If we had a significant amount of anti-hydrogen (but not enough to make a star), what would be the most complicated thing we could build out of it, and how would we go about doing so?

Edit: I give the condition "not make a star" because an anti-star would eventually be able to make other elements. Would we be able to make other elements, molecules etc if we had as much contained anti-hydrogen as we needed?

  • $\begingroup$ How much is "significant"? How are you measuring "complicated"? What assumptions are you making about how well we are able to manipulate the anti-hydrogen without annihilating it? A boring but correct answer to your question is that you can build anything out of 10 kg of anti-Hydrogen that you can build out of 10 kg of Hydrogen, in principle. Except, the 10 kg of anti-Hydrogen will immediately annihilate into a deadly burst of X-rays if it comes into contact with regular matter, so anti-Hydrogen is just a little harder to work with. $\endgroup$
    – Andrew
    Mar 16, 2022 at 17:43
  • $\begingroup$ @Andrew Yes that is the challenge. If you could get as much anti hydrogen in some contained area, how could we manipulate it without annihilating it. Complicated is intentionally somewhat vague, could we make other elements, molecules, technology, etc? $\endgroup$
    – Craig
    Mar 16, 2022 at 17:48
  • $\begingroup$ A similar question is physics.stackexchange.com/q/66800 $\endgroup$
    – Craig
    Mar 16, 2022 at 18:04
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    $\begingroup$ I think there's a few different issues wrapped up in what you're saying and it's not clear to me what you are most interested in. There's no fundamental physics obstruction to building anything you want out of anti-Hydrogen that you can out of Hydrogen. It's "only" a technological question to build the devices to put the anti-Hydrogen atoms together as desired. In principle we can make anything consistent with physics given enough time and money. If you want to know what practical constraints exist with current technology, see en.wikipedia.org/wiki/Antimatter#Antihydrogen_atoms $\endgroup$
    – Andrew
    Mar 16, 2022 at 20:26
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    $\begingroup$ I do not really see what this question have in common with the purported duplicate. It's an open-ended question from a new aspiring user who has more curiosity than training in physics, and that's a good thing (the former often leads to latter; the latter is mostly indifferent to the former). I answered what we, con. popular belief, can not do with a “significant” stash of anti-H, satisfying the question constraint (“less than a stellar mass”). Closer in order to a stellar mass, and the course of action recommended by the best physics is pray… And engineering physics is on topic, I believe. $\endgroup$ Mar 25, 2022 at 0:04

1 Answer 1


Of course, we should make an interstellar rocket propulsion drive from it!

It will yield an insane specific impulse, which I even won't try to estimate, because its magnitude won't fit in my poor head. Many, many orders of magnitude better than ~$10^2 s$ that the chemical engines produce, or even that of the best ion thrusters, reaching nearly $10^5s$. No wonder antimatter drives occur in SciFi so often!

But can we?

Besides obvious engineering problems to solve, namely how to keep the antifuel from contact with anything made of matter, there is a physics problem: how to steer the product of annihilation, the hard $\gamma$ momentum-carrying radiation, into an exhaust beam of the engine. It's not light which can be beam-formed with a parabolic mirror when the source is in its focus. The annihilation of a proton with an antiproton, the antihydrogen nucleus (we'll ignore electrons: they're too light, and you'll likely strip them off to keep your antifuel in an electromagnetic confinement; keep in mind that Coulomb repulsion in the gas of pure antiprotons will create a high pressure even at low density, making the storage kinda bulky), you get a few different kinds of particles, which will indeed contribute to your engineer's headaches. The reaction like $p+\bar{p} \to 2 \gamma$ would be nice to have, but it's not as simple as this. The immediate annihilation proceeds into one of three approximately equiprobable branches:

$$ p+\bar{p} \to \begin{cases} k\times\pi^0 && P \approx 1/3; (k \ge 1?)^*\\ \mu^{-}+\pi^{+}+\nu_\mu && P \approx 1/3\\ \mu^{+}+\pi^{-}+\nu_\mu && P \approx 1/3 \end{cases} $$

* I do not remember if there's one or more $\pi^0$ particles produced on this branch. Maybe only one if the reagent (anti)protons started at rest and only got kinetic energy from their mutual Coulomb attraction. I recall a number between 2 and 3, but that's from accelerator experiments, where the $p$ and $\bar p$ are already quite energetic.

