# Could anti-matter virtual particles be theoretically harvested from the quantum flux?

I was wondering if a vacuum created within a particular material in the shape of sphere could be used to harvest anti-matter from virtual particles? Within the vacuum, theoretically, virtual particles would form; is there a material that could be used to draw the electron towards it while it repels the anti-matter and forces it to the center of the sphere where it cannot come in contact with matter and be annihilated.

• My impression is that this is very similar to Hawking radiation, in which case your "particular material" is... a black hole. But I'll be the first to admit I don't know enough physics to be sure of that. – Cadence Oct 13 '19 at 1:20

I might actually make a second answer, for the first time, ever.

It has been theorised (or this New Scientist article on the paper) that a suffiently power laser pulse will have a strong enough electrical field to separate virtual particle-antiparticle pairs in the quantum vacuum, similar to the way that a black hole has a strong enough gravitational field to perform the same trick.

The theory suggests that once your laser gets up to about 5*1025W/cm2 (with an electrical field strength of ~1.32*1014V/cm) the power of the pulse will be rapidly depleted and you'll get a spray of electron/positron pairs that will then be amenable to more conventional electromagnetic collection and storage. This also means that you probably won't be able to get anything larger than an electron or positron out of the system, because of the challenges of developing even more power without it all being depleted by electron-positron generation. Note also that you cannot create neutron-antineutron pairs this way either, unlike the black-hole way, but this isn't necessarily a bad thing.

Whilst the production of particles will be technically more efficient than more boring old school means of generating positrons, the sheer difficulties of making such an outrageously powerful laser and the inefficiency of operating that laser will likely make the process uneconomical.

So there you go. Turns out that you may be able to use electrical fields to separate particle/antiparticle pairs, but the field strengths are so incredibly high that it may just be too awkward and expensive and inefficient to be worthwhile. It will at least be easier and safer than creating a million tonne black hole.

The only know mean to separate the components of virtual pairs forming everywhere is the event horizon of a black hole, according to the mechanism known as Hawking radiation

Hawking radiation is black-body radiation that is predicted to be released by black holes, due to quantum effects near the event horizon. [...] Hawking showed that quantum effects allow black holes to emit exact black-body radiation. The electromagnetic radiation is produced as if emitted by a black body with a temperature inversely proportional to the mass of the black hole. Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles.

Black holes are quite unpractical, since they have been shown to destroy any observer venturing too close to them.

As an additional remark, stacking charged particles in a point is not feasible: electrostatic repulsion would quickly spread the particles around, with easily predictable consequences in the case of anti-matter.

• On your additional remark: While like-charged particles do repel each other, they can still be held within a particular volume of space using an ion trap of some variety. In principle, antimatter ejected from a black hole (as part of its Hawking radiation, in addition to normal matter and photons) could be directed into and stored within such a device. – Someone Else 37 Oct 13 '19 at 5:18

L.Dutch is basically right, but I'll expand on their answer as there are a few important details that anyone hoping to harvest antimatter this way should know. Note that this involves maths and physics, and I'm not great at either. Take the numbers below with a small pinch of salt.

TL;DR: you'll need to make your own black hole, you can't get more stuff out of the quantum vacuum than the total mass the black hole started with, you can only extract stuff as fast as the black hole evaporates (which is either very slowly, or very dangerously quick), and a lot of the stuff coming out will be in the form of gamma rays, unstable and neutral particles and the kinetic energy of the stuff you do want, making it harder to catch. All in all, very difficult.

The energy (or temperature, if you like) of the Hawking radiation coming out of an evaporating black hole is related to its mass. The more massive the black hole, the cooler the radiation. This is important, because no particles can be spat out of a black hole that have a rest energy greater than the energy of the Hawking radiation.

This has a number of important knock-on effects.

The temperature of a black hole is related to its mass by this nice, simple equation (from the wikipedia page):

$$T = \frac{\hbar\,c^3}{\,8\,\pi\,G\,k_\text{B}\,M\,}$$

where $$ħ$$ is the reduced Planck constant, $$c$$ is the $$k_\text{B}$$ is the Boltzmann constant, $$G$$ is the gravitational constant, and $$M$$ is the mass of the black hole. You can simplify it to $$T \approx \frac{1.227\times10^{23}}{M}$$.

