# Why is there a mass defect in the nucleous?

1. Feb 1, 2010

### JK423

Can someone explain to me why is there a mass defect in the nucleous? That is, why the nucleous' mass is smaller than the sum of the masses of its constituents (nucleons)?

PS. Just to inform you about my background, im in the 4rth year in physics.
The funny thing is that i was first taught the mass defect in high school and now im in the 4rth year and still haven't understood it.

John

2. Feb 1, 2010

### JustinLevy

The quickest way to understand this is:
E = mc^2
If a system is at rest (its constituents can move, but the net momentum=0), the energy of the system is equal to the mass of that system.

So, because the nucleons are in a bound state in the nucleus, clearly the total energy for the system is less than the total energy of all those nucleons separately. Therefore the nucleus mass < the sum of the nucleon masses separately.

Does that help?

3. Feb 1, 2010

### JK423

So, if i understood correctly, the mass defect is clearly a quantum mechanical phenomenon.

Say you have two neutrons. They dont interact until they come so close to each other that the strong force will get them bound.
The energy levels of the new bound state are quantized. (E1,E2,..)
Initially neutron had energy E=mc^2. In which energy level (E1,E2,..) is going to get? I suppose to the first energy level En<mc^2 emmiting the rest energy (mc^2-En) as a photon.
1) ΔΕ=mc2 - En is the mass defect?
2) Am i correct about the mechanism that creates the mass defect?
3) If 1,2 are true, there is a mass defect because energy ΔΕ is emmited via photon emmision?

I hope i get it someday :P

4. Feb 3, 2010

Any help?

5. Feb 3, 2010

### inflector

I'm by no means an expert, but I've been looking at this specific issue for several years and I don't think anyone really knows the answer to this question. There are some models for how the nucleus works that give some hints but nothing that is even close to definitive, as far as I've seen. The best formula (i.e. the one that fits the data best) appears to be the semi-empirical mass formula:

http://en.wikipedia.org/wiki/Semi-empirical_mass_formula

The formula is a decent approximation but it doesn't explain the magic numbers and differs from observation in many ways. Further, it describes what we see rather than explains why we see it.

I'm interested in fusion and have been surprised at the lack of research into this particular problem as it seems to me to be quite fundamental to fusion itself since fusion energy comes from the energy captured because of the reduced per-nucleon binding energy as two lighter elements fuse to a heavier one. That energy is seen as the mass defect you ask about. It seems to me that if we understood how the nucleus is held together better, we might find some ways to make fusion happen easier.

There has been a lot of experimental observation of what happens but not much progress in learning why. Take a look at the binding energy graph from Wikipedia:

http://commons.wikimedia.org/wiki/File:Binding_energy.jpg

Why is there such a big jump in binding energy per nucleon between H and He, for example? Why does the per-nucleon binding energy of He exceed that of heavier elements Li, Be, and B and you have to go all the way to C before the binding energy of He is exceeded? The binding energy per nucleon increases until you get to Fe-56 (the most common highest-binding energy nuclide) or Ni-62 (the absolute highest binding energy nuclide). Why does it go down at these particular numbers? What gives the Fe-56 nuclide a higher binding energy than the other Fe isotopes? Ditto for Ni-62?

No one has answers to these questions, from what I've seen. Until we really understand the curve in the binding energy graph, I don't think we have enough information to adequately answer your original question.

6. Feb 3, 2010

### JK423

Wow.. i really wasnt expecting such an answer! I thought that it had a straightforward explanation that everyone knew about..
Im really gonna ask my proffessor abou it

7. Feb 3, 2010

### JustinLevy

I'm not sure what you mean by this. The only thing needed to have the system mass be < than the sum of the individual masses is a bound state (attactive force, so energy of system is LOWER if the particles are bound). You don't need quantum mechanics for this. You only need an attractive force.

None of the details of the energy levels, nor how the system settles into it (emitting photons, etc.) are important to understand the sign of this effect.

Well, either he misunderstood your question, or I misunderstood your question (or we both did).

Are you asking for the numerical value of the mass defect? Or are you just asking why the nucleus mass is smaller than the sum of the masses of its constituents (nucleons)?

The answer to the second is simple, as no messy details are needed.

