Curvature, Geodesics and Acceleration in GR

[nigelscott]

I am trying to get my head around curvature, geodesics and acceleration in GR. I've put together the following paragraph that attempts to describe qualitatively how I think these things play together.

In Newtonian mechanics, a freely falling object accelerating towards the Earth implies a force acting between the object and the Earth corresponding to gravity. In general relativity, however, the Earth and the object are considered to be both moving along geodesics (along straight lines with zero proper acceleration) and it is the convergence of these geodesics as a result of the curvature of the space-time that manifests as the coordinate acceleration that we see as the object falls.

My questions are:

1). Is this close to being correct?

2). Do and , if so, how do the converging geodesics actually constitute the acceleration?

3). How does one reconcile this with the plot for the situation where there is no orbital motion and the moon
just free falls toward the earth?
http://www.firstcoasttutors.com/images/curved_spacetime.jpg

 

[Nugatory]

1). Is this close to being correct?

Yes. You can even drop the "close enough" qualifier - it is correct.

2). If so, how do the converging geodesics actually constitute the acceleration

The converging geodesics in spacetime can appear as an acceleration in space. For an analogy, you could imagine two hikers starting one kilometer startung at the south pole and walking north along different lines of longitude... If you only consider only their east-west separation, you would see them move gradually apart, reach a maximum separation as they cross the equator, and then starting to move towards one another again. That is, they start with a positive relative velocity in the east west direction, this velocity goes down until it reaches zero at the moment they cross the equator, and then becomes negative - so even though they are both following on a locally straight line and subject to no forces in the east-west direction, their experience a change in velocity (acceleration!).

3). How does one reconcile this with the plot for the situation where there is no orbital motion and the moon
just free falls toward the earth?

That picture of the Earth making a dimple in a flat surface is very misleading, and if you search this forum for "rubber sheet analogy" you will find much invective (appropriately) hurled in its direction. This video by our own member A.T. is a much better explanation

 

[A.T.]

. How does one reconcile this with the plot for the situation where there is no orbital motion and the moon just free falls toward the earth?

See this recent thread regarding that picture:
https://www.physicsforums.com/threa...ace-thus-causing-gravity.803213/#post-5042220

The videos cover the straight radial fall you ask about.

 

[nigelscott]

Thanks to both of you. I assume the double covariant derivative of the distance between the 2 hikers with respect to the coordinate time (or is it the proper time?) gives the acceleration. The presence of second order differentials of the metric implies that the Ricci tensor must somehow be involved in the calculation. Also , I assume that it is possible for one of the hikers to be stationary. Would that be correct?

 

[stevendaryl]

Thanks to both of you. I assume the double covariant derivative of the distance between the 2 hikers with respect to the coordinate time (or is it the proper time?) gives the acceleration. The presence of second order differentials of the metric implies that the Ricci tensor must somehow be involved in the calculation. Also , I assume that it is possible for one of the hikers to be stationary. Would that be correct?

Not quite right. There are two different types of acceleration that you might be getting mixed up. If you drop a rock, it accelerates toward the ground. If you drop two rocks, they accelerate toward each other (they both are accelerating toward the center of the Earth, which means that they are getting closer together as they drop). The first type of acceleration is measured in Newtonian physics by the quantity \vec{g}, the acceleration due to gravity. The second type of acceleration is the so-called "tidal forces", which is due to the fact that \vec{g} varies from place to place, both in magnitude and direction.

The curvature tensor of General Relativity describes the tidal forces. It doesn't directly describe the acceleration due to gravity. The reason for this is that "acceleration due to gravity" is a subjective quantity; it depends on what reference frame you are using to measure it. For the reference frame of an observer in free-fall, the measured value of the acceleration due to gravity is zero (a person in freefall doesn't feel ANY gravity--it feels just like you are floating in space with no forces acting on you; well, if you ignore the wind and the sudden stop when you hit the ground). In contrast, tidal forces can be measured by anyone. If you are in freefall, and you release two rocks a foot apart, you won't see them accelerate downward, because relative to you, they're not going down, they're staying still. But you will still see tidal effects: the two rocks will be moving closer together as you all fall.

Getting back to your question: The "double covariant derivative of the distance between the 2 hikers" is a measure of the tidal forces, NOT the acceleration due to gravity.

 

[nigelscott]

OK. Thanks for the clarification. However, I am still a little confused. In the weak field approximation, coordinate acceleration along the geodesic is construed as the gravitational force (g I assume}. So now we have accelerations between geodesics (tidal forces) and accelerations along geodesics, which would also appears to be tidal. How do you reconcile the two?