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Small=high mass at quantum level, but big=high mass at classical level. Why? 
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#1
Feb2012, 06:35 PM

P: 10

At a classical physics level, physically big equates to big mass, but at the subatomic level, small seems to equate to big mass i.e. (short wavelength big mass relationship. "momentum=h/wavelength"). Any ideas why there is this complete contrast?



#2
Feb2112, 01:27 AM

P: 2,258

because all subatomic particles have (roughly) the same angular momentum (ħ)
http://en.wikipedia.org/wiki/Bohr_model so radius is inversely proportional to mass 


#3
Feb2112, 04:09 AM

Mentor
P: 16,353

If you don't know the answer, it's not necessary to reply. 


#4
Feb2112, 04:49 AM

P: 435

Small=high mass at quantum level, but big=high mass at classical level. Why?
In classical physics you have an intuition that all objects have constant density. So bigger size with the same density yields bigger mass.
In quantum physics "density" is not constant. You rather have some constant amount of something (aether, waves) and you squeeze it. The classical intuition would be that the mass of a squeezed body remains constant, but from special relativity you get that the energy of the body gets higher. And higher energy means higher mass. Those two approaches can be considered at the same time. You get then the balance between classical and quantum physics. This yields the definition of Planck mass and the Bekenstein bound. 


#5
Feb2112, 07:00 AM

Sci Advisor
P: 4,603

E=Ne X=Nx so bigger N means both bigger E and bigger X. 


#6
Feb2112, 08:28 AM

P: 10




#7
Feb2112, 09:42 AM

P: 194

Becuase on a quantumlevel, the universe isn't intuitive?



#8
Feb2112, 10:51 AM

Sci Advisor
P: 4,603

More precisely, it is not enough to have many particles. In addition, these particles must be FERMIONS (which protons and neutrons in the nucleus are), so that you cannot put all them at the same place. If you have BOSONS (e.g., photons), then you can put many of them in the same state, so with a fixed wavelength (fixed energy of one particle) the size of N bosons may not depend on the total energy of N particles. 


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