What is the reason for mass defect in a nuclide?

[Anonymous Vegetable]

I may be misguided here but to my understanding, separate nucleons have a higher mass altogether than the combined nucleus as the potential energy gained from being separated (in the field of the strong force) is being manifested in more mass. If this is true, and if it's not I'd like to know the reason, why would the nucleons retain this greater mass outside of the range of the strong force?

 

[CrazyNinja]

How are the individual nucleons gaining mass? The difference in mass manifests itself in the form of kinetic energy.

 

[Anonymous Vegetable]

How are the individual nucleons gaining mass? The difference in mass manifests itself in the form of kinetic energy.

Vibrational or of what 'form' of kinetic energy?

 

[CrazyNinja]

Well, vibrational energy is nothing but a combination of potential and kinetic energies. But no, the nuclides do not vibrate, they fly off. THAT form of kinetic energy. This is the principle used in nuclear reactors where fission occurs. This kinetic energy is transferred to water → converts to steam → runs turbines → electricity is generated.

 

[Anonymous Vegetable]

Well, vibrational energy is nothing but a combination of potential and kinetic energies. But no, the nuclides do not vibrate, they fly off. THAT form of kinetic energy. This is the principle used in nuclear reactors where fission occurs. This kinetic energy is transferred to water → converts to steam → runs turbines → electricity is generated.

So by that logic, if you hypothetically brought those nucleons to a rest and weighed them, their mass would be no greater?

 

[CrazyNinja]

No greater when compared to what? Do you want to compare :

1. The sum of their masses with the mass of the decayed nuclide; or
2. Their "moving" mass with "rest" mass?

 

[Anonymous Vegetable]

No greater when compared to what? Do you want to compare :

1. The sum of their masses with the mass of the decayed nuclide; or
2. Their "moving" mass with "rest" mass?

I think you may be misreading my question. The mass of a nucleus is less than the mass of its nucleons added together separately (I assume these are rest masses). So how can that extra mass be kinetic energy?

 

[CrazyNinja]

Okay I'm sorry. I was thinking in terms of decay because your question said "nuclides". My bad.

Yes, THAT mass defect turns into potential energy. So according to your question, if the nucleons did not retain this extra mass, where would it go?

EDIT: What other alternative do you suggest?

 

[Anonymous Vegetable]

Okay I'm sorry. I was thinking in terms of decay because your question said "nuclides". My bad.

Yes, THAT mass defect turns into potential energy. So according to your question, if the nucleons did not retain this extra mass, where would it go?

Oh so are you saying this potential energy is present when part of a nucleus?

 

[jtbell]

The difference in mass manifests itself in the form of kinetic energy.

No. The difference in mass comes about because of the work you have to do on the system in order to separate its components. If you could somehow "grab onto" the individual nucleons in a nucleus and pull them apart slowly, you would have to do work on them as you pull. When the nucleons are far enough apart that the binding forces are negligible, you stop. The mass of the system (the sum of the masses of the separated (and now stationary!) nucleons) is now greater than the mass of the original nucleus. The difference (the mass defect of the nucleus) equals the mass-equivalent (via E = mc²) of the work that you did.

[added: ah, I see you corrected yourself while I was typing.]

To AV: the increase in mass comes from the work done by whatever separated the nucleons.

 

[CrazyNinja]

@jtbell ... Yup, you are right. I read the question wrong. My bad.
@Anonymous Vegetable ... yes, it is only present when it is part of the nucleus. jtbell's explanation will help you out.

 

[Anonymous Vegetable]

No. The difference in mass comes about because of the work you have to do on the system in order to separate its components. If you could somehow "grab onto" the individual nucleons in a nucleus and pull them apart slowly, you would have to do work on them as you pull. When the nucleons are far enough apart that the binding forces are negligible, you stop. The mass of the system (the sum of the masses of the separated (and now stationary!) nucleons) is now greater than the mass of the original nucleus. The difference (the mass defect of the nucleus) equals mass-equivalent (via E = mc²) of the work that you did.

[added: ah, I see you corrected yourself while I was typing.]

To AV: the increase in mass comes from the work done by whatever separated the nucleons.

I may just be an idiot but I still can't normalise the fact that pulling them apart gives them more mass.

 

[jtbell]

The individual nucleons actually don't change their masses. The key thing here is that generally, the mass of a system of particles does not equal the sum of the masses of the particles. The mass of a system of particles includes (a) the masses of the component particles, (b) the mass-equivalent of the system's potential energy, and (c) the mass-equivalent of the kinetic energies that the component particles have when the system as a whole is stationary.

