Motion in curved space-time
Let’s consider an interesting little puzzle. We have two identical clocks, $A$ and $B$, sitting together on the surface of the earth as in Fig. 42–19. Now we lift clock $A$ to some height $H$, hold it there awhile, and return it to the ground so that it arrives at just the instant when clock $B$ has advanced by $100$ seconds. Then clock $A$ will read something like $107$ seconds, because it was running faster when it was up in the air. Now here is the puzzle. How should we move clock $A$ so that it reads the latest possible time—always assuming that it returns when $B$ reads $100$ seconds? You say, “That’s easy. Just take $A$ as high as you can. Then it will run as fast as possible, and be the latest when you return.” Wrong. You forgot something—we’ve only got $100$ seconds to go up and back. If we go very high, we have to go very fast to get there and back in $100$ seconds. And you mustn’t forget the effect of special relativity which causes moving clocks to slow down by the factor $\sqrt{1-v^2/c^2}$. This relativity effect works in the direction of making clock $A$ read less time than clock $B$. You see that we have a kind of game. If we stand still with clock $A$ we get $100$ seconds. If we go up slowly to a small height and come down slowly we can get a little more than $100$ seconds. If we go a little higher, maybe we can gain a little more. But if we go too high we have to move fast to get there, and we may slow down the clock enough that we end up with less than $100$ seconds. What program of height versus time—how high to go and with what speed to get there, carefully adjusted to bring us back to clock $B$ when it has increased by $100$ seconds—will give us the largest possible time reading on clock $A$? Answer: Find out how fast you have to throw a ball up into the air so that it will fall back to earth in exactly $100$ seconds. The ball’s motion—rising fast, slowing down, stopping, and coming back down—is exactly the right motion to make the time the maximum on a wrist watch strapped to the ball. Now consider a slightly different game. We have two points $A$ and $B$ both on the earth’s surface at some distance from one another. We play the same game that we did earlier to find what we call the straight line. We ask how we should go from $A$ to $B$ so that the time on our moving watch will be the longest—assuming we start at $A$ on a given signal and arrive at $B$ on another signal at $B$ which we will say is $100$ seconds later by a fixed clock. Now you say, “Well we found out before that the thing to do is to coast along a straight line at a uniform speed chosen so that we arrive at $B$ exactly $100$ seconds later. If we don’t go along a straight line it takes more speed, and our watch is slowed down.” But wait! That was before we took gravity into account. Isn’t it better to curve upward a little bit and then come down? Then during part of the time we are higher up and our watch will run a little faster? It is, indeed. If you solve the mathematical problem of adjusting the curve of the motion so that the elapsed time of the moving watch is the most it can possibly be, you will find that the motion is a parabola—the same curve followed by something that moves on a free ballistic path in the gravitational field, as in Fig. 42–19. Therefore the law of motion in a gravitational field can also be stated: An object always moves from one place to another so that a clock carried on it gives a longer time than it would on any other possible trajectory—with, of course, the same starting and finishing conditions. The time measured by a moving clock is often called its “proper time.” In free fall, the trajectory makes the proper time of an object a maximum.  Let’s see how this all works out. We begin with Eq. (42.5) which says that the excess rate of the moving watch is \begin{equation} \label{Eq:II:42:13} \frac{\omega_0gH}{c^2}. \end{equation}  Besides this, we have to remember that there is a correction of the opposite sign for the speed. For this effect we know that \begin{equation*} \omega=\omega_0\sqrt{1-v^2/c^2}. \end{equation*}  Although the principle is valid for any speed, we take an example in which the speeds are always much less than $c$. Then we can write this equation as \begin{equation*} \omega=\omega_0(1-v^2/2c^2), \end{equation*}  and the defect in the rate of our clock is \begin{equation} \label{Eq:II:42:14} -\omega_0\,\frac{v^2}{2c^2}. \end{equation}  Combining the two terms in (42.13) and (42.14) we have that \begin{equation} \label{Eq:II:42:15} \Delta\omega=\frac{\omega_0}{c^2}\biggl(gH-\frac{v^2}{2}\biggr). \end{equation}  Such a frequency shift of our moving clock means that if we measure a time $dt$ on a fixed clock, the moving clock will register the time \begin{equation} \label{Eq:II:42:16} dt\biggl[ 1+\biggl(\frac{gH}{c^2}-\frac{v^2}{2c^2}\biggr) \biggr], \end{equation}  The total time excess over the trajectory is the integral of the extra term with respect to time, namely \begin{equation} \label{Eq:II:42:17} \frac{1}{c^2}\int\biggl(gH-\frac{v^2}{2}\biggr)\,dt, \end{equation}  which is supposed to be a maximum. The term $gH$ is just the gravitational potential $\phi$. Suppose we multiply the whole thing by a constant factor $-mc^2$, where $m$ is the mass of the object. The constant won’t change the condition for the maximum, but the minus sign will just change the maximum to a minimum. Equation (42.16) then says that the object will move so that \begin{equation} \label{Eq:II:42:18} \int\biggl(\frac{mv^2}{2}-m\phi\biggr)\,dt= \text{a minimum}. \end{equation}  But now the integrand is just the difference of the kinetic and potential energies. And if you look in Chapter 19 of Volume II you will see that when we discussed the principle of least action we showed that Newton’s laws for an object in any potential could be written exactly in the form of Eq. (42.18). 