Energy released from annihilation

[Einstein's Cat]

Hypothetically, say there were two bodies, both with a mass of 1kg, and these bodies were to anhilliate as one body was an anti-version of the other body then would the energy released equal the speed of light (due to E= mc squared; the bodies are stationary) or twice the speed of light? In other words, is the energy released from anhilliation the sum of the energy of each particle that anhilliates?

 

[DaveC426913]

The energy released is determined by the amount of mass that is converted, regardless of objects. If 2kg of mass is converted to energy, then ... well ... 2 kg of mass is converted to energy. It doesn't matter what flavor of mass.

Note, BTW, that there are many ways that things can be converted into energy without it having to be a particle-anti-particle collision. eg, when fusion occurs, two hydrogen atoms combine to form one helium atom, and there is a little bit of mass left over. That little bit of the mass exits the reaction as energy.

BTW, don't forget that squared in there. The amount of energy released from total annihilation of 2kg of mass is 2*c².

 

[Einstein's Cat]

The energy released is determined by the amount of mass that is converted, regardless of objects. If 2kg of mass is converted to energy, then ... well ... 2 kg of mass is converted to energy. It doesn't matter what flavor of mass.

Note, BTW, that there are many ways that things can be converted into energy without it having to be a particle-anti-particle collision. eg, when fusion occurs, two hydrogen atoms combine to form one helium atom, and there is a little bit of mass left over. That little bit of the mass exits the reaction as energy.

BTW, don't forget that squared in there. The amount of energy released from total annihilation of 2kg of mass is 2*c².

Cheers from your help

 

[mathman]

When a particle and anti-particle annihilate and photons result, the photons are often paired, going at the speed of light in opposite directions.

 

[vanhees71]

In fact, you cannot annihilate two particles to just one photon, because you must obey energy-momentum conservation and the on-shell conditions, i.e., energy and momentum are related by the mass of the particles. The most simple tree-level diagrams for, say, electron-positron annihilation is $e^+ + e^- \rightarrow \gamma + \gamma$. It's a good exercise to work out the corresponding kinematics of this 2->2 process in some detail, using the Mandelstam variables.