 

[Anonymous Vegetable]

The individual nucleons actually don't change their masses. The key thing here is that generally, the mass of a system of particles does not equal the sum of the masses of the particles. The mass of a system of particles includes (a) the masses of the component particles, (b) the mass-equivalent of the system's potential energy, and (c) the mass-equivalent of the kinetic energies that the component particles have when the system as a whole is stationary.

But in this case the system of particles (the nucleus) actually has less mass than the separate pieces added together?

 

[jtbell]

Yes. Note that the potential energy of a bound system is negative. So the total energy of a bound system is less than the total energy of the unbound system (the separated components). Therefore the mass of the bound system is less than the mass of the unbound system.

 

[Anonymous Vegetable]

Yes. Note that the potential energy of a bound system is negative. So the total energy of a bound system is less than the total energy of the unbound system (the separated components). Therefore the mass of the bound system is less than the mass of the unbound system.

This may sound silly, if it helps I'm still in school, but the potential energy being negative in a bound system is kind of a revelation to me, I was completely unaware.

 

[CrazyNinja]

No prob bro. How old are you? I am 17, and fairly speaking, it feels weird the first time too. Don't worry, you will get used to it.

 

[Anonymous Vegetable]

No prob bro. How old are you? I am 17, and fairly speaking, it feels weird the first time too. Don't worry, you will get used to it.

And out of curiosity, can I ask you to elaborate on vibrational being a combination of kinetic and potential?

 

[CrazyNinja]

It would be easier for me if you could tell me how old you are.

 

[Anonymous Vegetable]

It would be easier for me if you could tell me how old you are.

Similar in age to you. But UK so perhaps different syllabuses.

 

[Anonymous Vegetable]

Similar in age to you. But UK so perhaps different syllabuses.

I can apply all of this and what it means mathematically, that wasn't the problem. I just want to make sure that I really get it frankly, otherwise it all feels a bit worthless.

 

[nikkkom]

Any bound system has a mass deficit relative to the case where you pull it apart. Be it Earth-Moon system, or hydrogen atom, or neutron star. (Neutron star is a particularly spectacular case: its mass is 20% less than it "should be" if you would multiply one free neutron mass by the number of its neutrons!)

 

[Anonymous Vegetable]

Any bound system has a mass deficit relative to the case where you pull it apart. Be it Earth-Moon system, or hydrogen atom, or neutron star. (Neutron star is a particularly spectacular case: its mass is 20% less than it "should be" if you would multiply one free neutron mass by the number of its neutrons!)

That's quite impressive thank you for that.

 

[nikkkom]

Any bound system has a mass deficit relative to the case where you pull it apart.

To add to this: there is an apparent counter-example: proton's mass is larger that masses of its quarks.

This is resolved as follows: the concept of "free quark mass" is ill-defined. You can't pull proton apart to a state where you have three quarks with large spatial separations (say, 1 meter).

What is meant by "quark mass" (for light quarks u,d,s) is roughly "by how much lighter nucleons would become if quark's mass would be zero?". This is not the same as what we think of when we speak about e.g. electron mass.

If you prefer, you can think that "true" free quark mass is infinite. Then proton is not an exception: "this bound system also has a mass deficit".

 

[Anonymous Vegetable]

To add to this: there is an apparent counter-example: proton's mass is larger that masses of its quarks.

This is resolved as follows: the concept of "free quark mass" is ill-defined. You can't pull proton apart to a state where you have three quarks with large spatial separations (say, 1 meter).

What is meant by "quark mass" (for light quarks u,d,s) is roughly "by how much lighter nucleons would become if quark's mass would be zero?". This is not the same as what we think of when we speak about e.g. electron mass.

If you prefer, you can think that "true" free quark mass is infinite. Then proton is not an exception: "this bound system also has a mass deficit".

Again that's very interesting and thank you however it doesn't really clear too much up hahaha.

 

[Anonymous Vegetable]

So to just sum up the problem, I've heard two explanations.
1. The nucleons gain mass from the work done on them to separate them.
2. The negative potential energy they have as a bound state in the nucleus effectively subtracts mass.
So either these are both the same thing told in different ways or... I'm not quite sure. I'm sorry if everyone feels like they're talking to a brick wall hahaha.

 

[nikkkom]

Both explanations say the same thing.

 

[Anonymous Vegetable]

Both explanations say the same thing.

So would you mind running through the concept of work done becoming mass? I know you may be repeating but I just want to be clear.

 

[nikkkom]

Work done on the system increases its energy. Energy and mass are equivalent.

 

[Anonymous Vegetable]

Work done on the system increases its energy. Energy and mass are equivalent.

And this work done would be in what form? If that's a valid question.

 

[vanhees71]

In relativistic physics there is no conservation of mass, and if I write mass I always mean what's called "invariant mass" not "relativistic mass", which is a misnomer for the energy of the system divided by $c^2$, but only energy and momentum conservation.

