When applying DC to a neon lamp, only the negatively-charged electrode glows:
But… why? The electrodes are the same shape, so the electric field around them should be the same shape, and the gas should break down in the regions at which the electric field strength is above some threshold, which seems like it would be symmetrical. Is there a difference in threshold between positive and negative coronas? If so, do both sides light up at high enough voltage? Or maybe only one type of corona is possible in neon since it's a noble gas? If it contained air would it glow at both electrodes?
Do neon signs work in a different manner, since they have a long region of glowing gas, rather than just glowing near the cathode?
Best Answer
The asymmetry comes from the different masses of electrons and neon ions (neon ions are about 36000 times heavier).
This mass asymmetry results in different cross sections for the excitation of neon atoms by electrons and ions. There are some plots of this here (figure 1a for electrons, figure 1b for ions). The interesting processes of excitation occur above around 10eV for electrons and 100eV for ions.
There is a very cool Java simulation of discharges: http://phet.colorado.edu/en/simulation/discharge-lamps . It covers electron excitation, the atomic structure of neon, acceleration and excitation cross-sections very well, and demonstrates why the glow can be localized (it glows at the point that electrons have been accelerated to the necessary energy), and why it is asymmetric (all electrons start from the cathode and can gain energy on the way to the anode).
I hope this answers why only one side can glow. I wish I could explain exactly why the glow is next to the cathode in the picture, but I would just be guessing.