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In his lecture notes on General Relativity (2nd page of the pdf, labelled page 7) and in the lecture itself, Scott Hughes puts forward a thought experiment to justify the existence of gravitational redshift. The thought experiment, as I understand it, goes as follows:

A rock with mass $m$ is dropped from a tower with height $h$. The rock is dropped into a device that converts it into a photon in such a way as to perfectly conserve energy. Therefore, this photon is created with $E_{bottom} = m + mgh = \hbar\omega_{bottom}$ at the bottom of the tower (units of $c=1$).

This photon is directed back to the top of the tower, where it will enter another device that converts it back into a rock in a way that perfectly conserves energy. Therefore the energy of the new rock will be $E_{top} = \hbar\omega_{top}$

If $\omega_{top} = \omega_{bottom}$, the new rock will have mass $m_{new} = m + mgh$. Dropping this rock into the device at the bottom would then create a photon of with energy $E_{bottom2} = m + mgh + mgh$ and so forth. Energy is not conserved and we have invented a free, boundless energy supply.

Therefore, $\omega_{top}$ must be lower than $\omega_{bottom}$ by exactly enough to remove the extra $mgh$ and preserve conservation of energy.

I have a question about this argument.

It seem to me that when the new rock is created at the top of the tower with energy $E_{top2} = m + mgh$ that nothing amiss is happening. A rock with mass $m$ and potential energy $E_{pot} = mgh$ is created. So each new photon created at the bottom continues to have the same energy $E = m + mgh$.

Why do we need the photon to redshift and lose energy here? To clarify my question is: "Does this thought experiment really require gravitational redshift to avoid violating more fundamental physics? Or can't we simply say that the photon does not change frequency, the new rock is created with mass $m$ and potential energy $mgh$ and everything lines up nice and tidy?"

Doing some research on this, it seems like this argument is often presented not as the redshift balancing out energy conservation, but that without the redshift, you would have 'perpetual motion.' This is also not convincing for me. Isn't it just a sequence of rocks being dropped?

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3 Answers 3

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it seems like this argument is often presented not as the redshift balancing out energy conservation, but that without the redshift, you would have 'perpetual motion.' This is also not convincing for me.

Perpetual motion in a theory is not a fatal blow to that theory. Practically perpetual motion is not possible because ideal conditions cannot be achieved. In principle Newtonian gravity also allows a perpetual motion for a 2 body system.

Energy is not conserved and we have invented a free, boundless energy supply.

Energy is not conserved is not a very big problem. In special relativity potential energy is not properly defined. In general relativity energy itself is not properly defined. But in certain systems energy can be defined and is conserved like in eternal Schwarzschild black hole since there will be time translation symmetry. But in many cases it cannot be defined like for example in FLRW metric.

The main problem is your assumption that infinite amount of energy can be extracted. It is not physically possible.

Also that derivation is not completely rigorous, it is just meant to be an intuitive introduction.

A better answer is using Schwarzschild metric (for example see 9.4 in General Relativity An Introduction for Physicists).

Another better approach is using equivalence principle. You should imagine a lift going at an acceleration $g$ wrt an inertial reference frame. Then a photon is released at the bottom of the lift. According to the inertial reference frame the velocity of light is $c$ towards above and is constant. Frequency will be constant in this frame. Now find the doppler shift in the reference frame of the lift (consider instantaneous inertial frames travelling at the same velocity as the lift). It doesn't need understanding of Schwarzschild metric. It also doesn't use things which are related to Quantum Mechanics (like $E=\hbar\omega$) and non relativistic ideas like potential energy.

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  • $\begingroup$ My question must not have been clear. I am well aware that producing infinite energy is not physically possible. My degree and field of research was statistical mechanics. My point is that it seems to me that this example fails to show a scenario that (erroneously) produces infinite quantities of energy unless we have gravitational redshift. $\endgroup$
    – Metropolis
    Jun 23, 2021 at 14:17
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"...Or can't we simply say that the photon does not change frequency, the new rock is created with mass $m$ and potential energy $mgh$ and everything lines up nice and tidy?"

No, this is not possible, because if EM energy of the radiation packet did not change while coming up to height $h$, this would mean the potential energy of that radiation packet in gravity field increased without energy decrease anywhere else, and we would already have problem with energy conservation.

Suppose that is what happens; at some point, we have radiation packet at height $h$ with total EM energy $mc^2 + mgh$. We can arrange things in such a way that this energy is spent to heat up a rock of negligible mass placed at height $h$, and thus it is made into a hotter rock with mass

$$ m\left(1 + \frac{gh}{c^2}\right). $$

This rock is more massive than the rock we started with, at the same height.

This whole system has greater total energy than the system we started with.

