Recent content by stvn

Ah yes thanks very much for your pointers - here is the rest of the solution for anyone else who needs some help: The x component of momentum (in the meson frame) before the decay is zero. The x component of momentum (in the meson frame) after the decay is equal to the sum of the momentums of...

Ah so in the frame moving with the meson the momentum is zero before the decay. After the decay momentum is h/λ_1 + h/λ_2. So this implies λ_1 = -λ_2, this doesn't seem right?

Sry for delay! So the energy of the photons would be E=hf_1 and E=hf_2, now I am assuming that f_1 does not have to equal to f_2, but that hf_1 + hf_2 = 135eV as per conservation of energy in this frame. I am not sure what to do next, maybe do the same in the lab frame?

Homework Statement A meson with rest mass 135 MeV has kinetic energy 1 GeV. It decays into two photons. 1 photon moves in the direction of motion, 1 is in the opposite. What are the energies of the photons? Homework Equations E=m_0c^2 E=hf λ'(\pi) = λ(\frac{(1+\beta)}{(1-\beta)})^{0.5} λ'(0)...