation of light, $ds=0$.

Hence, it is natural to interpret the motion of an object in a geometrical way as follows. Instead of talking about particle moving in space and following a spatial curve called trajectory, we regard this particle as an object in 4-dimensional spacetime (technically we call it manifold). You can imagine the time axis $t$ and usual 3D space which we denote $E^3$. Now, to each point of the time axis we attach one space $E^3$. This way we obtain a 4D spacetime whose points are called events.

The point is that while in the 3D space particle can be at rest or follow some trajectory, in the spacetime it is not so. The very fact that particle's own clocks (i.e. those at rest w.r.t. the particle) always measure particle's proper time implies that we have to replace the notion of trajectory in space by a worldline in spacetime. Let me illustrate it on a simple picture (I drew it without tooo much care, sorry for the quality)

Suppose that in 3D some particle is moving along the circle. Circle is the trajectory of the particle in space. In the figure shown the particle orbits in the $(x,y)$plane, the axis $z$ is suppressed. At the beginning, the time is $t=0$ and we mark this in the spacetime by a point (in my figure a ball) with coordinates
$(t,x,y)=(0,r,0)$
where $r$ is the radius of the circle.

At later time $t>0$ particle is already shifted along the circle by some angle. But now we must mark the position in spacetime at the point lying at different slice of the spacetime, i.e. lying at the slice corresponding to given time $t$ - it will lie above the plane where the ball is plotted and the coordinates are
$(t,x,y)=(t,r\cos\omega\,t,r\,\sin\omega\,t)$
where $\omega$ is the (angular) frequency of the motion of the particle.

To conclude:
1. Event is not a point in space, but in a spacetime and carries the information about the position of the particle but also about the time it happened.​
2. As the particle orbits with velocity vv (in fact, v=ωrv=ωr), in space, each position of the particle is represented in spacetime by an event lying at different time slice of the spacetime.​
3. All events (tt, position at time tt) together form a curve n spacetime called worldline.
4. Trajectory of our paticle is a closed circle, while its worldline is an infinitely long curve (called helix in this case, somewhat different kind of a spiral). Trajectory is depicted in my figure by a dashed circle, the worldline is the solid line.​
5. Although trajectory can be also a single point in space (if the particle is at rest), it must always move in the time direction in spacetime (upwards). In the case when the partice is at rest or moves with constant velocity, its worldline will be a straight line in spacetime but a single point in space.​
I know this is quite trivial but I am slowly getting to my point regarding gravity. What I wanted to emphasize is that in spacetime particle always moves along the worldline. In fact, in some sense the speed of any object in spacetime is always equal to the speed of light. I will not explain this issue now because I don't want to digress even more, I can explain on request.

If we have just space with particles which do not interact in any way (electromagnetically or gravitationally), the worldlines of particles will be simply straight lines whose slopes will be given by the velocities of particles. These velocities will not change, because there are no forces and hence no interactions. So, the helical motion depicted in the figure above cannot occur if there is no interaction. But you can imagine that our ball from the figure is, say, the Earth and in the center of the circle is the Sun which attract the Earth through gravitational force and therfore it curves the trajectory and, consequently, the worldline of the Earth.

Nevertheless, without interaction, particles follow straight worldlines. In usual spatial geometry the line from AA to BB is the shortest path connecting points AA and $B$. In spacetime geometry, on the contrary, it is the longest path. Have a look at the spacetime interval ds2ds2 given above. Unlike the Pythagorean theorem, coordinates and time enter with opposite signs, that's the reason for this unusual feature of spacetime geometry. Although it is important, in what follows I will not discuss whether line is shortes or longest distance and I will simply call it extremal (either minimum or maximum) like it is usual in mathematics.

In order to introduce more convenient terminology, we call an extremal worldline connecting points AA and BB a geodesic. Geodesic is "as straight curve" as possible. In special relativity the geodesics are really straight lines. But imagine a surface of a sphere (ball) and choose two points. What is the shortest curve connecting them? It is certainly not a line because such a line would lie inside the sphere, not on its surface:

The shortest path connecting AA and $B$ is in fact an arc lying on the surface:

This arc is one of geodesics on the sphere. Whether the geodesic is the shortest or longest path depends on the actual geometry.

