Is Proton Mass Really Reduced in Helium Nuclei?

AI Thread Summary
The discussion centers on the fusion of deuterium nuclei into helium nuclei and the implications for proton mass. It is argued that the mass of protons decreases when they fuse, leading to a decrease in energy, contrary to the expected increase in energy from fusion. However, participants clarify that the total mass of a nucleus is a property of the entire system, not the sum of individual nucleons, and that the strength of the nuclear and electrostatic forces remains unchanged. The conversation highlights common misconceptions about mass and energy in nuclear fusion. Ultimately, the thread suggests that the underlying misconceptions have been addressed, indicating a need to conclude the discussion.
avito009
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I am no Einstein but I would like to expand my knowledge and share it.

When two deuterium atoms fuse together they become a helium nuclei. Now deuterium nuclei contains 1 proton and 1 neutron. When the deuterium nuclei fuses to form a helium nuclei. Helium nuclei contains 2 protons, and 2 neutrons. Now the mass of the proton in the deuterium nuclei is reduced. This means that the individual mass of the proton when it was in deuterium nuclei is greater than the mass of the proton in helium nuclei.

So here you may say that the mass difference of the proton would have caused a proportional increase in energy in the helium nuclei. If you believe in E= MC2. But in reality it is the opposite.

What actually happens is that when deuterium nuclei fuse the energy in the atom decreases and so there is a corresponding decrease in proton mass. This means that the strong nuclear force and the electrostatic force are less stronger in the fused nuclei. (Not the other way round).

Is this correct?
 
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avito009 said:
What actually happens is that when deuterium nuclei fuse the energy in the atom decreases and so there is a corresponding decrease in proton mass. This means that the strong nuclear force and the electrostatic force are less stronger in the fused nuclei. (Not the other way round).

I think the correct conclusion would be that what actually happens is that when deuterium nuclei fuse the energy in the atom increases and so there is a corresponding decrease in proton mass in the nuclei. This means that the strong nuclear force and the electrostatic force are more stronger in the fused nuclei. More energy lesser mass.

Is this proper?
 
You are discussing a topic which has a substantial number of books and general summaries around it. When fusion occurs, energetic particle(s) are ejected. Naturally, there is conservation of total mass/energy through each step. E=mc^2 does apply.

Not sure there are any general or popular misconceptions about this.
 
avito009 said:
I think the correct conclusion would be that what actually happens is that when deuterium nuclei fuse the energy in the atom increases and so there is a corresponding decrease in proton mass in the nuclei. This means that the strong nuclear force and the electrostatic force are more stronger in the fused nuclei. More energy lesser mass.

Is this proper?

No. The total mass of the nucleus is a property of the nucleus as a whole. You can't pick out the individual nucleons and assign them masses that will add to the total mass - all you can say is that one configuration of two neutrons and two protons has a different mass than another.
 
Another way to put it is that the mass of a system of particles does not generally equal the sum of the masses of the individual particles. (Here by "mass" I mean what physicists generally mean by "mass" nowadays, which is often called "rest mass" in popular literature and sometimes "invariant mass" by physicists; but not the "relativistic mass" that you often see in popular literature)
 
avito009 said:
This means that the strong nuclear force and the electrostatic force are less stronger in the fused nuclei. (Not the other way round).

Is this correct?

No, the strength of the forces involved does not change.
 
avito009 said:
I am no Einstein but I would like to expand my knowledge and share it.
...
If you believe in E= MC2. But in reality it is the opposite.

What actually happens is that...

This thread is pushing the edge of the rule about personal theories. Now that the underlying misconception has been explained by several posters, we should close it.
 
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