Paths of Objects in GR: Acceleration vs Constant Velocity

[Swapnil]

In words, how would a path of an object moving in a straight line with a constant velocity differ from the path of an accelerating object moving in a straight line according to Einstein's picture of curved space-time?

 

[lightarrow]

If you already know that the object moves in a straight line in both cases, then...it moves in a straight line in both cases! The path is the same!

However, concerning space-time's shape around the object, I think it's not even bent, if the acceleration is constant, but it's only an intuitive idea I have in this moment; I really would like to know something more about it, because I don't have, practically, any knowledge of general relativity.

 

[selfAdjoint]

In words, how would a path of an object moving in a straight line with a constant velocity differ from the path of an accelerating object moving in a straight line according to Einstein's picture of curved space-time?

A unaccelerated body would fillow a geodesic (no literally straight lines in a curved geometry!) The path would look locally straight in three dimensions and it would also be linear in time.

An accelerated body's path would not be a geodesic; although it might look the same in three dimensions, it would bend in the time direction.

 

[Severian]

no[t] literally straight lines in a curved geometry!

Isn't a geodesic the definition of a straight line in a curved geometry?

 

[HallsofIvy]

No. A straight line is a geodesic in geometries that admit them but, in general, a geodesic in not a straight line. You may be thinking that geodesics have the properties of straight lines (shortest distance between two points) and some people do use the terms "straight line" for a general geodesic but, in my opinion, that's an unfortunate abuse of terminology.

 

[Swapnil]

A unaccelerated body would fillow a geodesic (no literally straight lines in a curved geometry!) The path would look locally straight in three dimensions and it would also be linear in time.

An accelerated body's path would not be a geodesic; although it might look the same in three dimensions, it would bend in the time direction.

I see, but I don't "see." Is there a more visual way to see the difference between the paths of the two objects. Maybe by an analogy or something...

 

[pervect]

If you follow your nose, without turning your head, you are following a geodesic path.

This is a less technical way of saying that a geodesic curve "parallel transports" a vector (represented in this case by a line from the middle of your head to the tip of your nose) that is intitally tangent along a path (pointing in the direction that you are walking) so that it remains tangent along the path.

There are some more technical explanations in http://math.ucr.edu/home/baez/gr/outline2.html which explain this in more detail to make it more precise.

A geodesic also usually extremizes an action function - in the case of space-time, a geodesic extremizes proper time. This usually means that a geodesic path has a longer proper time than any neighboring path.

 

[selfAdjoint]

I see, but I don't "see." Is there a more visual way to see the difference between the paths of the two objects. Maybe by an analogy or something...

Well let's see. In order to simplify, let's drop the curvature stipulation and consider what happens in "flat" Minkowski spacetime; and to make it more visual, let's consider a reduced dimension picture in which the spatial part is only two- and not three-dimensional, and time is represented by the third dimension.

So consider an unaccelerated body moving from (0,0) to (4,3) in time 1 (using whatever compatible units you like). The distance, by Pythagoras in this 3-4-5 triangle in the plane is 5 units. and the 4-displacement by Minkowki is \sqrt{-t^2 + s^2} = \sqrt{-1 + 25} = 2\sqrt{6}, measured along a straight line in the t-x-y solid. The straightness is the spacetime geometric fact that corresponds to the dynamic fact of no acceleration.

Now suppose the body is accelerated along the same line with constant acceleration 2 (i.e. 2 space units per time unit squared). So in the same span of time 1 its speed will increase from the original 5 to 10, and in the t-s plane the equation of its path will be t^2 - 2s^2, a parabola. The curvature of the path in spacetime is the geometric fact that corresponds to the presence of an acceleration.

Does that help any?