The speed of clocks in a gravitational field
Now we want to use the principle of equivalence for figuring out a strange thing that happens in a gravitational field. We’ll show you something that happens in a rocket ship which you probably wouldn’t have expected to happen in a gravitational field. Suppose we put a clock at the “head” of the rocket ship—that is, at the “front” end—and we put another identical clock at the “tail,” as in Fig. 42–16. Let’s call the two clocks $A$ and $B$. If we compare these two clocks when the ship is accelerating, the clock at the head seems to run fast relative to the one at the tail. To see that, imagine that the front clock emits a flash of light each second, and that you are sitting at the tail comparing the arrival of the light flashes with the ticks of clock $B$. Let’s say that the rocket is in the position $a$ of Fig. 42–17 when clock $A$ emits a flash, and at the position $b$ when the flash arrives at clock $B$. Later on the ship will be at position $c$ when the clock $A$ emits its next flash, and at position $d$ when you see it arrive at clock $B$. The first flash travels the distance $L_1$ and the second flash travels the shorter distance $L_2$. It is a shorter distance because the ship is accelerating and has a higher speed at the time of the second flash. You can see, then, that if the two flashes were emitted from clock $A$ one second apart, they would arrive at clock $B$ with a separation somewhat less than one second, since the second flash doesn’t spend as much time on the way. The same thing will also happen for all the later flashes. So if you were sitting in the tail you would conclude that clock $A$ was running faster than clock $B$. If you were to do the same thing in reverse—letting clock $B$ emit light and observing it at clock $A$—you would conclude that $B$ was running slower than $A$. Everything fits together and there is nothing mysterious about it all. But now let’s think of the rocket ship at rest in the earth’s gravity. The same thing happens. If you sit on the floor with one clock and watch another one which is sitting on a high shelf, it will appear to run faster than the one on the floor! You say, “But that is wrong. The times should be the same. With no acceleration there’s no reason for the clocks to appear to be out of step.” But they must if the principle of equivalence is right. And Einstein insisted that the principle was right, and went courageously and correctly ahead. He proposed that clocks at different places in a gravitational field must appear to run at different speeds. But if one always appears to be running at a different speed with respect to the other, then so far as the first is concerned the other is running at a different rate. But now you see we have the analog for clocks of the hot ruler we were talking about earlier, when we had the bug on a hot plate. We imagined that rulers and bugs and everything changed lengths in the same way at various temperatures so they could never tell that their measuring sticks were changing as they moved around on the hot plate. It’s the same with clocks in a gravitational field. Every clock we put at a higher level is seen to go faster. Heartbeats go faster, all processes run faster. If they didn’t you would be able to tell the difference between a gravitational field and an accelerating reference system. The idea that time can vary from place to place is a difficult one, but it is the idea Einstein used, and it is correct—believe it or not. Using the principle of equivalence we can figure out how much the speed of a clock changes with height in a gravitational field. We just work out the apparent discrepancy between the two clocks in the accelerating rocket ship. The easiest way to do this is to use the result we found in Chapter 34 of Vol. I for the Doppler effect. There, we found—see Eq. (34.14)—that if $v$ is the relative velocity of a source and a receiver, the received frequency $\omega$ is related to the emitted frequency $\omega_0$ by \begin{equation} \label{Eq:II:42:4} \omega=\omega_0\,\frac{1+v/c}{\sqrt{1-v^2/c^2}}. \end{equation}  Now if we think of the accelerating rocket ship in Fig. 42–17 the emitter and receiver are moving with equal velocities at any one instant. But in the time that it takes the light signals to go from clock $A$ to clock $B$ the ship has accelerated. It has, in fact, picked up the additional velocity $gt$, where $g$ is the acceleration and $t$ is time it takes light to travel the distance $H$ from $A$ to $B$. This time is very nearly $H/c$. So when the signals arrive at $B$, the ship has increased its velocity by $gH/c$. The receiver always has this velocity with respect to the emitter at the instant the signal left it. So this is the velocity we should use in the Doppler shift formula, Eq. (42.4). Assuming that the acceleration and the length of the ship are small enough that this velocity is much smaller than $c$, we can neglect the term in $v^2/c^2$. We have that \begin{equation} \label{Eq:II:42:5} \omega=\omega_0\biggl(1+\frac{gH}{c^2}\biggr). \end{equation}  So for the two clocks in the spaceship we have the relation  \begin{equation} \label{Eq:II:42:6} (\text{Rate at the receiver})= (\text{Rate of emission}) \biggl(1+\frac{gH}{c^2}\biggr), \end{equation} 
