- #1

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**4**m0c^2? Thanks.

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- Thread starter coregis
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- #2

CarlB

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If you're talking about photon to electron/positrons, then the reason is because electrons and positrons are always produced in pairs (conservation of electric charge, for instance). A triplet production includes one electron/positron that has so little momentum that it is missed.

The Feynman diagrams on page 4 of this paper will show what is going on:

http://arxiv.org/PS_cache/hep-ph/pdf/9909/9909323.pdf [Broken]

Carl

The Feynman diagrams on page 4 of this paper will show what is going on:

http://arxiv.org/PS_cache/hep-ph/pdf/9909/9909323.pdf [Broken]

Carl

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- #3

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m(e)c2+hv=m(b+)c2+m(b-)c2+E(kinetic energy of participating electron on left side of equation)

Now assume that elctron was at rest so m(e)c2=_m(0)c2 and E= mc2 and also they (triplets)are more probabilistic to take equal E(K)

so above eq becomes

m(0)c2+hv=3mc2

and per conservation of momentum we have

hv/c=3mv=3mBc (B=v/c)

solving above 2 eq in B gives

(1-B2)1/2+3B=3 and that gives B=4/5

=give m=m(0)*5/3

now putting m value in momentum eq we get

hv=3*5/3*m(0)*4/5*c2=4m(0)c2

And this is due to conservation of momentum of photon as without third partical either e or nucleus this process is impossible as photon momentum is not conserved and as elctron rest mass is small it need more energy than pair production do, to physically this process to occur.

now i suppose that it will clear the threshold requirment

- #4

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how did you find the value of B, i mean please show how B=4/5. I couldnt understand that step!

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