Question about Gravity and curvature of space time

[A.T.]

in the case where I drop a rock, is "gravitational acceleration" the coordinate acceleration of the rock in my rest frame, or the proper acceleration I feel that the rock doesn't?)

Your proper acceleration is due to electromagnetic repulsion. Gravitational acceleration is just coordinate acceleration in General Relativity.

 

[PeterDonis]

Gravitational acceleration is just coordinate acceleration in General Relativity.

I agree this is the most natural interpretation (and it still makes the term "gravitational acceleration" superfluous), but I don't think the GR literature is consistent on this point. For example, I've seen the term "gravitational acceleration" used to mean the proper acceleration necessary to hold an object stationary (with respect to the timelike Killing vector field of the spacetime).

 

[TrickyDicky]

I don't think that use is standard.
I know wikipedia is not always right but this excerpt looks right to me:"The "acceleration of gravity" ("force of gravity") never contributes to proper acceleration in any circumstances, and thus the proper acceleration felt by observers standing on the ground is due to the mechanical force from the ground, not due to the "force" or "acceleration" of gravity. If the ground is removed and the observer allowed to free-fall, the observer will experience coordinate acceleration, but no proper acceleration, and thus no g-force."

 

[A.T.]

I don't think the GR literature is consistent on this point.

If you talk just about the magnitude g as "gravitational acceleration", it obviously applies to both:
- downwards coordinate acceleration of free faller relative to hover
- upwards proper acceleration of hover

I agree with you that "gravitational acceleration" is not a third type of acceleration, just the name for the magnitude that can apply to both types above.

 

[TrickyDicky]

If you talk just about the magnitude g as "gravitational acceleration", it obviously applies to both:
- downwards coordinate acceleration of free faller relative to hover
- upwards proper acceleration of hover

I agree with you that "gravitational acceleration" is not a third type of acceleration, just the name for the magnitude that can apply to both types above.

You mean the three are actually the same up to sign?

 

[A.T.]

You mean the three are actually the same up to sign?

I mean that two listed vectors have the same magnitude, which is sometimes called "gravitational acceleration" or short "g".

 

[Dale]

You mean the three are actually the same up to sign?

For certain objects in certain coordinate systems in certain spacetimes, I would say "yes", but not in general.

 

[Christine88]

Actually, just the last two are enough; "gravitational" acceleration is just a particular case of one of the other two. (And there could be confusion over which one: in the case where I drop a rock, is "gravitational acceleration" the coordinate acceleration of the rock in my rest frame, or the proper acceleration I feel that the rock doesn't?)

That's where I get confused. Gravitational and coordinate acceleration. Or is it proper and coordinate acceleration? Or proper and gravitational acceleration? I guess I understand proper acceleration.

 

[A.T.]

Or is it proper and coordinate acceleration?

That's the two types of acceleration there are in general. Gravitational acceleration can refer to either of the two types in specific cases.

 

[Dale]

Or is it proper and coordinate acceleration?

Yes. Proper and coordinate acceleration are the two correct technical terms for the two different kinds of acceleration.

 

[Fek_]

Gravity is "special". It is an absolutely democratic interaction: it accelerates in same way all the bodies; more precisely, gravitational and inertial mass "accidentally" coincide (this does not happen for any of the other fundamental interactions, electro-weak and strong). Then, if you are in free fall, you won't measure any acceleration on you or in bodies close to you. In free fall, locally there is no gravity (that's why things in Space Station appear weightless). If you are in a closed box in free fall, you can't determine that you are free falling in a gravitational field with any experiment (till the moment you splash on the body's surface, of course :) ). This simple observation has far reaching consequences: the einstein theory of gravitation.
In this picture, gravity is space-time curvature induced by a space-time mass-energy distribution. Gravity is therefore a tensor. It is described by the 2nd rank einstein's tensor, which is proportional to the energy-momentum tensor of the mass-energy distribution.
As a consequence, a test particle will move "freely" in the curved space-time following the space-time geodetic between initial and final positions. This is defined by the 2nd order covariant derivative with respect to the proper time of 4-position to be zero. For an observer not moving with the test mass (eg not free falling), the motion will appear deviating from rectilinear, uniform motion (just as if you are in a rotating frame) because of curvature, that is he will invoke an acceleration (technically, affine connection of the coordinates makes the role of a "potential").
Matter tells the spacetime how to curve, spacetime tells the matter how to move.
In the case of "weak" gravitational fields (as Earth's or sun's field), one can restore from GR field equations the classical formalism of the Newtonian theory (eg lagrangian or hamiltonian). On the other hand, in strong fields as around compact astrophysical objects (neutron stars, black holes, say the schwarschild solution) the Newton theory fails. As if you try to describe the hubble's law or the acceleration of the universe's expansion (the cosmological constant).
Gravity is much more than a simple potential. Imho :-)

