Particle Annihilation: Electron-Positron & Photon Energies

pyo
Messages
3
Reaction score
0
Hey i am doing some studying and when it comes to the subject of nuclear decays it says that when an electron and positron annihilate at rest they produce two photons with equal energys. Can someone try to explain to me why the photons have equal energies and how come there must be two photons produced instead of one in this case? Is the conservation of momentum the answer here?
 
Last edited:
Physics news on Phys.org
Yep, its 4-momentum
 
Yes, since the initial system (electron and positron at rest) had no momentum, the final system must also have no momentum. Clearly this means that the pair of photons must be emitted in opposite directions and with equal magnitude for their momenta. Now, how is the energy of a photon related to its momentum?
 
Great thanks !
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top