In the case of electromagnetic field (EF) the interaction is explained by Maxwell's theory. The charged source produces the EF which propagates at the speed of light through the space and once it reaches our particle, it acts on it and exerts a force.

Einstein's ingenious idea was that gravity is of totally different origin. Einstein postulated that if the only interaction is gravity, particles always move along the geodesics - extremal worldlines. It agrees with SR, where we saw that inertial particles always follow straight worldlines. But if the Earth orbits about the Sun and follows a helix, and by assumption worldline is always geodesics, it means that the helix in the case of orbiting particle is a geodesic. But we saw that the shape of geodesic depends on the geometry. There is only one conclusions: gravitating bodies affect the geometry around them and they deform straight lines in curved geodesics. That's what we mean by "spacetime is curved by the matter".

So, to finally answer the question:
• Gravity is a force in classical sense: it causes an acceleration of objects measured by the observers.​
• It is not a usual force in the sense that the interaction is not due to some kind of field propagating between interacting objects but due to deformation of the geometry.​
To make at least small comment on your gravitons.
kfielder said:
Summary: Are gravitons consistent with general relativity?
To make at least small comment on your gravitons.

• EM field can exist in the flat spacetime (where geodesics are straight lines) because it is made of quanta called photons and EM interaction is realized by he exchange of these quanta. Everything is extremey well descibed by the so-called quantum field theory (QFT). There are many reasons to expect that gravity must also exhibit quantum features and thus be described by some kind of QFT. The question is: if gravity = geometry, what should be gravitons then? Some particles like photons? Nobody interested in quantum gravity has any doubts that the actual geometry is discrete. It is similar to condensed matter: piece of metal looks like a continuum of metalic substance, but in fact it is made of atoms that form a crystal lattice. We propose (but not just us, there are many similar approaches, but we relate our model to black hole information paradox) that gravitons should be regarded as emergent quasi-particles. If you know something about condensed matter physics, you probably heard the word phonon. Each atom of the lattice oscillates irregularly. But when you take into account the complicated interaction between all atoms of the lattice, you find a new, ordered behavior of the whole system - that we call emergence. The oscillations of the full lattice can be described similarly to EM field in QFT like a collection of particles, in this case they are called phonons. Phonons are not real, you cannot trap a single phonon, but you can understand the properties of metal like heat capacity, resistance, etc. in terms of phonons, their propagation and collisions. That's very beautiful. In our opinion, graviton should be understood similarly as a quasi-particle emerging from complicated ineraction between quantum gravitational degrees of feedom. If you are interested, you can read our paper. It is freely accesible on arXiv.org > gr-qc > arXiv:1704.00345, published version is on https://doi.org/10.1016/j.aop.2017.10.018, but if you don't have the access, I attach the paper to this post.​

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#### Dale

Mentor
I often hear gravity described as a force, including speculation that gravitons are its messenger particle, but is that consistent with general relativity?
Gravity can be described as a force even in general relativity, although it is not often done. When you choose to describe gravity as a force in general relativity then you also describe things like the centrifugal and Coriolis forces as forces.

Gravity shares a lot of characteristics with the centrifugal and Coriolis forces:
Proportional to mass
Vanish in an inertial frame
Not detected by accelerometers

Mathematically, they are also very similar. Specifically, they all arise from what are called the Christoffel symbols. These symbols describe the fields corresponding to the inertial forces, e.g. the Newtonian gravitational field, and in some of Einstein’s early writings he used this concept heavily.

The reason that this is not often done in GR today is because these Christoffel symbols are not tensors. They are therefore not covariant, but depend strongly on your coordinate system. We value frame invariant quantities more now so this description has become less prevalent.

#### Dale

Mentor
Hence, gravitational force is a true force, it accelerates objects.
The centrifugal force is also a true force by that criteria.

I have no objection to simply calling it a “force”, but if you must add a qualifier then “true force” is not appropriate. You could call it an “inertial force” (my preference) or a “fictitious force”.

