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Certain particles come together to form a nucleus, this causes a certain amount of energy to be released in the form of photons, the mass of the nucleus is less than the sum of the masses of its components, it is said that a certain mass has been converted into energy. If this itself is considered as a closed system, the energy of the resulting photons is considered as part of the mass of the system, which means that the mass of the system is conserved and is equal to the sum of the initial components. In the same phenomenon, it has been described in the first place that there is a loss of mass and in the second that the mass is conserved. It doesn't seem consistent. Is there something wrong with this reasoning?

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  • $\begingroup$ Photons are massless. $\endgroup$
    – DanDan0101
    Sep 20 at 19:51
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    $\begingroup$ If you consider the nucleus alone: loss of mass. If you consider the overall closed system: conservation of mass. No inconsistency - only different systems. $\endgroup$ Sep 21 at 8:32
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Indeed, the whole idea of “converting” mass into energy by $E=mc^2$ is a popular misconception. (Note that $E=mc^2$ applies in the center of momentum frame, so I will assume that from here on). Since energy of an isolated system is conserved $E$ must be constant, so there is no loss of mass and no gain of energy. The mass and the energy are both constant and they are related by $E=mc^2$ at all times.

What is more correct is that a system of particles with some given energy can be converted in to other particles with the same total energy, and therefore the same system mass. So, a slow moving electron and positron can be converted into two photons. The photons have an energy of 511 keV each, so that implies that the electron and positron each had 511 keV of rest energy.

Similarly, since the electron and positron together had 1 MeV/c^2 mass so also the system of the two photons have 1 MeV/c^2 mass. This means that the mass of a system is more than the sum of the masses of the pieces of the system. Mass of an isolated system is conserved, not the sum of the masses of the parts.

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    $\begingroup$ Since energy is conserved E must be constant, so there is no loss of mass and no gain of energy. The mass and the energy are both constant and they are related by E=mc2 at all times. I'm now more confused than before reading this answer: you make it sound like the nuclear mass defect doesn't exist! en.wikipedia.org/wiki/Nuclear_binding_energy $\endgroup$
    – Gert
    Sep 20 at 20:32
  • $\begingroup$ @Gert the nuclear mass defect is not a violation of conservation of mass or conservation of energy. The mass/energy that leaves the nucleus does so in the form of the nuclear decay products. The mass and energy of the system of all decay products is equal to the mass and energy of the original nucleus. $\endgroup$
    – Dale
    Sep 20 at 20:37
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    $\begingroup$ It's the sentence The mass and the energy are both constant that is ambiguous, for when read in a certain way it reads "the mass is constant and the energy is constant". But it's the sum of $m$ and $E$ that is constant. $\endgroup$
    – Gert
    Sep 20 at 20:44
  • $\begingroup$ @Gert you seem to have fallen prey to the popular misconception. The mass of an isolated system is constant. And the energy of an isolated system is also constant. The “read in a certain way” was the intended reading. Yes, the sum is also constant, but the actual conservation law is stronger than that. Energy is conserved $\endgroup$
    – Dale
    Sep 21 at 10:36
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    $\begingroup$ I owe you an apology and a +1. As chemists we are trained mainly to look at open systems (either for the extraction of energy or for systhesis of desirable matter), not closed ones. A bit of reading has made me fully understand that in isolated systems both mass and energy are conserved. Mea culpa. $\endgroup$
    – Gert
    Sep 22 at 17:58
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I've always found "binding energy" to be an unhelpful term.  It just exists for chemists and nuclear scientists to do their book-keeping.

Energy is possessed by particles. It does not exist on its own. And mass is simply a property that energy has when it is confined.

Look up "photon in a box" and you can read about how if you were to confine massless photons in a massless box, that box would still start exhibiting properties like inertia, i.e. it would have mass. Atoms are essentially nothing but this.

So it's not a question of "mass being converted to energy," although it's said this way as a shorthand. It's really "confined energy being released as a particle's kinetic energy."   To your question, mass is not conserved except approximately in macroscopic chemical reactions and physical processes. Energy is.

What happens in nuclear fusion reactions for example is:

(Deuteron) + (Deuteron) = (Helium) + (some photons, electrons, and neutrinos)

The particles of light and matter in that last term carry away energy X. Helium has lower energy than the two Deuterons combined did (by an amount X), which is why Helium doesn't spontaneously decay back into two Deuterons. It doesn't have enough energy to do so, unless something else (like for instance some other photon or matter particle with at least kinetic energy X) comes and crashes into it. That's why in this latter case they call X the Binding Energy of Helium, because it's the energy you have to add to get it to break into something else. And that also means you can't talk about binding energy without specifying the reaction you are talking about. The energy to break He into 2 Deuterons is much less than the energy to break it into 2 protons and 2 neutrons, which is itself MUCH less than the energy to break the quarks apart into pairs. So it's binding energy "with respect to" a certain process the atom could undergo, not an inherent thing every Helium atom has.

Hope that is helpful.

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  • $\begingroup$ Minor comment to the answer (v3): the reaction “deuteron + deuteron $\to$ helium + photons + electrons + neutrinos” doesn’t conserve charge. The electrons and neutrinos are created in the weak-interaction step $\rm pp\to d$. $\endgroup$
    – rob
    Sep 29 at 6:02
  • $\begingroup$ Major comment to the answer (v3): your statement “[energy] does not exist on its own” is not really the case. Consider this argument for energy storage in the gravitational field, or the rules in electromagnetism relating field strength to energy density. $\endgroup$
    – rob
    Sep 29 at 6:07
  • $\begingroup$ Good point! Hopefully my point was made anyway without the reaction eqn being completely literal. As for the energy storage in fields, that is an interesting point. I wonder, however, if it could be argued that the fields don't actually have that energy unless there is a particle or mass there to interact with it? $\endgroup$
    – RC_23
    Sep 29 at 13:30
  • $\begingroup$ A glib response is that the tool is called “quantum field theory,” not “quantum particle theory.” The relationship between the photon and the electromagnetic field has a lot in common with the relationship between the electron and “the electron field.” Stating the details convincingly depends a lot on how much QFT you know, and doesn’t fit in a comment in any place. But as a rule, you’re better off thinking of fields as the cornerstone of modern physics, and particles as an interesting collective phenomenon that fields engage in frequently enough that it’s interesting. $\endgroup$
    – rob
    Sep 29 at 20:47
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Mass is not conserved; energy is conserved. You shouldn't count the energy of the photons (which are massless) or the kinetic energy of the products or reactants as part of the mass (note, the concept of 'relativistic mass' is deprecated).

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The mass isn't lost, just separated from the nucleus when the energy was released. The mass is still somewhere flying through the universe.

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  • $\begingroup$ Could down-voter please explain? This is the most accepted explanation by the scientific community for where the extra energy from nuclear reactions comes from. If the entire scientific community is wrong, where do you think it's from? $\endgroup$ Sep 21 at 17:43

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