1 MV electrode limited by vacuum pair production?

1. Jul 22, 2011

johne1618

Hi,

Someone told me that a potential of 10^6 volts is not stable because it would cause spontaneous electron-positron creation from the vacuum.

Is this true?

I thought one could reach higher potentials than 1000000 volts.

2. Jul 22, 2011

ZapperZ

Staff Emeritus
That is definitely not true. I work with an RF field inside a cavity that has a peak field of 80 MV/m. No e-p pairs there!

What would be more of an issue is vacuum breakdown causing arcing.

Zz.

3. Jul 22, 2011

Dickfore

How is 80 MV/m the same as potential difference of 1 MV?

4. Jul 22, 2011

ZapperZ

Staff Emeritus
Not directly, but at least within the length of the cavity of roughly 10 cm, you have a potential of the order of MV.

Zz.

5. Jul 22, 2011

Dickfore

@ZapperZ:
But, the order of magnitude is not what the OP is concerned in this particular case. the rest energy of an electron is 0.511 MeV, thus the minimal energy for producing a pair is 1.022 MeV.

@OP:
Saying the value of the potential at a point has a particular value does not make any sense until you define a reference point where the potential is known. Consequently, what is physical is potential difference.

My thoughts:

I would say that electric fields are more important than potential differences. Namely, suppose a virtual pair is produced, with each particle travelling at speed $v$. We have borrowed an energy:
$$\Delta E = \frac{2 \, m \, c^{2}}{\sqrt{1 - v^{2}/c^{2}}}$$
and we had better return it in time of the order:
$$\Delta t \sim \hbar / \Delta E$$
In this time (ignoring acceleration of the charged particles for simplicity), the two particles will cover a distance)
$$\Delta x = v \, \mathrm{cos}{\theta} \, \Delta t$$
If the positron moves along the field and the electron opposite of the field, then the electric forces would gain energy:
$$2 \, e \, K \, \Delta x$$
If this energy is greater than or equal to the borrowed energy, pair creation might become possible. Combining everything together, we get:
$$2 \, e \, K \, v \, \frac{\hbar}{\Delta E} \gtrsim \Delta E$$
$$K \gtrsim \frac{(\Delta E)^{2}}{2 \, e \, \hbar \, v} = \frac{2 (m c^{2})^{2}}{e \, \hbar \, c} \, \frac{1}{v/c (1 - v^{2}/c^{2})}$$
The function
$$\frac{1}{x(1 - x^{2})}$$
has a local minimum for $x_{0} = 1/\sqrt{3}$ which is $3 \, \sqrt{3}/2$. Thus, a minimal electric field that would create pair production is:
$$K_{\mathrm{min}} = 3 \, \sqrt{3} \, \frac{(m c^{2})^{2}}{e \, \hbar \, c} = 3 \times 1.73 \, \frac{(0.511 \, \mathrm{MeV})^{2}}{1 \, \mathrm{e} \times 197.4 \, \mathrm{MeV} \cdot \mathrm{fm}} = 6.9 \times 10^{-3} \, \frac{\mathrm{MV}}{\mathrm{fm}} = 6.9 \times 10^{18} \, \frac{\mathrm{V}}{\mathrm{m}}$$
This field is very strong! In fact, it would take a distance of only 0.14 pm to generate a 1 MV potential difference.

Last edited: Jul 22, 2011
6. Jul 22, 2011

ZapperZ

Staff Emeritus
Er... I didn't realize I had to be verbose in the explanation.

The TM01 mode field has an axial electric field. This means that in a pillbox-type cavity, the E-field has its highest field in the center of the cavity. An 80 MV/m field means that the POTENTIAL DIFFERENCE in a 10 cm field across the pillbox is 80MV/m *0.1m = 8 MV!

Now, is this clear enough, or did I leave something out? Oy vey!

Zz.

7. Jul 22, 2011

Dickfore

I edited my post.

8. Jul 22, 2011

Dickfore

I searched for 'pair production electric fields' and found this on Wikipedia:

It seems the term 'decay of false vacua' is associated to the phenomenon we are contemplating. There is a PF thread about it.