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1 MV electrode limited by vacuum pair production?

  1. Jul 22, 2011 #1
    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. jcsd
  3. Jul 22, 2011 #2

    ZapperZ

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    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.
     
  4. Jul 22, 2011 #3
    How is 80 MV/m the same as potential difference of 1 MV?
     
  5. Jul 22, 2011 #4

    ZapperZ

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    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.
     
  6. Jul 22, 2011 #5
    @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 [itex]v[/itex]. We have borrowed an energy:
    [tex]
    \Delta E = \frac{2 \, m \, c^{2}}{\sqrt{1 - v^{2}/c^{2}}}
    [/tex]
    and we had better return it in time of the order:
    [tex]
    \Delta t \sim \hbar / \Delta E
    [/tex]
    In this time (ignoring acceleration of the charged particles for simplicity), the two particles will cover a distance)
    [tex]
    \Delta x = v \, \mathrm{cos}{\theta} \, \Delta t
    [/tex]
    If the positron moves along the field and the electron opposite of the field, then the electric forces would gain energy:
    [tex]
    2 \, e \, K \, \Delta x
    [/tex]
    If this energy is greater than or equal to the borrowed energy, pair creation might become possible. Combining everything together, we get:
    [tex]
    2 \, e \, K \, v \, \frac{\hbar}{\Delta E} \gtrsim \Delta E
    [/tex]
    [tex]
    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})}
    [/tex]
    The function
    [tex]
    \frac{1}{x(1 - x^{2})}
    [/tex]
    has a local minimum for [itex]x_{0} = 1/\sqrt{3}[/itex] which is [itex]3 \, \sqrt{3}/2[/itex]. Thus, a minimal electric field that would create pair production is:
    [tex]
    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}}
    [/tex]
    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
  7. Jul 22, 2011 #6

    ZapperZ

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    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.
     
  8. Jul 22, 2011 #7
    I edited my post.
     
  9. Jul 22, 2011 #8
    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.
     
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