#### PeterDonis

Mentor
Gravity is a force in classical sense: it causes an acceleration of objects measured by the observers.
It causes coordinate acceleration in particular coordinate charts. It does not cause coordinate acceleration in other charts. And it does not cause proper acceleration (acceleration you can feel) at all. As I pointed out in an earlier post, objects moving solely under the influence of gravity are weightless, in free fall. That is why GR attributes "gravity" to the geometry of spacetime instead of to a force.

#### Martin Scholtz

Gold Member
It causes coordinate acceleration in particular coordinate charts. It does not cause coordinate acceleration in other charts. And it does not cause proper acceleration (acceleration you can feel) at all. As I pointed out in an earlier post, objects moving solely under the influence of gravity are weightless, in free fall. That is why GR attributes "gravity" to the geometry of spacetime instead of to a force.
Thank you for your comment. I probably understand what you want to say and you are right. But first, I wrote that it is a force in classical sense and I explained in detail what is meant by that. We talk about two different accelerations. You are talking about an observer who is already moving along a geodesic which by definition has vanishing proper acceleration. I was saying that gravity is a force in the sense that for a given observer who observers some other object gravity acts as a force because it implies mutual acceleration. This is not a coordinate effect (anyway, any statement depending on coordinates is objectively useless). The choice of the observer is not about choosing the coordinate system but about choosing an orthonormal frame.

But the fact that the two observers, although both following geodesics, have non-zero mutual acceleration is coordinate-independent thing and it happens only in the presence of curvature. Hence: no gravity (flat spacetime) implies zero mutual proper acceleration of two geodesic observers, gravity (curved spacetime) implies non-zero mutual acceleration. This is coordinate independent thing so, in this sense, gravity qualifies as a force.

#### PeterDonis

Mentor
a given observer who observers some other object gravity acts as a force because it implies mutual acceleration
This is ambiguous: it could mean, for example, someone standing on the surface of the Earth watching a rock fall, or it could mean tidal acceleration, which you describe later on in your post (see below).

In the case of someone standing on the surface of the Earth, the force that causes the mutual acceleration of the someone and the rock is not gravity; it's the surface of the Earth pushing up on the person.

two observers, although both following geodesics, have non-zero mutual acceleration is coordinate-independent thing and it happens only in the presence of curvature.
This is tidal acceleration, and since both observers are in free fall, there is no force involved in the GR sense. It's simply spacetime geometry--as you say, tidal acceleration is a sign that spacetime is curved.

One could talk of "tidal force", but once again, the force involved is not gravity, but internal forces inside objects that cause stresses due to parts of the object not moving on the geodesic paths that they would move on if the object had no internal forces and every part of it could move independently.

#### Martin Scholtz

Gold Member
The centrifugal force is also a true force by that criteria.

I have no objection to simply calling it a “force”, but if you must add a qualifier then “true force” is not appropriate. You could call it an “inertial force” (my preference) or a “fictitious force”.
Thank you for the reaction. I don't necessarilly have to add a qualifier :) I did it in order to emphasize that classicaly force is something causing acceleration. Even in GR gravity has this property. I elaborated on this a bit in the post 11 which was a reaction on similar criticism in the post #9 by PeterDonis.

#### Dale

Mentor
I don't necessarilly have to add a qualifier :) I did it in order to emphasize that classicaly force is something causing acceleration.
Do you also claim that the centrifugal and Coriolis forces are "true forces"? If you do then that is a highly unusual claim, and if not then the qualifier is the wrong qualifier to use for gravity even by your own standards.

But the fact that the two observers, although both following geodesics, have non-zero mutual acceleration is coordinate-independent thing and it happens only in the presence of curvature. Hence: no gravity (flat spacetime) implies zero mutual proper acceleration of two geodesic observers, gravity (curved spacetime) implies non-zero mutual acceleration. This is coordinate independent thing so, in this sense, gravity qualifies as a force.
That is specifically "tidal gravity", not just "gravity". By this standard the usual uniform 9.8 m/s^2 gravitational field near the surface of the earth does not qualify.

#### Martin Scholtz

Gold Member
This is ambiguous: it could mean, for example, someone standing on the surface of the Earth watching a rock fall, or it could mean tidal acceleration, which you describe later on in your post (see below).

In the case of someone standing on the surface of the Earth, the force that causes the mutual acceleration of the someone and the rock is not gravity; it's the surface of the Earth pushing up on the person.

