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Free falling & geodesics

  1. Apr 13, 2009 #1
    http://www.youtube.com/watch?v=8MWNs7Wfk84&feature=PlayList&p=858478F1EC364A2C&index=2" , Edmund Bertschinger is talking about Einstein's field equations .
    during the lecture , under the title of : "Gravity as sapcetime curvature (GR viewpoint) " , he wrote :
    "Freely falling bodies move along spacetime geodesics ....."
    this statement made me confused.
    what does this actually mean ???
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Apr 13, 2009 #2

    A.T.

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    It means that they advance straight forward trough space-time. Here a visualization:
    http://www.relativitet.se/spacetime1.html
     
  4. Apr 13, 2009 #3
    thanks
    but i didn't get the point yet
    what does this has to do with geodesics ???
     
  5. Apr 13, 2009 #4

    A.T.

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    Geodesics are just a generalization of straight lines. If you advance straight ahead trough space-time, your space-time path (world line) is a geodesic.
     
  6. Apr 15, 2009 #5
    forgive my stupidity , but i don't understand
    let me put the question in another form ,
    why do things fall to the surface of the earth in straight lines according to GR ???
    (and straight lines are not geodesics in curved space time)
     
  7. Apr 15, 2009 #6
    You should see it the other way around. In curved space time the notion of a straight line generalized to that of a geodesic.

    In the absence of curvature a free test particle moves along a geodesic, which in this case corresponds to moving with a constant velocity along a straight line.

    In a curved background the particle still moves along a geodesic. Only now this geodesic (and the motion itself) no longer correspond to "straight lines" and constant velocity (if there even is such a thing as a straight line in a curved background!).

    For instance, on the earth the motion of a free falling test particle can be along a vertical line, with an accalerating velocity. The statement then is that the movement is along a geodesic.

    (Note that in the case of the earthe there are also geodesics which are not vertical. Satellites for instance are also "free falling objects". Their movement is also along a geodesic, but corresponds to an elliptic orbit.)
     
  8. Apr 17, 2009 #7
    I'll give this a bash.

    http://en.wikipedia.org/wiki/Schwarzschild_metric" [Broken] represents curved space-time in a vacuum in the following form (assuming that [itex]d\theta[/itex] and [itex]d\phi[/itex] equal zero)-

    [tex]ds^2 = \left(1-\frac{2M}{r} \right)dt^2 -\left(1-\frac{2M}{r}\right)^{-1}dr^2[/tex]

    where [itex]M=Gm/c^2[/itex] (often referred to as the gravitational radius) where G is the gravitational constant, m is the mass of the object, c is the speed of light, dt and dr are change in time and distance respectively and r is variable, reducing the closer you get to an object of mass.

    proper time would be represented by-

    [tex]dt_{shell}=\left(1-\frac{2M}{r} \right)^{1/2}dt[/tex]

    where dtshell represents the time dilation at a specific (coordinate) radius.

    and proper distance would be represented by-

    [tex]dr_{shell}=\left(1-\frac{2M}{r} \right)^{-1/2}dr[/tex]

    where drshell represents the distance inflation at a specific (coordinate) radius.

    Velocity is normally expressed as v=m/s (distance/time) and the velocity induced by spacetime curvature could be expressed as-

    [tex]v_{shell}=\frac {dr_{shell}}{dt_{shell}}[/tex]

    If we substitute the above equations for dtshell and drshell, we get the following equation-

    [tex]v_{shell}=-\left(\frac{2M}{r}\right)^{1/2}[/tex]

    negative because the object is moving away from the observer towards the source

    Multiply by c for m/s. Objects take the shortest path through spacetime so if it feels a time dilation, no matter how slight, to one side of it, it will tend towards the source of the time dilation (or curvature). v increases as r reduces which is in some way analogues with Newton's equation for gravity [itex]g=Gm/r^2[/itex] (technically, gravity is [itex]g=dr_{shell}\cdot Gm/r^2[/itex] but drshell is normally only included when calculating gravity for ultra-compact objects such as neutron stars and black holes and normally ignored for less dense objects such as planets). If you apply the above to Earth say, you'll notice that vshell for a free-falling object at the earths surface is -11.2 km/s which is the negative of the escape velocity expressed as [itex]v_{esc}=\sqrt(2Gm/r)[/itex] which means that an object that fell from rest at infinity would hit the Earth's atmosphere at ~11.2 km/s, 'moving along the curved spacetime geodesic' caused by Earth's mass. (Note: The above velocity represents an object free-falling from infinity only and doesn't account for an object falling from rest at a specific radius) In Minkowski space (i.e. flat space) the velocity (which would be the result of work done by an external source) would remain constant and there wouldn't be any acceleration.

    It's also worth noting that for an object free-falling from infinity, E/m=1, the energy required for an object to remain stationary in curved Schwarzschild space is-

    [tex]\frac{E_{shell}}{M}=\left(1-\frac{2M}{r}\right)^{-1/2}[/tex]

    so unless there's an input of energy, the object has to move along the curved spacetime geodesic.
     
    Last edited by a moderator: May 4, 2017
  9. Apr 17, 2009 #8

    A.T.

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    I will give you an analogy:

    If you stick adhesive tape nicely to a flat surface, without bends, you get a straight line (stripe). If you stick it to a curved surface (e.g. a vase) you get a geodesic. Locally the geodesic is still like a straight line with respect to the 2D-surface of the vase. The fact that the tape is curved within the embedding 3D-space, doesn't change the fact, that topologically it is always a straight line.

    To understand how this relates to the apparently curved paths of free fallers, you should check out this link:
    http://www.relativitet.se/spacetime1.html
     
  10. Apr 17, 2009 #9
    ok ,i'll tell what i have understood here
    space time is a 4-manifold
    the space time around the earth is locally flat , so the geodesics of freely falling bodies in this region are straight lines , so they fall in straight lines .
    but globally the sapce time around the earth is not flat , so the geodesics of freely falling bodies (e.g. the moon) are not straight lines , they are curves , so the moon orbits around the earth in this curve .
    did i get it right ????
     
  11. Apr 17, 2009 #10

    A.T.

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    I think you mix up things a bit:

    trajectory : the path of the object in space
    worldline : the path of the object in space-time
    geodesic : a locally straight path in a manifold
    line : special case of a geodesic in a flat manifold

    For a free falling object the worldline is always a geodesic. The trajectory is just the projection of this worldline onto the spatial dimensions, and can be curved or not.

    This picture shows how a vertically falling object moves straight in curved space-time (worldline is a geodesic):
    http://www.physics.ucla.edu/demoweb...alence_and_general_relativity/curved_time.gif
     
    Last edited: Apr 17, 2009
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