Is there Phase difference between voltages at end points of a long AC Power line I found this explanation somewhere:
Since wavelength=c/freq so in a AC power line of 50 Hz, wavelength=(3*10^8)/50 = 6000 Km, so voltage phase reverses after 6000 Km in a ac Power line.
Now this would have been correct if it was a 50 Hz E.M. Transmission line, but does the same occur in metal wire with A.C. ? Clearly the power transmission through the wire would not be through E.M. waves. I am also confused if a wave phenomenon does occur in a AC wire or not. If it occurs then how (in terms of flow of electrons) and at what speed/wavelength ?
 A: You are right that you should solve the Maxwell equations for air + metal wire, not just for air, but the solutions for the cylindrical metal wire + air give phase velocity component along the wire close to c (or at least of the same order of magnitude), as far as I understand (http://www.mathunion.org/ICM/ICM1924.2/Main/icm1924.2.0157.0218.ocr.pdf ) 
A: This might help understand why the current also obeys a wave equation. The voltage is essentially the gradient of the electric field. So if you have an oscillating voltage, you are trying to set up an oscillating electric field in the wire. The current is related through Ohm's law $\mathbf{J} = \sigma \mathbf{E}$. Now, you cannot have an instantaneous change in \mathbf{E} all through the wire -- there is a time delay corresponding to the time taken for the EM disturbance to propagate (in fact, we shouldn't be talking about voltage in the traditional sense any more, since this is not an electrostatics problem). So the "voltage" downstream on the conductor is different from that at the source. And after all, the electrons are driven by the changing electromagnetic field, so they will oscillate too.
I'm not sure if I'm making sense, but in summary, the thought that voltage is constant throughout the conductor is a hang-over from electrostatics, and this need not be the case when we have changing electromagnetic fields.
