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

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Consider natural [Gaussian units] 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
$$\vec{D}=\vec{E}.\tag{1}$$
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
$$\vec{D}=q^2\vec{E}.\tag{3}$$
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] 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. asked Sep 18
edited Sep 22

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