Fortunately, the neutral pions live only attoseconds and decay into a pair of photons, $\pi^0\to2\gamma$. They won't have time to travel very far. The other two branches are more problematic from the engineering point, as these particles are longer-lived and require some way to confine them if you want to keep the volume of the reaction zone compact; their relativistic energies extend their apparent lifetime. A quick guesstimate, based on comparing muons produced high up in the atmosphere from comparably energetic cosmic rays and detected on the surface, 50–80 km away, shows that “compact” here means “less than 100km in diameter.” And how to steer hard relativistic particles of exactly same mass but opposing charges flying in all directions ($\mu^{-}$/$\mu^{+}$ from the two bottom branches; same for $\pi^\pm$), I have only vague ideas. It's not impossible, but certainly a daunting physical engineering task.

There are further decays, but in the end, you'll get about a half of the total energy carried away by the $\nu_\mu$ neutrinos. Of the rest, about 2/3 end up as $\gamma$ photons and 1/3 as electron-positron pairs.

These electrons are your another engineering headache. It would be nice to annihilate them, too, but the catch is they are relativistic, with the kinetic energy that makes catching and steering them back to the reaction zone quite challenging. (Alternatively, you can usefully shoot them out as the engine exhaust, but steering them into the “nozzle” is hard for the same reason). We can make a back-of-the envelope for these electrons as this: 1/6 of the initial energy of the $p\bar{p}$, which is only their mass, ends up in $ee^+$, but the rest mass of a $p$ is $\approx\! 1800$ that of the $e$, and you get one $e$ from one $p$ (or anti). The energy of each of these $e$ will be $\approx\! 1800 \times 1/6 = 900\, m_e c^2$. That's a gobsmacking lot, to put it in the mildest way possible. If you want the actual velocity in the units of $c$, the relativistic relation $\gamma = 1/\sqrt{1-v^2} = 900$ yields $1-v=6\times10^{-7}$, written in the form which saves you counting the nines in $v=0.9999994\,c$. Bummer. It gets very hard.

The photons aren't a lesser headache. 1/3 of the total initial mass of the two protons ends up in only a few of them, let's take 6 just for a rough approximation, and assume their energies will be all the same. Each 3 $\gamma$ particles will have the 1/3 initial (rest) energy of the original proton, or 1/9 per photon. This yields an energy of about 100 MeV per photon. This is not something that can be reflected or refracted; hard gamma-radiation is anything but light as we usually think of it. I have no idea how to make these things exit the reaction chamber in the direction I want, not where they want. These are neutral, produce all sort of stuff when interact with matter, and push the reactor design task from “very hard” to “nearly impossible” on my tricky-o-meter.

But if you get a 10 kilo stash of antihydrogen,

please keep this thing far from the Earth!

There is not a scarcity of hard vacuum in deep space, but creating vacuum on Earth is another thing, and any neutral matter reaching your antistash suspended in the EM fields will produce the same 100 MeV gamma-radiation, the more of it the higher is the pressure of unevacuated gases. LHC has almost the best vacuum achievable on Earth, and it varies between $10^{-7}$ and $10^{-9}$ Pa. But that's still on the order of $10^{12}$ gas molecules, or, assuming that the contaminant is a mix of N$_2$ and O$_2$, $10^{13}$ antiproton-hostile protons per $m^3$ (no Coulomb attraction, though, so interactions are rather random). I won't go into calculations of how many meters of concrete should your antimatter storage bunker have; neither I'll be surprised if you arrive at a number of few tens. And that's only to protect from its inevitable $\gamma$-radiation; please make your vacuum and EM suspension systems at least quadruple-redundant...

Best of all, keep it in an orbit somewhere half-way to the Moon. The vacuum there is deep enough for antiproton storage: $\approx\! 10^{-9}$ Pa, mostly molecular and some radicalized atomic hydrogen. Let me leave it to you to calculate what would happen if your EM bottle suddenly loses power and pops open. I don't think you'll get anywhere close to the GRB level of radiation intensity: the cross-section of interaction between the $\bar p$ and the neutral H atom should be quite small.


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