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Mass of Proton 
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#1
Dec2612, 02:53 AM

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The mass of proton is 938mev/c2 where as mass of quarks in proton is 11 Mev/c2. Much of the reminder is gluon. But gluon does not have any mass. So how can it be gluon?
Also in neutron decay, one d Quark transform itself into u quark by absorbing / emitting W boson. in this case W boson is absorbed or emitted? i am trying to work out the equation UDD=UUD+e+ve in respect of Mass and charge. the U quark has mass 2.4Mev/c2 where as D quark has 4.8Mev/c2. So UDD = 2.4+4.8+4.8=12Mev/c2 where as UUD=2.4+2.4+4.8=9.6. E has mass of 0.511Mev/c2. There is a difference of 1.889 Mev/c2 of mass. What is this? in term of charge U has +2/3 where as D has 1/3. so UDD has no charge i can understand this. same is the case UUD which has +1 charge. so how come electron have 1 charge. where this charge comes from? the W boson which as emitted that later on transform into e+ve was either already there in UDD formation or absorbed and then emitted. If it was already there, why neutron is said to be composed of UDD? if it was absorbed first and then emitted, where does this W boson come from? 


#2
Dec2612, 06:12 AM

Sci Advisor
Thanks
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#3
Dec2612, 10:01 AM

Mentor
P: 11,928

See this Feynman diagram for the neutron decay (and forget the "space" dimension shown there, this is not very useful): A downquark emits (or absorbs, does not matter) a virtual W boson and becomes an upquark. The virtual Wboson decays into (or gets produced from, does not matter) an electron and an antielectronneutrino.
The Wboson is charged, charge is conserved at both interaction points and both interaction points are valid within the weak interaction. Concerning mass, you always have to consider the whole hadron, and as Bill_K pointed out this is dominated by QCD effects (for protons and neutrons). 


#4
Dec2612, 02:37 PM

Sci Advisor
P: 6,080

Mass of Proton
A major factor in the mass of the proton or neutron is the presence of virtual quarkantiquark pairs that come in out and out of existence.



#5
Dec2712, 12:19 AM

P: 2

thanks for the reply.



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