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GR starts with a principle called the Equivalence Principle, which states that gravity is indistinguishable from acceleration locally.

Here, what do we mean by gravity?

Case 1: Do we mean gravity as a force? But then what sense does it make to say "A force called Gravity is indistinguishable from acceleration locally"?

Case 2: Do we mean gravity as spacetime curvature? If this is the case, then doesn't it mean that General Relativity starts with the assumption of spacetime to be curved? But then, what sense does it make to say that solutions of EFEs predict spacetime to be curved, given that we had started with a principle that needs spacetime to be curved?

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9 Answers 9

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General relativity changes the language of Newtonian mechanics at a fundamental level, such that it is no longer completely straightforward to describe what a force is.

To see this, consider that Newton's second law involves the acceleration $$ m \frac{d^2 x^i}{dt^2} = F^i $$ where $x^i=(x, y, z)$ is a position vector in three-dimensional space, and $F^i$ is the force vector. In Newtonian mechanics, there is an absolute time that everyone agrees on. However, even in special relativity, this is no longer true, because of the relativity of simultaneity.

Now, in special relativity, there is a more or less straightforward solution. So long as we stick to inertial observers, we can define a quantity called the "four-acceleration", $A^\mu$, where now $\mu=(t,x,y,z)$ is a label for a four-dimensional vector. The four-acceleration reduces to the usual notion of acceleration in the small-velocity limit, and otherwise is related to the change of velocity of some object, according to an inertial observer. We can set the rest mass times four-acceleration equal to a generalization of the force called the four-force, and end up with a version of Newton's second law.

But now we need to take another step, to general relativity, and now we can no longer assume there are inertial observers in the same way as in special relativity. There are locally inertial observers -- in other words, over small regions of spacetime, you can pretend you are working with special relativity, just like you can pretend geometry is Euclidean so long as you only work on a small patch of the Earth where you can ignore the curvataure. But we want a formulation that will work for any observer, and relate different inertial observers.

The generalization of Newton's second law that includes the effect of non-inertial coordinates is the geodesic equation $$ \frac{d^2x^\mu}{d\tau^2} + \Gamma^{\mu}_{\nu\lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} = 0 $$ where $\Gamma^\mu_{\nu\lambda}$ are the Christoffel symbols, $x^\mu=(t,x,y,z)$ is a set of coordinates in four dimensional spacetime, and $\tau$ is proper time (a parameterization of a curve, such that $g_{\mu\nu} \frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}=-1$, where $g_{\mu\nu}$ is the metric tensor and I'm using a mostly plus sign convention). This equation can be rearranged to be $$ \frac{d^2x^\mu}{d\tau^2} = - \Gamma^{\mu}_{\nu\lambda} \frac{dx^\nu}{d\tau} \frac{dx^\lambda}{d\tau} $$ which is analogous to the force law we started with, and in an appropriate limit, the "$\Gamma$" term does reduce to the Newtonian force in the way you expect. So you could say that general relativity does contain an idea of a gravitational force on particles that is analogous to the Newtonian one.

However, from the more abstract perspective we need to even be able to talk about any of these quantities at all, the "$\Gamma$" term is part of what is called a "covariant derivative" and arises due to the fact that we need to consider non-inertial coordinates. Even without gravity, you can choose coordinates where this term is non-zero (essentially amounting to having fictitious forces in non-inertial frames). But gravity forces you to have this term, due to the presence of spacetime curvature. Therefore, it's also valid to say that spacetime curvature is the real reflection of gravity, and "force" is just an old-fashioned concept that is awkward to formulate in the new and more powerful language we need to use.

