Accelerating protons and antiprotons for pair production?

AI Thread Summary
The discussion centers on the definition of pair production, which is described as the conversion of a photon into a particle-antiparticle pair. A specific example is given where a proton collides with a stationary proton, resulting in the creation of an antiproton and additional protons. The question arises whether this reaction qualifies as pair production and if the definition should be adjusted accordingly. Additionally, the conversation touches on the conservation of charge in particle reactions, emphasizing that charge must always be conserved. Overall, the participants seek clarity on the nuances of particle interactions and definitions in physics.
Tangeton
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Okay so I have a problem with what my textbook is saying.

It defined pair production as a process in which a photon of electromagnetic energy is converted to a pair of particles.

But then it gave the discovery of an antiproton. Which was when a proton was accelerated to 6MeV and collided into a stationary proton, making one antiproton and 3 protons.

So, does this p + p --> p + p + antip process count as pair production? In which case, would the definition of pair production be a bit different?

Also, I found a website on which it says that ''sometimes, a pair of particles annihilates, but then the photon produces another pair of particles.'' When it comes the the first line, is this basically when a pair annihilates, but then the photon still has a lot of energy due to the extra kinetic energy of the particle and its antiparticle, meaning that when it passes close to a nucleus?

Edit: It also says that a antip + p = n + antin Is this besically an example of what I just wrote above (particles annihilate but then the photon produces another pair of particles)?

Sorry for lots of questions and thank you for your time.
 
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Tangeton said:
p + p --> p + p + antip

Does that conserve charge?
 
Vanadium 50 said:
Does that conserve charge?
oooo whoops it's p + p --> p + p + p + antip and well it doesn't look like it does conserve charge, but what has that got to do with anything? Says here it does happen so I am not trying to disprove it?
 
Last edited:
Tangeton said:
but what has that got to do with anything?

Charge is always conserved. If you write down a reaction that doesn't, it's wrong.
 
Vanadium 50 said:
Charge is always conserved. If you write down a reaction that doesn't, it's wrong.

Well yes I know but charge is conserved here. I just wrote the wrong equation by accident.
 

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