What is the percentage of useful energy do we get from matter-antimatter annihilation?

This is a theoretical question since we haven't made enough antimatter to try it in reality of course. But I am asking about the physics part in this.

Also, by "useful energy" I mean the energy we are able to use either as a heating source for something like a nuclear reactor, or as energy for an explosion like nuclear explosions.

If I am not mistaken, a large part of the energy we get from the annihilation is in the form of neutrinos, which we for some reason can't consider them useful energy. So now, if we subtract the energy of the neutrinos, is it safe to consider the rest as useful energy as I explained ?

Please try to be as simple as possible because I don't speak English very well.

• The reason the neutrinos can't be treated as "useful" is that they don't interact in your working fluid. Indeed most of those pointed 'down' will fly right through the planet undisturbed carrying that energy away. That energy is lost to your power plant. – dmckee Aug 10 '13 at 20:16
• – Ben Crowell Aug 11 '13 at 16:31

I found a interesting paper on the use of antimatter for rocket propulsion from NASA that addresses this subject. Your question has a pretty complicated answer.

I think you will find more formation than you wanted reading that article. In the report it says about an electron-positron collision:

His results indicate that the neutrinos carry off ~22% of the available pion energy (~1248 MeV) whereas the muons retain ~78%. The unstable muon, having an average energy of ~300 MeV, also decays (in ~6.2μs) into an electron, or positron, and two neutrinos as shown in Table 2. The energy appears to be about equally distributed among the three particles with the neutrinos carrying off ~2/3 of the available energy. Ultimately, the electrons and positrons can also annihilate yielding additional energy in the form of two 0.511- MeV gamma rays. The neutrinos are considered to be massless and move at essentially the speed of light. They are extremely penetrating and rarely interact with matter. Under vacuum conditions the various muon and electron neutrino particle–antiparticle pairs carry off ~50% of the available annihilation energy following a p–p reaction.

• So, to be more accurate, we should only consider ~50% of the energy released by the annihilation as useful energy. So now a kilogram of matter annihilating another kilogram of antimatter would only release 9×10^16 which is about 21.5 megatons of TNT. – Abanob Ebrahim Aug 10 '13 at 21:51
• @AbanobEbrahim only? That's a lot of energy for only 2KG of matter/anti-matter. But the down side is that is more anti-matter than what will ever exist in our life times. – user6972 Aug 11 '13 at 17:17

As I said in my answer Can antimatter be used as fuel for nuclear reactors? to your other question life is not that simple.

The estimate with neutrinos is for the free annihilation. If one wants to make a reactor out of this there are options to trap all charged particles before they decay and dissipate their kinetic energy in a medium, and ditto for all gammas from pi0 with ofcourse a different medium.

So, with an average of 5 pions per annihilation 2/3* 5*105MeV=350Mev cannot be recovered, the energy is lost in muon etc decays from charged pions when they become at rest( a bit can be gained from the muon track too). So the number of lost energy will be much less than 50% if the charged pions can be trapped and the energy of the pi0 photons converted to heat.

The problem is that with current technology one would need very strong magnetic fields to curl the charged pions. Calorimetry in high energy physics works by trapping all the energy of gammas, but these are large setups not designed for energy extraction as heat.

It is very probable that the cost of the technology needed to produce the antiprotons and create the magnetic fields to contain the charged products and capture the heat from the gammas will be much higher than the gain from the energy of annihilation no matter what the fraction. In my opinion the reactor cannot break even, i.e. give more energy than it eats up in its production, certainly not with the present technology.

So much for reactors. For explosions the energy loss in neutrinos holds, because the decays happen in flight and the energy is taken away by them. The technological problem is reduced to containing antimatter, and it is far from trivial, and I suspect equally cost/benefit negative as with a reactor.

• In 2006, Gerald Smith estimated \$250 million could produce 10 milligrams of positrons (equivalent to \$ 25 billion per gram); in 1999, NASA gave a figure of \\$62.5 trillion per gram of antihydrogen. The only way to get it cheaply might be to harvest it. It is estimated that 10 micrograms of antiprotons and 10 milligrams of positrons are locally contained within the Earth’s magnetosphere at any given time. And perhaps more in a planet with strong fields like Jupiter. – user6972 Aug 11 '13 at 17:29