# Free energy from the vacuum?

An electric field with an intensity of the Schwinger limit

$$E=\frac{m_e^2c^3}{e\hbar} \approx 10^{18} \hbox{ V/m}$$

is strong enough to create electron-positron pairs out of the vacuum by separating the particles with an apparent horizon through a process similar to the Unruh Effect.

Theoretically these electrons and positrons could separately leave the region of high electric field (through holes in the electrodes) and then later be brought back together to produce real energy through annihilation (see diagram below).

Would this be a way of obtaining "free" energy from the vacuum or does that energy come from the original electric field (see third paragraph of Schwinger_effect)? As I understand it quantum field theory predicts that the vacuum has an infinite energy density and therefore electron-positron pairs can be created continuously without limit.

I could imagine steady streams of electrons and positrons coming out of the high field region and being annihilated to produce photons as illustrated in the diagram below. The steady currents in the particle streams along with the static charges on the electrodes would produce a static electromagnetic field. Therefore the system (comprising charges, currents and local EM field) would be unchanging apart from the output of photons.

Hypothesis that electric field energy density produces electron-positron pair

The Compton wavelength $$\lambda_e$$ of an electron is

$$\lambda_e = \frac{\hbar}{m_e c}$$

The energy density of an electron $$\rho_e$$ is

$$\rho_e = \frac{m_e c^2}{\lambda_e^3}$$

The energy density in the electric field (ignoring factor of $$1/2$$)

$$\rho_E=\epsilon_0 E^2$$

If we equate $$\rho_e=\rho_E$$ we find the following expression for the electric field strength E

$$E=\frac{m_e^2 c^{5/2}}{\epsilon_0^{1/2}\hbar^{3/2}}=10^{16.5}\hbox{ V/m}$$

This is a completely different expression to the Schwinger limit described above. For example it doesn’t refer to the electron/positron charge.

Work done by field in a Compton wavelength equals rest mass energy of particle

At @lux’s suggestion I tried $$e E \times \lambda_e = m_e c^2$$ $$E=\frac{m_e^2 c^3}{e \hbar}$$

This is the Schwinger limit formula. However I don’t think this suggests that the particle energies are being supplied from the energy density of the electric field itself. Rather the particle charges pick up an energy equivalent to their rest mass as they move a Compton wavelength through the field.

• The energy comes from the applied electric field, assumed here to be constant and without incorporating depletion.
– nox
Commented May 14, 2022 at 17:20
• But if the charge on the electrodes producing the electric field doesn't change then I don't see how the electric field can lose energy. Commented May 14, 2022 at 17:22
• This is closely related to my comment about depletion. The E field is idealised, not at all physically realistic, as it is infinitely extended in space and time and constant!
– nox
Commented May 14, 2022 at 17:30
• remember that every electron-positron pair being produced is trying to reduce the electric field below the critical intensity, hence you need to pump more energy into your electric field to keep it constant. Commented May 14, 2022 at 18:20
• Response to the edited post - have you tried equating work done in a Compton wavelength to rest mass energy of electron and positron pair?
– nox
Commented May 15, 2022 at 0:02

As per energy conservation law (or "no free lunch theorem"),- whatever energy you'll put into a pair production - on annihilation it will return given energy back to the field, so that total energy density is the same : $$\gamma \to e^{^+} + e^{^-} + K \to \gamma$$ Of course electron-positron pair kinetic energies can vary due to vacuum fluctuations, and so output photon energy will vary too. However on average, you'll get your same input field density, so answer is NO,- you can't extract free energy from vacuum this way (or any other, cause it would break thermodynamic laws).