 \begin{equation} \label{Eq:II:42:6} \begin{pmatrix} \text{Rate}\\[-.75ex] \text{at the}\\[-.75ex] \text{receiver} \end{pmatrix}= \begin{pmatrix} \text{Rate of}\\[-.75ex] \text{emission} \end{pmatrix} \!\biggl(\!1+\frac{gH}{c^2}\biggr), \end{equation}  where $H$ is the height of the emitter above the receiver. From the equivalence principle the same result must hold for two clocks separated by the height $H$ in a gravitational field with the free fall acceleration $g$. This is such an important idea we would like to demonstrate that it also follows from another law of physics—from the conservation of energy. We know that the gravitational force on an object is proportional to its mass $M$, which is related to its total internal energy $E$ by $M=E/c^2$. For instance, the masses of nuclei determined from the energies of nuclear reactions which transmute one nucleus into another agree with the masses obtained from atomic weights. Now think of an atom which has a lowest energy state of total energy $E_0$ and a higher energy state $E_1$, and which can go from the state $E_1$ to the state $E_0$ by emitting light. The frequency $\omega$ of the light will be given by \begin{equation} \label{Eq:II:42:7} \hbar\omega=E_1-E_0. \end{equation}   Now suppose we have such an atom in the state $E_1$ sitting on the floor, and we carry it from the floor to the height $H$. To do that we must do some work in carrying the mass $m_1=E_1/c^2$ up against the gravitational force. The amount of work done is \begin{equation} \label{Eq:II:42:8} \frac{E_1}{c^2}\,gH. \end{equation}  Then we let the atom emit a photon and go into the lower energy state $E_0$. Afterward we carry the atom back to the floor. On the return trip the mass is $E_0/c^2$; we get back the energy \begin{equation} \label{Eq:II:42:9} \frac{E_0}{c^2}\,gH, \end{equation}  so we have done a net amount of work equal to \begin{equation} \label{Eq:II:42:10} \Delta U=\frac{E_1-E_0}{c^2}\,gH. \end{equation}  When the atom emitted the photon it gave up the energy $E_1-E_0$. Now suppose that the photon happened to go down to the floor and be absorbed. How much energy would it deliver there? You might at first think that it would deliver just the energy $E_1-E_0$. But that can’t be right if energy is conserved, as you can see from the following argument. We started with the energy $E_1$ at the floor. When we finish, the energy at the floor level is the energy $E_0$ of the atom in its lower state plus the energy $E_{\text{ph}}$ received from the photon. In the meantime we have had to supply the additional energy $\Delta U$ of Eq. (42.10). If energy is conserved, the energy we end up with at the floor must be greater than we started with by just the work we have done. Namely, we must have that \begin{equation*} E_{\text{ph}}+E_0=E_1+\Delta U, \end{equation*} or \begin{equation} \label{Eq:II:42:11} E_{\text{ph}}=(E_1-E_0)+\Delta U. \end{equation}  It must be that the photon does not arrive at the floor with just the energy $E_1-E_0$ it started with, but with a little more energy. Otherwise some energy would have been lost. If we substitute in Eq. (42.11) the $\Delta U$ we got in Eq. (42.10) we get that the photon arrives at the floor with the energy \begin{equation} \label{Eq:II:42:12} E_{\text{ph}}=(E_1-E_0)\biggl(1+\frac{gH}{c^2}\biggr). \end{equation}  But a photon of energy $E_{\text{ph}}$ has the frequency $\omega=E_{\text{ph}}/\hbar$. Calling the frequency of the emitted photon $\omega_0$—which is by Eq. (42.7) equal to $(E_1-E_0)/\hbar$—our result in Eq. (42.12) gives again the relation of (42.5) between the frequency of the photon when it is absorbed on the floor and the frequency with which it was emitted. The same result can be obtained in still another way. A photon of frequency $\omega_0$ has the energy $E_0=\hbar\omega_0$. Since the energy $E_0$ has the relativistic mass $E_0/c^2$ the photon has a mass (not rest mass) $\hbar\omega_0/c^2$, and is “attracted” by the earth. In falling the distance $H$ it will gain an additional energy $(\hbar\omega_0/c^2)gH$, so it arrives with the energy \begin{equation*} E=\hbar\omega_0\biggl(1+\frac{gH}{c^2}\biggr). \end{equation*}  But its frequency after the fall is $E/\hbar$, giving again the result in Eq. (42.5). Our ideas about relativity, quantum physics, and energy conservation all fit together only if Einstein’s predictions about clocks in a gravitational field are right. The frequency changes we are talking about are normally very small. For instance, for an altitude difference of $20$ meters at the earth’s surface the frequency difference is only about two parts in $10^{15}$. However, just such a change has recently been found experimentally using the Mössbauer effect.5 Einstein was perfectly correct.