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If you increase the h (=height), potential energy will be increased given by U=mgh.

Where does the energy go, into atoms?

mini
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2 Answers2

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I would say that the energy is in some sense "stored" in the gravitational field. When the gravitational force is acting on your mass, it is "transferring" the energy from the field to the falling object (namely, its kinetic energy). You have the same situation in electromagnetism, the field has an energy density and that energy is "transferred via the Poynting vector". (This is just a simple way to imagine things, but not totally "true").

Christoph
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Worldsheep
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  • and if one raises an object, against the gravity, one gives energy to the field? – Sofia Jan 14 '15 at 23:31
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    The issue with the energy density is not clear. In an electrostatic field you can bring as many charges as you wish. And if you let them free, the field will do mechanic work on all of them. So, is the field storing infinite energy? Potential energy is a property of both charges, not only of the field surrounding the object. No energy passes from the field to the body, the energy transforms from potential to kinetic. So, where the two bodies keep the potential energy? – Sofia Jan 14 '15 at 23:42
  • @Sofia: I do not understand your objection; as long as you only bring an arbitrary but finite amount of charges, the potential energy will be finite; anyway, if I recall correctly, the interpretation of potential energy as field energy (once you substract the self-energy of the charge) can be found in chapter 1 of Jackson's EM book – Christoph Jan 14 '15 at 23:56
  • @Christoph : I think that my objection is correct with all the due respect to the book. As an argument, let me remind you the Schrodinger equation of the electron in atom - for me is the easiest example. Which energy you write there as potential energy? Do you write $Q/(4 \pi \epsilon_0 r)$ (where $Q$ is the nucleus charge) ? Or do you write $Q e/(4 \pi \epsilon_0 r)$, (where $e$ is the electron charge)? So, the field is of both objects. – Sofia Jan 15 '15 at 00:03
  • @Christoph this is why, no matter how many charges you bring in the neighborhood of a central charge $Q$, there is potential energy for all of them. – Sofia Jan 15 '15 at 00:07
  • @Sofia: and the energy density of the electromagnetic field will have increased by the required amount; take 2 charge configurations A and B, integrate the difference of electromagnetic field energy density and you'll arrive at the potential energy – Christoph Jan 15 '15 at 00:23
  • @Christoph : yes, but from your answer one understands that the electric field and the potential are due only to the bigger charge. And that is not true. About Poynting vector, with the electrostatic field it is zero. I didn't understand what is doing in static fields. The electric or gravitational field is between the two charges and it stores the potential energy or consumes it for doing mechanical work. That's it. – Sofia Jan 15 '15 at 01:05
  • @Sofia: from your answer one understands that the electric field and the potential are due only to the bigger charge no, one has to consider the field generated by both charges; the potential energy comes from the mixed term when evaluating $|E_1+E_2|^2$ – Christoph Jan 15 '15 at 01:13
  • @Christoph I think that at this stage we think the same. The field and the potential is due to the two charges. – Sofia Jan 15 '15 at 01:17
  • @Sofia: yes, we're in agreement; I just did not get your objection until your final comment – Christoph Jan 15 '15 at 01:25
  • @Christoph now it's O.K. – Sofia Jan 15 '15 at 01:26
  • @Christoph : just you can remove the doubts, "the energy is in some sense "stored" in the gravitational field". No, it is stored, and that's it. But the "transferred via the Poynting vector. (This is just a simple way to imagine things, but not totally "true")." If it is not totally true why do you need this statement. Poynting vector is connected to flux. Which flux do you have here? The potential energy goes to kinetic. So, which flux? – Sofia Jan 15 '15 at 01:33
  • @Sofia: you'd have to ask Worldsheep about why he felt the need to qualify these statements; possibly, it has to do the GR (no energy density for gravity) and the non-uniqueness of the Poynting vector, but he may have been thinking about something completely different... – Christoph Jan 15 '15 at 01:53
  • @Christoph : I see. But if the answer is his, why I have this long talk with you? It's him that should have discussing with me. How does it happen that all the talk is with you? – Sofia Jan 15 '15 at 01:59
  • @Sofia: The reason could be that I have just woken up... Anyway, the "not totally true statement" was referred to the gravitational field... And I was indeed thinking about GR – Worldsheep Jan 15 '15 at 08:04
  • @Worldsheep : but it is futile. For this "woken up" you can loose points. There is no flux radiation there, all the potential energy passes to kinetic, and vice-versa. I mean, a moving charge in an electric field does produce a magnetic field and therefore the Poynting vector is not null, but it points toward the charge, not outwards (please check if I am right). – Sofia Jan 15 '15 at 10:25
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I think the clue is in the name - it's only potential energy, due to the position of an object in an external force field. The energy doesn't actually exist yet (although it can be converted into other forms of energy). It's not present in the molecules and it's not stored anywhere.

EDIT: Think I was a bit hasty on this one. $E=mc^2$ implies that the potential energy must exist, because mass exists. I'll think on it some more ...

Time4Tea
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    This is not right. Because of E=mc^2, the potential energy has an actual mass which with a fine enough scale can be measured. A 1 m on a side box sitting on earth with a 1 kg mass inside it at its top weighs slightly more than a 1 m on a side box sitting on earth with a 1 kg mass in it on its bottom. It weighs mgh/c^2 more where h is 1 meter and m is 1 kg. – mwengler Jan 14 '15 at 22:00
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    That's significant for the interneucleon force that holds protons and nutrons together. The "binding energy" is negative and does indeed affect the mass of the atom. That's why the sun shines! 1 He is lighter than the sum of its (unbound) parts. – JDługosz Jan 14 '15 at 22:04
  • Hmm ... thinking on this some more, maybe I was a bit trigger-happy on this one. $E=mc^2$ implies that all forms of energy must exist, because mass exists. @mwengler: have any experiments been conducted that have successfully measured that mass difference? It must be extremely tiny. – Time4Tea Jan 15 '15 at 00:06
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    @Time4Tea http://plato.stanford.edu/entries/equivME/#4 discusses measurements. Nothing as satisfying as weighing a compressed spring and comparing it to the weight of the uncompressed spring. – mwengler Jan 16 '15 at 01:40
  • Thanks for the link @mwengler. I've actually asked about this topic as a new question here. If you know the answer then feel free to post and get some easy points! ;-) – Time4Tea Jan 16 '15 at 05:10