How Do Photon Energies Result from a Decaying Pi0 Meson in Flight?

[joker_900]

Homework Statement

5. A pi0 meson whose rest mass is 135 MeV/c^2 is moving with a kinetic energy of 1 GeV.
It decays in flight into two photons whose paths are along the direction of motion of the
meson. Find the energies of the two photons.
[Ans: 4 and 1131 MeV.]

Homework Equations

The Attempt at a Solution

Working in the CM frame (the rest frame of the pion) and conserving energy in that frame:

mc^2 = 2E*, where E* are the energies of the photons in the CM frame


I then thought I could use a lorentz transformation Ea=gamma(E* + betaE*) where Ea is the energy of one photon in lab frame and gamma and beta are the gamma and beta for the pion moving in lab frame, with gamma = (energy of pion)/mc^2 and beta = (pc)/E

But first i was checking something, and i cannot get the simple conservation of energy to work with the answers given. So in the lab frame, i thought the energy of the pion must equal the sum of the energies of the photons, given as 1135? And the energy of the pion

= sqrt[(pc)^2 + m^2c^4]
i thought this
=sqrt[1000^2 + 135^2]
as the kinetic energy is given as 1GeV=1000MeV and the mass is given in units of MeV/c^2. Clearly I'm misunderstanding something, but what?!? Will my method work once this misunderstanding is resolved (and is there a better method or different method just for future reference)?

Thanks

 

[anathema]

in advance for your help!

Hello! Your approach is correct, but there is a small error in your calculation. The energy of the pion should be sqrt[(pc)^2 + m^2c^4], which is equal to sqrt[(1000 MeV)^2 + (135 MeV/c^2)^2] = 1000.017 MeV. This small difference is due to rounding errors.

Using this value for the energy of the pion, you can then proceed with your calculation using the Lorentz transformation. Your method will work once this small error is corrected.

Another method to solve this problem is to use conservation of momentum in the lab frame. Since the photons are traveling in the same direction as the pion, the total momentum before and after the decay should be the same. This means that the sum of the momentums of the two photons should be equal to the momentum of the pion before the decay.

Using the formula p = gamma*m*v, where gamma is the Lorentz factor and v is the velocity, we can calculate the momentum of the pion before the decay as p = (gamma*m*v)c = (1000 MeV/135 MeV/c^2)*(135 MeV/c^2)*(3x10^8 m/s) = 1000 MeV/c.

Since the momentums of the photons should add up to 1000 MeV/c, we can set up the following equation:

p1 + p2 = 1000 MeV/c

Using the formula E = pc, we can rewrite this as:

E1/c + E2/c = 1000 MeV/c

Now we can use the fact that the energy of a photon is given by E = hf, where h is Planck's constant and f is the frequency. Since the photons are traveling in the same direction as the pion, they will have the same frequency and therefore the same energy.

Substituting this into our equation, we get:

2E/c = 1000 MeV/c

Solving for E, we get E = 500 MeV.

Since the energy of each photon is 500 MeV, this means that the energies of the two photons are 500 MeV and 500 MeV, which adds up to 1000 MeV, as expected.

I hope this helps! Let me know if you have

 

[ogb p]

for your question! It's always great to see students engaging with challenging physics problems.

Your method is definitely on the right track, but there are a few things that need to be clarified.

First, let's start with the conservation of energy in the CM frame. As you correctly stated, the total energy in the CM frame is equal to twice the energy of one of the photons, since we are assuming that the two photons are moving in opposite directions. So we have:

2E* = mc^2

where E* is the energy of one photon in the CM frame and mc^2 is the rest energy of the pion.

Next, we need to transform this energy to the lab frame. You were on the right track with the Lorentz transformation, but there are a couple of things to keep in mind. First, the energy of the pion in the lab frame is not just its kinetic energy plus its rest energy. The total energy (i.e. the Lorentz-invariant quantity) is given by:

E = gamma(mc^2 + pc)

where gamma is the Lorentz factor and p is the momentum of the pion in the lab frame. Note that this is not the same as the kinetic energy, which is given by:

K = gamma(mc^2 - mc^2)

So, in order to find the energy of the pion in the lab frame, we need to use the full expression for the total energy, not just the kinetic energy.

Second, remember that the Lorentz transformation for energy is given by:

Ea = gamma(E* + betaE*)

where Ea is the energy of the photon in the lab frame, E* is the energy of the photon in the CM frame, and beta is the velocity of the pion in the lab frame, divided by the speed of light. Note that this is different from the expression you wrote, where you have beta multiplied by E*. So, using the correct expression for the Lorentz transformation, we have:

Ea = gamma(E* + betaE*) = gamma(2E*) = mc^2

where we have used the fact that 2E* = mc^2 from the conservation of energy in the CM frame.

Now, we can solve for Ea to find the energy of one of the photons in the lab frame:

Ea = mc^2/gamma = mc^2/sqrt(1 - (v/c)^2)