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Paths of Objects in GR

  1. Sep 22, 2006 #1
    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?
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  3. Sep 25, 2006 #2
    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.
    Last edited: Sep 25, 2006
  4. Sep 25, 2006 #3


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    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.
  5. Sep 25, 2006 #4
    Isn't a geodesic the definition of a straight line in a curved geometry?
  6. Sep 25, 2006 #5


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    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.
  7. Sep 25, 2006 #6
    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...
  8. Sep 25, 2006 #7


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    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.
  9. Sep 25, 2006 #8


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    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 [tex]\sqrt{-t^2 + s^2} = \sqrt{-1 + 25} = 2\sqrt{6}[/tex], 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 [tex]t^2 - 2s^2[/tex], 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?
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