The principle of general covariance says that a physical equation holds in a general gravitational field if 2 conditions are met: 1) it holds in absence of gravitation, that is agrees with the special relativity when the metric tensor equals the minkowsky tensor and the affine connection vanishes and 2) equation is generally covariant, that is it preserves its form under a general coordinates tranformation.
Principle of equivalence states that the effects of gravitational fields are indistinguishable by accelerated reference frames by any law of mechanics ("weak" principle) or by any law of physics ("strong" principle). One can be derived from the other and viceversa (eg see Weinberg, Gravitation and cosmology, principles and applic. of GR, Ch. 4).
They rely on the experimental fact that M_inertial = M_gravitational.
It should be stressed that general covariance has its meaning in its statement about the *effects* of gravitation, that a physical equation by virtue of its general covariance will be true in a gravitational field if it is true in absence of gravitation. It is not a principle of relativity as galileo's relativity or special relativity or gauge symmetries, but it is a statement about the effects of gravitation and nothing else. For example, in special relativity we pretend that, when passing from a frame to another with a coord transformation, the final result must *not* depend on the relative speed: it must not appear in the equations. This condition puts stringent limits on the possible transformations the have *physical* meaning. In GR there is *not* such a restiction. When passing from a general frame to another, new pieces appear in the field equations (through affine connection) that we interpret as gravitational fields.
Is it not true? Well, till now we can predict what will be the deflection of a photon in a gravitational field *and* the pulsar period decay *and* lensing *and* gravitational red-shift *and* AGN emissions *and* anything else by using GR. This makes GR "true".

 

[stefjourdan]

Ok now I think I get it. Because space and time are the same any object in space is already in motion through time?

Forget time and relativity, IMHO your problem is that you see (in space) some objects as 'moving' and other objects as 'still'. Then you understand the behavior of moving objects, but not the behavior of still objects (which 'start to move')

There are no such things, in the universe, as 'still objects' and 'moving objects'. You see a ball in your hand as a still object because you think you are still too, and the Earth is a still referential. Now you must realize that Earth is thrown in space at a very high speed. Seen from the sun, or even from the moon, or from any referential choosen in the universe apart Earth, the experience you describe is one involving only extremely fast moving objects.

Therefore what you interpret as a starting movement is only a slight modification of trajectory, something you reckon is easy to understand

Actually, the action of gravity on objects, acting like a force, is not easy to understand; a curvature of space is one way of seeing it, provided you realize the curvature doesn't belong to any of the three dimensions of space we see. The movement of objects happens in the three dimensional

A more difficult problem, seldom noticed, about gravity is the fact that it acts instantly. Newtonians calculations are right (provided the speed or mass of object are not extreme) and they predict correctly the future position of moving objects in gravitational field, and yet they are based on the idea that any mass attracts any other mass along the line connecting directly thoses two objects regarless of the distance between them !

 

[brotherbobby]

Hello all

I just joined this forum so forgive me for jumping right in but I have a question about Gravity and the curvature of space time that I can't get answer with a Google search. My question: though I understand that an object remains in orbit because of the curvature of space time and it is this curvature which is responsible for Gravity, but what causes an object that is stationary to fall toward the center of mass if nothing sets it in motion? Does the curvature of space give it a nudge? If so How? Why does a ball which is motionless in my hand fall if I let go of it without giving a push? I understand that if I set it into motion fast enough that it will fall around the Earth following the curvature of space but what makes it move toward center of mass if no force is acted on it?

Christine, imagine you're go to a place in empty space and just fix yourself there, far away from any nearby matter. What is your spacetime curve? A straight line, parallel to the time axis. Your space coordinate remains the same (because you remain at the same place) but your time coordinate continues to flow, since time flows.

