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This is a follow-up my previous question. Several of the answers indicated that the mass of a particle (atom, proton, etc.) increase with the "binding energy" of it's component particles - the energy needed to split them apart - which doesn't make sense to me. I'm convinced that, at least in an exothermic chemical reaction (where the product bond energies are larger) the product's particles will lose mass (proportionally to the heat dissipated) or at least have no change.

To use a larger-scale analogy, if an object, a "particle", is 100m above the Earth's surface, it has potential energy from gravity. As it falls, this energy is lost, converted into KE. Overall, the two "particles", the object and the Earth, end up with the total energy, and therefore the same total mass. There is no room for a "binding energy" that adds mass.

My reasoning is that this extends to particles, with electrostatic or nuclear forces taking the place of gravity. Potential energy of component particles becomes KE of the bonded particle, they end up with the same mass. Actually, if this KE is dissipated (as in a burning / nuclear fusion reaction) the particles should actually have more mass in their uncombined/unreacted state, thanks to their PE. Surely it isn't possible for the mass to increase without an external input of energy?

However, answerers of my energy in chemical reactions question said that:

the energy involved in the bonds is ... half of what we normally consider the "mass" of the proton - David Zaslavsky

and

potential energy of the chemical bonds do correspond to an increase of mass - Ben Hocking

So, how can this be, and where is my reasoning incorrect? What exactly is the binding energy (if not just the energy needed to break the bond), and where does it come from?

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Are you aware of the famous E=mc**2? When one goes to nuclear sizes, it becomes important. Have a look at en.wikipedia.org/wiki/Binding_energy –  anna v Jun 23 '11 at 20:13
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You assume wrong in your analogy, @myk-- if the "particle" has a mass of 1 kg, then the potential energy at 100m is ~981 Joules. That energy is part of the combined earth-particle system mass -- but we ignore it since it's ~10^-14 kg! –  wsc Jun 23 '11 at 20:25
    
@myk: I suppose I may have misspoke a little in my answer to your other question: half of the proton's mass is attributable to kinetic energy of the gluons, which is not technically "binding energy" as the term is usually used. I'll see if I can edit that. –  David Z Jun 23 '11 at 21:58
    
@wsc I realise that the additional mass is very small, but how it comes about is important. If the object falls, then just before it hits the ground, it has 981 J KE, so the system is still 10^-14 kg heavier. After it stops, that KE dissipates into the Earth as 981 J of heat/sound, and the system is still 10^-14 kg heavier. The size of the change is irrelevant, since I am challenging the idea that the falling could lead to a net increase in mass - here, it couldn't. –  one-more-minute Jun 23 '11 at 22:02
    
@anna Yes, I am. –  one-more-minute Jun 23 '11 at 22:12
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3 Answers 3

A bound system will indeed have a lower mass than it's constituent free parts. The binding energy is usually understood to be the energy it would take to separate the bound system to free constituent parts. Then we would say $M_{tot} = M_1 + M_2 - E_{bind}/c^2$ Or we could by convention make the binding energy a negative number and say it "increases" mass by a negative amount. I think people speaking a little loosely is what's causing the confusion.

If we are talking about a nucleus, then it is impossible because of color confinement to separate the constituent parts (quarks) to independent free particles, so what we call the "binding energy" in this case is not as clear.

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It looks like you are right, see David's comment on the question. It sounds like you have explained Ben's quote as well, it wasn't clear whether he was talking about an increase or decrease. –  one-more-minute Jun 23 '11 at 22:21
    
Note: I neglected to mention the other possibility of metastable bound states, which would indeed have a larger mass than their constituent parts, but would only be short lived. –  user1631 Jun 23 '11 at 23:21
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I'm convinced that, at least in an exothermic chemical reaction (where the product bond energies are larger) the product's particles will lose mass (proportionally to the heat dissipated) or at least have no change.

It would be helpful to have a more explicit reaction, like $A\Rightarrow B+C+energy$, because otherwise it's hard to define what kind of energy it is and then identify if it stays in the system or not. In the example I present, it is true that $m_A>m_B+m_C$. This is simple enough and I think everyone will agree on that. Now, if you are considering a chemical reactor (where the mass change is small anyway), then energy manifested as kinetic energy of $B$ and $C$ will still be measured as a part of the weight of the total system until the thermal energy is dissipated out.

A system that undergoes an exothermic reaction and then expels the released energy (through emission or conduction) will have a lower mass.

To use a larger-scale analogy, if an object, a "particle", is 100m above the Earth's surface, it has potential energy from gravity. As it falls, this energy is lost, converted into KE. As it falls, this energy is lost, converted into KE. Overall, the two "particles", the object and the Earth, end up with the total energy, and therefore the same total mass. There is no room for a "binding energy" that adds mass.

Ah ha! You have discovered an argument for photon emission from quantum mechanical state transitions.

Your example seems nonsensical precisely because there is no obvious path for released energy to leave the system. It is dissipated as thermal energy and the total mass of the system remains constant (although thermal energy contributes a negligible fraction anyway). In the case of elementary particles energy must be conserved and there is no obvious analog for a force like friction. In these cases where QM physics rules, transitions are allowed (nuclear excited states, electron orbital transitions) by the release of energy through photon emission.

Binding energy is generally negative

I hope it's obvious that the system of you + the Earth does, in fact, have binding energy. This would account for a very small mass difference between the two states you discussed assuming everything else is equal (and everything else is clearly not equal). But in general, new particles are most often formed through exothermic reactions (thermo 2nd law arguments apply here), and in all these reactions energy leaves the system, provided that you start with everything at rest and end with everything at rest and don't complicate it with things like thermal energy. That means that the mass change will be negative. Over time the universe likes to see these systems loose rest mass and dissipate that energy through high-entropy photon emission.

This means the Earth right now is lighter than all its constituents would be if they remained scattered throughout space. This is the case for most things in the universe.

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Maybe an easier way to see what is going on is to consider the bound state. What will you have to do to take it apart?

Since it's bound, separating the components means you're going to have to do work against a potential. That work is energy that you're adding to the system. And so, as you separate the particles, their energy increases and so does their mass.

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