Space-time gravity and acceleration

[ittiandro]

1. Hi !

 

, tI If, according to General Relativity, bodiesfall ( or,rather,move ) towards the Earth not due to a “ force” of gravity pulling down on them but simply because they naturally follow the geodesics of curved space-time and can therefore be viewed( I suppose) as moving in uniform motion, how come they are still seen as moving ( falling) at an acceleratingrate? Indeed their acceleration rate ( in the sense of an increase in velocity) is accurately measured at 9.8 m/s/s. My question may sound naïve to some, but may be I have reached the limits of conceptual understanding, without using maths, in which I have little familiarity beyond college algebra, long time ago ( I am 71 years old and my background is in philosophy). Thank you for what you can do to dispel my ignorance.

 

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[imatfaal]

Welcome to the forum ittiandro - I don't know if this is right but I think you need to make a sharper distinction between movement in 3-dimensional space (which is where we measure the 9.8ms-2) and 4-d spacetime. In 4d spacetime we are all following worldlines at the same speed moving in a complex of 3d space and time - each inertial body travels through spacetime on the geodesics that minimizes the spacetime interval. The 4 dimension geometry needs to be taken into account - you need to consider that time is a dimension and that travel is through all four dimension. This concept of 4-vectors and invariant intervals between events takes place in a smooth Euclidean geometry for Special Relativity ie Minkowski Space - and you must get your head around these ideas first. Then you can move on to the same ideas within a geometry that can be curved through the presence of mass and energy as expressed in General Relativity and represented by the Riemann tensor.

 

http://en.wikipedia.org/wiki/Minkowski_space

[michel123456]

(...)My question may sound naïve to some, (...)

 

Don't bother about that.

Welcome to the club.

 

Welcome to the forum ittiandro - I don't know if this is right but I think you need to make a sharper distinction between movement in 3-dimensional space (which is where we measure the 9.8ms-2) and 4-d spacetime. In 4d spacetime we are all following worldlines at the same speed moving in a complex of 3d space and time - each inertial body travels through spacetime on the geodesics that minimizes the spacetime interval. The 4 dimension geometry needs to be taken into account - you need to consider that time is a dimension and that travel is through all four dimension. This concept of 4-vectors and invariant intervals between events takes place in a smooth Euclidean geometry for Special Relativity ie Minkowski Space - and you must get your head around these ideas first. Then you can move on to the same ideas within a geometry that can be curved through the presence of mass and energy as expressed in General Relativity and represented by the Riemann tensor.

 

http://en.wikipedia.org/wiki/Minkowski_space

 

Is this an answer to ittiandro's question?

 

If, (...) bodies fall (...) simply because they naturally follow the geodesics of curved space-time and can therefore be viewed( I suppose) as moving in uniform motion, how come they are still seen as moving ( falling) at an accelerating rate?

 

The question is clear like fresh water.

Edited November 9, 2012 by michel123456

[michel123456]

So nobody here can answer ittiandro's question? (edit: I can't)

Only give neg. rep?

Edited November 19, 2012 by michel123456

[mireazma]

Related to the op question I can contribute with... some more questions ) like if gravity is not an intrinsic force of a body but is merely a curvature of spacetime, so pure geometry, (my) ignorant logic would dictate that a static body left in mid-air would remain static, as opposed to a moving object that would follow the curved spacetime around the Earth, for example. Where's the moving from, be it accelerated or uniform, after all?

[elfmotat]

Here's a way to think about it:

 

Say we place two small objects some distance away from each other so that they are at rest with respect to each other. Now say we graph the position (in one dimension) of these objects vs. time in flat spacetime. What would this graph look like? It would just be two parallel lines, as shown below.

 

 

 

But what happens if we draw the same graph on a curved surface (for example a sphere)? The result is shown below:

 

 

As you can see, the objects start out at rest with respect to each other (the lines start out parallel). But as the objects follow their straightest possible paths through spacetime (i.e. as they follow their respective geodesics) the distance between the objects appears to decrease over time, and the rate at which the objects approach each other increases over time (they are accelerated towards each other). Even though no forces are present and the two objects are simply following the most natural path through curved spacetime, they accelerate toward each other and eventually collide.

 

 

Now obviously a sphere isn't a realistic spacetime (every geodesic is a CTC), but it demonstrates the idea well.

[michel123456]

Great explanation!

 

How do you draw the situation at T2 in your spherical diagram? (where is the horizontal axis at T=2 ? }

[elfmotat]

Great explanation!

