Why does a body at rest move if Gravity is not a force?

In summary, according to general relativity, gravitational force is not a real force but rather a result of the curvature of space-time. A body in motion follows an "invisible rail" while a body at rest cannot remain at rest relative to a large mass without an external force. This is explained by the principle of maximal proper time, which shows that an object cannot maximize its proper time by remaining at rest relative to a large mass. The body is at rest with respect to a free-falling observer, and proper acceleration is not relative and has nothing to do with relative motion. In short, all motion is relative in the context of general relativity.
  • #1
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hello

according to GR gravitational force is not a real force rather a space time curvature
ok a body in motion follows a sort of invisible rails
why doesn't a body at rest remain at rest, then ?
 
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  • #2
zoltrix said:
why doesn't a body at rest remain at rest, then ?
It does.
 
  • #3
I supposed it was something like that
however I definitely understand why relativity of simultaneity can explain schrink of body in motion of special relativity
I think also to grasp the reason why the acceleration of a body in presence of a mass is caused by the space time curvature
is the body at rest with respect to a certain observer ?
which one ?
may you expand on that ?
 
  • #4
zoltrix said:
hello

according to GR gravitational force is not a real force rather a space time curvature
ok a body in motion follows a sort of invisible rails
why doesn't a body at rest remain at rest, then ?
This question gets asked a lot. The short answer is that all motion is relative. And, relative to a large mass an object cannot remain at rest without a force holding it in place.

Note also that in Newtonian gravity it's not enough to say that gravity is a force. You also need Newton's second law telling you how an object subject to a force moves.

Likewise with GR it's not enough to say spacetime is curved. You also need the equivalent of Newton's second law, which is the principle of maximal proper time. When you add that you find that an object cannot maximise its proper time by remaining at rest relative to a large mass.
 
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  • #5
zoltrix said:
hello

according to GR gravitational force is not a real force rather a space time curvature
ok a body in motion follows a sort of invisible rails
why doesn't a body at rest remain at rest, then ?
Is this a version of "why does a hovering object start to fall", inspired by the rubber sheet model of space? The problem is that this model describes the curvature of space which is pretty much negligible in all but the most extreme circumstances. The interesting curvature is in planes orthogonal to this, which is why a worldline curves towards the source of gravity - remember that it has timelike extent.
 
  • #6
zoltrix said:
I think also to grasp the reason why the acceleration of a body in presence of a mass is caused by the space time curvature
No. The point is that the object does not accelerate (unless acted upon by a force) in the sense of proper acceleration. If you consider an object "falling" to the ground, it is actually the ground that has proper acceleration, not the falling object.

zoltrix said:
is the body at rest with respect to a certain observer ?
which one ?
A free-falling observer, which is an observer upon which no external forces are acting.
 
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  • #7
zoltrix said:
why doesn't a body at rest remain at rest, then ?
A body "at rest" in the space coordinates is moving through time. That means everything is moving in the spacetime coordinates of relativity. A free-falling object that appears to be accelerating due to gravity is following a "straight path" geodesic (not accelerating) in the spacetime coordinate system.
 
  • #8
FactChecker said:
A body "at rest" in the space coordinates is moving through time. That means everything is moving in the spacetime coordinates of relativity.
The question is why it is moving in the conventional sense of its spatial coordinates changing with time. Not why its clock is ticking.
 
  • #9
Orodruin said:
If you consider an object "falling" to the ground, it is actually the ground that has proper acceleration, not the falling object.
If either of the falling object or the ground has proper acceleration, wouldn't both of them have more than zero of it? Isn't the ground also a slightly falling object wrt to the lesser-massed object that is hurtlingly falling toward it? Why the complete exclusion of proper acceleration for the object of lesser mass?
 
  • #10
sysprog said:
If either of the falling object or the ground has proper acceleration, wouldn't both of them have more than zero of it?
No. Proper acceleration is not relative and has nothing to do with relative motion. It is what is measured by an accelerometer attached to the object.

sysprog said:
Why the complete exclusion of proper acceleration for the object of lesser mass?
It's not a matter of "lesser mass", it's a matter of whether the object is in free fall or is being subjected to a force (which, as noted in the thread title, does not include gravity). An object on the Earth's surface and at rest relative to the Earth is subjected to a force: the Earth's surface is pushing up on it. A rock falling after it is dropped, or a spaceship in orbit, is not subjected to a force. That's why they have zero proper acceleration.
 
