# GR layman's question

1. Jun 4, 2005

### dodo

If the curve described by an object passing near a star is not due to a 'force of gravity', but to the space being curved by the star's mass, and the object is just following a 'straight' (geodesic) inertial path... then why would a still, not moving object begin to fall toward the star, if there's no force applied?

Thank you, and excuse the curiosity. :D

2. Jun 4, 2005

### Mortimer

A stationary object in a gravity field does not follow a geodesic path. Its path is curved (read "accelerated") in 4D from the perspective of a geodesic rest frame (the rest frame of a freely falling object). See attached picture.

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3. Jun 4, 2005

### dodo

Hmm... so, if it was still (under a g.field) was because a force was exerted on it; that force released, it followed its inertial trend. Neat. -- Thanks

4. Jun 4, 2005

### gonzo

I might be confused here, but it seems like Mortimer answered a different question. It seems to me that Mortimer is talking about an object that is stationary in a gravity field and why it is not "falling".

If I understand Dodo right, his question was more about something that troubled me also when I first learned about GR. Althougn, again I could be misunderstanding both of you.

If your initial question was what I think it was, Dodo, then I think the answer is that everything is always moving through spacetime. There's no such thing as an object that is stationary in spacetime.

5. Jun 4, 2005

### dodo

1) I have a little confusion on the meaning of '4D' - whether it means '3D space plus time' -- or '4D space' and 'curves' as analog to plane spherical vs. euclidean, and the apparent curves being the projection of that 4D into the little 3D of ours. Is it meant sometimes one and sometimes the other, or in GR it's always meant to be 3D+time?

2) And if it is 3D+time, then 'everything moving on spacetime' is a bit of a comfortable tautology... I wonder if the apparent curved trajectory of a comet around the Sun can take so much variation at the first 3 coordinates at the expense of a change in the 4th (if some sort of modulus of the velocity vector is meant to be constant along a geodesic). Unless that factor 'c' on the units of 4th axis makes the difference. Besides, a change in direction would also imply the presence of acceleration, wouldn't it?

3) Nevertheless, it's puzzling to find no stationary objects within a gravity field (except perhaps at its center). It's no 'pit' and things doesn't 'fall' on it. If it is possible to have an object at rest outside the field, why can't you find one inside? (Time motion notwithstanding here when speaking of 'at rest'.)

Last edited: Jun 4, 2005
6. Jun 4, 2005

### gonzo

4D usually means 3-space + time. But this isn't quite the meaningless idea you might first think, since relativity teaches us there are mathematical relations between time and space.

Here's one way to think of it. However, I have to warn you that some people don't like this analogy, since as with all analogies for complicated issues, it lends itself to some wrong conclusions.

You can sort of think of your movement through time plus your movement through space as a constant, in the sense that the more you move through space, the less you move through time, and the less you move through space the more you move through time. Please, note, however, that this does not imply a prefered frame of reference (which is why some people don't like this analogy). But it is a simple way to understand some of the ideas, and to show the deep connection between space and time.

I'm not sure I fully understand the questions in your second point. Maybe someone more qualified can figure them out and give you a better answer.

7. Jun 4, 2005

### pervect

Staff Emeritus
There isn't much difference between the two situations from a GR perspective. The initally moving object follows a geodesic, and so does an observer with an initial velocity of zero relative to the star.

Both geodesics accelerate towards the star in a cartesian coordiante system anchored to the star.

The main differences show up when the velocity of the object is very high (near that of light). In this case, the acceleration of the object has components due to both space curvature and space-time curvature, which result in the objects apparent acceleration being twice as great as the Newtonian formula suggest.

8. Jun 4, 2005

### dodo

Part of the intention of the original question focused on why should an still object begin to move, when no force is applied to it. I assume Newton's first law still holds.

Mortimer's post suggests that an stationary object in a gravity field, if it is not only momentarily stationary, must be the subject of a force (an astronaut's hand holding it?). Fine up to here.

Yet it is also suggested than no object can exist at rest in a gravity field (as it could elsewhere - it's not moving and nobody pushes it, so it keeps still). I wondered why.

9. Jun 4, 2005

### Garth

Dodo "I wondered why."
Because we are all 'time travellers'!

Do you want to travel into the future? Then sit back and wait! Nobody's pushing you!

