Could a high electric field pull particles out of the vacuum?

1. Aug 31, 2011

johne1618

If one had a high enough electric field could one pull electrons and positrons into existence out of the vacuum?

The field needs to be high enough to accelerate the virtual particles away from each other strongly enough so they don't get a chance to recombine.

If a particle has acceleration g then there is a horizon distance r behind it given by:

r = c^2 / g..........(1)

such that no light signal starting at a distance of r or greater could ever reach the accelerating particle. Thus no virtual partner particle at a distance r or greater could recombine with the accelerating particle. This gives us an uncertainty in the particle position that we can convert into a momentum and then into an energy using the uncertainty principle.

If we have the uncertainty principle:

Momentum * distance = Planck constant

using E = p c

E / c * r = h

E = h c / r

Substituting for r in the above equation using (1)

E = h g / c

as E = M c^2

then we have

M = h g / c^3

Thus if we provide an acceleration g using an electric field we can produce a pair of particles of mass M.

For electrons/positrons being accelerated by a field F we have:

m_e = (h / c^3) * (e F / m_e)

F = m_e^2 c^3 / h e

F = field = 10^17 Volts / metre

It seems to me that the rest mass energy of the particle pair doesn't come out of the applied electric field but rather from out of the vacuum itself. The applied field just provides the acceleration to separate the virtual particles. The extra energy supplied to the particles during their acceleration could in principle be recovered in the process of bringing them back to rest.

Is this right?

Last edited: Aug 31, 2011
2. Aug 31, 2011

Ajihood

Re: Could

You have to be careful when treating virtual particles, because the only reason they exist is due to the uncertainty principle associated with energy and time.

When you want these virtual particles to become real particles (Ie you can point to a location of space and say they exist) you need to provide their rest energy in some form otherwise there will be an energy deficit.

This is what Hawking Radiation is associated about, with these pair-annihilation events taking place near the event horizon of a Black hole. Sometimes, the gravitational pull will be greater on one of the pair, than the other, seeing one partner (for example the electron) pulled into the black hole. This results in the other partner (the positron in this case) not having a partner to annhilate with and over stay its cheating of energy that the uncertainty princple allows it to do. So where does the energy come from in this case as clearly we have created a particle form nothing? Theory tells us the mass of the black hole actually decreases in the exchange.

So with your theory, I dont like the way you have derived it, especially using E = pc. However, I would say if you provided an electric field with an energy density of the rest mass of an electron AND position (need to conserve charge - with the above BH, we black holes have charge so we dont have to worry about charge not being conserved). That is a very large electric field. So it is possible, just like it is in accelerators.

3. Aug 31, 2011

johne1618

Re: Could

Actually an electric field of 10^17 volts/metre itself does have enough energy in a cubic electron compton wavelength to produce an electron/positron pair.

So I guess the electron/positron pair would come out of the applied electric field after all.

4. Sep 4, 2011

Hahagui

This is exactly the Schwinger mechanism to polarize the vacuum and produce the electron-positron pair. Unfortunately, as a non-perturbative results from QED, no direct experimental test has ever been done. In 1997, Bula from SLAC carried out a similar experiment with intense lasers and high-energy electrons.

5. Sep 4, 2011

Dickfore

Last edited by a moderator: Apr 26, 2017
6. Sep 4, 2011

fleem

Re: Could

I believe it could produce pairs even if it were a weaker field, just with less probability.

7. Sep 6, 2011

johne1618

Perhaps the high electric field merely "pulls back the veil" allowing particle pairs to leap out of the vacuum?

Is this "free" energy?

8. Sep 6, 2011

Dickfore

no, if a particle-antiparticle pair forms in the electric field, the positive particle will move in the direction of the field and the negative particle will move in the opposite direction. If you had the field produced by two isolated parallel plates, this would tend to decrease the charge on each of the plates, decreasing the field strength between them. The energy of the pair is gained by decreasing the energy of the electrostatic field.

If the plates are connected to (a huge) voltage source, there would be the need for charge to flow to keep the field the same. Thus, the external voltage source would produce the work necessary for a particle-antiparticle pair to emerge.