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  • #11
sysprog said:
Why the complete exclusion of proper acceleration for the object of lesser mass?
It's not the lesser mass that makes the difference, it is the complete absence of non-gravitational (that is, "real") forces. The surface of the Earth is being pushed upwards by the material underneath it (otherwise it would freefall to the center of the Earth instead of being a surface) and it is that force that produces proper acceleration upwards. The free-falling object is not subject to any such force (until it hits the surface) so has no proper acceleration.
 
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  • #12
@PeterDonis, @Nugatory ##-## thanks for your explanations.
nugatory said:
The surface of the Earth is being pushed upwards by the material underneath it (otherwise it would freefall to the center of the Earth instead of being a surface) and it is that force that produces proper acceleration upwards.
Isn't that equally true of the surface of the object of lesser mass? I appreciate your illustration; however, wouldn't your example suggest that the proper acceleration of the surface of the lesser mass is non-zero?
 
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  • #13
zoltrix said:
hello

according to GR gravitational force is not a real force rather a space time curvature
ok a body in motion follows a sort of invisible rails
why doesn't a body at rest remain at rest, then ?
Note that locally even though the gravitational force is locally not a real force, it locally still is a fictitious (or inertial) force. Inertial forces certainly can and do produce coordinate acceleration in the reference frames where they exist.*

To get rid of inertial forces requires going to an inertial reference frame. The local frame attached to a free falling object is such a frame, and indeed free falling objects do not accelerate in such local inertial frames.

*added the qualifier “coordinate” see below
 
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  • #14
Dale said:
Inertial forces certainly can and do produce coordinate acceleration
The bolded qualifier I inserted above is important. Inertial forces cannot produce proper acceleration.
 
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  • #15
sysprog said:
Isn't that equally true of the surface of the object of lesser mass?
If that's the scenario you are interested in, yes. For example, if you're standing on the Moon, you have nonzero proper acceleration. But if you are at the Moon's center of mass, you don't. (Nor do you if you are at the Earth's center of mass.)

I don't think that is the kind of scenario the OP is asking about, though. I think the OP is asking about the simpler situation where only one body (such as the Earth) has non-negligible mass (i.e., enough mass to have measurable gravity).
 
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  • #16
sysprog said:
Isn't that equally true of the surface of the object of lesser mass?
In principle, yes, but it's incredibly small for anything smaller than a decent sized asteroid since gravity is very weak.
 
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  • #17
sysprog said:
appreciate your illustration; however, wouldn't your example suggest that the proper acceleration of the surface of the lesser mass is non-zero?
It is - indeed this will be true of anything that is not an idealized point particle with no surface or spatial extent.

It would be a good exercise to calculate the proper acceleration of the surface of, for example, a 10 kg iron cannonball in free fall. Compare this with the 9.8 meters per second per second of the surface of the Earth and you will see why we feel justified in treating all parts of the cannonball as if they are in free fall.
 
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  • #18
Orodruin said:
No. The point is that the object does not accelerate (unless acted upon by a force) in the sense of proper acceleration. If you consider an object "falling" to the ground, it is actually the ground that has proper acceleration, not the falling object.
OK the "Einstein's elevator" is well known
a good starting point but far away from being a comprehensive and exhaustive explanation ,in my opinion
Here in the old Europe material bodies use to fall down with an improper acceleration
The same in Australia, I suppose
where is such common ground ?
I think that the correct explanation come from your own words

"You did not take relativity of simultaneity into account."

such concept shout be expanded for a correct explanation
relativity of time and space does not depend on motion only
also mass and energy have an effect on the flow of time and space and cosequenti on velocity and acceleration
in other words also the "improper " acceleration should be a matter of "metrics" same as the apparent shrink of bodies in motion
 
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  • #19
This is one of the few occasions when relativity of simultaneity has nothing to do with the resolution.

Instead, it is a question for the curvature of spacetime and how that results in the surface of the Earth being stationary despite having non-zero proper acceleration.
 
