Stretching across China, from the Xinjiang region in the west to the province of Anhui in the east, there's a ±1,100 kV high-voltage direct-current transmission system. I'm not 100% sure what "±1,100 kV" means, but I think it means that the system uses two conductors, one with a voltage of +1,100 kV relative to ground and one with a voltage of -1,100 kV relative to ground.
Now, the rest mass of an electron is about $510\ \mathrm{keV}/c^2$. That got me thinking. Imagine if we had an electron and a positron somewhere near the positive conductor of the transmission line. If the electron were sucked into the line, and the positron were flung away, the total amount of kinetic energy the two particles would gain is 1,100 keV, right? However, my understanding of what rest mass means is that 1,020 keV is enough energy to create an electron and a positron. If I understand particle physics correctly (which I certainly don't), doesn't this mean that electron–positron pairs should be created near the transmission line and flung apart just like this, with the two particles containing about 80 keV of kinetic energy in total?
The way that Hawking radiation is sometimes described to laypeople like me is that a particle–antiparticle pair spontaneously appears near the event horizon of the black hole, and the kinetic energy produced when one of the two particles falls in is enough to "pay off the debt" and make the particles "real" (or something like that!). If that explanation is in any way vaguely accurate, then it seems like maybe the same thing should happen in the presence of a very strong electric field.
So, do high-voltage transmission lines like these really emit positrons, or am I totally off base here?
Best Answer
Your analysis doesn't make sense because the units don't match up. $1100 \, \text{kV}$ is not more than twice $510 \, \text{keV}/c^2$, because the two quantities can't be compared at all. It's like saying $4$ meters is twice as big as $2$ minutes.
It's indeed possible to create electron-positron pairs, but you need a tremendously large electric field, given by the Schwinger limit, $$E = \frac{m_e^2 c^3}{e \hbar} \sim 10^{18} \, \text{V}/\text{m}.$$ Power lines don't have electric fields anywhere near this big, and it's good that they don’t, because this is eight orders of magnitude higher than the field needed to rip the electrons off atoms.