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  Does vacuum displacement field imply space filled with Planck particle/antiparticle pairs?

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Consider natural [Gaussian units][1] such that $\hbar=c=1$ and the electron charge $e=\sqrt{\alpha}$ where $\alpha\approx1/137$ is the fine structure constant.

In Gaussian units the vacuum displacement field $\vec{D}$ is identical to the applied electric field $\vec{E}$ so that
Let us assume that in each small volume of space, $L^3$, there are charges $+q$ and $-q$ both with a mass $M$.

The vacuum displacement field, which is the vacuum polarization per unit volume, is given by
$$\vec{D}=q\times\frac{q\vec{E}}{M}\cdot T^2\times\frac{1}{L^3},\tag{2}$$
where the middle term is the separation distance of each pair of charges due to their acceleration in the electric field $\vec{E}$ during time $T$.

According to the quantum uncertainty principle in natural units $T = L = 1/M$ so that we find
Thus it seems that there must be masses $M$ with charges $q=+1,-1$ at each point in space in order to satisfy Eqn.$(1)$. The charge $q$ is $e/\sqrt{\alpha}=11.7e$.

Could the masses $M$ at each point in space be Planck masses with dimensions given by the Planck length $10^{-33}$ cm?

The Planck masses could have the fundamental charge $e$ if at the Planck energy the electromagnetic force is unified with the other forces so that $e=\sqrt{\alpha}=1$.

The pairs of Planck masses can be generated by positive-energy zero point modes with wavelengths down to the Planck length. If the Planck masses consist of matter and antimatter then according to the [Feynman-Stueckelberg interpretation][2] the antimatter can be considered as having negative mass. Thus the combined gravitational mass of each pair is zero so that space can remain flat.

  [1]: https://en.wikipedia.org/wiki/Gaussian_units
  [2]: https://en.wikipedia.org/wiki/Antiparticle#Feynman%E2%80%93Stueckelberg_interpretation

asked Sep 18, 2022 in Theoretical Physics by John [ revision history ]
edited Sep 22, 2022

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