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  • #20
zoltrix said:
Here in the old Europe material bodies use to fall down with an improper acceleration
The same in Australia, I suppose
where is such common ground ?
The curvature of spacetime does not explain local effects, like the fact that locally the force of gravity is an inertial force which disappears in local inertial frames. The curvature of spacetime does explain non-local effects like the fact that the surface of the Earth is not expanding even though it is accelerating outward at each point.

In flat spacetime, the fact that each point on a sphere has an outward proper acceleration would imply that the sphere would be expanding. But the spacetime around the Earth is not flat, and so that reasoning does not hold.
 
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  • #21
Dale said:
The curvature of spacetime does explain non-local effects like the fact that the surface of the Earth is not expanding even though it is accelerating outward at each point.
In the above could we think of the surface of the Earth as a continuous set of "hovering" observers in Schwarzschild spacetime ?
 
  • #22
cianfa72 said:
In the above could we think of the surface of the Earth as a continuous set of "hovering" observers in Schwarzschild spacetime ?
yes
 
  • #23
zoltrix said:
I supposed it was something like that
however I definitely understand why relativity of simultaneity can explain schrink of body in motion of special relativity
I think also to grasp the reason why the acceleration of a body in presence of a mass is caused by the space time curvature
is the body at rest with respect to a certain observer ?
which one ?
may you expand on that ?

Are you familiar with the concept of geodesics - and geodesic motion?

I would suggest re-interpreting your question in terms of "geodesic motion", mainly because it's rather unclear exactly and precisely what you mean by "a body at rest".

Let's take an example. An airplane is flying a parabolic trajectory which makes it's passengers weightless. We consider only the part of the airplane's trajectory in which the passengers on the airplane are "weightless".

Is the airplane "at rest"? I would say that would be an unusual statement. But the airplane is undergoing geodesic motion.

Now consider someone standing on the surface of the Earth. Said person is not undergoing geodesic motion. But you'd probably consider them to be "at rest", at least I'd assume you might. It's hard to be sure.

Now, consider a person on a train moving at a constant groundspeed around the equator. Is the train "at rest?" It would be possibly a bit unusual to say that the train was at rest, but not unheard of.

In terms of whether the motion is geodesic or not, the example of the train moving at a constant groundspeed along the equator may or may not be undergoing geodesic motion. If the train's velocity relative to the ground is such that the passengers on it are weightless, it is following a geodesic path. This requires one specific velocity of the train. If the train does not have that exact velocity, then it is not undergoing geodesic motion.

Finally, let's consider someone on Einstein's elevator. They are not in the presence of any significant mass - the elevator is assumed to have negligible mass. Is the person standing on the elevator "at rest?" It's unclear. But they're not undergoing geodesic motion.

Having talked about "bodies at rest" and the more precise concept of "bodies undergoing geodesic motion"geodesic motion, we can now talk about gravity and curved space-time.

In the flat space-time of special relativity, two bodies undergoing geodesic motion have zero relative acceleration. We'd say they have a constant relative velocity.

In curved space-time, in the presence of mass, two bodies undergoing geodesic motion will not, in general, have zero relative acceleration.

And that's mostly all there is to gravity, the fact that two bodies undergoing geodesic motion have a non-zero relative acceleration. This is formalized by the "geodesic deviation equation" of General relativity. See for instance the wiki article https://en.wikipedia.org/wiki/Geodesic_deviation, though I'm not sure how clear it will be. The relative acceleration betwen two initially parallel geodesics can be used to define the Riemann curvature tensor, called the Riemann tensor and usually denoted by R, which appears on the right hand side of the geodesic deviation equation:

$$D^2 \frac{x^a}{d\tau^2} = R^a{}_{bcd} u^b u^c \delta(x^c} $$.

The Riemann tensor here is the mathematical entity R.

There's one final clarification to be made. The notion of "geodesic motion" actually applies precisely only to a point particle, not an extended body. If the body is small enough, the distinction isn't too important. But for a large enough body, it becomes significant.