If GR were the end to the story, we would probably just say that gravity is spacetime curvature, and "force" is a word that applied to an old formulation of physics, and can be given meaning in the new and better formalism, but only in an ugly way. However, since we ultimately expect to be able to unify gravity with the other forces of nature, it seems like there should be some unified picture where we can talk about the forces in the standard model and gravity in the same language. Whether that requires us to reformulate the standard model (and quantum mechanics) in a more geometric language, or reformulate general relativity in a more "quantum interaction" based language, or something else, remains to be seen.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Nov 27 at 17:48
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The idea of gravity as a force comes from the fact that it makes objects deviate from straight paths in flat spacetime. Following Newton's laws, this means that a force acts on them. However, in general relativity, the idea of a "straight line" is generalized to a geodesic on a manifold, meaning curves of minimal length connecting two points. Note that in flat space, these are exactly straight lines. In general relativity, the notion of a force still exists, but instead of measuring deviation from a straight path, it measures the deviation from geodesic motion. The force you feel on Earth which you interpret as gravity pulling you down is instead the Earth pushing up against your body, keeping you from performing geodesic motion into the Earth's core. Indeed, you stop feeling this force when in free fall, since you are now performing geodesic motion. This is what Einstein noticed as well when formulating the theory.

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  • $\begingroup$ But this doesn't answer my question. In GR, is Gravity a force? (Tendency of objects to follow) curvature of spacetime? Or something else? Can you first give me a single word answer, i.e. Case 1, Case 2, something else. Then elaborate? $\endgroup$
    – user247935
    Commented Nov 25 at 17:04
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    $\begingroup$ @Ekalabya No it’s not a force, at least in GR. In GR (and in some QM theories) “gravity” is a name we use to describe an effect that appears to put force on bodies but really is just the bodies following “straight lines” where the underlying manifold is curved. A “force” describes something that causes a body to deviate from those “straight” geodesics. $\endgroup$ Commented Nov 25 at 17:05
  • $\begingroup$ Your answer essentially means that in GR, by Gravity we mean tendency of objects to follow curvature of spacetime i.e. case 2. Isn't it? $\endgroup$
    – user247935
    Commented Nov 25 at 17:13
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    $\begingroup$ @Ekalabya that is the case. Famously, this is usually stated as "Matter tells space how to curve, space tells matter how to move" $\endgroup$
    – paulina
    Commented Nov 25 at 17:55
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    $\begingroup$ @Swarupananda There's no logical fallacy, you're just mistaking the equivalence principle to take a stand on whether gravity is a force or not, which it doesn't. It is a physically testable thing to say "This body under acceleration will act the same if it was in a uniform gravitational field".. well at least to some extent. There is no such thing as a uniform gravitational field, but experiments deal with all kinds of practical limitations anyway, that's not the main issue. The issue is that the equivalence principle makes a testable prediction without having to refer to force nor curvature $\endgroup$
    – Amit
    Commented Nov 26 at 21:25
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I think that such a question can only be answered by taking into account a historical perspective. Since the Equivalence principle, in its original form, in fact preceded the full theory of general relativity, one can only partly speculate and try to glean from the various written papers and other sources by Einstein, whether he was already aiming to reformulate the view of gravity from being considered a force, as we say that Newton took it to be, to become a geometrical property of spacetime which is our modern viewpoint today.

My reading of history, and I am certainly no expert, is that in the relatively (pun unintended) early days, circa 1907, after the publication of the special theory of relativity, Einstein was simply interested to work out the consequences of applying the same physical postulates of relativity to accelerating reference frames.

One of his insights was that indeed, a uniformly accelerating frame can be taken to be as indistinguishable from a frame that's in the presence of a uniform gravitational field. He was able to work out a few consequences from that insight alone, for example the slowing down of clocks in a gravitational field, in 1907, still 8 years before the publication of the final GR papers. However, during that time and in particular within this very paper, there is no talk of "curvature", and no particular reference to "geometry of spacetime", etc. I am inclined to believe that at this time, Einstein still didn't form the view that gravity was such a radically different phenomenon from, for example, the electric and magnetic forces he knew about.

One of the best Einstein biographies is the one by Abraham Pais, "Subtle is the lord...". In it, page 210, he more or less gives his view on when Einstein formed his conviction about the relation of geometry and gravitation:

In August 1912, Einstein and his family arrived back in Zurich. On the tenth of that month they were officially registered as residents of an apartment at Hofstrasse 116. Some time between August 10 and August 16, it became clear to Einstein that Riemannian geometry is the correct mathematical tool for what we now call general relativity theory. The impact of this abrupt realization was to change his outlook on physics and physical theory for the rest of his life. The next three years were the most strenuous period in his scientific career.