Since this does not correspond to our beliefs about energy conservation, something had to go wrong in the assumptions. The most important wrong assumption is that energy of an EM radiation packet has the same value when emitted at height 0, as when it comes to height $h$. Instead, it must be that its energy at higher height is lower.

This can be derived more rigorously using special relativity to analyze accelerated motion. But this "proof by contradiction" is the easiest argument to see why it is necessary.

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  • $\begingroup$ My confusion is, in each of these examples we are always saying the energy from the photon is entirely converted into mass. That seems like it wouldn't be the case. If we simplify and just image a photon with energy E that is converted into a rock in a device at some height h, I posit that rock wouldn't have mass = E/c^2. I posit that rock would have mass E/c^2 - mgh. Some of the photon's energy would be converted to mass, some would be converted to the rock's potential energy. $\endgroup$
    – Metropolis
    Jun 24, 2021 at 16:02
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    $\begingroup$ I understand your proposition. It is wrong as I've explained above. In other words, when the radiation packet is at height $h$, all of its EM energy $E_{\text{EM}}$ at that height is available for heating the rock at that height. None of this energy is needed to increase gravitational potential energy of the system planet-rock; the whole system already has this gravitational potential energy $\frac{E_{\text{EM}}}{c^2}gh$ before the conversion, because the EM energy is localized at a small region at height $h$. $\endgroup$ Jun 24, 2021 at 23:22
  • $\begingroup$ " the whole system already has this gravitational potential energy 𝐸EM𝑐2𝑔ℎ before the conversion, because the EM energy is localized at a small region at height ℎ." Do you mean to say the photon has a gravitational potential energy? If so, that seems like a circular argument. If we assume a photon has gravitational potential energy, then as it move upward in a g field, it has to lose energy from another source and that can only come from its frequency. I think the idea here is to try to demonstrate that it must have gravitational potential energy. Not use that fact as part of the argument. $\endgroup$
    – Metropolis
    Jun 25, 2021 at 16:12
  • $\begingroup$ And thanks for trying to help me through this. I'm certain the confusion is entirely mine. $\endgroup$
    – Metropolis
    Jun 25, 2021 at 16:12
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    $\begingroup$ The photon concept is problematic here for other reasons (quantization of radiation energy has nothing to do with GR and can't possibly be relevant for explaining gravitational red shift, quantum theory and GR are different theories with no accepted unification). But the same kind of argument works rather well with classical EM wave packet; since EM energy of a wave packet is localized, its presence at some height means there is associated contribution to gravitational potential energy, and thus as the wave packet goes upwards, it loses EM energy in favour of gravitational potential energy. $\endgroup$ Jun 25, 2021 at 16:38
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Every cycle the machine at the bottom measures a bigger amount of energy arriving than what top people measured to be the amount of energy that they sent towards the bottom.

Then the bottom people send photon, the measured energy of which the top and the bottom measure to be the same.

So therefore top measures larger amount of photon energy than what the measured energy of the rock was.

From a bigger amount of measured energy a rock with a bigger measured mass can be made.

$$ m_{newrock} = m_{oldroc} + m_{oldrock}gh /c^2$$

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  • $\begingroup$ So this is my question. Why is the new rock more massive? Isn't it simply the case that the new rock has mass m and potential energy mgh? I mean to say: if a photon with E = m + mgh gets to the device that makes rocks, when it makes a rock that newly created rock will have potential energy and mass, both. If it's created with mass m, it will be created with potential energy mgh. No new mass or energy is created in this cycle. $\endgroup$
    – Metropolis
    Jun 23, 2021 at 20:58
  • $\begingroup$ @Metropolis Does a falling rock gain energy? Well here it's assumed that it does. If rock's energy at upper position is E=m+mgh then after gravity has done work on said rock its energy E=m+mgh+mgh........ You can actually exchange an upper pebble for a lower boulder. $\endgroup$
    – stuffu
    Jun 23, 2021 at 21:17
  • $\begingroup$ @Metropolis Or maybe in this thought experiment it was so that ascending photon did not lose any kind of energy but gained potential energy. And a falling rock had constant energy? The rock gets bigger anyway, no matter which alternative we choose. $\endgroup$
    – stuffu
    Jun 23, 2021 at 21:34
  • $\begingroup$ No, a falling rock does not gain energy. Potential energy becomes kinetic energy, but conservation holds. The rock's total energy is still equal to m + mgh at the bottom. It would just be in the form m + 1/2mv^2. But it's just easier not to calculate v, since it's an unneeded step. $\endgroup$
    – Metropolis
    Jun 23, 2021 at 21:51
  • $\begingroup$ An ascending photon doesn't gain potential energy. At least, not classically, as it's massless. Moreover, the claim here is that it would lose energy in the form of gravitational redshift. $\endgroup$
    – Metropolis
    Jun 23, 2021 at 21:52

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