However, you can only fix yourself at that point in space because you are away from any nearby matter. Or in other words, the geometry of the spacetime is flat. On the other hand, if you were nearby some massive object, you could not do this. Your spacetime path in this new geometry will no longer be a straight line. It will be some curve - as time varies, your position will also vary. The path will determine how you move.

It is the new geometry due to the massive body that needs to be used to find your path - x(t). You can know how to do this when you learn the mathematics of general relativity - but mostly, einstein's equation (to determine the geometry) and given the geometry, the geodesic equation (to find the path).

 

[Nugatory]

This thread is open again after some amount of cleanup.
If I seem to have done someone's ideas a serious injustice, please PM me.

 

[Mike442]

Why does a ball which is motionless in my hand fall if I let go of it without giving a push?

The reason you don't have to give the ball a push is because it already has a pushing force (gravitational force) acting on it due the curvature of space-time (this is what gives it weight). However, your hand is applying an equal and opposite force upward on the ball that prevents it from falling. Once you remove your hand from under the ball, the ball accelerates downward in accordance with F=ma. The gravitational force vector acts perpendicular to the lines of curved space-time and toward the center of the earth.

I understand that if I set it into motion fast enough that it will fall around the Earth following the curvature of space but what makes it move toward center of mass if no force is acted on it?

I believe you are confusing the mechanics of escape velocities and orbits with the gravitation force vector which pushes objects toward the center of the earth. The reason the Earth does not fall into the Sun is because the Earth's orbital velocity of 67,000 miles per hour around the Sun results in an outward force on Earth in accordance with F= m (v^2) / R. This outward force offsets (balances out) the inward gravitational force vector acting toward the center of the Sun and keeps Earth in its orbit and following the lines of curved space-time.

 

[A.T.]

The reason you don't have to give the ball a push is because it already has a pushing force (gravitational force) acting on it due the curvature of space-time (this is what gives it weight). However, your hand is applying an equal and opposite force upward on the ball that prevents it from falling.

That is basically the Newtonian explanation, with "curvature of space-time" randomly thrown in. In General Relativity there is no "gravitational force", that opposes the force of the hand. That's why the ball experiences proper-acceleration upwards, when held in the hand.

Once you remove your hand from under the ball, the ball accelerates downward in accordance with F=ma.

That is again the Newtonian explanation. In General Relativity there is no force acting in free fall. That why the ball experiences zero proper-acceleration when falling.

 

[A.T.]

You see a ball in your hand as a still object because you think you are still too, and the Earth is a still referential. Now you must realize that Earth is thrown in space at a very high speed

The OP specifically asks for an explanation, based on the reference frame, where an object is initially stationary. Changing the reference frame to one where it already moves is not really addressing the question, but rather avoiding it.

Actually, the action of gravity on objects, acting like a force, is not easy to understand; a curvature of space is one way of seeing it,

General Relativity describes gravity as curvature of space-time, not just space. Even when you change the reference frame to one where the ball already moves, you will not get the right quantitative result considering only spatial curvature.

 

[Nugatory]

The OP specifically asks for an explanation, based on the reference frame, where an object is initially stationary. Changing the reference frame to one where it already moves is not really addressing the question, but rather avoiding it.

I think you are misunderstanding stefjourdan here - I read his as post as trying to explain that there is no such thing as absolute rest.

 

[A.T.]

I read his as post as trying to explain that there is no such thing as absolute rest.

I understand that part. But it doesn't change the fact, that there are frames where the falling object is initially at rest. And that's where the OP wants gravity due to curvature of space-time explained.

Talking about some other frames, where Earth and ball are moving very fast is not addressing this, and gives the wrong impression that the movement of Earth and ball in some arbitrary frame is relevant for their gravitational attraction.

Advising the OP to "forget time" and consider only spatial curvature is similarly misleading. Even in the frames where the object already moves, you still need to consider space-time, not just space curvature.

 

[jerromyjon]

I think one more piece is required to fully comprehend "why": Gravity is virtually indistinguishable from acceleration through space which nicely removes the time factor for simplified comprehension. Imagine standing in a rocket in empty space on a floor perpendicular to acceleration. If you hold a ball, your hand is accelerating it and if you let go it stops accelerating with you and the floor accelerates to reach it. I hope that helps!

I don't know why I wrote time, I meant gravity...