 

How do you draw the situation at T2 in your spherical diagram? (where is the horizontal axis at T=2 ? }

The x-axis at any time t is a great circle perpendicular to both geodesics.

[michel123456]

The x-axis at any time t is a great circle perpendicular to both geodesics.

So all x axis intersect.

[elfmotat]

So all x axis intersect.

Yes. They intersect at the left and right "poles" of the sphere.

[michel123456]

Yes. They intersect at the left and right "poles" of the sphere.

What is the physical meaning of this?

[elfmotat]

What is the physical meaning of this?

There isn't any. They're coordinate singularities that can be removed by diffeomorphisms. And, like I said, a sphere isn't a realistic spacetime in the first place.

[michel123456]

So you say that along the T axis there is a physical interpretation (that objects get closer to each other at an accelerating rate) but that along the X axis there is no physical interpretation. I disagree on that.

IMHO the interpretation is that objects far away happen to coincide in time (they are at the intersection of time axis. It also means the time rate diminishes the same way distance diminishes.

 

i agree that a sphere is not a realistic spacetime but it shows that when something "weird" happens in space (like 2 parallel lines intersecting), a similar "weird" phenomenon must happen in time.

Edited December 10, 2012 by michel123456

[elfmotat]

So you say that along the T axis there is a physical interpretation (that objects get closer to each other at an accelerating rate) but that along the X axis there is no physical interpretation. I disagree on that.

IMHO the interpretation is that objects far away happen to coincide in time (they are at the intersection of time axis. It also means the time rate diminishes the same way distance diminishes.

 

i agree that a sphere is not a realistic spacetime but it shows that when something "weird" happens in space (like 2 parallel lines intersecting), a similar "weird" phenomenon must happen in time.

 

No, it's physically meaningless. For example, consider the coordinate transformation [math]x \rightarrow x+k[/math]. Now the "poles" (place where the x-axes intersect) have been shifted a distance k. This is just a coordinate change; nothing physical occurred, yet the poles are now at different locations.

[michel123456]

The x-axis at any time t is a great circle perpendicular to both geodesics.

 

There must be something wrong here.

Can a second great circle be perpendicular to both geodesics ?

[elfmotat]

 

There must be something wrong here.

Can a second great circle be perpendicular to both geodesics ?

 

Yes, you're right. The x-axis great circles aren't perpendicular to the geodesics after t=0.

Edited December 10, 2012 by elfmotat

[michel123456]

Edited December 10, 2012 by michel123456

[ittiandro]

Elfmotat’s comment is right on : parallel lines drawn on a flat space seem to converge when drawn on a curved surface.

Similarly, I have to infer, the paths of two bodies falling in a curved space along their geodesics seem to converge towards the center of the Earth, but in reality this convergence is only apparent and there is really no deviation from straight line motion caused by a “force” of gravity. It is only a question of different geometries of space. I fully agree on this. This explanation seems, however, to address only one aspect of acceleration, i.e. the change in direction. It does not address the other aspect I referred to in my initial thread: the change in velocity.

 

Indeed, in this regard, my question still remains : if there is no longer a " force" of gravity acting on bodies and making them deviate from the straight line of uniform motion, what explains that( falling) bodies when moving towards the Earth along the geodesics still undergo an acceleration (they change their velocity) at a rate of 9.8 m/s^2 ?

 

G.R. holds that gravity is stronger near the Earth . Since it is not a force, but is due to the curvature of spacetime, do we have to conclude that the acceleration of a falling body as it approaches the Earth is due perhaps to a steeper curve in the space near the Earth? But why so? Such a notion would reintroduce gravity as a force, wouldn’it?

 

Can anybody cast some more light on this?

 

Thanks

 

Ittiandro

[ittiandro]

Elfmotat’s comment is right on : parallel lines drawn on a flat space seem to converge when drawn on a curved surface.

Similarly, I have to infer, the paths of two bodies falling in a curved space along their geodesics seem to converge towards the center of the Earth, but in reality this convergence is only apparent and there is really no deviation from straight line motion caused by a “force” of gravity. It is only a question of different geometries of space. I fully agree on this. This explanation seems, however, to address only one aspect of acceleration, i.e. the change in direction. It does not address the other aspect I referred to in my initial thread: the change in velocity.

 

Indeed, in this regard, my question still remains : if there is no longer a " force" of gravity acting on bodies and making them deviate from the straight line of uniform motion, what explains that( falling) bodies when moving towards the Earth along the geodesics still undergo an acceleration (they change their velocity) at a rate of 9.8 m/s^2 ?