The answer to the first however is indeed incredibly complicated. It was a huge feat that people recently calculated the mass of the proton from QCD using lots of clever math and lots and lots of brute forcing computer calculations. Calculating something like the energy of multiple nucleons bound together from "first principles" is still a long ways off. There are many different emprical methods people have used to try to explain the nucleon interactions though. As inflector mentioned, these are not entirely satisfactory.

So while it may be easy to understand why the nucleus mass is less than the sum of the nucleons individually, it would be incredibly difficult to try to explain the numerical value of that difference.

8. Feb 3, 2010

### Staff: Mentor

To put it another way: the protons and neutrons in the nucleus stick together. To pull them apart, you have to do work on them, that is, put energy into the system. This energy manifests itself as an increase in the mass of the system, via E = mc^2. (assuming the particles start and end up stationary, so we don't have to deal with kinetic energy.) This part is simple and well-understood.

What is not as yet well-understood is how to calculate the exact amount of work necessary from first principles, i.e. the fundamental equations of the strong interaction between the quarks that make up the protons and neutrons.

9. Feb 4, 2010

### JK423

Justinlevy you're talking about it like its something "usual", and by "usual" i mean classical! I dont think mass defect is a classical phenomenon so i named it quantum mechanical. If you have classical masses that get bound gravitationally (like the sun with the planets), youre not gonna see any mass defect! Even there you have negative potential energy!
However due to energy conservation this negative energy is equalized by the kinetic energy of the bodies. So in classical physics we cannot talk about bound states like they are something "special". They are "special" only in quantum mechanics because many strange things happen when two systems get bound and these are 1)quantization of energy and 2)minimum non zero energy.
(@jtbell als How can you speculate that particles end up stationary and are bound?
Again an example of Classical physics. A comet goes by the sun and it happens to get bound. If you discard its kinetic energy of the initial and the final state, then youll end up finding a huuuuge mass defect (if you consider to use E=mc2 like u do here with the nucleous).

If i misunderstood anything youve said, explain to me.
Till now, im quite convinced that mass defect:
1) is a QM phenomenon only - no classical analogue
2) has something to do with the fact that two quantum systems (proton & neutron) get bound for the reasons i explained above.

10. Feb 4, 2010

### Prathyush

I think Justinlevy is correct
We can define mass defect as the the extra energy in a bound system over constituents alone. It is not restricted to only quantum effects. It is only based on mass energy equivalence which is due to special relativity.

I see nothing special in quantum mechanics.

The is extra mass when ever bound states exists. Electromagnetic field will have mass. The mass defect for the hydrogen atom will be small ~ev compared to nucleons ~Gev.
In case of gravitational mass due to gravitation potential energy was discussed recently in this thread https://www.physicsforums.com/showthread.php?t=371845.

In case of the nucleons the effect is very important due to the strength of the field. The mass defects are ~MEV compared to ~1GeV of nucleons.
Important point is that in all systems(classical or quantum) a small increase in mass will happen due to interaction. Only in case of the nucleons the effect becomes important and measurable.

Last edited: Feb 4, 2010
11. Feb 4, 2010

### ansgar

No there is mass defect in the solar system as well... (i.e. gravitational bound systems)

12. Feb 5, 2010

### JK423

So, its explained by the fact that the particles' fields carry some of their energy -->converted to mass?
Anyway, i havent study this carefully so ill wait for my proffessor to explain to me (unfortunately i didnt find him today).
But i have another question regarding all these:
If the initial energy of the system is E=m1+m2 (c=1) and the final E'<E, then we are missing some energy here. The final system has less energy. Where did the energy difference go if there is no photon emmission as you say and you dont need quantum mechanics?

13. Feb 5, 2010

### inflector

The "energy difference" in the final energy for fusion reactions is found in the kinetic energy of the fusion products.

For reactions with two products the energy is divided in inverse proportion to the masses of the products. For example: the fusion of D and T results in a 4He (alpha particle) with 3.5 MeV of kinetic energy and a neutron with 14.1 MeV of kinetic energy.

For reactions with three or more products, the energy is divided between the products but not necessarily in predictable ways.

