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Annihilation - matter/antimatter

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jnorman
#1
Jul2-09, 01:31 PM
P: 308
sorry for this elementary question. when an electron (matter) and a positron (anitmatter) collide, they annihilate, releasing high energy photons/gamma rays - correct?

so, when particle pairs are created within the quatum foam, which pop into and out of existence within the planck time period, where does the annihilation energy go?

and, antimatter is not the same as a negative energy particle, correct? what is a negative energy particle?

thanks.
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HallsofIvy
#2
Jul2-09, 02:20 PM
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PF Gold
P: 39,352
My understanding is that what happens is that the energy produced when two particles annihilate in the quantum foam is that it almost immediately goes into the production of two new particles. It is not a "given" that there exist "negative energy particles". I believe that they was originally postulated as "anti-particles" but I don't know what their status is now.
benk99nenm312
#3
Jul2-09, 02:24 PM
P: 302
I believe the energy you're talking about is what we refer to is vaccuum energy. The non-zero expectation value of energy in a vaccuum implies that even nothing has energy. If I'm wrong, please correct me.

However, I can tell you that antimatter is not the same as negative energy matter. Antimatter has positive mass, as I found out once :).

jnorman
#4
Jul2-09, 03:34 PM
P: 308
Annihilation - matter/antimatter

well, then where does the "negative energy particle" come from which supposedly accounts for BH evaporation?
DaveC426913
#5
Jul2-09, 04:15 PM
DaveC426913's Avatar
P: 15,319
Quote Quote by benk99nenm312 View Post
Antimatter has positive mass, as I found out once :).
This highly-intriguing and imagination-provoking statement (I'm picturing a Family Guy cut-away scene) will probably be much less intriguing if you actually explain it. Best entertainment value if you don't.
Bob S
#6
Jul2-09, 04:16 PM
P: 4,663
In the bubble diagram (virtual pair production) for photons, the total mass (energy) uncertainty is dE=2m0c2 (where m0 = electron rest mass.
The Uncertainty Principle allows a dE for a duration dt = hbar/dE.

If we write hbar = λbarm0c
where λbar is the reduced electron Compton wavelength, then

dt = hbar/dE = λbar/2c, or

c dt = λbar/2 is the length of the bubble = 1/2 reduced electron Compton wavelength = (1/2)3.86 x 10-11 cm.

Thus the dt is the time it takes for light to travel ~ (1/2 pi) x electron Compton wavelength.

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
− √ .
benk99nenm312
#7
Jul2-09, 04:30 PM
P: 302
Quote Quote by jnorman View Post
well, then where does the "negative energy particle" come from which supposedly accounts for BH evaporation?
The negative energy particle comes from 1 of 2 virtual particles in pair production. If a virtual pair of particles comes in, then one of them has negative energy in order to conserve the total energy of the pair (=0), I believe. The negative energy particle could be the antimatter or matter particle.

Where's George (Jones)?


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