Going Further: 10.2 Another look at
Gravitational Redshift
In the examples so far we have assumed that the equivalence principle holds, and so we have computed the shift in wavelength of light in a gravitational acceleration based upon what we expect on an accelerating rocket ship. We will now find what the shift will be for a photon on Earth if we fire it from the bottom of a tower to the top. We will again use the equivalence principle to accomplish this.
One way to state the equivalence principle is that an accelerated frame is indistinguishable from a frame where there is a gravitational acceleration, at least for small scales. Another way to state it is as follows: On small scales, freely falling frames are equivalent to the inertial frames of special relativity. So if we are in a frame that is freely falling in a region with a gravitational acceleration, then we expect all our measurements to be the same as we would have in an inertial frame. We will now follow this reasoning through.
Imagine we have a colleague who directs a laser beam upward from the bottom of a tower to the top, where we are located. Further imagine that our colleague measures some wavelength for the beam, call it λ0—it might be 420 nm, like in the worked example from the activity. What wavelength will we observe at the top of the tower? To answer this question, we make a rash decision and step off the tower platform at the instant the beam is fired upward.
According to the equivalence principle, once we leave the support of the top of the tower and begin to fall, we are in an inertial frame because we are only under the influence of gravity. The light is also freely falling, influenced only by gravity. We neglect the effects of the air, as usual.
For a moment imagine another observer, next to us but traveling upward at a speed v. From the point of view of such an observer, the wavelength of the light would be increased by an amount given by the speed multiplied by the time required for the wave to go by the observer, which we will call T. This is because the observer would move slightly in the direction of the light wave as it moves rapidly upward. For the tiny shift in wavelength, we may write:
\begin{equation} \Delta\lambda=vT \end{equation}
We can rewrite the time period, T, in terms of the frequency, f, of the light using the relation T = 1/f. Then we have:
\begin{equation} \Delta\lambda=\frac{v}{f} \end{equation}
We can also rewrite the shift in wavelength using the relation ∆λ = λ–λ0, where λ0 is the unshifted wavelength of the light (the wavelength in the frame of the source) and λ is the wavelength we measure. Then we have:
\begin{equation} \lambda-\lambda_0=\frac{v}{f} \end{equation}
And now, since we know that the product of wavelength and frequency is the speed of the wave, or (λ = c/f), we have:
\begin{equation} \frac{c}{f}-\frac{c}{f_0}=\frac{v}{f} \end{equation}
If we rearrange this we can write:
\begin{equation} \frac{f}{f}-\frac{f}{f_o}=\frac{v}{c} \end{equation}
Which simplifies as follows:
\begin{equation} 1-\frac{f}{f_o}=\frac{v}{c} \end{equation}
Now we can solve this for the frequency, f, measured by the observer:
\begin{equation} \frac{f}{f_o}=1-\frac{v}{c} \end{equation}
Or...
\begin{equation} f=f_0\left(1-\frac{v}{c}\right) \end{equation}
So an observer moving upward at the instant we step off the tower would measure a shift of this size for the light’s wavelength. Of course we are not moving upward; we have just stepped off the tower and are beginning to fall to its base. However, if a friend of ours remains on the tower, as we begin to fall, it looks to us as though our friend is moving upward. Since we are in free fall, and therefore in the inertial frame of the also freely falling light, it must also look to the light as if our friend is moving upward relative to its own frame of reference. And just as in the case of the rocket, the velocity we attain downward is very small in the short time needed for the light to reach the top of the tower: this means that the frame of the top of the tower is only moving slowly relative to the free-falling frame of the light.
We can calculate our speed when both the acceleration due to gravity (g) and the height of the tower (H) are “small,” and thus determine the speed of our friend’s reference frame. We realize that the time we have been falling is just the height of the tower divided by the speed of light: t = H/c. Our speed is therefore v = gt = gH/c. This gives us the following expression for this new frequency:
\begin{equation} f=f_0\left(1-\frac{gH}{c^2}\right) \end{equation}
We have found exactly the expression we got before for the accelerating rocket. In this example we explicitly used the symbol g for Earth’s gravity. Of course it is valid not only on Earth, but on any other objects for which our assumption of low gravity and small H are valid.