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I've looked into how the binding energy functions and that increasing the particles' potential energy or their distance from the other particles increases the mass of the system. My question is why this is, why do we measure a heavier particle just because of the additional energy?

For more clarification, I've heard that binding energy and the mass of the fundamental particles (from the Higgs field) are the cause of most of anything's mass, but from what I know, energy isn't normally measured to be mass (we have to calculate the binding energy from the difference in mass since the binding energy can't be directly measured). Why does the binding energy affect mass, how would the strong force make other particles harder to accelerate?

  • What about this related question https://physics.stackexchange.com/q/11515/. –  Aug 02 '17 at 02:15
  • Are you asking the abstract philosophical question of why $E = mc^2$ at all? If not, please be more specific. – CR Drost Aug 02 '17 at 02:33
  • Though the question is is similar, the answers focus more on the fact that everything does has binding energy, not on why that energy is measured as mass. – Jonathon K Aug 02 '17 at 02:46

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In classical mechanics mass is a conserved quantity. Every element has its specific weight, attributed to mass through the classical F=mg, m the weight g the acceleration of gravity , measured with scales to quite a great accuracy. The hypothesis was that measuring the weight of a pure substance one had the mass of the atom.

Then came the quantum revolution , the attribution of a specific number of protons and neutrons to specific substances, and it was found that mass is not a conserved quantity. That is what the binding energy curve really tells us.

The difference in mass between the sum of a proton and a neutron mass when they bind to make a deuteron is what the binding energy is about.

The case of the deuteron, the combination of a proton and a neutron, is easy to visualize. It was first noticed in the 1930's that if gamma rays of at least 2.22457 MeV irradiated deuterium then deuterons would dissociate into free neutrons and protons. Later it was noticed that if slow neutrons were brought in contact with protons deuterons would form, accompanied by the emission of gamma rays of 2.22457 MeV energy.

As, when a proton and neutron bind into a deuteron there is electromagnetic radiation released, carrying off energy as a gamma ray, it is evident that classical mechanics and classical electrodynamics is not adequate to describe these physical observations.

The conundrum is solved by special relativity, the mathematics of which is simple , but is revolutionary in the way nature in the microcosm is described.

The particles and nuclei masses are organized under what is called the "invariant mass"

invarmass

Each matter conglomerate can be uniquely characterized by its invariant mass, the deuteron mass for example is its invariant mass, and the difference with the sum of constituent neutron and proton masses is the binding energy released as a gamma at the creation of the deuteron, the energy and the mass concept are united.

The E=mc^2 well known formula is an outcome of this , where the m is velocity dependent and complicates the simple concepts . It is not used by particle physics.

I've heard that binding energy and the mass of the fundamental particles (from the Higgs field) are the cause of most of anything's mass

This is a misunderstanding. The mass given to elementary particles by the Higgs field is the mass given in this table of the standard model

elempart

only these small quark masses enter in the proton mass, which is very much larger. The rest of the mass of the proton comes from the strong interactions and the four vectors added up of all the quarks and gluons composing the proton. . The invariant mass of the proton is fixed to the given number.

In order to understand what is really happening with the nuclear table one has to study the mathematics of special relativity and quantum mechanics .

anna v
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  • +1, but the difference between mass conservation is really a relativistic/nonrelativistic divide. (Classical or quantum w/ Galilean symmetry => mass is conserved. Classical or quantum w/ Lorentz symmetry => mass is not conserved.) –  Aug 02 '17 at 05:16
  • @anna Why is the gamma ray absorption/emission not adequately described by classical mechanics? is there a problem with kinetic energy turning into radiation (gamma rays)? or do you mean that: 1) we see a mass loss 2) only gamma rays are emitted 3) so where did the mass go? – Reinstate Monica Apr 10 '18 at 20:39
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It's because $E=mc^2$.

A different way to phrase this question would be, why doesn't the binding energy of atoms compared to ions, or of molecules compared to atoms, affect their mass? And the answer is that it also does, which starts to tell you something interesting about the strong nuclear force.

If you produce some neutrons essentially at rest, and allow them to interact with protons essentially at rest, the radiation that's released as they form deuterium is about $\rm2.2\,MeV$. That's surprisingly large: the mass of the neutron and proton are each about $940\,\mathrm{MeV}/c^2$, so the energy that's released is equivalent to about 0.1% the mass of the entire system. It's right at the threshold where you could discover connection between mass and binding energy by measuring everything to three significant figures. And that's not an unusual scale. At the other end of the periodic table, when you fission uranium by adding a neutron (which works best if the neutrons are essentially at rest) the typical energy released in a fission is about $\rm200\,MeV$ --- but there are a couple hundred nucleons in a uranium nucleus, so we're still at the level of exchanging about one MeV per nucleon to reconfigure how a nucleus is arranged.

(The neutron-proton system is a counterexample to your statement that "binding energy can't be directly measured"; there the transition from the unbound state to the bound state involves emission of a single photon, whose energy is related to its wavelength in a straightforward way. The deuteron is less massive than a free proton and a free neutron; the helium-3 nucleus is less massive than a free deuteron and a free proton; and so on up the table of isotopes.)

From a chemist's perspective, imagine you have free electrons and free protons and permit them to interact. They'll also emit radiation, suggesting that each de-ionization releases about $\rm13.6\,eV$. But the hydrogen atom's mass is dominated by the proton's mass, which is still roughly $940\,\mathrm{MeV}/c^2 \approx 940\,000\,000\,\mathrm{eV}/c^2$. If you wanted to observe the effect on mass due to the electrical interaction that turns a hydrogen ion into a neutral hydrogen atom, you'd need to measure all the masses involved to ten significant figures. That's a much taller order. And it continues to get harder if you consider molecules, where the intrinsic masses are larger and the binding energies are smaller.

The connection from here to the Higgs mechanism is surprisingly tortuous. The Higgs is mostly responsible for the electron mass. However, unlike nuclei compared to nucleons (and atoms compared to ions, and molecules compared to atoms), the three "valence quarks" that make up a proton or a neutron are less massive than the particle they compose. Within the nucleon, then, there are several competing things happening. First there are an indeterminate number of quarks with finite rest mass due to the Higgs mechanism. (It's always true that the number of quarks minus the number of antiquarks is three, but the number of quark-antiquark pairs is not fixed.) Second, the interactions among these quarks force each to carry a large amount of kinetic energy --- a kinetic energy that's much larger than the energy equivalent of the rest masses of the quarks, but that's hard to quantify because, again, the number of quarks and antiquarks inside a nucleon is not well-defined. This internal kinetic energy increases the effective mass of the system. Third, there's an additional energy barrier that must be overcome for any of the constituents of a nucleon to escape. That's the "binding energy" of a nucleon, but its definition is much murkier that the equivalent for nuclei or atoms or molecules, where the number of constituents is fixed.

You could in principle start to measure binding energies among quarks in the same way you measure binding energies among nucleons: by running reactions like \begin{align} \gamma + p &\leftrightarrow n + \pi^+ \\ \gamma + n &\leftrightarrow p + \pi^- \end{align} However, things get tricky quickly for reasons which eventually boil down to "you can't say how many quarks there are in a proton."

rob
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