High School Query about Pound-Rebka like experiment

[etotheipi]

I was helping (or at least attempting to help...) my sister with this question but ended up getting confused about one part myself! The idea is that you have a mass $m$ at a height $h$ which you drop, and on contact with the ground it is converted into a photon of equivalent energy (frequency $f$) that is shot back up to the original height before its new frequency $f'$ is measured.

The first way I did it was to take the mass as its own system, then its initial energy is just $mc^2$. On falling to the ground $mgh$ of work is done by gravity so its energy increases to $mc^2 + mgh$, and this then becomes the $hf$ of the photon (i.e. $mc^2 + mgh = hf$). If the energy of the photon at the original height is then $hf'$, we see that this must equal the initial energy so $mc^2 = hf'$ and we finally obtain $mgh = h(f-f')$. This is then sufficient to calculate $f'$ if we know $h$ and $f$, and agrees with the answer on the mark scheme.

However, I tried then attempting it by considering the whole system (including the Earth). The initial energy of the configuration is $mc^2 + M_E c^2 + mgh$, which is of course also the energy of the system just when the mass hits the ground. Once the photon is created, I reasoned that the total energy is now expressed as $hf + M_E c^2$ (because we're not converting any of the Earth's mass), and then the final energy once the photon reaches the top again is $hf' + M_E c^2$.

This second approach becomes very problematic to me at least. If we are considering the system as a whole, it's total energy in all four snapshots should be equal. The only resolution is that $f = f'$, which is evidently wrong. I am sure there is an oversight in the second method on my part, but I can't see it. I have the feeling that my treatment of the rest energy of the Earth is questionable. I wondered if anyone could clarify? Thank you!

 

[PeterDonis]

The first way I did it was to take the mass as its own system, then its initial energy is just $mc^2$.

Doing this means treating the mass as an open system, which can have work done on it, so its total energy is not conserved, and similarly you are treating the photon as an open system, which can have (negative) work done on it. Which is fine, but seems to have led you into confusion when you try the second way. See below.

I tried then attempting it by considering the whole system (including the Earth).

If you're working the problem in a frame in which the Earth is at rest, the Earth's energy never changes, so it drops out of the calculation. That seems to be what you are doing.

However, when you talk about "considering the whole system", what you actually appear to mean is that you are now including the potential energy due to the Earth's gravitational field, instead of just modeling the Earth's gravity as something outside the system that does work on it. That means the total energy is conserved throughout the process; all that is happening is that some potential energy is getting converted to kinetic energy and then back to potential energy. But that means you also have to treat the photon as having potential as well as kinetic energy.

The initial energy of the configuration is $mc^2 + M_E c^2 + mgh$,

Yes.

which is of course also the energy of the system just when the mass hits the ground

Yes, but now the potential energy $m g h$ has been converted to kinetic energy $\frac{1}{2} m v^2$ of the mass.

Once the photon is created, I reasoned that the total energy is now expressed as $hf + M_E c^2$

Yes, where $h f = m c^2 + \frac{1}{2} m v^2$.

and then the final energy once the photon reaches the top again is $hf' + M_E c^2$.

No, it's $h f' + M_E c^2 + U$, where $U$ is the potential energy of the photon. And we must have $U = m g h$ in order for total energy to be conserved.

The problem is that we don't have any convenient way of explicitly representing the potential energy of a photon in the math, because we think of potential energy as $m g h$ and a photon is massless so $m = 0$. The only place the potential energy of the photon shows up is in the gravitational redshift, i.e., in the fact that $h \left( f - f' \right) = m g h$. But that doesn't mean photons don't have potential energy; it just means it's easy to forget about it, so you have to be extra careful not to leave it out of your analysis.

I have the feeling that my treatment of the rest energy of the Earth is questionable.

No, that part's fine; as noted above, the Earth's energy never changes, it's just $M_E c^2$ all through the calculation.

 

[Ibix]

You're just repeating your original calculation but with $M_Ec^2$ added, no?

When the mass is at the top, it has energy $mc^2$ and the gravitational field has energy $mgh$. When the mass is at the bottom it has energy $mc^2+mgh$ and the gravitational field has zero energy. The mass is converted to radiation, $mc^2+mgh=hf$. Then the light climbs back up losing energy $mgh$ to the gravitational field and redshifting to $hf'=mc^2$. You can just repeat that, adding "Earth has energy $M_Ec^2$" after each sentence.