14. Feb 5, 2010

### Orion1

mass defect...

$$m_2 = (Z m_{p} + N m_{n})$$ - unbound system calculated nuclear mass
$$m_1$$ - experimentally measured nuclear mass

Mass defect:
$$\Delta m = (m_2 - m_1)$$

Binding energy:
$$E_B = \Delta m c^2$$

In nuclear physics, the semi-empirical mass formula (SEMF), sometimes also called Weizsäcker's formula, is a formula used to approximate the mass and various other properties of an atomic nucleus.

In the following formulae, let A be the total number of nucleons, Z the number of protons, and N the number of neutrons.
The mass of an atomic nucleus is given by
$$m_1 = m_2 - \Delta m = (Z m_{p} + N m_{n}) - \frac{E_{B}}{c^{2}}$$

Where $m_p$ and $m_n$ are the rest mass of a proton and a neutron, respectively, and $E_B$ is the binding energy of the nucleus. The semi-empirical mass formula states that the binding energy will take the following form:
$$E_{B} = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z-1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^{2}}{A} + \delta(A,Z)$$

Binding energy:
$$E_B = \Delta m c^2 = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z-1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^{2}}{A} + \delta(A,Z)$$

Each of the terms in this formula has a theoretical basis, as explained by reference 2.

Reference:
http://en.wikipedia.org/wiki/Binding_energy#Mass_defect"
http://en.wikipedia.org/wiki/Semi-empirical_mass_formula" [Broken]

Last edited by a moderator: May 4, 2017
15. Feb 5, 2010

### JK423

Yes i know about that. I was talking specifically for the proton-neutron binding, which gives as the Deuteron. The reaction is: p+n-->D+gamma.
There is always a gamma ray radiated when they bound! Accoring to the post of Orion1, this gamma ray "carries" the missing energy -->Binding energy--->The missing mass.
So, i think my current understanding of the mass defect is correct..
Its purely a quantum mechanical phenomenon....

16. Feb 5, 2010

### JustinLevy

No. I don't understand where the miscommunication here is, as multiple people have said it several ways.

Let's try this:
1] Do you agree the mass defect is given by the following?
$$E_{binding} = \Delta m c^2$$

2] Do you agree E = mc^2 is a classical result of relativity?

No one ever said there can't be photon emission.
What I said was:
"None of the details of the energy levels, nor how the system settles into it (emitting photons, etc.) are important to understand the sign of this effect."

From very general arguments (including E=mc^2, which is a classical argument relying Poincare symmetry (which still holds in QM)), if you have to do work to separate the constituents, then the mass of the bound system will be less than the sum of the constituents alone. The details of the Lagrangian (be it classical or QM, or how energy is transferred), are immaterial to understanding the sign of this effect.

17. Feb 6, 2010

### JK423

1] Ofcourse i agree. I dont have a problem with the binding energy concept. I just dont understand the origin of the mass defect. In classical physics there is also the concept of binding energy, like for example when you are on earth you must have a minimum velocity if you want to escape earth's gravitational field. In this case, the gravitational potential represents the binding energy, and the minimum velocity represents the work we have to do in order to separate the two objects (Earth and spaceship)
2] Yes i agree...

In order to clear this out, can you describe me with detail the mechanism that creates the mass defect? Why is the classical bound system 'Earth-Sun' supposed to have less mass?
I agree that we have to do work to separate the constituents but i dont think that this is relevant to the mass defect. If we do work in order to put a sattellite in orbit, then we have to do work in order to unbind it. So? Conservation of energy holds the whole time. Why do you need mass defect and E=mc2? Where does the mass defect come from in that case???

18. Feb 6, 2010

### ansgar

since E = mc^2 let us compute the Energy for the system sun + earth:

E(sun + earth) = M(sun)c^2 + M(earth)c^2 + BindingEnergy(gravity)

now Masses are per defenition positive definite units, and binding energy negative:

E(sun + earth) = M(sun)c^2 + M(earth)c^2 - absValue{ BindingEnergy(gravity) }

But since E(sun + earth) = M(sun+earth)c^2

we see that M(sun+earth) < M(sun) + M(earth)

Did you follow?

----

but we are putting the satelite in orbit FROM earth, potential energy is defined as when you put something at a position taking it from infinite radius....