This is tidal acceleration, and since both observers are in free fall, there is no force involved in the GR sense. It's simply spacetime geometry--as you say, tidal acceleration is a sign that spacetime is curved.

One could talk of "tidal force", but once again, the force involved is not gravity, but internal forces inside objects that cause stresses due to parts of the object not moving on the geodesic paths that they would move on if the object had no internal forces and every part of it could move independently.
OK, I think we can agree, it makes no sense to repeat everything again. After all, whether we call it a force or just interaction or even differently is just a matter of terminology. My logic was to explain the classical notion of force and that even in GR and its geometrical formulation, gravity can cause relative acceeration. First on the level of mutual acceleration in 3D space but I emphasized that unlike classical physics, this acceleration is relative. Then in the reaction on your post, I considered indeed 2 geodesic observers and in 4D it again makes sense to consider the deviation vector and its 2nd covariant derivative which is a mutual acceleration of two geodesics - tidal forces, as you write. Hence, for me gravity is a force from various points of view.

Could you, please, explain what you mean by "force in the GR sense"? I don't critisize the term I am just not sure what you mean. And finally, when we talk about tidal forces and Jacobi deviation equation, why do you say "One could talk of "tidal force", but once again, the force involved is not gravity, but internal forces inside objects that cause stresses due to parts of the object not moving on the geodesic paths...." We can talk about 2 test point particles where no internal forces are present, particles are moving along geodesics, still there is non-vanishing relative acceleration (in 4d sense).

In any case, thank you for an interesting discussion.

#### Nugatory

Mentor
Gravity is the force in a classical sense. By force we mean the acceleration, i.e. the change of velocity.
Gravity is indeed a force in classical physics, but to avoid the criticisms from @Dale and @PeterDonis above you will have to be a bit more precise about what that means:

Newton’s first law defines an inertial frame. Newton’s second law defines force, not just as acceleration but as acceleration in an inertial frame.

Thus the Newtonian definition is based on coordinate acceleration. The distinction between proper and coordinate acceleration is irrelevant to this definition; what matters is that there is coordinate acceleration in an inertial frame. Gravity as a real force (a falling object has coordinate acceleration in an inertial frame) but centrifugal force is not (produces coordinate acceleration only in the non-inertial rotating frame).

General relativity (more cleanly, IMO) treats all coordinate acceleration as a mere convention and defines force in terms of proper acceleration. That definition doesn’t change the interpretation of the classical fictitious forces, but it does exclude gravity as a force.

#### PeterDonis

Mentor
the Newtonian definition is based on coordinate acceleration
And on a different definition of "inertial frame" from the GR one. Under the Newtonian definition, someone standing at rest on the surface of the Earth is at rest in an inertial frame. Under the GR definition, they aren't.

#### pervect

Staff Emeritus
Summary: Are gravitons consistent with general relativity?

Hi all, another newbie question for you. I often hear gravity described as a force, including speculation that gravitons are its messenger particle, but is that consistent with general relativity? I thought relativity implies it's more like rolling down a hill than being pulled by a force.
General relativity, being a purely classical theory, does not incorporate "gravitions".

The usual approach to General relativity is based on geometry, which is what I think you mean by "like rolling down a hill".

There are some proposals that attempt to describe gravity in a non-geometrical way, for instance Straumann's "Reflections on Gravity", https://arxiv.org/abs/astro-ph/0006423

Straumann said:
A pedagogical description of a simple ungeometrical approach to General Relativity is given, which follows the pattern of well understood field theories,
However, if you study GR seriously, i.e. if you get a textbook on the topic and start to read it, the textbook will undoubtedly use the geometrical description.

Straumann's approach is certainly interesting, and it is an illustration that there are often more than one approach to a problem. It's unclear if Straumann's approach exactly replicates all the predictions of GR, particularly in the case of black holes. I do not believe he claims that it does, though it's been a while since I've read the paper.

A couple of interesting quotes from the paper

The idea of this alternative approach is to describe gravity - in close analogy to electrodynamics - by a field theory onflatMinkowski spacetime.
Now I come to a conceptually important point: we shall see that the flat Minkowski metric is not observable.
These two snippets might give you some insight into Straumann's proposal - if you move away from actual, physical rulers and clocks to ones that you "make up" - these clocks and rulers are made up in the sense that they are not observable - you can avoid having to deal with some of the intricacies of curved geometry.

