Gravitation in General Relativity.

1. May 26, 2013

klen

Consider a man standing inside lift which is free from all external influences (gravitational or otherwise). Suddenly someone begins to pull the lift with some acceleration. Now if the man releases a ball he sees it falls onto the floor. Since he sees the lift at rest he concludes that there is a gravitational force acting on the ball downwards. But a man standing outside in an inertial frame sees it is because of the acceleration of that lift which caused the ball to fall.

But consider two balls resting on a table inside this accelerated lift the man inside the lift would say that there is gravitational force acting between the balls.

How would the man in the inertial frame explain this force?

2. May 26, 2013

Staff: Mentor

With gravitational attraction between the two objects, too, as this is the easiest explanation.

The person in the lift would not conclude that he is in a gravitational field - he cannot know this. All he knows is that some external force acts on him. This could be the surface of a planet, or a string somewhere in space.

3. May 26, 2013

Staff: Mentor

by the gravitational force between the two balls you mean their Newtonian gravitational attraction $\frac{Gm_{1}m_2{2}}{r^2}$ where r is the distance between the balls on the table and the two m's are their respective masses?

If so... The quick answer is that in the thought experiment both observers completely ignore that attraction because it is unmeasurably small compared with the effect of the acceleration or gravitational force that is moving the balls towards the floor of the elevator.

The slightly longer answer is that the equivalence principle only applies locally; here an informal definition of "local" would be something like "close enough that the acceleration/force is in the same direction for all observers". If you have two balls massive enough that their gravitational attraction is not negligible, the direction of the forces will be very different between the two observers (and even for the observer inside the elevator, depending on where he stands) so we can't treat this as a local situation.

Now, if you could position an ant in a very small sealed compartment located somewhere between the two masses... That could be a small enough region to apply the equivalence principle, give the ant the same experience whether he's feeling gravity from the two masses or being accelerated through empty space.

Last edited: May 26, 2013
4. May 26, 2013

Naty1

You mean gravitational 'attraction'.....whether it is a 'force' or not depends on the model one uses to interpret the attraction. Hence the different terminology in the two prior posts.

5. May 26, 2013

pervect

Staff Emeritus
I would say that the man inside the lift and the man in the (approximately) inertial frame would both explain the "gravitational force" as being due to space-time curvature.

At least if you're trying to understand GR. There isn't any particular problem with describing the attraction as being due to "gravitational force", but it isn't in the spirit of GR. There certainly isn't any motivation to explain the "force" differently depending on what frame one is in.

Of course, talking about space- time curvature implies that the frame isn't _really_ an inertial frame - because of the distortion in the frame caused by the masses.

This may or may not be particularly helpful, without the mathematics. I'd say it isn't all that helpful without the math, but I"m not sure what sort of answer you're trying to get.

To throw in some possibly unfamiliar math, the Christoffel symbols determine the motion of the balls via the geodesic equation. I won't go into the details, but perhaps a very quick sketch will answer yoru question.

If it were not for the table, and the balls were following geodesics, the sum of all the terms of the geodesic equations would be zero. With the presence of the table, the motion of the balls departs from a geodesic, due to the upwards force that the table exerts on the balls. So the sum of the terms of the equation is equal to the applied force, said force being applied by the table to the balls.

The curvature of space-time generates Christoffel symbols (as well as the acceleration of the elevator). These Christoffel symbols are the mathematical entity which describes the balls attraction. In the simplest case (a frictionless table) the balls slide towards each other in order to satisfy the geodesic equation, as perturbed by the table. The table exerts an upwards force on the balls, but being frictionless the table doesn't impede their slow slide towards each other.

6. May 27, 2013

Naty1

pervect posts :
I would only add 'the same spacetime curvature'.....but what that means is not so straightforward to explain....as least not for me.

It occurs to me you might be familiar with spacetime in special relativity[ SR]. It's described as flat spacetime...Minkowski spacetime as an example. Yet objects in relative motion 'distort' space and time in SR...We call that time dilation and length contraction. And objects in SR can move in curved worldlines...that is, curved paths in spacetime. So there IS a type of 'curvature' in SR, but it is not gravitational in nature. In other words, bodies in relative motion may curve spacetime but not in a gravitational way. *

Gravity and curvature are not such simple topics as to often succumb to clearly understood one line explanations. In the following, gravitational curvature is sourced from the [Einstein] stress energy momentum tensor.

PeterDonis:

[I did not record the poster on that one]

and sort of an ad hoc concept:

If the kinetic energy of a fast moving object contributed to its gravitational strength, it could turn into a black hole by going fast enough. In other words, rapid motion may curve spacetime but that is not gravitational space time curvature.]

Finally, The stress-energy-momentum tensor is the source of gravity in general relativity. As it has been explained to me in these forums, that gravitational source is calculated in the frame of the object....not in the frame of a distant observer.

Which is 'gravitational' and which is just 'space time curvature' depends on your definition.

* It was explained to me elsewhere in these forums some years ago that you can picture world line [paths] curves in SR as you would curves on a flat graph paper. When gravitational curvature is involved, as in GR, the graph paper itself on which the curved worldlines are drawn is itself curved. It is this latter type curvature that is captured in the Einstein mathematics pervect refers to in his post.