Consider again the example of the train running around the equator such that the passengers are weightless. If you look at the problem more precisely, the notion that I'm calling weightless, and also calling "geodesic motion", applies exactly only at a point. (In terms of space-time, it applies to a worldline.) For instance, a point on the floor of the train might be undergoing geodesic motion, while a point on the ceiling would then have the wrong velocity (the train is assumed to be rigid), and the point on the ceiling would not be precisely following a geodesic. But if the train isn't too tall, the "weight" (aka proper acceleration) of a point (worldline) on the ceiling would be small enough to be ignored.
 
  • #24
pervect said:
If the train's velocity relative to the ground is such that the passengers on it are weightless, it is following a geodesic path. This requires one specific velocity of the train.
You mean the right velocity like the velocity needed for a spaceship to free-fall in orbit around the Earth, I believe.
 
  • #25
PeterDonis said:
If that's the scenario you are interested in, yes. For example, if you're standing on the Moon, you have nonzero proper acceleration. But if you are at the Moon's center of mass, you don't. (Nor do you if you are at the Earth's center of mass.)

I don't think that is the kind of scenario the OP is asking about, though. I think the OP is asking about the simpler situation where only one body (such as the Earth) has non-negligible mass (i.e., enough mass to have measurable gravity).
I was inquiring regarding this remark from @Orodruin:
Orodruin said:
If you consider an object "falling" to the ground, it is actually the ground that has proper acceleration, not the falling object.
I guess I was curious about the minuscule comparative mass "not" having "proper acceleration" -- It seems to me that if the Earth has it, then the less massive object must have it too, albeit neglibly, but not non-existently.
 
  • #26
sysprog said:
It seems to me that if the Earth has it, then the less massive object must have it too, albeit neglibly, but not non-existently.
See post #17 from @Nugatory.
 
  • #27
PeterDonis said:
See post #17 from @Nugatory.
In my view, that post asserts negligibility reasonably, but doesn't assert non-existence.
Nugatory said:
It is - indeed this will be true of anything that is not an idealized point particle with no surface or spatial extent.
That seems to assert existence . . .
Nugatory said:
It would be a good exercise to calculate the proper acceleration of the surface of, for example, a 10 kg iron cannonball in free fall. Compare this with the 9.8 meters per second per second of the surface of the Earth and you will see why we feel justified in treating all parts of the cannonball as if they are in free fall.
. . . that seems to assert negligibility.
 
  • #28
sysprog said:
that post asserts negligibility reasonably, but doesn't assert non-existence.
True. So what?

I think you are taking statements about "zero proper acceleration" much too literally. Nobody has made a dogmatic assertion of absolute non-existence anywhere in this thread. For many purposes, "negligible" and "non-existent" are equivalent in practical terms, so there is no point in trying to excessively police our statements to make sure we always leave open the possibility of something being present but negligible.
 
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  • #29
PeterDonis said:
An object on the Earth's surface and at rest relative to the Earth is subjected to a force: the Earth's surface is pushing up on it.
Isn't the object's surface pushing its own 'up', however weakly, against the Earth, even though the object is of so much lesser mass?
 
  • #30
sysprog said:
Isn't the object's surface pushing its own 'up', however weakly, against the Earth, even though the object is of so much lesser mass?
By Newton's Third Law, obviously the object's surface pushes on the Earth's surface with the same force that the Earth's surface pushes on the object's surface. But that has nothing to do with the object's self-gravity.

The effect of the object's self-gravity (if we ignore the fact that it's negligible in this scenario) will not be to "push" back on the Earth but to make the distribution of internal stresses inside the object, due to its response to the Earth's gravitational field, very slightly different than they would be if we only took into account non-gravitational interactions between the atoms in the object.
 
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  • #31
PeterDonis said:
True. So what?

I think you are taking statements about "zero proper acceleration" much too literally. Nobody has made a dogmatic assertion of absolute non-existence anywhere in this thread. For many purposes, "negligible" and "non-existent" are equivalent in practical terms, so there is no point in trying to excessively police our statements to make sure we always leave open the possibility of something being present but negligible.
I think that when we're saying things in a manner which suggests that we are discussing them 'in principle', we should distinguish beween 'is not present' and 'is for most practical purposes negligible' ##-## the mass of a small object compared to that of the Earth may be negligible for many purposes, but I think that it's not always legitimately disregardable ##=## e.g. when we are accounting for why the object remains attached to the Earth ##-## it's because the Earth and the object are pushing against each other; it's not only because the Earth is pushing against the object.
 
