1 MV electrode limited by vacuum pair production?

[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.

 

[ZapperZ]

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.

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.

 

[Dickfore]

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.

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

 

[ZapperZ]

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

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.

 

[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 traveling 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.

 

[ZapperZ]

But, of the order of is not what the OP is concerned. the rest energy of an electron is 0.511 MeV, thus the minimal energy for producing a pair is 1.022 MeV.

Furthermore, 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.

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.

 

[Dickfore]

I edited my post.

 

[Dickfore]

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

Another example is pair production in very strong electric fields, sometimes called vacuum decay. If, for example, a pair of atomic nuclei are merged together to very briefly form a nucleus with a charge greater than about 140, (that is, larger than about the inverse of the fine structure constant), the strength of the electric field will be such that it will be energetically favorable to create positron-electron pairs out of the vacuum or Dirac sea, with the electron attracted to the nucleus to annihilate the positive charge. This pair-creation amplitude was first calculated by Julian Schwinger in 1951.

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