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Pair production  conservation of momentum VS conservation of energy 
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#1
Jan1313, 09:01 AM

P: 205

Allover the web i am only seeing a statement similar to this:
"Pair production is not possible in vaccum, 3rd particle is needed soWell noone out of many writers shows, how to prove this matematically. So this is what interests me here. First i wanted to know if pair production really cannot happen in vacuum, so i drew a picture and used equations for conservation of energy and conservation of momentum to calculate energy of a photon ##(h \nu)## needed for pair production. It turns out ##h\nu## is different if i calculate it out of conservation of energy or conservation of momentum. And even more! It can never be the same because equallity would mean parts ##v_1 \cos \alpha## and ##v_2 \cos \beta## should equall speed of light ##c##. Well that cannot happen. Below is my derivation. CONSERVATION OF ENERGY: $$ \scriptsize \begin{split} W_{1} &= W_{2}\\ W_f &= W_{e^} + W_{e^+}\\ h\nu &= W_{ke^} + W_{0e^} + W_{ke^+} + W_{0e^+}\\ h\nu &=\left[m_ec^2 \gamma(v_1)  m_ec^2\right] + m_ec^2 + \left[m_ec^2 \gamma(v_2)  m_ec^2\right] + m_ec^2\\ h\nu &=m_ec^2 \gamma(v_1) + m_ec^2 \gamma(v_2)\\ h\nu &=m_ec^2 \left[\gamma(v_1) + \gamma(v_2) \right]\\ \end{split} $$ CONSERVATION OF MOMENTUM: ##y## direction: $$ \scriptsize \begin{split} p_{1} &= p_{2}\\ 0 &= p_{e^} \sin \alpha  p_{e^+} \sin \beta \\ 0 &= m_e v_{1} \gamma(v_{1}) \sin \alpha  m_e v_{2} \gamma(v_{2}) \sin \beta\\ &\text{If $\boxed{\alpha = \beta} \Longrightarrow \boxed{v_1 = v_2}$ and:}\\ 0 &= 0 \end{split} $$ ##x## direction: $$ \scriptsize \begin{split} p_{1} &= p_{2}\\ \frac{h}{\lambda} &= p_{e^} \cos \alpha + p_{e^+} \cos \beta \\ \frac{h \nu}{c} &= m_e v_{1} \gamma(v_{1}) \cos \alpha + m_e v_{2} \gamma(v_{2}) \cos \beta\\ h \nu &= m_e c \Big[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Big] \end{split} $$ Alltogether: Because momentum in ##y## direction equals 0 (holds for some combinations of ##\alpha, \beta, v_1, v_2##) whole momentum equals just the momentum in ##x## direction. So if i add them i get: $$ \scriptsize h \nu = m_e c \Big[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Big] $$ From this i can conclude only that i cannot sucessfully apply conservation of energy and conservation of momentum at the same time and therefore pair production in vacuum cannot happen. QUESTION1: Why do writers state that 3rd particle is needed so that conservation of momentum holds? What if conservation of momentum holds and conservation of energy doesn't? How can we say which one holds and which one doesnt? QUESTION2: Do writters actually mean that if a 3rd particle is included we can achieve ##h \nu## to match in both cases? QUESTION3: Can someone show me mathematically how this is done? I mean it should right? 


#2
Jan1313, 10:01 AM

Sci Advisor
Thanks
P: 3,412

2) yes. 


#3
Jan1313, 11:59 AM

Sci Advisor
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P: 2,311

First of all you have to define which process you really mean. I don't understand you picture. What's "razpad"?
Usually under pair production one understands the process [tex]\gamma + \text{nucleus} \rightarrow e^+ + e^ +\text{nucleus}.[/tex] Of course both energy and momentum are conserved in this reaction. What you may have read is the correct statement that a singlephoton decay in the vacuum is impossible due to energymomentum conservation. 


#4
Jan1313, 12:26 PM

Sci Advisor
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P: 3,412

Pair production  conservation of momentum VS conservation of energy



#5
Jan1313, 01:13 PM

P: 205

$$ \begin{split} h\nu &=m_ec^2 \left[\gamma(v_1) + \gamma(v_2) \right]\\ h \nu &= m_e c \Big[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Big] \end{split} $$ 


#6
Jan1313, 01:29 PM

P: 205




#7
Jan1313, 02:05 PM

Mentor
P: 11,573

Your inequality shows that the process is not possible without a nucleus.
You can look at the timereversed process, too: With electron+positron, there is an inertial system with a fixed center of mass. This is not true for a single photon. With a nucleus, it is a bit lengthy to calculate it in an analytic way. The general idea is that a nucleus is heavy compared to the electrons  it can gain significant momentum with little energy (##E \approx \frac{p^2}{2m}## (edit: fixed prefactor)). Therefore, you are "allowed" to ignore momentum conservation if you consider photons and electron/positron only: the nucleus will get the momentum difference. 


#8
Jan1313, 02:38 PM

P: 205

If we use center of mass system we get ##p_{1} = p_{2} = 0## out of conservation of momentum. I can't help myself with this but could i help myself with a conservation of energy? 


#9
Jan1313, 03:23 PM

Mentor
P: 11,573

Without proton (=as you calculated), it cannot be done. There is a similar effect in classical collisions: For elastic collisions, you have both energy and momentum conservation. For inelastic collisions, energy is still conserved (of course, it is a very fundamental law), but a part of the energy can go to heat or other forms of energy you don't care about. Therefore, for (perfectly) inelastic collisions, energy conservation is ignored, and momentum conservation only is considered. Let's look at an actual example with a nucleus: We have an incoming photon with an energy of 1277 keV and a momentum of 1277 keV (using c=1), approacing a ^{208}Pb nucleus (m=194 GeV) and performing pair production there. Assume that the angle between electron and positron is 0 (easier to calculate): Electron+positron fly in xdirection with v=0.6c (giving gamma=1.25). Their summed energy is 2.5*511keV = 1277 keV, and their summed momentum is 1277 keV * 0.6 = 766.5 keV (with c=1). The ^{208}Pb nucleus (m=194 GeV) moves in xdirection with a momentum of 510.5 keV and a kinetic energy of 0.00067 keV. As you can see, its energy is neglibile  you could dump basically every amount of momentum you like into it, and it would not make a difference. Therefore, energy conservation is relevant, momentum conservation is not (unless you want to get the velocity of the nucleus after the collision). 


#10
Jan1313, 03:40 PM

P: 205

Thank you soooo much!
There is only one more thing. Does anyone know of a good center of mass = center of energy system animation or. good picture for pair production. 


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