Why Must Electron and Positron Have Equal Angles in Annihilation?

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In electron-positron annihilation, the angles of emission must be equal due to conservation of momentum. When an electron and positron collide, they produce photons, and the momentum before and after the collision must remain balanced. Since both particles have the same mass, their velocities and angles of emission are inherently linked to maintain this balance. The requirement for equal angles ensures that the total momentum vector remains constant. Understanding this principle is crucial for grasping the dynamics of particle interactions in quantum theory.
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Hi everyone,
I'm currently studying a bit of quantum theory and I came across this paragraph: When an electron collides with a positron, it moves with an angle theta_1 with the normal. Ditto for the positron, except that it has an angle theta_2. But the two angles must be equal. Why? Conservation of momentum.
Well, that was all the explanation given, and I don't understand the reasoning behind it, so could someone explain it to me? It would be very helpful during my self-study. (See the attached diagram)

Thank you!
 

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I can't formulate it for you mathematically, but since they have the same mass, wouldn't you EXPECT them to have the same angle?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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