Pion decays into gamma rays. Find the energy and angle

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SUMMARY

The discussion centers on the decay of a pion with a rest energy of 135 MeV into two gamma rays while moving at a velocity of 0.97c. The user applies the principles of conservation of relativistic energy and momentum, calculating the Lorentz factor (γ) as approximately 4.11. This leads to an energy calculation for the gamma rays of about 555.32 MeV. The user is advised to switch to the pion's rest frame to simplify the analysis of the decay process.

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  • Basic calculations involving relativistic mass-energy equivalence (E=γmc²).
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Hi all,

I was having trouble with this problem and hoping that someone could help me with it.

A pion has a rest energy of 135 MeV. It decays into two gamma rays, bursts of electro magnetic radiation that travel at the speed of light. A pion moving through the laboratory at v = 0.97c decays into two gamma rays of equal energies, making equal angles θ with the direction of motion. Find the angle θ and the energies of the two gamma rays. (Hint: gamma rays are electromagnetic radiation with E = pc.)

I am pretty sure I will be using conservation of relativistic energy and momentum but I am still confused.

I know that E=γmc^2 and P=γmv

I find gamma 1/sqrt(1-(0.97c)^2/c^2)≈ 4.11

I know that mc^2= 135MeV

With this I get E=(4.11)(135MeV)≈ 555.32MeV

For P I get (4.11)m(0.97c)=3.99mc≈ 538.66MeV/c

I am a bit unsure what to do next with this.
 
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This post should be moved to "Special & General Relativity" subforum. It's a simple exercise in special relativity, and doesn't really require knowledge of particle physics.
 
Boost yourself into the pion's rest frame, then back again afterwards?
 

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