In order to appreciate what happened in August 1912, it is essential to know that before his arrival in Zurich Einstein had already concluded that the description of gravitation in terms of the single scalar c-field of the Prague days had to go and that a new geometry of physical space -time was needed. I am convinced that he arrived in Zurich with the knowledge that not just one but ten gravitational potentials were needed. This opinion is based on some remarks in Einstein’s papers; on a study of all the letters from the period March- August 16, 1912, which are in the Einstein archives in Princeton; and on recollections by myself and by Ernst Gabor Straus, Einstein’s assistant from 1944 to 1948, of conversations with Einstein.

So, I think the answer to your question is simply that the various sources you've probably came across and introduce the equivalence principle while still taking gravity as a "force", are simply following along the footsteps of Einstein himself as he took them towards the final theory of general relativity. Of course, Einstein himself had to make a lot more twists and turns on that path, and what we're learning about is a kind of an abridged version of what really happened.

I think that all this really means is that the equivalence principle isn't really as fundamental as we perhaps take it to be. In an alternative universe, I can imagine someone working out the theory of GR without postulating such a principle. It is only because we didn't have a geometrical view of gravity, that we needed to pass through that stage, i.e. notice that gravitational mass is identical with inertial mass, then from that hypothesize that gravity is similar to a fictitious force that arises in a uniformly accelerated frame, etc. But if we could see right away that gravity is different, the matter of the identity of gravitational and inertial mass would be a matter of course, and the equivalence principle would not be a necessary stepping stone.

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  • $\begingroup$ My point is on the conclusion of spacetime being curved. Is that really a conclusion of GR? $\endgroup$
    – user247935
    Commented Nov 26 at 18:51
  • $\begingroup$ I'm not sure what you mean by conclusion. A physical theory makes predictions. The mathematics of GR, which results in the predictions of this theory, implies that spacetime is curved, yes. (Then again, someone may come along and find a theory that makes the same predictions, without curvature of spacetime being part of the description). $\endgroup$
    – Amit
    Commented Nov 26 at 19:03
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GR starts with a principal called Equivalence Principal, that Gravity is indistinguishable from acceleration locally. Here, what do we mean by Gravity?

The equivalence principle doesn’t really care if you’re talking about gravity or other “forces” or what have ye. Quoth the equivalence principle Wikipedia article, the Einstein equivalence principle states that “the outcome of any local, non-gravitational test experiment is independent of the experimental apparatus' velocity relative to the gravitational field and is independent of where and when in the gravitational field the experiment is performed.” Essentially: under a uniform gravitational field (which is understood in the context of GR as a manifold of uniform curvature), we have Lorentz and positional invariance for all physical processes: it doesn’t matter what our velocity vector is relative to anything and it doesn’t matter what our position vector is relative to anything. “Local” means that the process happens in a small region, i.e. one where the gravitational field is roughly constant.

What the equivalence principle is often taken to mean is that you cannot distinguish between sitting on the ground in a box and sitting in a 1-$g$-accelerating rocket in a box; that is, inertial and gravitational mass are exactly equivalent.

To answer your questions: in GR, we say that gravity is an effect of spacetime being curved, and is thus not a “force” in the traditional sense; you’re not being forced in any direction by any other body, you’re following a “straight line” that happens to be on a curved manifold; a true “force” would be something that makes your trajectory deviate from these geodesics (paulina’s answer has a good explanation of this).

Now,

If gravity is spacetime curvature, then doesn't it mean that General Relativity starts with the assumption of spacetime to be curved?

GR does not make that assumption. The presence of gravitational “forces” implies that spacetime is curved in some way, but you are implicitly assuming that there are such fictitious forces, which is not a necessary premise of GR. A common example is a gravitational wave propagating in otherwise-flat space. That doesn’t require any significant “gravity”, and aside from the G-wave there’s no spacetime curvature, but it’s still a very interesting GR problem that is undergoing active research.

