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Yes, that's my view too.. *Ouch*But when counting on the mass in this sphere it seems to give you a sudden 'rest mass' as those vectors get canceled out?
So yeah, it confuses me in many ways?==You wrote "zero total momentum of the atoms" Lightarrow. I have to be thick because when I read a statement like that I do not see individual atoms 'bouncing around' each one with its own momentum intact, instead I see a Bose Einstein condensate. And neither can I see how they are thought to take each other out practically? All of them will at some time reach the wall, and each one of them will still have a momentum? So why do we use that definition, it doesn't seem to describe the situation?
When we say a thing like their vectors taking each other out, magically transforming momentum to so called 'rest masss' or 'proper invariant mass' aka 'matter' I still want to understand why. Looking at the math doesn't tell me a thing there Lightarrow.
If your experiment were possible then sooner or later you would start creating electrons and protons.
I think i can see how you think Lightarrow, but doesn't the movements inside that sphere take each other out statistically, meaning that the 'bouncing' is on all parts of the inner 'surfaces' which to me seems to indicate that there won't be any noticeable 'new inertia' to the sphere if you tried to move it?Where do I go wrong?(And don't you tell me it's due to my misspent youth:)
Are you thinking of it in terms of energy=mass?But if you consider the situation inside and look at the individual 'photons'. They all carry a momentum and it is that you will see be converted into 'force' here, their 'energy' notwithstanding. I mean, even if each one of the photons represent an energy that at some level of concentration may express itself as mass, isn't it the momentum that deliver the 'push' here, as all photons are intrinsically massless?
The momentum isn't any rest mass?