The rest energy of an electron or positron is 510KeV and for a proton or antiproton about 932MeV with equivalent temperatures of 6*109K (hotter than the centre of the sun and our hottest fusion reactor projects) and 1013K respectively. The equipment for grabbing escaping particles, which will come out in all directions, must fully englobe the black hole and survive this rather hostile environment.

Once your black hole reaches those temperatures it'll start spitting out lots of the appropriate kinds of matter and antimatter particles... you'll get some at lower temperatures, but they won't exactly gush out. So, a black hole with a mass of about 20 billion tonnes will spit out things no heavier than an electron, and a black hole with a mass of about 11 million tonnes will spit out things no heavier than an antiproton, but the stuff it does spit out will include electrons and positrons, as well as protons and antiprotons and a whole soup of other particles and EM radiation (lots of gamma rays), most of which you won't want. Many of those particles will be neutral, so you can't catch them. Fifty percent of the escaping neutral particles will be antiparticles, and antineutrons flying out will react violently with your capture mechanisms. Many of the other particles will be unstable and will decay into all sorts of stuff, but mostly gamma rays (which will further scorch your equipment) and neutrinos (which are wasted mass, from your point of view).

The next important fact is that black holes evaporate over time, as their mass-energy is carried away by this Hawking radiation. The evaporation time of a black hole with mass $$M_0$$ is given by

$$t_\mathrm{ev} \approx 8.41092 \times 10^{-17} \;M_0^3$$

You don't actually have to worry about these electron or proton-emitting black holes evaporating any time soon... the lighter, hotter proton-emitting hole will take over a million years to evaporate. A black hole that spits out protons now would have been less than a stellar mass when the universe was born, which raises interesting questions about how that hole formed in the first place. No-one has spotted such a thing... that doesn't mean that they don't exist, but the chances are you'll need to make it yourself with regular matter. Good luck!

But here's the important thing: you can't get more stuff out via Hawking radiation than you put in. Really, what you're doing is converting millions or billions of tonnes of regular matter into energy and a 50/50 mixtures of particles and antiparticles via the means of a black hole. Whilst you might technically be harvesting those particles from the quantum vacuum, you can't even get more out than the mass you put in to start with. So there you go. The answer to your question is therefore "yes, kinda."

There's one final (probably) thing to be gathered from all of the above... if you can't get out more than you put in, and you can't get it out faster than the hole evaporates, and an 11 million tonne black hole takes nearly 4 million years to evaporate, your average yield is going to be a few tonnes-equivalent of mass energy every year (less than a tenth of a gram per second). Lighter holes evaporate faster, so it'll be lower than this to start with. How low?

Well, initial power generated from a black hole of mass $$M_0$$ is given by

$$P \approx 3.56345\times 10^{32}(1/M_0)^2$$

which for the electron-producing hole is a mere 830kW, equivalent to about 1019 electrons per second. The proton producing hole has a rather more fierce initial output of 2.77TW, equivalent to about 1022 protons per second. That's about a third of the average yield... about 3 hundredths of a gram per second. Of a mixture of particles and antiparticles and things you don't want. Assuming 100% production of desired particles (which won't happen... you'll get all sorts of stuff in there, and lots of EM radiation). For an initial input of some tens of billions of tonnes, if you can create a black hole without wasting any mass. It starts to sound like making a dyson swarm is the sensible and easy alternative, and you don't get to say that very often.

If you crank up your black hole initial power output to a toasty petawatt (effective temperature, 2*1014K), you'll get a tonne-equivalent of matter and energy out every day. A black hole a mere half-million tonnes in mass will provide this output. It'll evaporate in a little over 500 years, and every day it'll release a little more power and hence a little more matter. Eventually your equipment will melt and after a while the hole will go bang spectacularly, releasing about a million megatonnes equivalent as it finally evaporates. You might think you could forestall that by pushing mass and energy back into the hole... only a) there's a lot of radiation coming out of the hole that'll make it difficult to get anything in, and b) and 10-19m across, its a pretty small target to hit. Honestly, the whole plan seems like a big "don't do that".