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    $\begingroup$ GR is a theory of gravity. The purpose of GR was to combine SR with Gravity. But to do so, one must assume that Gravity exists. $\endgroup$
    – user247935
    Commented Nov 25 at 17:34
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    $\begingroup$ It actually does not make that assumption. In GR our foremost assumption is that our spacetime can be modeled as a 3+1-dimensional Lorentzian manifold, and all the effects of whatever have ye come from our choice of a metric on that manifold, which does not necessarily have to be one that produces what we normally think of as gravity. $\endgroup$ Commented Nov 25 at 17:49
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    $\begingroup$ The equivalence principle, too, does not actually depend on anything gravitational. It just posits Lorentz invariance for processes, and sometimes (local) positional invariance. This also leads to gravitational and inertial mass being equated if you consider the case of acceleration due to a rocket vs gravity. $\endgroup$ Commented Nov 26 at 2:58
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    $\begingroup$ Like I said - GR is a general theory of the way physics behaves in curved spacetime, which does not necessarily include gravity, and which has SR as a limiting case as spacetime becomes flat (Minkowski). Like I said in my post, the equivalence principle does not actually state anything about gravity in particular (neither does GR), but gravity and the properties thereof arise as consequences of equivalence and GR. "SR" is me taking GR and saying "We start with an everywhere-flat Minkowski space...", and gravity is me taking GR and saying "Consider the following black hole metric...". $\endgroup$ Commented Nov 26 at 3:11
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    $\begingroup$ @Ekalabya no GR is not a theory of gravity. Is a theory of locally inertial observers. It differs from SR because SR is a theory of GLOBALLY inertial observers. GR expands the theory of SR by dropping the assumption of global inertiality. Now gravity comes into play as a SPECIAL case of locally inertial observers. But GR is a theory made to describe all types of locally inertial observers, not only the ones associated to gravity. $\endgroup$
    – LolloBoldo
    Commented Nov 26 at 13:32
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Both.The famous experiment on the elevator with a accelerometer concludes that by switching between non-inertial frames we perceive gravity differently so gravity<-->spacetime curvature.

If your frame of reference is g m/s^2 towards a object which has a gravitational pull on objects then gravity dissapears but if you are in any other frame of reference you can feel it VERY much.

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  • $\begingroup$ Are you suggesting to introduce Force Curvature duality for Gravity? $\endgroup$
    – user247935
    Commented Nov 26 at 2:56
  • $\begingroup$ No as I have said it depends on which frame of reference you are considering. $\endgroup$ Commented Nov 26 at 8:16
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    $\begingroup$ @Ekalabya It is not how it works. Physics look at the world, see something and name it. Gravity is just "something", as well as force or curvature of space. "Gravity" is just a label that we put on this "something". "Force" is another label on something else, as well as "Curvature". We see "gravity" affecting reality. We say that the effect is similar to "force", and we say that the effect is similar to "curvature". So the classical theory interprets "gravity" as "force", while GR interprets "gravity" as "curvature". None of them can definitely establish what gravity is. $\endgroup$ Commented Nov 26 at 13:24
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    $\begingroup$ ...and nobody actually cares what it is as long as our equations help us to predict gravity's effects on the rest of world. "Gravity is gravity". $\endgroup$ Commented Nov 26 at 13:33
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    $\begingroup$ @RootGroves I think user13964273 is making a different point which can be put more concisely as: distinguish between phenomena and the theories that we use to describe them. Gravity is a phenomenon. Curvature is part of a theoretical description. Hence, it is a categorical error to ask if they are identical $\endgroup$
    – Amit
    Commented Nov 26 at 20:07
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Both perspectives are equivalent - you can use either to model gravity, and you will find that both result in the same equations governing the dynamics.

To better understand this, let us walk through the content of the weak equivalence principle step-by-step

  1. We make the observation that in a uniform gravitational field, any acceleration due to gravity experienced by particles can be eliminated by a transformation to an accelerating reference frame. So, in this case, the effect of gravity is equivalent to that of an accelerated reference frame.

Note: I purposefully use the phrase "effect of gravity" here instead of "force of gravity" so as to not bias myself to either the force or curvature perspective for analysing gravity.

