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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?

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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").

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  • $\begingroup$ and if one raises an object, against the gravity, one gives energy to the field? $\endgroup$
    – Sofia
    Jan 14, 2015 at 23:31
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    $\begingroup$ 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? $\endgroup$
    – Sofia
    Jan 14, 2015 at 23:42
  • $\begingroup$ @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 $\endgroup$
    – Christoph
    Jan 14, 2015 at 23:56
  • $\begingroup$ @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. $\endgroup$
    – Sofia
    Jan 15, 2015 at 0:03
  • $\begingroup$ @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. $\endgroup$
    – Sofia
    Jan 15, 2015 at 0:07
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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 ...

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    $\begingroup$ 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. $\endgroup$
    – mwengler
    Jan 14, 2015 at 22:00
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    $\begingroup$ 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. $\endgroup$
    – JDługosz
    Jan 14, 2015 at 22:04
  • $\begingroup$ 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. $\endgroup$
    – Time4Tea
    Jan 15, 2015 at 0:06
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    $\begingroup$ @Time4Tea 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. $\endgroup$
    – mwengler
    Jan 16, 2015 at 1:40
  • $\begingroup$ 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! ;-) $\endgroup$
    – Time4Tea
    Jan 16, 2015 at 5:10

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