Is free space a 'dielectric'?

Question

The physical constant $\epsilon_0$ is usually called "permittivity of free space" or "vacuum permittivity". Griffiths says:

I dislike the term, for it suggests that the vacuum is just a special kind of linear dielectric, in which the permittivity happens to have the value $8.85 \space 10^{-12}\space \rm {C^2\over Nm^2}$

But later on, in a comment, he specifies:

In quantum electrodynamics, the vacuum itself can be polarized, and this means that the effective (or “renormalized”) charge of the electron, as you might measure it in the laboratory, is not its true (“bare”) value, and in fact depends slightly on how far away you are!

So? Can free space be treated as a legit dielectric or not? And by the way, what does it mean "depends slightly on how far away you are"? I thought the elementary charge was not a relative value...

Answer 1

In a simplified picture, which often tends to give the wrong idea but may serve to illustrate some aspects, every photon spends some time as a combination of a virtual electron plus its antiparticle, the virtual positron, which rapidly annihilates each other shortly thereafter. The combination of the energy variation needed to create these particles, and the time during which they exist, fall under the threshold of detectability expressed by the Heisenberg uncertainty relation, $$\Delta E· \Delta t \geq \hbar$$

In effect, the energy needed to create these virtual particles, $\Delta E$, can be "borrowed" from the vacuum for a period of time, $\Delta t$, so that their product is no more than the reduced Planck constant, $\hbar \approx 6.6×10^{−16}$ eV·s. Thus, for a virtual electron, $\Delta t$ is at most $1.3×10^{−21}$ s.

While an electron-positron virtual pair is in existence, the Coulomb force from the ambient electric field surrounding an electron causes a created positron to be attracted to the original electron, while a created electron experiences a repulsion. This causes what is called vacuum polarization. In effect, the vacuum behaves like a medium having a dielectric permittivity more than unity. Thus the effective charge of an electron is actually smaller than its true value, and the charge decreases with increasing distance from the electron. This polarization was confirmed experimentally in $1997$ using the Japanese TRISTAN particle accelerator. Virtual particles cause a comparable shielding effect for the mass of the electron.

Answer 2

The medium has a relative permittivity >1 and can be treated as a dielectric. So yes to your first question.

The vacuum can be polarised, the polarisability depends on the applied field which itself is a function of distance.

Therefore, the field you might measure in order to determine the charge of an electron will vary with distance due to the different amounts of polarisation of the vacuum at the different distances.

The charge you are measuring in the lab is not the exact charge rather it is an effective charge.

Answer 3

Firstly, $\epsilon_0$ is an artifact of using SI system of units. In many other system of units $\epsilon_0=\mu_0=1$ and both permittivities are dimensionless, whereas permittivities of other materials are measured in respect to that of vacuum.

Electric permittivity, along with fields $\mathbf{D}$ and $\mathbf{H}$, is an object that belongs to the electrodynamics of continuous media, i.e., macroscopic description of electrodynamics phenomena. One should not really discuss such macroscopic quantities on a microscopic level, alongside a discussion of the electron mass renormalization by virtual particles. The difficulties that Griffiths points out in the second passage cited are precisely the reason why this should not be done.

The dependence of how far you are seems to be a reference to relativity, since quantum electrodynamics is in essence relativistic.