  1. Next, we notice that in a non-uniform gravitational field, the above observation does not hold due to tidal effects. But, if we focus only on a small region of space-time, then no tidal effects occur, and we can still eliminate the effect of gravity by going to an accelerated reference frame, just like the previous point. So, this means that locally, the effect of gravity is indistinguishable from acceleration (at least for particle motion). This is the weak equivalence principle.

Note: So, your statement "A force called Gravity is indistinguishable from acceleration locally" is not entirely accurate, as it is biased towards the force perspective (Case 1). This statement holds even in the curvature perspective (Case 2), and is better phrased as "The effect of gravity is indistinguishable from acceleration locally".

  1. Finally, we see that in a non-uniform gravitational field, particles move on curved paths. So, we can see that the effect of gravity is to cause the particle to move on a curved path.

Collecting all these observations, we see that we can describe the effect of gravity in two ways -

  1. We can treat gravity like an external force which causes particles to move on curved paths. This external force causes particles to deviate from the straight-line motion (geodesic motion) they would have had if gravity was not there.
  2. We can treat gravity as the curvature of space-time. In this perspective, the particles do not experience an external force, but their motion is still curved as the geodesics of space-time are now curved (due to the curvature of space-time). Note that this perspective exists only because of the weak equivalence principle.

Note: Thus, we see that if we take the second perspective, then we do assume that gravity is modelled by curved space-time. Exactly what this curvature is, is given by the EFEs. Additionally, if there is no gravity, then this perspective indicates that space-time should be flat, which is also what the EFEs tell us.

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"Gravity" usually means a force. In GR, it's vogue to say "gravitation", which mean all of gravity.

For example, if your coordinates are defined by the horsey on which you sit (on a merry-go-round spinning at ${\bf \Omega}$), you can do Newtonian calculations provided you include external forces:

$$ {\bf F}_g = m\vec g $$ $$ {\bf F}_c = -m{\bf \Omega} \times ({\bf \Omega} \times {\bf r})$$ $${\bf F}_C = -2m{\bf \Omega} \times \dot{{\bf r}} $$

Traditionally: the last two go away if you change from carousel-fixed to carousel-inertial coordinates, because inertial forces are a result of non-inertial coordinates. This is basic 3D Euclidean stuff.

So in theme-park coordinates, you're revolving (and rotating) at ${\bf \Omega}$ (which are mainted b a centripetal force from the ride's structure) with an external force described by Newton:

$$ {\bf F}_g = m\vec g $$

(actually, since the Earth is spinning, we need to that once more, but let's ignore that for now).

Thanks to GR, the remaining force, "gravity", goes away if we let the carousel be in free-fall. In 4D curved geometry, the Earth's surface is applying a constant force to the ride, and it is accelerating at ${\bf g}$ away from the geodesic.

So in weak field local GR, all gravitation is a result of choosing the wrong coordinates.

Now I said "weak field" because gravity is non-linear. We can't just put a black hole on the ride an expect my analysis to make any sense.

Let's imagine a horizon bisects the ride. Then a force analysis is useless, and everything is due to $g_{\mu\nu} \ne \eta_{\mu\nu}$ (even if $G_{\mu\nu}=0$).

So that's your answer: gravitation is always the curvature of the metric; however, in weak fields (that being the scale of the problem is much larger than the Schwarzschild radius of the masses in the problem) you can treat gravitation as forces.

If you choose inertial coordinates, then you can get by with external fields ${\bf E}_g$ and ${\bf B}_g$ that cause a force:

$$ {\bf F}_g = m\Big({\bf E}_g + {\bf v}\times 4{\bf B}_g\Big)$$

(see: https://en.wikipedia.org/wiki/Gravitoelectromagnetism)

Now if you choose non-inertial coordinates, then you need to include centrifugal and Coriolis terms too; however, they can be included in the above ${\bf E}_g$ and ${\bf B}_g$. That is: all inertial forces are manifestations of gravitation, and gravitation is described by the metric.

tl;dr Physics doesn't care about your choice of coordinates, nor whether you describe it as a force, when possible.

