How do geodesics explain two identical balls thrown up at the different speeds?

Question

As stated in the title, two identical balls, both thrown directly upward, but at different speeds. The slower ball will reverse direction at a lower height than the faster ball. But the curvature of spacetime that they are passing through would be nearly identical.

How do geodesics explain these two paths?

Another way of looking at the same issue: Two identical balls dropped, but from different heights. Both balls travel straight down but hit the ground at different velocities. Each ball should be passing through the same spacetime curvature at the point of impact. Since neither ball experienced a force or acceleration and their motion is purely a product of the curvature of spacetime, why are they traveling a different speeds at impact?

Answer 1

Two worldlines starting at the same event in spacetime but having different velocities are going in different “directions” in spacetime, even if they are going in the same direction in space. So their geodesics are different.

Remember, most worldlines here on Earth are nearly parallel, since they travel at very small fractions of c relative to each other. So the extent in the time domain is much larger than in the spatial domains.

For a sense of scale, consider a baseball pitch. It travels about 18 m in about 0.5 s, for a velocity around 37 m/s which is 1.2 e-7 c. During the flight it travels 18 m in space and 1.5 e8 m in time. This means there is a spacetime "angle" of about 1.2 E-7 radians between the earth and the baseball, which is "nearly parallel". Now, if we consider the height, a baseball goes up about 0.3 m during the flight. If we approximate that as a spacetime circular segment with a chord length of 1.5 e8 m and a height of 0.3 m, then we find a radius of curvature of 9.4 e15 m, which compares closely with c^2/g of 9.2 e15 m (the discrepancy being due to the approximation as a circle). So, because the baseball pitch is so long in time, the small difference in direction in spacetime can lead to the observed difference in the Newtonian path.

Answer 2

You have already got some good answers. But I would like to add some more concrete and visual explanations for convincing you. This can be done with Newtonian mechanics and elementary geometry only, but without the advanced mathematics usually used in general relativity.

Here is a table with some trajectories of balls thrown up with velocity $v$, reaching maximum height $h$, and total time of flight $T$. Using Newtonian mechanics (with $g=10\text{ m/s}^2$) we have $$h=\frac{v^2}{2g} \tag{1}$$ and $$T=\frac{2v}{g} \tag{2}.$$

\begin{array}{c|c|c|c|c} v & h & T & cT & R \\ \hline 1\text{ m/s} & 0.05\text{ m} & 0.2\text{ s} & 6\cdot 10^7\text{ m} & 9\cdot 10^{15}\text{ m} \\ \hline 10\text{ m/s} & 5\text{ m} & 2\text{ s} & 6\cdot 10^8\text{ m} & 9\cdot 10^{15}\text{ m} \\ \hline 100\text{ m/s} & 500\text{ m} & 20\text{ s} & 6\cdot 10^9\text{ m} & 9\cdot 10^{15}\text{ m} \\ \hline \end{array}

We need to analyze the trajectory not in 3-dimensional space, but in 4-dimensional space-time (with $ct$ as the 4th coordinate, $c=3\cdot 10^8\text{ m/s}$ being the speed of light). Therefore the table above also contains the values of $cT$.

Our goal is to find the curvature radius $R$ in 4-dimensional space-time. We can draw the curved trajectory in space-time ($z$ versus $ct$).

Of course this drawing is not to scale. Because of the huge value of $cT$ the curved trajectory at the top would be much less curved, and the center of curvature would be much further down. But it is enough for analyzing the geometry.

Applying Pythagoras' theorem to the right triangle we get: $$\left(\frac{cT}{2}\right)^2+(R-h)^2=R^2$$ or $$\left(\frac{cT}{2}\right)^2+R^2-2Rh+h^2=R^2$$ The $R^2$ cancels out. We can neglect the small $h^2$ and resolve for $R$: $$R=\frac{(cT)^2}{8h} \tag{3}$$ The calculated values of this curvature radius $R$ are also shown in the table above. It turns out, the radius $R$ is huge (nearly one light year) and (may be as a surprise) completely independent of the trajectory. To understand how this comes about, we can insert $h$ and $T$ from (1) and (2) into (3). We find $$R=\frac{c^2}{g} \tag{4}$$ This relation makes evident: The curvature of the trajectory in 4-dimensional space-time depends only on gravity $g$ and nothing else. Especially it is independent of the initial velocity $v$.

Answer 3

You must take into account the time dimension. You are assuming that since both balls where thrown upward from the same location on the same gravitating mass that they will follow the same geodesic. This is not true. You must take into account the time dimension (in addition to space). When you do this, you see that they do not follow the same path in space-time. That is, it takes less time for the ball thrown faster to reach a certain height. The other ball will take longer to reach this point (or may never reach this point because it may have already reached its maximum height and may be falling downward). Therefore, the two balls follow different paths in space-time and therefore their motion is determined by two different geodesics.

Answer 4

The curvature of the spacetime through which the balls are moving is identical, but the balls set off through it in different directions. Whereas in space alone they appear to be heading in the same direction, in four dimensions they are taking different paths. To see this, simply plot two dimensions, z and t, using everyday units of m and s- you will see that the balls are launched in quite different directions through spacetime. If you use the unit ct for the time axis, the difference in the angles of launch of the two balls will then seem tiny, but they are moving at such a huge speed using those units that the overall effect on the trajectory is exactly the same.