# Sum of its parts?

1. Apr 1, 2005

### Omexus

I used to be an avid reader on the subject, however with age I have lost some of the finer details. I was thinking back upon what I learned the other day and I was wondering if I had recalled correctly that a particles mass is less than the sum of its parts. Can anyone here tell me if this is true or not? Also are there any theories or explainations as to why this is so? (If it is true of course :) )

Any help would be much appreciated.

2. Apr 1, 2005

### DaveC426913

Isn't the missing mass hiding in the energy used to hold the components together?

3. Apr 1, 2005

### dextercioby

No,no,usually morethan the sum...There are 2 types of particles:fundamental (v.SM) and composite particles.Since i think you refer to the latter,the sum of the masses of the constituents is less than the mass of the composite particle.The common examples are with baryons and the quarks/antiquarks inside...The theory explains the phenomenon very well,however the calculations are very difficult and obviously only approximate...

Daniel.

4. Apr 1, 2005

Check out :

regards
marlon

5. Apr 4, 2005

### Meir Achuz

The mass of a bound state is usually less than the sum of the masses of its components because of the negative binding energy.
For instance, the mass of a deuteron (a proton-neutron bound state)
is 2.23 MeV less than the sum of the p and n masses.
The same would hold true for the hydrogen atom (a p-e bound state)
but the 13.6 eV binding energy is too small to be noticed in the masses.

6. Apr 4, 2005

### marlon

Yes but it is the other way around for baryons or mesons versus the constituent quarkmasses

marlon

7. Apr 5, 2005

### Meir Achuz

Some constituent quark models have the sum of quark masses greater than the baryon mass, and behave as above. There are others, for which this isn't true. The quark-quark force differs from the Coulomb or nuclear force by having a linear repulsive confinement force (like a rubber band) that prevents the quarks from escaping. This gives a positive potential energy.
In any event the sum of the constituent's kinetic and potential energies equals the mass of the composite object.
This also happens in heavy nuclei that are unstable to alpha decay.
The positive Coulomb PE makes the nucleus [A,Z] heavier than the sum
of the [A-4,Z-2] nucleus and an alpha particle (helium nucleus).
The heavy nucleus can have a long lifetime, but eventually decays through "quantum tunneling". It would be stable in classical mechanics.

8. Apr 5, 2005

### marlon

Well, this is only valid in the long range...For higher energies, the Coulombic potential becomes predominant.

In what models is the sum of the quark-mass bigger then the baryon-mass. I have never heard of those ?

'Normally' the baryons-mass is BIGGER then the sum of the quarkmasses and this is certainly the case with the POSITIVE linear confinement potential that can be described in the dual Abelian Higgs-model (you know, with them dual superconductors allĂ  Landau Ginzberg). The extra mass comes from virtual quark anti-quarkpairs that exist in the baryon due to the HUP. The expectation value of these dynamical quarks (the virtual pairs i talked about) is non-zero when calculated non-perturbatively (ie in low energy QCD), so the vaccuum has a non-zero energy value. This is a basic requirement for such virtual pairs to exist. In the vacuum-state, the linear potential is dominant, so you see ?, there is something fundamentally wrong with what you stated here.

Baryons are totally different then atomic nuclei...

marlon

Last edited: Apr 5, 2005
9. Apr 5, 2005

### marlon

The decay-processes are not to be compared with baryon-behaviour. Alpha decay is a QM-thing, certainly not a QFT-thing...There won't be no virtual particles here...Only beta-decay is a QFT-thing...Besides, keep in mind that both positive and negative beta decay are very different because one is a particle decay mode, the other is a nuclear decay mode.

marlon

10. Apr 5, 2005

### Meir Achuz

You have also confused my simple answer.
This is my final post on this thread.

11. Apr 5, 2005

### marlon

:rofl: :rofl: :rofl:

Don't be so insulted i was just correcting your false statements...If i hadn't done that, then somebody else would have...

regards
marlon...

And this is NOT my final answer :rofl:

12. Apr 28, 2005

### CarlB

Bjorken and quark masses

Here's James D. Bjorken's notes on quark mass, with the assumption that they are very heavy, with references to the literature (1979):
http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-2366.pdf

I find this more acceptable than the standard model as it is compatible with what we know would happen if we had three free quarks and combined them to make a proton. Of course there is a binding energy and of course the mass would decrease. Therefore the free quarks must weigh more than the proton.

The fact that the standard model assumes light quark masses reminds me of the basic problem with the standard model, the cancellation of infinities. In this case, it is the infinite binding energy that is being cancelled.

Saying that a quark bound inside some complex object has a light effective mass makes sense, but the fact that we can't separate out the quark doesn't make its mass low any more than my inability to weigh a battleship with a postal scale makes the battleship massless just because it is always floating.

In short, I say that when the standard model talks about light quark masses, they are really talking about light effective quark masses for quarks bound in hadrons.

Carl

P.S. I knew about Bjorken's work only because I'm giving a talk at the Pheno2005 APS meeting on the subject of the centauro events, which is what interested Bjorken. I don't know if there is more recent work along the quark mass line.