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  • $\begingroup$ What I understood, your answer is case 2. But then can you answer the logical fallacy that I posed in my original question? $\endgroup$
    – user247935
    Commented Nov 26 at 2:54
  • $\begingroup$ I am surprised by your distinction between the terms "gravity" and "gravitation". I don't think this is correct. $\endgroup$ Commented Nov 27 at 4:01
  • $\begingroup$ @ColinMacLaurin I learned it directly from Kip Thorne, when he was a mere pre-Nobel demi-god of gravitation, so I'm going with it ;-). $\endgroup$
    – JEB
    Commented Nov 30 at 5:49
  • $\begingroup$ Thorne is a leading expert, but it is still a surprising distinction. Do you have a reference? Or, if your communication was verbal, what was the context? Perhaps the terminology is specific to gravitomagnetism, weak fields, or Thorne himself. I am used to people including centrifrugal effects in "surface gravity" e.g. for Earth. But in most contexts in GR, people take gravity (or gravitation) to mean curvature alone. (Though personally I suspect a convention of inertia should play more of a role.) $\endgroup$ Commented Dec 1 at 7:28
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It is usually said that in Newtonian gravity, gravity is a force, but that in GR, gravity is not a force and is explained by curvature. However this is a false dichotomy. Instead, gravity is explained by curvature when we use a modern geometrical description -- that is, a manifold with connection $\nabla$ -- of either theory. (A connection tells you how to take derivatives. And this is needed to define acceleration.) Such a formulation of Newtonian gravity is known as Newton-Cartan theory.

What is gravity? The usual answer of "spacetime curvature" seems excellent to me. However this is partly a language choice. After all, the obvious next question is: what is curvature? There is a formulation of GR known as teleparallel gravity. It has exactly the same physical predictions, from what I am told, but involves no curvature but only torsion. So you can call it curvature, or call it torsion, but more importantly here (in physics), what are the physical observables? The foremost example offered is typically geodesic deviation. Roughly, it means that if you have a small ball of freely falling test particles, then its volume and/or shape will distort over time, due to gravity.

I see no reason why the version of the Equivalence Principle you cite, about gravitation and acceleration being indistinguishable (with certain nuances), wouldn't hold similarly for Newtonian gravity. So what is the difference between the theories? Newtonian gravity is non-relativistic, and also changes in the gravitational field propagate instantaneously. GR is relativistic, and changes in gravity propagate only at the speed of light. Finally, I have not discussed the convention of "gravitational" inertia (or whatever it should be called; this will be unfamiliar to many readers, but I mean the gauge freedom reviewed in Malament 2012 §4.2). This would modify some of my statements.

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  • $\begingroup$ Note: ColinsCosmos site cannot be reached. $\endgroup$
    – Wookie
    Commented Nov 28 at 19:30
  • $\begingroup$ thank-you Wookie. This one still works: cmaclaurin.com $\endgroup$ Commented Dec 1 at 7:40
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View of an object/observer in space

The movement of an object in space depends exclusively on the distribution of celestial bodies and other masses and its own instantaneous speed (velocity and direction). The motion trajectory will be a curvilinear one, depending on the local change of the gravitational potential along its trajectory.

The observer will not notice this curvilinear trajectory without a view of the surrounding celestial bodies, because no acceleration can be detected. If the observer in the windowless spaceship is caught by the Earth's gravitational field, he will have no idea of the approaching catastrophe until the impact (neglecting the atmospheric braking effect).

TL;DR Since no acceleration acts on the observer along the geodesic path, no force acts on him. In GR, there are geodesic paths clearly determined by position and speed and no force. And it becomes more complicated when the object itself has a significant mass. For example, the Earth and the other planets have an influence on the (empirically confirmed) wobbling motion of the sun. Therefore, when describing the gravitational potential of a section of the sky, one should refer to the movement of light as a basis. Its photons are massless and all have the speed c; therefore, their geodesics are also uniquely determined. But be careful here too, c is a local quantity from the point of view of a third observer. At a higher gravitational potential (for example near a black hole), c is smaller than at a lower potential. TL;DR

View of an observer on earth

An object falling to earth gains speed and kinetic energy. F=m*g describes the process exactly and g is the acceleration acting on an object.

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