  • #32
PeterDonis said:
By Newton's Third Law, obviously the object's surface pushes on the Earth's surface with the same force that the Earth's surface pushes on the object's surface. But that has nothing to do with the object's self-gravity.

The effect of the object's self-gravity (if we ignore the fact that it's negligible in this scenario) will not be to "push" back on the Earth but to make the distribution of internal stresses inside the object, due to its response to the Earth's gravitational field, very slightly different than they would be if we only took into account non-gravitational interactions between the atoms in the object.
The object's mass pushes back against the push of the Earth, and both the small object and the Earth have non-zero 'self-gravity', and if we're explaining, I think that we should in this matter not disregard the difference between 'zero', and 'non-zero but much smaller, and for most purposes negligible, but still for some purposes effectual'.
 
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  • #33
cianfa72 said:
In the above could we think of the surface of the Earth as a continuous set of "hovering" observers in Schwarzschild spacetime ?
Yes. A set of such hypothetical observers is called a congruence. The congruence for the surface of the Earth is the same as the congruence for a sphere of hovering Schwarzschild observers, e.g. outside an equal mass black hole.
 
  • #34
sysprog said:
the mass of a small object compared to that of the Earth may be negligible for many purposes, but I think that it's not always legitimately disregardable e.g. when we are accounting for why the object remains attached to the Earth it's because the Earth and the object are pushing against each other; it's not only because the Earth is pushing against the object.
You evidently have not done the calculations that @Nugatory suggested in an earlier post. The fact that the object remains "attached to" the Earth has nothing to do with the object's self-gravity. The fact that the Earth and the object are pushing against each other with equal force, as I said in my previous post, is due to Newton's Third Law, and has nothing to do with the object's self-gravity. The ultimate reason why the trajectory of the object and the piece of the Earth it is pushing on are stable is the spacetime curvature due to the Earth; it has nothing to do with the (negligibly small) spacetime curvature due to the object.

sysprog said:
The object's mass pushes back against the push of the Earth, and both the small object and the Earth have non-zero 'self-gravity', and if we're explaining, I think that we should in this matter not disregard the difference between 'zero', and 'non-zero but much smaller, and for most purposes negligible, but still for some purposes effectual'.
The bolded part is false (see above). What you are actually doing here is giving a very good illustration of why it is not a good idea to include negligible quantities like the object's self-gravity in this discussion: it invites the misconception that you have confused yourself into.
 
  • #35
PeterDonis said:
You evidently have not done the calculations that @Nugatory suggested in an earlier post. The fact that the object remains "attached to" the Earth has nothing to do with the object's self-gravity. The fact that the Earth and the object are pushing against each other with equal force, as I said in my previous post, is due to Newton's Third Law, and has nothing to do with the object's self-gravity. The ultimate reason why the trajectory of the object and the piece of the Earth it is pushing on are stable is the spacetime curvature due to the Earth; it has nothing to do with the (negligibly small) spacetime curvature due to the object.The bolded part is false (see above). What you are actually doing here is giving a very good illustration of why it is not a good idea to include negligible quantities like the object's self-gravity in this discussion: it invites the misconception that you have confused yourself into.
I think that in this instance calling something that is "negligibly small", "nothing", is occlusive of the reality that an object affected by gravity cannot be massless (outside of black holes bending light, depending on whether light is regarded as massless).

I think that it's a disagreement; not a misconception. I didn't say that I thought that the object's 'self-gravity' had anything to do with its attachment to the Earth; that is accounted for by the mass of the object and of the Earth; but if it were 'massless', there would be no Third Law equal pressure holding it to the Earth.

I inquired as to why when explaining something in principle, a choice is made to exclude consideration of a comparatively small quantity, and I think that I was given good reason and good explanation for that by you and by @Nugatory, and I in this instance respectfully disagree that the mass of the smaller object should be neglected, even if it is quantitatively comparatively negligible, because I think that the qualitative fact of the non-zero-ness of the mass of the small object should be part of a full explanation.
 
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