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Small=high mass at quantum level, but big=high mass at classical level. Why?

by SteveinLondon
Tags: energy, mass.
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SteveinLondon
#1
Feb20-12, 06:35 PM
P: 10
At a classical physics level, physically big equates to big mass, but at the sub-atomic 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?
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granpa
#2
Feb21-12, 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
Vanadium 50
#3
Feb21-12, 04:09 AM
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Quote Quote by granpa View Post
because all subatomic particles have the same angular momentum (ħ)
No they don't.

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

haael
#4
Feb21-12, 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.
Demystifier
#5
Feb21-12, 07:00 AM
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Quote Quote by SteveinLondon View Post
At a classical physics level, physically big equates to big mass, but at the sub-atomic 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?
It is not so much classical vs quantum, but rather one vs many particles. For one quantum particle, smaller length x means more energy e, as you said. But if you have MANY (say N) such small particles at DIFFERENT positions, then total energy and total length scale like
E=Ne
X=Nx
so bigger N means both bigger E and bigger X.
SteveinLondon
#6
Feb21-12, 08:28 AM
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Quote Quote by Demystifier View Post
It is not so much classical vs quantum, but rather one vs many particles. For one quantum particle, smaller length x means more energy e, as you said. But if you have MANY (say N) such small particles at DIFFERENT positions, then total energy and total length scale like
E=Ne
X=Nx
so bigger N means both bigger E and bigger X.
But if you use DeBroglie's wave/particle formula on a large object, say a rock, you get momentum=h/wavelength, so a big rock, at the same speed as a little rock, has bigger momentum yet smaller wavelength - yet it's made up from more than one particle.
Hobin
#7
Feb21-12, 09:42 AM
P: 194
Becuase on a quantum-level, the universe isn't intuitive?
Demystifier
#8
Feb21-12, 10:51 AM
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Quote Quote by SteveinLondon View Post
But if you use DeBroglie's wave/particle formula on a large object, say a rock, you get momentum=h/wavelength, so a big rock, at the same speed as a little rock, has bigger momentum yet smaller wavelength - yet it's made up from more than one particle.
Yes, but it is incorrect to apply the DeBroglie's wave/particle formula on an object consisting of many particles. For example, the nucleus of an atom is very small (much smaller than the atom itself), and yet a heavier nucleus is bigger than a lighter one. That's because the nucleus consists of many particles.

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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