 

[Christine88]

I understand that part. But it doesn't change the fact, that there are frames where the falling object is initially at rest. And that's where the OP wants gravity due to curvature of space-time explained.

Talking about some other frames, where Earth and ball are moving very fast is not addressing this, and gives the wrong impression that the movement of Earth and ball in some arbitrary frame is relevant for their gravitational attraction.

Advising the OP to "forget time" and consider only spatial curvature is similarly misleading. Even in the frames where the object already moves, you still need to consider space-time, not just space curvature.

Thank you for adding this. I feel vindicated somewhat. The other posters addressed other interesting issues but not really my question directly so I kind of gave up figuring I was out of my league.

 

[stevendaryl]

Thank you for adding this. I feel vindicated somewhat. The other posters addressed other interesting issues but not really my question directly so I kind of gave up figuring I was out of my league.

To me, it helps to visualize curved spacetime by thinking about simpler analogies. For example, consider the surface of a globe, where east-west represents spatial separation and north-south represents time. An object just sitting do nothing is still traveling in time--the time coordinate is increasing. On the globe analogy, that means that all objects that are sitting still in space (not traveling east-west) are still traveling north (forward in time). But as two objects travel northward along two lines of longitude, the spatial distance between them decreases (till you get to the north pole, where all lines of longitude come together). So objects traveling in time (northward) get closer together in space (east-west separation) due to the spacetime being curved.

 

[A.T.]

Thank you for adding this. I feel vindicated somewhat. The other posters addressed other interesting issues but not really my question directly so I kind of gave up figuring I was out of my league.

In think the opposite is the case. The question you raised, is the typical question of an intelligent person, who is rightly confused by the misleading rubber-sheet analogy. Some of the responses were typical for people who uncritically swallow this analogy, because they are happy with a mere superficial feeling of understanding something. The other extreme are people who see the flaws of the analogy, and wrongly conclude that the theory itself is flawed.

 

[PeterDonis]

The reason you don't have to give the ball a push is because it already has a pushing force (gravitational force) acting on it due the curvature of space-time (this is what gives it weight). However, your hand is applying an equal and opposite force upward on the ball that prevents it from falling.

The first part of this is not correct; the effect of spacetime curvature by itself does not cause the ball to feel weight. If the ball were moving purely due to the curvature of spacetime (i.e., if your hand were not applying a force to it), it would be weightless. What causes it to feel weight is the force your hand exerts to prevent it from falling.

Some purists (such as myself) would not even call the effect of spacetime curvature a "force", precisely because it does not cause the ball to feel weight; we reserve the term "force" for something that is actually felt as weight (like the force your hand exerts on the ball). But others allow the term "force" to be used to describe the effect of spacetime curvature on the ball, at least in the weak field approximation that applies to this discussion; this allows you to account for the ball being motionless, in a frame in which the ground is at rest, by saying that it has two equal and opposite forces acting on it. But you have to be careful to keep in mind that one of these "forces" is not felt as weight.

 

[m4r35n357]

Just testing some words here. Is it fair or even accurate to say that in the GR analysis the force of your hand on the ball is not balanced by the ball (specifically contrary to the Newtonian spatial "equilibrium" of the ball's weight and your hand's reaction to it), causing the ball to be accelerated away from it's geodesic? In which case there is no equilibrium and no reaction force on your hand, and the thing you feel as weight is just the inertial resistance of the ball to this acceleration.

Hope that is clear . . .

 

[A.T.]

Is it fair or even accurate to say that in the GR analysis the force of your hand on the ball is not balanced by the ball (specifically contrary to the Newtonian spatial "equilibrium" of the ball's weight and your hand's reaction to it), causing the ball to be accelerated away from it's geodesic?

Yes, that's what proper acceleration is: deviation from a geodesic worldline.

In which case there is no equilibrium and no reaction force on your hand,

No, that's wrong. If you apply a force to the ball, the ball applies an equal but opposite force to your hand. Whether your force on the ball is balanced by other forces on the ball is irrelevant for this.

and the thing you feel as weight is just the inertial resistance of the ball to this acceleration.

What you feel is the force applied to you. The ball's "inertia" relates the net force on the ball, to the ball's acceleration, not to your feelings.

 

[m4r35n357]

No, that's wrong. If you apply a force to the ball, the ball applies an equal but opposite force to your hand. Whether your force on the ball is balanced by other forces on the ball is irrelevant for this.

