curved space/ spacetime
The GR equ. tells us that a test particle will follow a geodesic line in spacetime, which is not a geodesic line in space. Usually space is flat, but that does not imply the geodesic line of a test particle in space is a straight line (right?). Since if I throw a ball the gravity of earth makes the ball go a parabolic line.
So why go balls curved trajectories in the presence of earth gravity? What does GR say about little balls in earth's gravity? Can we speak here about space or spacetime curvature that causes its parabolic trajectory? I know my thinking went somewhere utterly wrong here. Could someone clearify? Thanks 
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Suppose you throw a ball so that it lands 100 metres away, after rising to 2.5 metres altitude before falling to the ground ~ 1.5 seconds later. To visualise the trajectory with the time dimension to scale, the time duration would be equal to the distance to the Moon (1.5 light seconds away). So, visualise your parabola stretched out 100 metres in one direction, 2.5 metres in another and ~ 384,000,000 metres in the other, it is pretty straight! The Earth's gravitational field only deviates from flat spacetime by a factor of around 7x10^{10} ([itex]\frac{GM}{rc^2}[/itex]). I hope this has helped. Garth 
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Formally, one can do the analysis with "only" the calculus of variations. We say that the path an object takes makes the proper time an extremum. Suppose we have a particle following a trajectory r(t) in (for example) the Schwarzschild metric. Then we can write (using geometric units, in which c=1 for convenience) [tex] d\tau^2 = g_{00} dt^2  g_{11} dr^2 [/tex] [tex] \tau = \int \sqrt{ g_{00}  g_{11} \left( \frac{dr}{dt} \right)^2} dt [/tex] where g_00 and g_11 are functions of r. For weak fields and slow velocities, we can ignore the effects of g_11 entirely, and assume that it is unity. Thus we are taking into account only g_00, which represents the fact that clocks tick at different rates at different altitudes. We then have a basic problem in the calculus of variations http://mathworld.wolfram.com/CalculusofVariations.html i.e. we are extremizing the intergal [tex] \int L(r,\dot{r},t) dt \hspace{.5 in} \dot{r} = \frac{dr}{dt} [/tex] with [tex] L(r,\dot{r},t) = \sqrt{g_{00}(r)  \dot{r}^2} [/tex] which imples that the falling body obeys the EulerLagrange equations [tex] \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) = \frac{\partial L}{\partial r} [/tex] You can solve this in detail if you like (for the specific examle you'll need to look up how g_00 depends on r by looking up the Schwarzschild metric). T he only points I want to make are very general ones, that the "force" term, [itex]\partial L/\partial r[/itex], results from the fact that g_00 is a function of r, and is in fact proportional to [itex]\partial g_{00}/\partial r[/itex], and that the momentum term is very close to the SR form for momentum, with some small corrections for the metric coefficient g_00. You can also include the effects of g_11 if you're really ambitious, and show why they are small for low velocities/weak fields. 
Thanks, gentlemen. You are all great people.
And did I mention recently how much I like this site? 
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By the way, particles follow geodesics of [bextremal[/b aging, not maximal. Pete 
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[QUOTE=JesseM]Is this is basically the same as noting that in SR an object will follow a curved trajectory in an accelerated coordinate system, and then also noting that the equivalence principle of GR means you can explain the observations of any accelerated observer in terms of a gfield?[/qupte]That is exactly right. The answer to the question posed had nothing to do with curved spacetime.
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Pete 
I should probably note yet again, that "curvature", especially in scare quotes, does not necessarily mean a nonzero Riemann curvature tensor.
Speaking losely, it can apply to any situation where the metric coefficients vary with position. For an example of this usage, see for instance MTW pg 187, chapter 7, section 3. The title of this chapter is "Gravitational redshift implies spacetime is curved". Reading this section, it is clear that the "curvature" being described in this chapter does necessarily mean a nonzero curvature tensor. MTW, and many other authors, use the term "curvature" to describe any situation where the metric coefficients vary with position. In fact gravitational redshift does occur in situations where the Riemann curvature tensor is zero, such as the Rindler metric of an accelerating observer (also sometimes called a uniform gravitational field). One can either believe that several wellrecognized texts are "wrong" and go to great efforts to "correct" them, or one can believe that "curved spacetime" does not always mean "a nonzero curvature tensor". I have chosen the later course, pmb has chosen the previous course. 
