A How do you convert the Pound–Rebka redshift into time variation?

[ARChohan]

TL;DR Summary

Converting Pound Rebka experiment result redshift to time variation per second, and other questions.

Title:
How do you convert the Pound–Rebka result into time variation per second, and do you need the base frequency?

Post Body:
The Pound–Rebka experiment measured a fractional frequency shift of
Δf/f = –5.13 × 10⁻¹⁵
from a vertical displacement of 22.5 m in Earth’s gravitational field.
How do you convert this number into a measure of time variation per second—for example, how much more or less proper time accumulates per second across that distance?
Do you need to know the base frequency (like the photon’s frequency used in the experiment), or can the fractional shift be directly interpreted as a rate of time difference per second?
I’m looking for:
• A clear explanation of whether the base frequency is needed
• And what the actual value of the time variation per second is—in the correct unit, whether femtoseconds, attoseconds, or something else
Follow-up Question:
Would the proper-time difference per second between two stationary atomic clocks placed 22.5 m apart vertically (one at the emitter height, one at the absorber height) be identical to the fractional frequency shift observed in the Pound–Rebka experiment?

Or, put differently:

Does the gravitational redshift measured by a photon traversing the gravitational gradient exactly match the difference in clock rates measured by comparing two clocks placed at either end of that same gradient?

 

[Dale]

The Pound–Rebka experiment measured a fractional frequency shift of
Δf/f = –5.13 × 10⁻¹⁵
from a vertical displacement of 22.5 m in Earth’s gravitational field.
How do you convert this number into a measure of time variation per second—

Just multiply it by 1 second. So $\rm{-5.13 \ 10^{-15}} = \rm{-5.13 \ fs/s}$

 

[A.T.]

Does the gravitational redshift measured by a photon traversing the gravitational gradient exactly match the difference in clock rates measured by comparing two clocks placed at either end of that same gradient?

Yes. Instead of measuring the frequency change of a continuous light beam, you could also send light pulses on every clock tick (for example 1s), and measure at what rate those ticks arrive. For clocks at relative rest this process must give you directly the effect of gravitational/potential-difference time dilation.

 

[Ibix]

Would the proper-time difference per second between two stationary atomic clocks placed 22.5 m apart vertically (one at the emitter height, one at the absorber height) be identical to the fractional frequency shift observed in the Pound–Rebka experiment?

That's essentially what the Pound-Rebka experiment was. Modern atomic clocks usually use caesium as their frequency source, but that's an engineering detail in this context.

 

[ARChohan]

Just multiply it by 1 second. So $\rm{-5.13 \ 10^{-15}} = \rm{-5.13 \ fs/s}$

Thank you for that. I appreciate it. Since I don't have a calculator capable of calculating exponents, I must also ask how much time dilation would that be per 24 hours?

 

[Ibix]

Since I don't have a calculator capable of calculating exponents

You have a computer in front of you right now. Even if you don't have and cannot install a scientific calculator app you can search "scientific calculator online". And the calculation could be done by hand in under a minute anyway.

 

[A.T.]

Since I don't have a calculator capable of calculating exponents

In the 80s teachers told us we must learn written calculations by hand, because we might not always have a calculator available, which turned out to be complete nonsense.

Whatever you use to connect to the internet, almost certainly has a scientific calculator. You might just need to switch the calculator mode to scientific. You can google how to do that on your device.