# Are 40 Kelvin interstellar gas clouds evidence of virtual positron and real electron annihilation

Given that Hawking radiation includes separation of a virtual electron-positron pair, then why cannot it be the case that in the vicinity of a real electron it annihilates with the virtual positron?

The net effect must surely be that two real photons result and that for charge conservation etc the virtual electron must now become real (and displaced and moving slightly). The energy of the two photons of the annihilation event is also now manifest and will be spread around 40 Kelvin (the equivalent temperature of an electron rest mass). Conservation of energy is of course broken and should be evident.

Spectra of interstellar gas clouds do range across 40K and given their thin nature and that they sit in the vast heat sink of empty space then it is not clear to me why they sustain their temperature, i.e. they ought to rapidly cool. If this is the case then it may be evidence that the real electron and virtual positron annihilation is the source of energy that sustains their temperature and that energy is generated/released from that of the vacuum!

• I edited the question to make it more reader-friendly, but still I must say that it has several issues. It looks like you could get more acquainted with how the notion of pair production and annihilation works (in "mainstream" quantum physics, before going to cosmological cases). – Helen May 13 at 19:34
• Without specifics Helen then I would guess that your issues relate to virtual particles being entirely mathematical artifacts to make the QFT maths work. My primary contention with that is that if an event horizon can make them real then why cannot electron annihilation? – RhysDefinitiveFrameworks May 13 at 19:56
• Keep in mind that room temperature (about 290 kelvin) corresponds to an energy of $40\,\mathrm{meV}$ and the anihilation energy of $e^+ + e^-$ is $2 \times 511 \,\mathrm{keV}$ (about five orders of magnitude higher). – dmckee May 14 at 18:02
• The virtual particle explanation of Hawking radiation is just a heuristic. Please see math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… – PM 2Ring May 15 at 3:18

the difference between virtual and real particles is that real particles have incoming and outgoing legs associated to asymptotic states. Laws of conservation (on-shell dynamics) applies to incoming and outgoing legs.

Any process involving an initial state with a single real electron implies an incoming leg. Conservation of charge implies that the total charge of outgoing legs for that process must also equal -1. Ignoring other hadron and lepton transitions, that implies that outgoing states must be either one electron, or two electrons and one positron, or three electrons and two positrons, ... or $$N$$ electrons and $$N-1$$ positrons

Conclusion: You cannot have real electrons be annihilated with virtual positrons

• Hello Lurscher, the initial and final particle would be the same, i.e. one electron but with the (dubious) result that it would have gained momentum matched by the resultant momentum of the two photons. If a virtual particle of a pair can become real in Hawking radiation then exactly what prevents the same in the case I described and would that then not apply to prevent Hawking radiation? – RhysDefinitiveFrameworks May 13 at 20:03

No, they are not. In addition to the reason in @lurscher’s answer, the temperature equivalent to the mass of an electron is not 40 K. It is 6,000,000,000 K!

The interstellar medium (ISM) has been well studied for the past fifty years. There are models of the ISM that invoke less exotic heating mechanisms than Hawking Radiation. The Wikipedia page on the ISM provides a good rundown on both heating and cooling mechanisms.

The interactions between the different phases of the ISM are complex, and there is still a lot to be understood. But it's clear that the gas is not just sitting isolated out in deep space. It is part of the dynamic system of the entire galaxy. The fact that the coolest parts have not cooled below 10's of Kelvin is not a mystery that needs solving.