Pair production

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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..
 

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  • #2
nrqed
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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..
there are several to prove that. The most intuitive and quick way is to consider the center of mass frame of the electron-positron pair. In that frame, the total three-momentum is zero. But it's impossible for a single photon to have zero three-momentum. Therefore the reaction is impossible, in that frame three-momentum can't be conserved.
 
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"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 travelling 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.
 
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nrqed
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"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 travelling 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.
No problem.

The point is that if you have two massive particles, it is always possible to find a physical center of mass frame (in the sense that it is moving at a speed less than c so you can actually boost yourself to that frame).
 
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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. travelling 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?
 
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nrqed
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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. travelling 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?
The point is that if you have pair production, you can find a physical frame (moving below the speed of light) where the total three-momentum of the electron and positron is zero.

If you have a single photon, you can NOT find a frame where the three-momentum of the single photon is zero. I mean a physical frame, one which travels at a speed smaller than c.

Therefore, it's impossible for for a single photon to convert into an electron-positron pair.
QED

The point is to show that a single photon cannot convert into an electron positron pair , right?
 
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yeah I totally understand what you've said and it makes sense intuitivly.

I was just wondering why you can't have a frame travelling AT c (probably a silly question)
 
  • #8
nrqed
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yeah I totally understand what you've said and it makes sense intuitivly.
Ah, ok.
I was just wondering why you can't have a frame travelling AT c (probably a silly question)
Well, you can define a frame travelling at c. It's just irrelevant to the question you asked here.

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.
 
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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.
i see what you're talking about:)
 
  • #10
nrqed
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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. travelling 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?

Iknow you said that you had understood my point but the second paragraph seemed to say that you had not. My point was that the CMF of the e+e- pair was NOT a frame travelling at c.


Just to make sure things are clear.

Regards
 
  • #11
pam
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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..
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.
 
  • #12
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thanks pam - that's a good explanation.
 

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