Originally posted by: f95toli
Originally posted by: Born2bwire
The potentials A and V are not really physical, but they allow us to solve a lot problems much more easily than if we strictly worked with E and B. For example, voltage allows us to work with the vector E as a scalar. In addition, voltage is the integral of the electric field over distance. One of the nice results of this is that the voltage is only dependent upon the field at the end points of the curve of integration. So if we take the potential difference, V(a)-V(b), then it does not matter how we draw the curve of integration. And in real life, you always take the potential difference, V(a)-V(b).
I think you are confusing a few issues here. You are right that the vector potential is not "physical" (it is not an observable) and therefore gauge dependent.
However, voltage IS an observable since it is related to potential energy.
A particle with the charge 1 Coloumb held at a electric potential 1V will have potential energy of 1J. A more "physical" picture is that voltage determines the distribution of energies of the quasiparticles with respect to the Fermi surface (in a semi-classical picture applying a voltage essentially "tilts" the Fermi-level). Also, using second quantizaion you can sometimes define a voltage operator in terms of creation and annihilation operators of the electric field.
Morever, voltage also happens to one of the corners of the metrological triangle (resistance. current, voltage) and therefore forms the basis of the SI-system. If we ever manage to close the triangle (at the moment we can not measure current directly with good enough precision, we should be able to do that in a few years) we will be able to define most SI units using voltage.
We can measure voltage with almost ridiculus precision using the Josephson AC effect which relates frequency and voltage (the conversion factor is 2*e/h=483 GHz/mV). Since we can measure frequency with high precision we can therefore also measure voltage with the same precision.