 

G.R. holds that gravity is stronger near the Earth . Since it is not a force, but is due to the curvature of spacetime, do we have to conclude that the acceleration of a falling body as it approaches the Earth is due perhaps to a steeper curve in the space near the Earth? But why so? Such a notion would reintroduce gravity as a force, wouldn’it?

 

Can anybody cast some more light on this?

 

Thanks

 

Ittiandro

[elfmotat]

Indeed, in this regard, my question still remains : if there is no longer a " force" of gravity acting on bodies and making them deviate from the straight line of uniform motion, what explains that( falling) bodies when moving towards the Earth along the geodesics still undergo an acceleration (they change their velocity) at a rate of 9.8 m/s^2 ?

 

My analogy explains this as well. Notice how, as the particles move forward in time, the distance between the particles decreases at an increasing rate (i.e. they are accelerated toward each other). This is purely a consequence of the geometry of the particular spacetime - no forces are required.

 

 

 

G.R. holds that gravity is stronger near the Earth . Since it is not a force, but is due to the curvature of spacetime, do we have to conclude that the acceleration of a falling body as it approaches the Earth is due perhaps to a steeper curve in the space near the Earth? But why so? Such a notion would reintroduce gravity as a force, wouldn’it?

 

The curvature certainly changes as you get closer/farther from a large mass. Why would this imply gravity is a force?

[J.C.MacSwell]

If you jump off a cliff, you accelerate at 9.8 m/s/s with respect to the Earth, but do not accelerate at all with respect to your natural path in space time.

 

If you decide not to jump, you do not accelerate with respect to Earth, however, due to the continuing force on your feet at the top of the cliff, you continue to accelerate, deviating from your natural path in space time.

[michel123456]

 

My analogy explains this as well. Notice how, as the particles move forward in time, the distance between the particles decreases at an increasing rate (i.e. they are accelerated toward each other). This is purely a consequence of the geometry of the particular spacetime - no forces are required.

 

 

 

 

The curvature certainly changes as you get closer/farther from a large mass. Why would this imply gravity is a force?

(bolded mine)

The "increasing rate" will become obvious only when the time axes are showned. That's why I asked about the position of time axis for T=2.

[ittiandro]

 

My analogy explains this as well. Notice how, as the particles move forward in time, the distance between the particles decreases at an increasing rate (i.e. they are accelerated toward each other). This is purely a consequence of the geometry of the particular spacetime - no forces are required.

Yes, I now see this acceleration, but it is relative to the other body , because the two geodesics come closer and closer at an accelerating rate and so do the two bodies moving along them. However if we have only one body, instead of two, moving along its geodesic, its acceleration (= change of velocity) is in relation to what? AlI I can see is a steeper curvature towards the end of the path, where it would intersect the geodesic of the other moving body, IF IT WAS THERE. But it is not there. So, again what does acceleration mean in this case?What is its point of reference?

 

 

 

 

The curvature certainly changes as you get closer/farther from a large mass. Why would this imply gravity is a force?

This would answer my original question only if we posit TWO bodies moving parallel to one another along their curved spacetime geodesics, but not in the case there was only one body, as already queried in the preceding section.

[elfmotat]

However if we have only one body, instead of two, moving along its geodesic, its acceleration (= change of velocity) is in relation to what? AlI I can see is a steeper curvature towards the end of the path, where it would intersect the geodesic of the other moving body, IF IT WAS THERE. But it is not there. So, again what does acceleration mean in this case?What is its point of reference?

 

Exactly! Unless you have some reference point to compare your motion to you won't be able to tell that you're "accelerating" due to gravity. If you were placed in a room where you were weightless, you wouldn't be able to tell whether you were free-falling towards a massive body or just floating around in empty space. There's no local test you can perform which can distinguish the two!

 

It's not until you start resisting gravity that you can actually measure your "acceleration." I.e. if you're standing on the Earth an accelerometer will read "9.8 m/s²." If you jump off a cliff, the accelerometer will read zero during your fall.

[michel123456]

 

Exactly! Unless you have some reference point to compare your motion to you won't be able to tell that you're "accelerating" due to gravity. If you were placed in a room where you were weightless, you wouldn't be able to tell whether you were free-falling towards a massive body or just floating around in empty space. There's no local test you can perform which can distinguish the two!

 

It's not until you start resisting gravity that you can actually measure your "acceleration." I.e. if you're standing on the Earth an accelerometer will read "9.8 m/s²." If you jump off a cliff, the accelerometer will read zero during your fall.

 

bolded mine: that should ring a bell.