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Are the Electron & Positron Created from a Photon Scattered at the Same Angles? 
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#1
Nov612, 02:44 PM

P: 22

Hello. Firstly, I hope this is the right subforum for this question. Here's a quick question that I have.
After an electronpositron pair production, do the angles that the paths of the electron and the positron make with the original path of the photon have to be equal? Another way to ask this is, do the velocities of the created electron and positron have to be equal? If they do have to be equal, I'd really like to know how this can be proved. Thanks. 


#2
Nov612, 07:50 PM

Sci Advisor
P: 5,451

try energy and momentumconservation



#3
Nov712, 07:15 AM

P: 22

Okay, let the photon's movement be in the +x direction. Then, if the resultant particles have different scattering angles with the horizontal, the one with the greater angle should have greater momentum, so that the sum of the ycomponents of their momenta will be 0.
Some of the photon's momentum in the +x direction will be transferred to a nucleus, so I'm not sure if we can express the conservation of the xcomponent of momentum with an equation here. We still have the energy conservation. Assuming the speeds are much smaller than the speed of light, we can use the nonrelativistic momentum formula [itex]\bf{P} = m\bf{v}[/itex]. So the relationship between the velocities will be linear ([itex]v_1sin(\theta1) = v_2sin(\theta2)[/itex]). Then we can just write [itex]\frac{1}{2}mv_1^2 + \frac{1}{2}m(v_1\frac{sin(\theta_1)}{sin(\theta_2)})^2 = hν2mc^2[/itex]. I'm just asking this becase I've heard some people say that the two angles have to be equal, but none of the things I have said above require that. Is there something I am missing? If they indeed do have to be equal, how would we show it? I'd be glad if anyone could explain it briefly or just point me to a relevant website. 


#4
Nov712, 12:24 PM

P: 107

Are the Electron & Positron Created from a Photon Scattered at the Same Angles?
They should be equal in magnitude but opposite in direction, because of symmetry. Note: that one of the products is particle while the other antiparticle plays no role here. What matters to this problem is the mass.



#5
Nov712, 05:11 PM

Sci Advisor
P: 5,451

The initial state is a photon γ and a target T. The final state is an electronpositron pair e^{} and e^{+}
The initial 4momenta are (with c=1) [tex]p^\mu_\gamma = (p_\gamma,\vec{p}_\gamma)[/tex] [tex]p^\mu_T = (M,0)[/tex] The final 4momenta are [tex]p^\mu_{e^} = (\sqrt{m^2+\vec{p}_{e^}^2},\vec{p}_{e^}) = (\sqrt{m^2+\vec{p}_{e^}^2},p_x,\vec{p}_\perp) [/tex] [tex]p^\mu_{e^+} = (\sqrt{m^2+\vec{p}_{e^+}^2},\vec{p}_{e^+}) = (\sqrt{m^2+\vec{p}_{e^+}^2},p_x,\vec{p}_\perp) [/tex] The y and zcomponents (perpendicular) add to zero b/c of momentum conservation. One immediately sees that the angles must be identical, and one can calculate this explicitly [tex]\cos\alpha_{\gamma,e^\pm} = \frac{\vec{p}_\gamma\,\vec{p}_{e^\pm}}{p_\gamma\,p_{e^\pm}} = \frac{p_\gamma\,p_x}{p_\gamma\,p_{e^\pm}} = \frac{p_x}{p_{e^\pm}} [/tex] 


#6
Nov812, 06:07 AM

P: 22




#7
Nov812, 09:34 AM

Sci Advisor
P: 5,451

The μ=0 component is the energy; the μ=1,2,3 components are the usual 3momentum; forget about the μ=0 component, you don't need it for the scattering angle



#8
Nov1012, 07:22 AM

P: 22




#9
Nov1012, 08:01 AM

Sci Advisor
P: 5,451

I see.
1) Start with different values p_{x} and p'_{x} which results in different energies (!) for electron and positron and try to find a contradiction with energymomentumconservation. or 2) Start in the c.o.m. frame with vanishing total momentum. The initial momenta (in x and x direction) are carried by photon and the target and add up to zero. So the total momentum in xdirection is zero. Now make a Lorentz transformation from c.o.m to rest frame of M; the Lorentz trf. introduces one velocity so both particles must have same momentum in x direction. 


#10
Nov1012, 08:59 AM

P: 22

That's what I was trying to do in my second post, but I couldn't come to the conclusion that the two angles have to be equal. I'll try to make it clearer what I did there so you can point out what I'm missing.
Say the photon is moving in the +x direction and the resultant particles move in the xy plane, for simplicity (so no zcomponent of momenta). Let's say the electron is scattered at an angle [itex]\theta_1[/itex] and the positron at [itex]\theta_2[/itex]. By the conservation of momentum in the y direction, [itex]\gamma_1mv_1sin\theta_1=\gamma_2mv_2sin\theta_2[/itex] [itex]\gamma_1v_1sin\theta_1=\gamma_2v_2sin\theta_2[/itex] The condition for the xcomponents of the momenta to be different is: [itex]\gamma_1mv_1cos\theta_1\neq\gamma_2mv_2cos\theta_2[/itex] [itex]\frac{\gamma_1v_1}{\gamma_2v_2}\neq\frac{cos\theta_2}{cos\theta_1}[/itex] which is satisfied as long as the two angles are different because: [itex]\frac{\gamma_1v_1}{\gamma_2v_2}=\frac{sin\theta_2}{sin\theta_1}\neq\fra c{cos\theta_2}{cos\theta_1}[/itex] Also, some of the momentum of the photon will be transferred to a nucleus, so the initial momentum of the photon will be greater that the sum of the xcomponents of the final momenta, but as far as I know, we can't write an exact equality for it: [itex]\frac{hν}{c}>\gamma_1mv_1+\gamma_2mv_2[/itex] [itex]hν>\gamma_1mv_1c+\gamma_2mv_2c[/itex] And for the energy conservation, the energy transferred to the nucleus will be insignificant compared to the momentum transferred, so: [itex]hν=\gamma_1mc^2+\gamma_2mc^2[/itex] (approximately) It's clear that [itex]hν=\gamma_1mc^2+\gamma_2mc^2>\gamma_1mv_1c+\gamma_2mv_2c[/itex] so there is no contradiction with what we've found using the energy conservation and the momentum conservation in the x direction. I don't see anything here that violates one of the conservation laws if the angles are not equal. It is even easier to see that it should be allowed if we let the speeds be much less than c, so [itex]\gamma_1=\gamma_2=1[/itex]. That way, there will be a linear relationship between the speeds, and you can easily determine the speeds for a given [itex]ν[/itex] and two angles you pick, as I explained in my second post. Now the CoM is moving in the +x direction with respect to the frame where M is at rest. Let's call this speed [itex]v[/itex]. Then using the Lorentz transformation, the speeds in the x direction [itex]u_1[/itex] (of the electron) and [itex]u_2[/itex] (of the positron) in the rest frame are: [itex]u_1=\frac{u'+v}{1+\frac{u'v}{c^2}}[/itex] [itex]u_2=\frac{u'+v}{1\frac{u'v}{c^2}}[/itex] which cannot be equal unless [itex]u'=0[/itex] or [itex]v=c[/itex]. [itex]u'=0[/itex] is the case where the particles' scattering angles are the same in the rest frame. 


#11
Nov1612, 04:15 PM

P: 22

Innocent bump.



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