In 4D space-time the length of the 4-velocity vector of any particle, yourself included, U, is the same as that of light, 1 in sensible units, (in which light speed is unity; one light year per year). So let go of your object and it will travel into the future, however, the space-time that it is travelling in is curved so the direction it goes in is deflected, not by a force acting on it but by the curvature of the surface on which it is moving, the curved space-time continuum.

So, as it travels through time, its straight line trajectory on that curved surface, its geodesic (similar to the great circle on the earth's surface) draws it closer and closer to the centre of the Earth, which is also travelling along such a geodesic through space-time.

The object and the Earth converge. Back in the 3D world + time we see the object fall to the floor.

I hope this helps.

Garth

Last edited: Jun 4, 2005
10. Jun 4, 2005

### dodo

Hmmm! Thanks to everybody's patience. Now it is more clear.

I guess the kind of drawings that show a weighty ball deforming a plane, elastic surface down, are terribly misleading in the end. They send the mental image that it is SPACE that curves. Perhaps a more accurate (though simple) example is to imagine two objects on the Earth's equator, and forced to move steadily North (latitude representing time). In this example, space is only 1-dimensional (longitude), and while nobody is pushing the objects IN SPACE (along longitude), they do join at the north pole.

11. Jun 4, 2005

### pervect

Staff Emeritus
That's exactly the right picture. It curves space-time (in this case 1d space+1d time) rather than space.

12. Jun 6, 2005

### da_willem

Just wondering how this concept follows from the equations of GR. I guess from the geodesic equation:

$$\frac{D p^{\mu}}{d \tau } = m( \frac{d^2 x^{\mu}}{d \tau ^2}+\Gamma ^{\mu} _{\nu \sigma} \frac{d^2 x^{\nu}}{d \tau ^2} \frac{d^2 x^{\sigma}}{d \tau ^2})=0$$

this follows from the nonzero connection coefficients. For if they are zero (i.e. an Euclidean space) this just says there is no change in velocity (the spatial part of the velocity 4-vector). Now in a curved space the connection coefficients are nonzero and a constant velocity can yield a change in velocity by the second term.

Last edited: Jun 6, 2005
13. Jun 6, 2005

### pervect

Staff Emeritus

14. Jun 6, 2005

### pmb_phy

I think you're a bit confused. Tidal force and Spacetime curvature are not seperate things. They are identically the same thing in differerent languages. However a body in free-fall in a uniform g-field experiences a gravitational force but there is no spacetime curvature in such a field.

Here is the derivation of pervects equation
http://www.geocities.com/physics_world/gr/geodesic_deviation.htm

Pete

15. Jun 6, 2005

### pmb_phy

If a particle is in a gravitational field such as the Earth and it's not moving then there is a non-zero 4-force acting on the particle.

It seems rather easy to concieve of a particle at rest in a curved spacetime and which follows a geodesic. Hold two spherical masses apart. Place a particle at the exact center along a line connecting the two centers of the mass spheres. The spacetime curvature there is non-zero but the gravitational force is zero.

See pretty picture at
http://www.geocities.com/physics_world/gr/grav_field.htm

The situation I speak of is when a point particle is at x = 0. An arbitrary object with a finite size would have tidal forces exerted on it and may not stay at rest and may also not follow any geodesic.

Pete

16. Jun 6, 2005

### dodo

Pete,
I can understand the usefulness of the same thing being expressed in two languages -- in this context, I was asking for the translation of Newton's 1st law, from the 'force' language to the 'non-force' one. Which I got. (I apologize for the inexact expression of my query, though.)

Your second post sets, IMO, an interesting observation. Is it possible, at that curved spacetime, for two lines to have different [punctual] curvatures and yet be 'parallel' (in the sense that the objects seem not to move, relative to each other, in space)? A first line passing through the center of one of the masses (and forward through time), and a second line through that 'Lagrange point' and forward to time, will always have the same spatial distance (as measured by their first three components, at equal values of the fourth), yet spacetime is curved differently at both. Curious.

In euclidean 2d, two lines with different slopes are never parallel, something that breaks up when stepping to euclidean 3d (until you change 'parallel' by 'coplanar', mutatis mutandis). I assume a similar relaxation of constraints occurs with GR's spacetime.

Last edited: Jun 6, 2005
17. Jun 6, 2005

### pmb_phy

Dodo - There are some interesting points I'd like to make when I'm feeling up to par. One is a quote about GR by Eddington. The other is the notion of "GR" as "geometry" by Einstein, who dislikes that interpretation, as I do.

Pete