To put it another way, if we describe the actual geometry of physical clocks and rulers, we inevitably come up with the conclusion that the underlying geometry is curved. We can deal with this directly, or we can instead start redefining our basic idea of clocks and rulers in an attempt to avoid having to learn about curved geometries.

So - the short version is that non-geometrical proposals to describe gravity do exist in the literature, but the standard approach to gravity is geometrical.

To highlight some of my concerns about black holes, the question is whether the "fields" that Straumann is envisioning as existing in a flat Minkowskii space acutally have a sensible solution for black holes. That's not something that he really explores in his paper. So if you are interested in black holes, such as the one that we are observing in the center of the galaxy, and via the gravitational waves that we detect with Ligo, Straumann's approach is probably not the best approach to use. You'd be better off learning something about curved geometries.

At the layman level, I'd suggest starting to learn about non-flat geometries by studying spherical trignometry. A general treatment of curvature in n dimensions is just too hard to describe at a layman level, but some insight can be gained by studying the simplest cases.

#### Dale

Mentor
I think we can agree
While I have no objection to calling gravity a force, I do not agree with the qualifier “true”.

#### 1977ub

Which forces remain which are unqualified?
While I have no objection to calling gravity a force, I do not agree with the qualifier “true”.
Which forces remain that are not in need of any qualifier? If they are mediated by exchanges of particles, what do we gain by referring to them as forces?

#### PeterDonis

Mentor
Which forces remain that are not in need of any qualifier?
Any force you can feel is a force that is not in need of any qualifier.

If they are mediated by exchanges of particles
We are way, way above this level of modeling. We are just talking about ordinary classical forces. We are not talking about how the things we experience in ordinary classical ways as forces (the things we can feel as forces) are produced way down at the level of quantum field theory. That level is (a) off topic for this thread, and (b) off topic for this forum (it belongs in the quantum forum).

#### Dale

Mentor
Which forces remain that are not in need of any qualifier?
No forces need a qualifier. My objection is to using an inappropriate qualifier. Of course, using an appropriate qualifier conveys more information than not using a qualifier, but using an inappropriate qualifier is simply wrong.

#### pervect

Staff Emeritus
Two of my objections to calling gravity a force are:

1) It doesn't transform as a force. Forces are vectors, with well-known transformation rules. "Gravity" doesn't follow those rules. This can cause conceptual errors when one insists on forcing gravity into the mold of a "force".

2) Certain elements of gravity just don't fit well into the conceptual model of a force.

2a) Example: "gravitational time dilation". This is a consequence of gravity, but the fact that gravity has any effect whatsoever on time doesn't come from the "force" model at all. It arises naturally from the usual space-time curvature model, however.

There is no other "force" that affects time in this way. It's an important effect, one that is significant even for high precision timekeeping on the Earth. The "force" model will totally fail to predict, explain, or allow one to think about this effect.

2b) Example: "Spatial curvature". Consider the "extra" deflection of light above the Newtonian prediction by massive bodies. Trying to fit this "extra" deflection into a force model does not work well. In some particular coordinate system, one might be able to interpret the "extra deflection" as due to some "velocity dependent force". But this idea is a bit of a kludge - it doesn't hold up well when one takes it seriously enough considers the details of how this "force" would transform with a change of coordinates.

#### Dale

Mentor
"gravitational time dilation". This is a consequence of gravity, but the fact that gravity has any effect whatsoever on time doesn't come from the "force" model at all. It arises naturally from the usual space-time curvature model, however.

There is no other "force" that affects time in this way.
The centrifugal force does.

#### PeterDonis

Mentor
The centrifugal force does.
More precisely, all "pseudo-forces" can be viewed as manifestations of spacetime geometry the same way the "force" of gravity can, so they all produce the same spacetime geometry effects.

#### A.T.

More precisely, all "pseudo-forces" can be viewed as manifestations of spacetime geometry the same way the "force" of gravity can, so they all produce the same spacetime geometry effects.
Coriolis?

"Is Gravity a Force?"

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