Electron/Positron Annihilation

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In electron-positron annihilation, the energy of the resulting photons equals the total energy of the colliding particles, which is the sum of their rest mass and kinetic energy. The confusion arises from interpreting "each photon" as referring to a single photon rather than the two photons produced in the collision. Since momentum must be conserved, two photons are necessary to balance the system's momentum, especially when the particles collide head-on. The initial assumption of a 2:1 energy ratio for the photons is incorrect; instead, each photon carries energy derived from the total energy of both particles. Understanding this clarifies the energy distribution in the annihilation process.
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I recently read a textbook that stated without explanation "When a positron collides head on with an electron, the energy of each photon is the sum of one particle's rest and kinetic energy." So E = m_o*c^2 + K. However my question is, why isn't the energy of each photon twice the mass and kinetic energy of each particle since the ratio is 2:1? I know it's the incorrect answer, but that's just what I intuitively think. Any help?
 
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Von Neumann said:
I recently read a textbook that stated without explanation "When a positron collides head on with an electron, the energy of each photon is the sum of one particle's rest and kinetic energy." So E = m_o*c^2 + K. However my question is, why isn't the energy of each photon twice the mass and kinetic energy of each particle since the ratio is 2:1? I know it's the incorrect answer, but that's just what I intuitively think. Any help?

The total energy of the two photons is the same as the sum of the mass and kinetic energy of the colliding electron positron pair. Two photons, two particles. What's this 2:1 ratio of which you speak?
 
The way it's worded, to me it sounds as though one photon is created as the result of the pair colliding. Therefore the total kinetic energy of one photon being twice the energy of one colliding particle.
 
Von Neumann said:
The way it's worded, to me it sounds as though one photon is created as the result of the pair colliding. Therefore the total kinetic energy of one photon being twice the energy of one colliding particle.

One photon could not be created. There is zero momentum total if they collide head on. It couldn't conserve momentum and energy. You need two. I think "each photon" doesn't mean each single photon created by the collision. It means each of the two photons created by the collision.
 
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Oh right! Thanks!
 
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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