Due to their low mass, electron/positron pairs should be relatively easy to produce in the form of virtual pairs in the vacuum. During their short lifetime, these particles should interact with photons and diffuse them. Shouldn't this phenomenon affect the images of the very distant quasars and shouldn't we correct for this effect the cosmic background ?
Virtual particles don't have a well-defined mass; they describe the polarization of the vacuum, which is a ``fancy'' way of describing a refraction index. A curved spacetime can be shown to have a refraction index, therefore the description of the propagation of electromagnetic waves (or photons) in a curved spacetime takes this automatically into account. And indeed it is in this way that the properties of the spacetime geometry of the Universe can be-and are-deduced from studying the cosmic background.
Producing pairs would mean describing ``dielectric breakdown'' of the background, on which the electromagnetic waves (or photons) propagate. Even for electron-positron pairs the required electric fields aren't ``small'', however. Such effects are taken into account, for instance, in understanding the results of such experiments: https://en.wikipedia.org/wiki/Alpha_Magnetic_Spectrometer
Stam Nicolis I was referring more precisely to vacuum fluctuation induced pairs and their diffusion of the photons via the electromagnetic interaction, not the gravity.
Vacuum polarization doesn't affect the diffusion of photons in flat spacetime, precisely because the spacetime is flat. And it is a standard exercise that only an external field (that, therefore, implies that the spacetime isn't flat) can produce real pairs, a single photon can't produce a real electron-positron (or any charged particle-antiparticle) pair in flat spacetime, just from energy-momentum conservation.
It is possible to induce dielectric breakdown of the vacuum by an external field and produce pairs; this is known as the Schwinger effect. The electric field necessary to do so, however, can't be produced, already for electron-positron pairs: https://javierrubioblog.files.wordpress.com/2016/09/report_rott.pdf
But, once more: ``vacuum fluctuation induced pairs'', in flat spacetime, don't ``diffuse'' photons. Flat spacetime has unit refraction index, even when quantum fluctuations are taken into account. This just expresses the property that interactions don't break global Lorentz invariance.
Thank you for your very precise and thorough answer. I understand then that these virtual particles only appear in special conditions but not in the vacuum.
electron/positron are just theory, they never been observed inside/outside of atom. they are just made up thing to make model of atom, that is all.
Thus, what is electron?? what is particle ?? they are just fancy name for nothing.
study them, find them, then make a comment on it.
regards,
Javad Fardaei Particles are the base of the standard view of quantum mechanics, so we can't deny their existence without good arguments. They could be emergent properties of quantum fields, deriving from the conservation laws.
No, due to energy-momentum conservation laws.
If your gamma quantum deflects due to Compton scattering on a virtual electron or positron, there must be a momentum exchange. The pair will then be unable to reannihilate into vacuum again. But can it instead become a real pair? The answer is no, too, because the 3-particle final state (e+, e- and gamma) has a space-like 4-momentum, whereas the original photon was light-like.
Miccola Bondarenco Thank you for your straightforward answer. I was not aware of this rule. But it rises the question : how can the photon "know" that the electron is a part of a virtual pair ?
A virtual electron is generally off mass shell, i.e., its momentum and energy are not interrelated. Yet they are anticorrelated with those of the virtual positron. But for any event to happen, all the final particles must be on mass shell. This is what cannot be fulfilled, unless the photon scatters strictly forward. In the latter case, its phase can change, but 4-momentum does not, i.e., there is no diffusion you ask about.
Thank you all for your willingness to enlighten curious non-specialists.
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