19. Feb 6, 2010

### JK423

Is this the right way to think?
What you say is that if somehow you apply the same force on both Earth and Sun, then their center of mass would move like it had mass M<M_earth+M_sun.
This idea is not supported by Newtonian mechanics!
I took -as an example to test it- bodies of mass m and connected them together with a spring (oscillators). With your logic the mass of the system should be larger than the sum of the masses of the individuals since the potential is positive.
However, if you apply a force on both of them (put that force in F=ma) you will see that the center of mass behaves like it had a mass M=M1+M2 and not larger..

Note: At least i understood the logic you were pointing me :)

Last edited: Feb 6, 2010
20. Feb 6, 2010

### Raghnar

Because is relativistic Mechanics.

The effects is not measurable in slightly bond systems (gravitational force is far the weakest of the force), become easly observable in nuclear system because the bonding force come to be appreciable (in the order of 0.8% because ~8MeV of binding force and 1GeV of mass). The effects becomes crucial e.g. in subnuclear system, where the binding force determines that 3 little components (quarks) of weight in the order of 3~10 MeV make a bulky nucleon with a GeV mass.
But not because of the quantum entity of these, only because the strong force that is really strong (and act at little distances, so can't bond classic entities).

21. Feb 6, 2010

### JK423

Well i thought that in the non relativistic limit the two theories should have identical results.. In the classical examples we surely are in the non relativistic limit..

22. Feb 6, 2010

### Raghnar

As I said, if the force is small (like Gravity) the two gives identical results.
The binding energy of Sun+Earth is

U= GmM/r = 2*10^30*6*10^24*6.6*10^-11/(1.5*10^11)=5.3*10^33 J = 0.6* 10^16 kg*c^2

That, evenif big, is 10^-15 times the mass of the system, so practically unmeasurable and give identical motion.

I hope to have did the right math ;)

23. Feb 6, 2010

### Prathyush

I think the point JK423 is making is that if you have 2 objects of mass M1 M2 objects at a distance r. We are observing their combined gravitational effects at a distance R.
Given R>>r. We need to calculate the effective potential due to the two.

Say the effective potential is
$$V=\frac{GM_{eff}}{R}$$

If there was no corrections due to Binding energy i.e c>>1 we expect
$$M_{eff}=M1+M2$$
Indeed this is the case of Newtonian gravity

If there were effects due to self energy of the system then we can expect an extra factor due to the mass defect.
$$M_eff=M1+M2+\frac{G*M1*M2}{Rc^2}$$

We need to some how derive this(or similar) formula from first principles of GR or disprove its existence. I don't understand GR properly and its not clear to me how such a formula can come about.

24. Feb 6, 2010

### JK423

Well im not really used to the relation E=mc2 and how its used. Ive only used it in high energy particle collissions where there is no potential.
I read in the book of W. Rindler that every kind of energy contributes to the mass.
And by that he means the following. If m is there rest mass, then the particle's energy is:
E=mc2+K+V+Q+.... [1]
where K,V,Q,... are the various forms of energy (kinetic,potential,heat,...). In the case of two masses that interact gravitationally we have the kinetic and potential energy present.
E=mc2+K+V
Now we can re-use the relation of energy-mass equivalence for the system:
E=m'c2, where m' is the mass of the whole system: m'=m+(K+V)/c2.

Just to note a few things:
1)One reason it took me so long to understand was that i had not realized that the energy of a bound system is negative.. That is: E=K+V=|K|-|V|<0, cause |V|>|K|.
2)The second is that i hadnt realized that the relation E=mc2 can be used so generally (I'm talking about equation [1] from Rindler's book).
3)And last, the thing i did realize now is that mass is almost never equal to the rest mass due to various interactions.. even if the difference is infinitesimal. Due to equation [1], i now understand why in nuclei reactions the mass defect or the mass excess (in the case of quarks) is so large, and why when two nucleons bind there is mass defect and where 3 quarks bind there is mass excess (That has to do with the potential's form)

All these were known to me before but i hadnt 'realize' them.
(I hope you all understand what im trying to say)

I thank everybody for your effort to try and explain to me..

Last edited: Feb 6, 2010