# Confusion regarding geodesic of thrown ball - curved or Cartesian coordinates?

I'm confused trying to understand what's happening in terms of spacetime geodesics when a ball is thrown and its trajectory plotted, height against time to give a parabola.

I read (from more than one source) that although the curve is a parabola in three (spatial) dimensions “we must recognize that a massive body like the Earth actually curves the coordinate system itself, so that rather than following a curved path in a flat (Cartesian) coordinate system, the ball actually follows a minimum-distance path, or geodesic, in a curved coordinate system.” In other words, the problem here is “mapping curved space-time onto a flat piece of paper.”

OK. But then I also read (again from more than one source) that, as an example of the weakness of the Earth's gravitational field, if the ball is in the air for, say, 1 second, using $ct$ units this is equivalent to $3\times10^{8}\,\mathrm{m.}$ One author then talks about plotting this curve on an ordinary $ct/x$ spacetime diagram saying “in this case the spacetime is practically flat, and thus the geodesic is very close to a straight line.”

OK again, but aren't these two positions contradictory? If you aren't allowed to plot curved spacetime using Cartesian coordinates in the first example, how come you are in the second? And vice versa of course.

Thanks

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Your second example explains why the first one, though theoretically correct does not apply for sizes of ball throwing and some tens of meters or even kilometers or so.

Maybe you can understand when you think as an analogue the map of the earth. When the whole globe is mapped, there are no parallel lines and plots of land are trapezoids; there are geodesics that have to do with the fact that the earth is round, not with any space distortion. When we come to city streets though, and the plot of our house, the curvature of the earth is so small that it can be ignored completely, as we can draw parallel lines to a very great accuracy, and still be on a horizontal plane, ignoring the underlying curvature due to the shape of the earth.

Another way of looking at it is that we use water to define the horizontal, but water as a whole ocean has a curvature, invisible to the accuracy of our measurements at small lengths.

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Thank you, though I find your first sentence a little ambiguous. Are you saying it's a valid approximation to plot the ball's path using Cartesian coordinates because (1) it's moving relatively slowly and/or (2) is in a weak gravitational field and/or (3) is not being thrown very far? So if I was throwing the ball at a velocity near the speed of light on the surface of a neutron star, for example, I couldn't plot its path using Cartesian coordinates but would need to take into account spacetime curvature and use curved coordinates - whatever they are? –  Peter4075 Apr 26 '12 at 7:25
yes, it needs a very strong gravitational field for space distortions to appear in meter distances. Gravitational distortion in the gravity field of the earth is taken into account in calculating satellite orbits, for example. –  anna v Apr 26 '12 at 11:22

Your confusion is legitimate--- how can we use cartesian coordinates when the Earth's gravitational field is distorting the geometry? If you were to use the falling ball trajectory to define a notion of "straight line", then the parallel lines wouldn't stay parallel--- they would fall down toward each other on opposite sides of the Earth.

In this case, and in all near static weak-field Newtonian cases, you do the following weak field valid approximation. You stipulate that there are good coordinates x,y,z where the metric is approximately diagonal, but the straight lines are not geodesics.

Then you define the Newtonian gravitational potential $\phi(x)$, which at any point x is the sum over all masses m at position y of $Gm\over|x-y|$. Using the Newtonian potential, the metric tensor is

$$ds^2 = -(1-{2\phi\over c^2})dt^2 + (1+{2\phi\over 3c^2})(dx^2 + dy^2 + dz^2)$$

The result is a very small perturbation to a flat metric, and it is self-consistent with General Relativity to leading order in the velocities of the objects, it's a legitimate description of slow-moving matter. If you ignore the space part of the metric, this is Einstein's 1907 approximation, which only had the rate at which clocks tick vary from point to point, and was the most useful half-way station to the full theory of General Relativity.

The most important thing is the time-time component of the metric, the rate at which clocks tick. This changes as the Newtonian potential changes, clocks go faster as you go up.

The condition of geodesic motion states that objects thrown on the Earth's surface maximize their proper time spent in the air with fixed endpoints. This means that they go up to where time ticks faster, then curve down to where time is slow. The geodesic motion reproduces Newton's law in response to a potential.

The choice of near-diagonal metric defines a preferred coordinate system, and in general, of course, you can't do it. But it works in the Newtonian limit, so we can use approximately rectilinear coordinates on Earth without contradiction.

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