vertices said:why can't a gamma ray photon -->positron+electron? My notes simply say this would violate momentum conservation (you need a recoiling nucleus), but why? I mean, the photon WOULD have momentum (=E/c), so the sum of momentums of the positron and electron would just have be E/c..
vertices said:"But it's impossible for a single photon to have zero three-momentum"
well, the photon COULD have zero three momentum if the CMF was traveling at a speed 'c'. I take it this isn't allowed? My knowledge of SR isn't amazing shall we say:(
thanks for your reply, again nrqed.
vertices said:just did a little recap on SR:
from wikipedia:
# The Principle of Invariant Light Speed - Light in vacuum propagates with the speed c (a fixed constant) in terms of any system of inertial coordinates, regardless of the state of motion of the light source.
SO you can pick a reference frame in which the photon IS at rest, thus having ZERO momentum (ie. traveling at c relative to the lab frame, if you will). This IS also the CMF for e+ and e-.
What am I not understanding clearly?
Ah, ok.vertices said:yeah I totally understand what you've said and it makes sense intuitivly.
Well, you can define a frame traveling at c. It's just irrelevant to the question you asked here.I was just wondering why you can't have a frame traveling AT c (probably a silly question)
nrqed said:As long as you don't ask questions like "If I am in that frame, what will the photon look like" and so on. So yes, you can define a frame moving at c but there is nothing you can do with it, no application, no gedanken experiment. So it's pretty much useless.
vertices said:just did a little recap on SR:
from wikipedia:
# The Principle of Invariant Light Speed - Light in vacuum propagates with the speed c (a fixed constant) in terms of any system of inertial coordinates, regardless of the state of motion of the light source.
SO you can pick a reference frame in which the photon IS at rest, thus having ZERO momentum (ie. traveling at c relative to the lab frame, if you will). This IS also the CMF for e+ and e-.
What am I not understanding clearly?
Without worrying about frames, E^2-p^2 is an invariant that must be the same before and after an interaction. E^2-p^2>0 for the electron+positron, while E^2-p^2=0 for a photon.vertices said:why can't a gamma ray photon -->positron+electron? My notes simply say this would violate momentum conservation (you need a recoiling nucleus), but why? I mean, the photon WOULD have momentum (=E/c), so the sum of momentums of the positron and electron would just have be E/c..
Pair production is a phenomenon in which a photon of sufficient energy can spontaneously create an electron and a positron (antiparticle of the electron) in the presence of a nucleus or an electric field.
This is due to the conservation of energy and momentum. A single photon does not have enough energy to create a massive particle such as an electron and a positron.
The minimum energy required for pair production is equivalent to the combined rest mass energy of the electron and the positron, which is approximately 1.02 MeV (million electron volts).
The excess energy is converted into the kinetic energy of the created electron and positron. This kinetic energy can be calculated using the conservation of energy and momentum equations.
Yes, pair production can occur in a vacuum. However, it requires the presence of a nucleus or an electric field to conserve energy and momentum. In the absence of these, the created electron and positron will quickly annihilate each other, producing two photons with the same energy as the original photon.