# Do extremely high-voltage power lines emit positrons?

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.

doesn't this mean that electron–positron pairs should be created near the transmission line and flung apart

To understand this from a semiclassical view, the key point is that pair creation from vacuum is limited by the uncertainty principle $$\Delta H\, \Delta t \lesssim \hbar$$ (using $$H$$ for energy to avoid confusion with electric field $$E$$). You have noted that if energy to create the pair is "borrowed", it seems that enough energy can be gained from the electric field to "pay it back" and make the pair "real".

However, the energy has to be paid back within a specific amount of time, of order $$\hbar/m_e c^2$$, not just "eventually". The longest distance an electron and positron can separate in this time is $$\hbar/m_e c$$, the Compton wavelength, which is the inherent quantum "fuzziness" in an electron's position. Thus, the energy that can be gained from the electric field $$E$$ before the clock runs out is $$e\hbar E/m_e c$$. This must cover the borrowed energy of order $$m_e c^2$$. Thus, creating real pairs from vacuum requires $$E \sim \frac{m_e^2 c^3}{e \hbar},$$ as noted by knzhou.

A different potential mechanism that has been mentioned is acceleration of a preexisting free electron in the field, leading to a collision with sufficient energy to create a real positron. But this is unlikely given the medium present (air).

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?

Correct so far.

doesn't this mean that electron–positron pairs should be created near the transmission line and flung apart just like this[?]

No, because the hypothetical electron and positron don't have the energy until they've been acted on by the field for some time. The possible electron-positron pair that hasn't been created yet doesn't have any energy, because it hasn't been acted on by the field.

Now, if your wires were surrounded by vacuum, and you dropped an electron in near the negative wire, it would be be accelerated toward the positive wire. By the time it got there, it would have enough kinetic energy that if it collided (speaking very crudely) with some other appropriate particle, it might conceivably create some third particle with ~2000 keV of energy (based on a quick Wikipedia exploration it seems the most likely particle you could make this way would be an x-ray photon, but don't ask me for any more detail than that --- I'm not by any means a particle physicist).

But real-world power lines aren't surrounded by vacuum, so none of this can happen because any electron being moved around by the field between the two wires will be constantly interacting with the molecules of the air and losing kinetic energy in small steps as it moves, rather than all at once in a particle creating event.