What you feel is the force applied to you. The ball's "inertia" relates the net force on the ball, to the ball's acceleration, not to your feelings.

OK, my problem here is that I'm associating an equal reaction force with equilibrium, so I think I see my mistake; my hand and the ball are still in (spatial) equilibrium even thought we are both in motion (through spacetime). So the reaction force makes sense again. I keep having moments like these . . . pardon me.

 

[A.T.]

I'm associating an equal reaction force with equilibrium

That is the usual confusion of 2nd and 3rd Newtons Law. The 3rd Law equal but opposite forces are on different objects and don't imply equilibrium for either of them.

See also:
https://www.lhup.edu/~dsimanek/physics/horsecart.htm

 

[Bob Walance]

Hello all

I just joined this forum so forgive me for jumping right in but I have a question about Gravity and the curvature of space time that I can't get answer with a Google search. My question: though I understand that an object remains in orbit because of the curvature of space time and it is this curvature which is responsible for Gravity, but what causes an object that is stationary to fall toward the center of mass if nothing sets it in motion? Does the curvature of space give it a nudge? If so How? Why does a ball which is motionless in my hand fall if I let go of it without giving a push? I understand that if I set it into motion fast enough that it will fall around the Earth following the curvature of space but what makes it move toward center of mass if no force is acted on it?

- - -

For me, to understand the basics of General Relativity (GR), I really had to put aside all my previously-conceived notions of position, movement, and acceleration because they are all "relative" concepts.

The only absolutes that I can really accept (now) is whether an object has a net external force on it (that is, it is existing non-inertially) or that object has no net external force on it (that is, it is existing inertially). Those qualities can always be measured as an absolute - and that cannot be said for position or any change in position (with time) for any object on its own. Position and change in position (with time) can only be assigned as relative quantities between two (or more) objects.

Note that some people say that an object is "travelling inertially" or "travelling non-intertially". To me, the word "travelling" is a relative term so I prefer not to use it.

In GR, whether the Moon is in orbit around Earth (in your first case), or the ball is stationary (the moment after it has been released) above the Earth (your second case), both the Moon, the Earth (including your mass), and the ball (assuming it's not touching you) have no external forces on them. That is, the Moon, the Earth, and the ball (in this case) are all intertial.

So, what causes the Moon and the Earth to travel relatively as they do? What causes the ball and the Earth to approach each other as soon as the ball has been released (in your second case)? Well, that is determined by the initial conditions (relative positions and velocities of the objects in question) and the shape (curvature) of spacetime in their vicinity. The Einstein Field Equations give the exact answers.

The other interesting thing about GR is what happens, in your second case, when the ball is in contact with either you or the Earth. When this is happening, you, the ball, and the Earth are part of one larger single object. This new bigger object (you+ball+Earth) exists intertially (i.e., no external forces on it as a whole). However, if you examine pieces of this new bigger object, you, the ball, and the chunk of Earth that doesn't include you and your ball now DO have a net external force on them. In fact, assuming that your mass and the ball's mass are small compared with the rest of the Earth, the direction of the net force on you and the ball is pointing away from the center of mass of the Earth+you+ball as a whole. In fact, this is the case for you right now. The only net force on your body is on your butt, and that force is pointing away from the center of mass of the Earth. This is what Einstein is talking about in his Equivalence Principle.

To answer your second question - to give an answer to "Why does a ball which is motionless in my hand fall if I let go of it without giving a push?", look to the Equivalence Principle. When you're holding on to the ball, there is a net force (upwards) on your feet, and there is also a net force upwards on the the ball. You and the ball have no relative motion. As soon as you let go of the ball, the net force on that ball goes to zero (ignoring air resistance). So now, your feet have a net force (pointing upwards) and the ball has no net external force on it. Of course, there will now be relative motion between you and the ball. Try this experiment when you and your ball are on a rocket ship with its engine turned on. You will get the equivalent result - just as Einstein predicts in his Equivalence Principle.

Bob

 

[jerromyjon]

Don't forget you and the ball are pushing towards the Earth a tiny bit.
As you release the ball you push slightly less at your feet and as well as fluid resistance or aerodynamics, which is flowing through the air, also a pressure gradient to traverse, and some slight torque involved in the rotation of the Earth propelling it forward rotational direction as it approaches the axis a.k.a. angular momentum.