Principle behind conduction

1. Nov 20, 2013

aaaa202

I am studying some simple models of how conduction arises on a microscopic level. A central idea is the following:
Look at two leads attached by a conducting channel. Initially the system as a whole is in equilibrium which especially means that the chemical potential μ is the same everywhere. Now the crucial idea is: Apply a voltage V across the two leads. This lowers the energy levels in one lead respect to the other keeping their chemical potentials separated by qV, which then is what gives rise to conduction.
I have underlined what I think to be the crucial step that I don't understand. To lower the potential means physically that we somehow connect one lead to a collection of negative charge (i.e. the negative pole of a battery) and the other lead to a collection of positive charge (the positive pole of a battery). The electrons in the lead connected to the negative pole will then have a potential relative to the electrons in the lead connected to the positive ones.
But physically how do we go from this observation to chemical potentials being shifted. The chemical potential is a variable that maximizes entropy for a system which can exchange particles, so more a less something which controls the amount of particles in a system.

2. Nov 20, 2013

Staff: Mentor

Not sure if this answers your question, but I found it so I figured I'd post it in case it does.
From wiki: http://en.wikipedia.org/wiki/Chemic...l.2C_external.2C_and_total_chemical_potential

3. Nov 21, 2013

DrDu

You must be careful which potential to use.
You have
$dU=TdS-Udq +\mu dn$
Hence mu is the change of energy with entropy S and charge q being kept constant.
However you are shifting charged electrons with charge ne, so you have
$dU=TdS +(\mu-eU) dn$
The expression in the brackets is known as electrochemical potential and has to be equal for two systems which may exchange electrons.