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Pete 
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First, consider the analgous situation in 2dimensional spaces (not spacetimes). For points p and q in flat R^2, a geodesic (i.e., a straight line) is a local minimum (in terms of length) for curves that run from p to q. For point p and q in curved 2dimensional spaces (i.e., positive definite Reimannian manifolds), it is not always the case that a geodesic that joins p and q is a local minimum in terms of length. As an example, consider S^2, the 2dimensional surface of a 3dimensional ball, and, for concreteness, take the surface to be the surface of the Earth. Take p to be the north pole and q to be Greenwich England. The shortest route from the north pole to anywhere is along the appropriate line of longitude, and the shortest route from the north pole to Greewich is south along the 0 line of longitudeprime (prime meridian). This route is a geodesic and a local minimum for length. Call it the short geodesic. There is, however, another geodesic route that starts at the north pole and ends at Greenwich. From the north pole, go south along the 180 line of longitude to the south pole pole and then north along the 0 line of longitude from the south pole to Greenwich. This route is also a geodesic, but clearly is not a local minimum for length. Call this the long geodesic. Taken together, the short and long geodesics comprise the unique great circle that runs through both the north pole and Greenwich. How can the diffence between the short geodesic and the long geodesic be characterized? First consider the long geodesic. Any geodesic that starts the north pole, and that is infintesimally close to the long geodesic crosses the long geodesic before it gets to Greenwich, i.e., at the south pole. (Note: these "close" geodesic start at he north pole, but do not go through Greewich). The south pole is what is called a conjugate point. Now consider the short geodesic. Any geodesic that starts at the north pole, and that is infinitesimally close to the short geodesic does *not* cross the short geodesics between the north pole and greenwich. The short geodesic does not have any conjugate points. This gives the required characterization: a geodesic between any point p and q of a 2dimensional space is a local minimum for length if and only if the geodesic has no conjugate points. Something similar is true for spacetime: a timelike geodesic between events p and q is a local maximum for proper time if and only if there are no conjugate points (to p or q) between p and q. See Wald for a proof. This is very much related to the singularity theorems of Penrose and Hawking, as the focusing nature of gravity (assuming an appropriate energy condition) often causes conjugate points to occur. Regards, George 
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You shouldn’t expect so much from GR & spacetime they say very little about parabolic curves of balls, or Galileo’s cannonballs, on earth. GR affects compared to the predominantly Newtonian effects are just too small. First  in the ideal case the paths are NOT parabolic – they are elliptical as they attempt to establish an elliptical orbit around the center of the earth. Of course the surface of the earth gets in the way but before it hits it is basically on an orbital path. Second – We never get to see the “ideal” because AIR gets in the way. Due to the air resistance the horizontal speed is slowly but only partly reduced all the way to infinity. Maximum speed in the vertical is also limited by terminal velocity. Since the horizontal speed is never 100% eliminated the curve only approaches a vertical line. Thus it’s parabolic. Guys, you don’t have to create the most complicated answers, just a little insight to a proper concept is so much simpler. I’m sure Occam would agree. 
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Your example of crossing orbits is another example of two geodesics which may have different proper times associated with them and each geodescic is one for which the proper time is an extremal. This means that if you vary either path just a tad then the proper time would be either slightly higher or slightly lower than the one on the geodesic. If you chose alpha as an index to paramaterize worldlines which are close to the geodesic then plotter the proper time vs the index alpha then you'd see that those values of alpha for which the curve has a zero derivative corresponds to a geodesic. The curve doesn't have to have a maximum there. It could have a minimum of a point of inflection. Pete 
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