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An unknown particle with rest energy of $1 \text{ MeV}$ is traveling with a speed of $0.6c$ along the $x$-axis in our rest frame when it decays into two photons, also travelling along the $x$-axis. What are the energies of the photons in $\text{MeV}$?

I've been trying to learn special relativity recently and came across this question, but all I really know is basic length contraction and time dilation. Would solving this use similar concepts and how should I go about it?

physicsaficionado
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  • Length contraction and time dilation won't really help in this problem. As a first step [which will be refined later], how would you approach this using non-relativistic physics? What have you tried? – robphy May 25 '20 at 03:05
  • I think conservation of momentum would apply if thinking of non-relativistic physics, but since photons are massless I don't know how to approach the situation. – physicsaficionado May 25 '20 at 03:14
  • thinking in terms of four momentum vectors simplifies such problems http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html – anna v May 26 '20 at 03:54

2 Answers2

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You need to apply conservation of energy & momentum.

Based on your comment it seems the difficulty is that "photons are massless". They are indeed massless, but they can still carry momentum. Once you know the momenta of the photons, it's easy to convert to energy via $E = pc$.

Allure
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  • I see that momentum is equal to Planck's constant divided by the wavelength, but I do not know how to determine the wavelength. I know the photons travel at $c$, but I don't have their frequency to find wavelength. – physicsaficionado May 25 '20 at 03:42
  • @physicsaficionado there's also conservation of energy which you can use. – Allure May 25 '20 at 03:45
  • So if I use $E^2 = (mc^2)^2 +(pc)^2$ for the photons, it boils down to $E = pc$, but I do not have $p$. I don't know what else to do with the photons. As for the original particle, I have its rest energy, but also not its $E$ or $p$. I do not know where the original particle's speed belongs in this picture, but I think that's a key insight I am missing at the moment. – physicsaficionado May 25 '20 at 04:02
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    There are formulas for the initial massive particle’s energy and momentum in terms of its rest mass and velocity. See https://en.wikipedia.org/wiki/Mass_in_special_relativity#The_relativistic_energy-momentum_equation – G. Smith May 25 '20 at 04:18
  • I used these and got that the initial energy of the particle is $1.25mc^2$ and its initial momentum is $0.75mc^2$. I tried to relate this to the photons by setting that energy equal to $2pc$ because there are two photons, but now I am mixed up and do not know how to get the energy of the photons. – physicsaficionado May 25 '20 at 05:12
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    You are nearly there! The two photons do not have the same energy (in the lab). One goes forwards and one goes backwards. Their individual energies are the same as their momenta so, if your numbers are right, $p_1+p_2=1.25 mc$, $p_1-p_2=0.75 mc$. – RogerJBarlow May 25 '20 at 08:18
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You know that $m = 1 MeV/c^2$. Your calculation of the initial total energy of 1.25 MeV is correct. The initial momentum should be 0.75 MeV/c. You can now find 2 equations for the momenta of the 2 photons, one equation from conservation of energy, the other equation from conservation of momentum.

Note that energy is a positive scalar, but momentum is a vector. In this one dimensional problem, momentum can be positive or negative (or zero). So to calculate photon energy from momentum you need to use the magnitude of the momentum, that is $E=|p|c$.

In the rest frame of the initial particle, the momentum is obviously zero. So in that frame the sum of the photons' momenta must also be zero to conserve momentum.

Hopefully, you now have enough clues to solve this.

PM 2Ring
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