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Hall effect

 
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Oct21-12, 08:00 AM   #18
 
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Hall effect


You need to learn how "effective mass" is define. It is the 2nd derivative of the dispersion curve. It means that the curvature of the dispersion matters, and if it changes, then the "sign" on the effective mass changes.

Zz.
 
Oct21-12, 08:05 AM   #19
 
Quote by ZapperZ View Post
...The effective mass in metals... has nothing to do with... "acceleration and gravity", because such interactions are non-existent in the standard Hamiltonian of materials! When was the last time you see gravity in the Hamiltonian for a metal or a semiconductor?...
I dare to claim that mass does have to do with acceleration.

And please feel free to introduce gravity in a Hamiltonian each and every time you're interested in gravity's effects, that is, when you don't want to neglect it.
 
Oct21-12, 08:18 AM   #20
 
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Quote by Enthalpy View Post
I dare to claim that mass does have to do with acceleration.

And please feel free to introduce gravity in a Hamiltonian each and every time you're interested in gravity's effects, that is, when you don't want to neglect it.
I introduce BOTH "gravity and acceleration" because that is part of what you wrote. It is to emphasize that gravitational effects are irrelevant in practically ALL of condensed matter phenomena. Your "invitation" to introduce such a thing is meaningless. Where, for example, is gravity needed in the Hamiltonian for superconductivity? Does the exclusion of gravity affects the accurate description of this phenomenon? What about the description of the band structure of a semiconductor? Where is gravity needed here?

Zz.
 
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