about the Pound-Rebka experiment and the answer to a second year undergrad student' s question 7 months ago

Question

The answer tells us about a specific subtraction of the doppler shift so as to obtain the net gravitational redsift. Can anyone explain all this thing about that subtraction?

Answer 1

This shows a schematic illustration of the Pound-Rebka experiment:

At the bottom we have $^{57}$Fe source that emits gamma rays upwards with an energy of 14.4 keV. The frequency of the gamma ray is $3.48 \times 10^{18}$ Hz, but let's just call this $\nu_0$ to avoid messing around with figures.

The gamma ray travels upwards, and as it travels it is red shifted by Earth's gravitational field so when it reaches the detector its frequency has changed to $\nu_1$, where $\nu_1 < \nu_0$. Calculating the new frequency $\nu_1$ involves general relativity, but as long as the distance $h$ isn't too large we can use a convenient approximation called the weak field limit:

$$ \nu_1 \approx \left(1 - \frac{gh}{c^2}\right)\nu_0 \tag{1} $$

The idea of the Pound-Rebka experiment was to blue shift the original frequency $\nu_0$ to a new frequency $\nu_2$ given by:

$$ \nu_2 = \frac{\nu_0}{1 - \frac{gh}{c^2}} \tag{2} $$

If we take this value f $\nu_2$ and plug it into equation (1) we get:

$$ \nu_1 \approx \left(1 - \frac{gh}{c^2}\right)\nu_2 = \left(1 - \frac{gh}{c^2}\right)\frac{\nu_0}{1 - \frac{gh}{c^2}} = \nu_0 $$

So the point is that the Doppler shift and the gravitational time dilation cancel each other out and the gamma ray reaches the top detector with its frequency unchanged.

If we move the source upwards with a velocity $v$ then provided the velocity $v$ is much less than $c$ the new frequency $\nu_2$ is given by the equation for the Doppler shift:

$$ \nu_2 = \left(1 + \frac{v}{c}\right) \nu_0 \tag{3} $$

If we take this value for $v_2$ and plug it into equation (1) we get:

$$ \nu_1 \approx \left(1 - \frac{gh}{c^2}\right)\left(1 + \frac{v}{c}\right)\nu_0 $$

So to make $\nu_1 = \nu_0$ we just need:

$$ \left(1 - \frac{gh}{c^2}\right)\left(1 + \frac{v}{c}\right) = 1 $$

And with some rearrangement we get:

$$ v = \left(\frac{\frac{gh}{c^2}}{1 - \frac{gh}{c^2}}\right)\,c $$

In the Pound-Rebka experiment this velocity $v$ turns out to be about 0.7 mm/sec. So what the experiment did was to show that if you moved the source upwards at a steady 0.7 mm/sec the Doppler shift exactly balanced out the gravitational red shift and the gamma rays reached the detector with the frequency unchanged.