I think where you've confused yourself is forgetting about the energy in the field. In general that's a simewhat problematic concept, but it works well enough here.

 

[etotheipi]

Ah, thank you both I see my error. That's pretty silly of me! I must have done exactly as @PeterDonis stated and mistakenly took the $U$ of the photon to be zero.

 

[etotheipi]

I think where you've confused yourself is forgetting about the energy in the field. In general that's a simewhat problematic concept, but it works well enough here.

Slight digression; by pure coincidence I happened to be reading about something similar earlier and I was under the impression that we can't define an energy density for a gravitational field, at least not in the same way that we can with electric fields e.g. $\frac{1}{2}\epsilon E^2$? Is that what you mean when you say it's a problematic concept?

 

[Ibix]

Is that what you mean when you say it's a problematic concept?

Yes. There isn't a generally accepted definition of "energy of the gravitational field" outside of a few specific circumstances, as you seem to have read.

For gravitational fields that don't change over time it turns out that you can get something very like Newtonian gravitational potential energy (as usual, it's a bit different as you get close to the event horizon). So a standard conservation of energy approach works in this case.

 

[etotheipi]

I went away to do some reading about potential energies of photons but lots of explanations seem to me to be quite suspect.

A common one seems to define some sort of mass $m_{ph}=\frac{E_{ph}}{c^2}$ so that $U = -\frac{GMm_{ph}}{r} = -\frac{GMhf}{rc^2}$. This might well be a correct relation, but I don't know if I follow the logic. Specifically, I have no idea how to interpret $m_{ph}$. Normally you would just have $m=0 \implies E_{ph} = pc$. But even if for some reason they are using $E_0 = mc^2$, then I'm pretty sure $E_{ph}$ is not a rest energy (!), so I can't see how it's valid.

So by process of elimination, I arrive at the conclusion that this $E_{ph} = m_{ph}c^2$ is just an arbitrary definition (with no relationship to the relativistic dispersion relation) that happens to give a correct answer for potential energy which would otherwise need to be calculated with a much more complicated GR method.

I wondered whether this is a correct conclusion? It's difficult for me to gauge what is correct just from hits from google (I have noticed that for introductory GR especially there is a lot of pseudo-science out there!). I suspect to understand it fully I'd need to go further into the weeds of gravitational time dilation.

 

[PeterDonis]

I was under the impression that we can't define an energy density for a gravitational field

In general, we can't, no. But for the particular case we are considering, we can; it's just the gravitational potential energy you are familiar with. (The technical reason this works in the particular case we are considering is that, for our particular case, the spacetime is stationary.)

I went away to do some reading about potential energies of photons but lots of explanations seem to me to be quite suspect.

Yes, they are. There isn't a consistent general way to define potential energy for a photon in a Newtonian framework.

To really get at what is going on, you need to use the machinery of General Relativity for stationary spacetimes. For such spacetimes, you can define a quantity called "energy at infinity", which is a constant of geodesic motion; this is the conserved total energy (not counting the energy of the Earth) of both the photon and the falling object, and can be broken down into kinetic and potential energies in the way we have been doing it. (There are plenty of technical complexities lurking under this apparently simple explanation, of course.)

 

[Ibix]

So by process of elimination, I arrive at the conclusion that this $E_{ph} = m_{ph}c^2$ is just an arbitrary definition (with no relationship to the relativistic dispersion relation) that happens to give a correct answer for potential energy which would otherwise need to be calculated with a much more complicated GR method.

Newtonian physics is the low-speed, weak-field limit of relativistic physics. Since we mean "travelling a lot slower than light" when we say "low-speed", light is never going to be a comfortable fit in Newtonian physics.

In this static case, the fully relativistic way of doing things is entirely consistent with defining a gravitational potential and letting light exchange energy with it. But I don't think you can really derive that from Newtonian gravity. You can certainly make a plausibility argument for gravitational redshift from Newtonian potential (if you repeatedly cycle the experiment you initially described you could extract free energy if there were no redshift) but you are right to be very suspicious of anything talking about "potential energy of a photon" without going through proper relativistic arguments. Actually, a decent rule of thumb is that anything that uses "photon" and "gravity" in the same sentence should be treated with suspicion unless it's an actual quantum field theory text.