If you have some protons and neutrons at rest, the total energy is $E_0=(Z m_p+(A-Z) m_n)c^2$, where $Z$ is the number of protons and $(A-Z)$ the number of neutrons ($A$ is called the number of nucleons). Now when these nucleons bind together to a nucleus, there is some binding energy, $-E_B$, which is negative, and it's freed somehow, e.g., by electromagnetic radiation. So the energy of the nucleus at rest is
$$E_A=E_0-E_B$$,
but by definition for the nuclus at rest you have the invariant mass of the nucleus
$$m_A=E_A/c^2.$$
in other words the mass of the nucleus is smaller than the sum of the masses of the nucleons by the amount $E_B/c^2$.

The same holds true for any other system too. E.g., suppose you have a capacitor with capacitance $C$ and let it's mass be $m_0$ if it's uncharged. Now if you charge it by connecting it to a battery of voltage $U$ the electromagnetic energy stored in this capacitor is
$$E_{\text{em}}=\frac{C}{2}U^2,$$
and this amount of energy you must provide from the battery. Consequently the mass of the charged capacitor is a bit larger than when it's uncharged, i.e., you have
$$m_{\text{charged}}=m_0+\frac{C}{2c^2} U^2.$$

The case of nucleons as bound states of quarks is much more involved and not fully understood. The reason is that the strong interaction is confining, i.e., we never observe free quarks and gluons but only color-charge neutral objects, the hadrons (baryons as bound states of three quarks and mesons as bound states of a quark and an anti-quark). The bound state is a very complicated system of bound quarks and gluons, and about 98% of the proton mass comes somehow from the dynamics of the quark- and gluon-quantum fields. As I said, it's not fully understood, but we have very convincing reasons to believe in this picture, because there are socalled lattice-QCD calculations of the hadron-mass spectrum. The idea behind this is to model space and time as a discrete lattice of points and evaluate the socalled Euclidean path integral of QCD approximately on big computers. With this you can evaluate the masses of the hadrons from QCD. The socalled current-quark mass of u- and d-quarks is only some $\mathrm{MeV}/c^2$ (which is due to the famous Higgs mechanism of the electroweak theory), but nevertheless the mass of the proton comes out at the observed physical value of about $940 \text{MeV}/c^2$. So it must be the dynamics of the quantum fields via the strong interaction that generates almost the entire mass of the nucleon.

 

[jtbell]

And this work done would be in what form?

Any form of work. When object A does work on object B, energy is transferred from A to B.

 

[maline]

AV, I fully agree with your puzzlement. It seems very odd that energy can be negative, given that energy manifests itself in concrete ways. Mass is basically a "symptom" of energy (the mass of a system is 1/c^2 times the energy as measured by an observer moving along with the center of mass) and spacetime curvature is also proportional to energy. As far as I know, the total energy in any region is never negative (someone please correct me if I'm wrong on this).
Furthermore, the idea of negative energy means that the energy of a system does not really limit how much energy you can extract from it! As long as you create enough negative energy within the system, you can pull out a corresponding amount of useful positive energy. Of course there are only so many ways to bind things, so this isn't really unlimited, but what does "zero energy" really mean? I say no one knows!
This is kind of how things are in physics. We find all these beautiful equations that form consistent systems & describe all the things we observe. But when- or if- we try to think about "what it means", things can be very, very cloudy. Some physicists consider the questions "meaningless philosophy" and strongly discourage them. I encourage you- even if you never get answers, keep right on asking!p.s Vanhees91, with all due respect, I think you could be a bit gentler on a high-school kid. The main result of a detailed answer full of unfamiliar terms will be to scare kids away from the forum...

 

[Anonymous Vegetable]

p.s Vanhees91, with all due respect, I think you could be a bit gentler on a high-school kid. The main result of a detailed answer full of unfamiliar terms will be to scare kids away from the forum...

No offence but being 'slightly' patronising is more likely to scare people away and I think people expect answers to be of a high standard if they ask them. I appreciate what you're saying but frankly, any help is gratefully received. Just because he put some maths in there doesn't make it a demanding answer, proton numbers and atomic mass numbers are not new and a concise formula can be quite enlightening.

 

[nikkkom]

As far as I know, the total energy in any region is never negative (someone please correct me if I'm wrong on this).
Furthermore, the idea of negative energy means that the energy of a system does not really limit how much energy you can extract from it! As long as you create enough negative energy within the system, you can pull out a corresponding amount of useful positive energy. Of course there are only so many ways to bind things, so this isn't really unlimited, but what does "zero energy" really mean? I say no one knows!

Zero energy, as far as General Relativity is concerned, is a patch of flat Minkowski space.