# Confusion over binding energy

1. Jan 23, 2010

Nucleons are less massive when in the nucleus than when they are separated,atoms are less massive than the particles which make them up,these differences being explained in terms of binding energy.Does this mean,for example, that the electron and proton both lose mass when they come together to form the hydrogen atom?If so I thought the belief was that particles such as electrons have a mass which is invariant.What am I missing?

2. Jan 23, 2010

### jimmy neutron

Mass isn't invariant: It is the total mass+energy that is conserved.

As well, I'm not sure if you're aware of the well known formula from special relativity that relates the mass of a particle to its speed v: m = m0/sqrt( 1 - v^2/c^2)?

3. Jan 23, 2010

Thank you jimmy.I think I did not explain myself properly.The invariant mass I was referring to was m0 (as in the equation you quoted).In this equation it is assumed that m0(often referred to as the rest mass) is invariant.Taking the hydrogen 1 atom as an example we need to input approximately 13.6eV of energy to separate the proton and electron.Mass/ energy is conserved but the energy input seems to result in the particles gaining more rest mass.How can these mass changes be explained if rest mass is invariant?

4. Jan 23, 2010

Staff Emeritus
The conceptual problem is not something that has anything to do with particles or relativity or even quantum physics. This goes all the way back to Physics 101: energy (kinetic or potential) is not a property of individual objects; it is a property of configurations of objects.

Saying one or more objects gains or loses mass based on the energy of a configuration of several objects misassigns this, and is almost certain to lead to misunderstandings and confusion.

5. Jan 23, 2010

I agree but take the example of the simplest configeration I can think of,the hydrogen atom.The total mass energy of the atom is that contained within the rest masses of the particles the kinetic energy they carry and the potential energy of the system.When the atom is ionised by the input of energy we might expect the total mass energy to increase and published data seems to show that this is the case.The data tells us that the mass of the hydrogen atom is less than that of its component parts.

6. Jan 26, 2010

### torquil

Yes, the rest mass of the bound state that we call the hydrogen atom, when considered as a unit, is less than the sum of the rest masses of its individual constituents. This comes about because the potential energy between the constituents contributes negatively to the rest energy of the bound state when considered as one unit. There is also kinetic energy as you mention, which contributes positively, but it is smaller in absolute value than the potential energy, so the net results is a negative difference in rest mass / rest energy.

If this were not the case, the bound state would not appear. It appears because it is energetically favourable.

Torquil

7. Feb 15, 2010

Thank you torquil.Please correct me if I'm wrong but you seem to be agreeing that the absolute rest mass of the proton plus electron when separated is greater than that when they are bound together in the hydrogen atom.This in fact is what the published data seems to show the mass difference being equal to the binding(ionisation)energy.
So is the data interpreted incorrectly or could it be that rest mass is not invariant?What am I overlooking.

Last edited: Feb 15, 2010
8. Feb 15, 2010

### JK423

I havent done any field theory but from what i have understood till now:
When you have a gravitational interaction, or Coulomb interaction, the energy -that is the negative potential for a bound system- is stored in the field.
So, maybe when you write down the proton's energy (in hydrogen):
E=mc2+K+V
the negative V potential that you insert is actually the field's energy. So you are not considering just the proton but the proton+field. Which means that its not the rest mass of proton that decreases but the system's proton+field.
Im not sure about the above, just speculating and give you a hint

9. Feb 15, 2010

### Prologue

How do you come to the conclusion that REST mass is not invariant in a MOVING system? I am not talking net movement but internal movement.That is the error and was pointed out already by Vanadium. You are trying to attribute a 'rest' mass to the atom, which is fine I guess if you only look at the atom. But, don't try and pull it apart or else you run into the exact problem that you have with nuclei. The sums of the parts (at rest) don't equal the mass of the unit when bound (which means the constituent parts are moving and NOT at rest).

Also, bound states don't have the same energy as a non bound but at rest configuration. This is by definition but also has a practical interpretation. In order for an electron to move in that close to a proton and orbit it, it must release energy in some form. This is due to the fact that the force balancing the inward pull is the inertial force, and this force would be too great if energy were stored in kinetic energy of the electron and proton. In other words, to be captured the electron has to be moving slower than what it would be if just released from rest at infinity. The energy that is missing must be cast off in the form of photons. I suppose that if you could somehow wrangle these photons that were carrying the energy, and make them somehow stuck to the atom, then it would 'weigh' the same as the constituent parts (photon, electron, proton).

Last edited: Feb 15, 2010
10. Feb 16, 2010

### JK423

I dont think it has to do with motion here. Motion increases mass, not decrease it. And if you think of another configuration you can achieve mass defect without any motion at all. For example, "take" the electron and leave it at some distance from the proton. Before it moves at all, it'll have its mass reduced because of the negative potential. (Im thinking a bit classically here but relativity also holds for classical bodies)
I really think it has to do with the idea that the negative energy is stored in the field as i said at my last post, without being totally sure about that..

11. Feb 16, 2010

### torquil

We have Me, the rest mass of the electron, and Mp, the rest mass of the proton. These are invariant quantities, and correspond to the total energy of each particle when at rest in vacuum.

The rest mass of the hydrogen atoms is the total energy of it when it is located at rest in a vacuum. For the hydrogen atom, at rest means that its centre of mass momentum is zero, and its total spin is zero.

It still has internal movements of course, that contribute positively a value E_k to the total energy, but also negative electric energy E_p that contributes negavitely. The sum of all of these is the hydrogen rest mass:

Mh = Me + Mp + E_k - E_p.

By solving the Schrödinger equation, we find that E_p will be larger than E_k, and thus it is a bound state with Mh < Mp + Me. This difference is the binding energy (the energy needed to take the bound state apart).

Mh is the rest mass of hydrogen, is invariant for motions of the hydrogen that don't influence the internal degrees of freedom of the hydrogen atom, and is related to its energy and momentum in the usual way: Mh^2 = E^2 - p^2.

Torquil

12. Feb 16, 2010

### Prologue

Your point is completely valid, I was just cautioning against comparing apples to oranges. The second paragraph I wrote sums up what I thought of this particular situation. If you add up all of the energy, it will 'weigh' the same as the non bound state. So, I suppose I should have extended that list a little to be (proton, electron, field) -> (proton, photons, electron, field). Those two systems would 'weigh' the same. Good point.

13. Feb 16, 2010

Thank you everybody.
In summary the view expressed here seems to be that for the hydrogen atom:

Mp+Me-Be=Mh

(Mp=proton mass,Me=electron mass,Mh=mass of atom,Be=binding energy mass)

The equation,although specific for hydrogen,can be generalised and summarises how I have always understood binding energy to be. I wanted confirmation that the equation is correct and generally accepted and I have been getting that confirmation as a result of getting replies from certain organisations I have approached(I am still waiting for some replies) and from the discussion that has been going on here.
If one bears in mind the methods by which absolute atomic mass(as distinct from relative atomic mass) is measured,there is something rather odd and interesting about the equation.

Last edited: Feb 16, 2010
A Confusion regarding the $\partial_{\mu}$ operator Jul 24, 2017