What are the energies of the photons in a meson decay?

[stvn]

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

$E=m_0c^2$
$E=hf$
$λ'(\pi) = λ(\frac{(1+\beta)}{(1-\beta)})^{0.5}$
$λ'(0) = λ(\frac{(1-\beta)}{(1+\beta)})^{0.5}$

The Attempt at a Solution

I obtained the speed of the meson from:

$\gamma = (135 + 1000) / 135 = 8.407$
To give beta = 0.9929

Then I have used the beta formaulae to obtain:

$λ'(\pi) = λ \times 16.755$
$λ'(0) = λ \times 0.05968$

I think the ratio of these should somehow link to the ratio of the energies but I am not sure how!

Many Thanks

 

[vela]

In the rest frame of the meson, what is the energy of the photons?

 

[stvn]

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?

 

[vela]

Your analysis isn't quite correct. You got the conservation of energy right. What does conservation of momentum imply?

Once you have figured out what happens in the center-of-mass frame, you can Lorentz-transform those results to the lab frame.

 

[stvn]

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?

 

[vela]

Remember that momentum is a vector.

 

[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 the photons which is hf_1 + (-hf_2) since the photons are traveling in the + and - direction on the x axis:
So we have 0 = hf_1 - hf_2 or 0 = h(f_1 - f_2).

This implies f_1 = f_2 ie the frequencies of the photons are the same.

Also energy is conserved (in the meson frame) during the collision. Before the energy is 135Mev (the meson is at rest in this frame) After the decay the energy is the sum of the photon energies which is hf_1 + hf_2 (note energy is a scalar).

Now f_1 = f_2 so we have E = 2hf or f = 135mev/2h = 1.63x10^22 Hz. This is the frequency of the photons (gamma rays!) 1/1.63x10^2 = 6.135x10^-23 This is the wavelength of the photons.

We can use the doppler formulae to obtain the wavelengths in the lab frame:

lambda(pi) = 6.135x10^-23 x [(1+0.9929)/(1-0.9929)]^0.5 = 1.0278x10^-21
lambda(0) = 6.135x10^-23 x [(1-0.9929)/(1+0.9929)]^0.5 = 3.6619x10^-24

These are the wavelengths in the lab frame.

E = h/lambda so:
E of photon 1 = 6.626x10^-34 / 1.0278x10^-21 = 6.447x10^-13 Joules
E of photon 2 = 6.626x10^-34 / 3.6619x10^-24 = 1.809x10^-10 Joules

To convert to eV:
6.447x10^-13 / 1.6 x10^-19 = 4029375 = 4 MeV
1.809x10^-10 / 1.6 x10^-19 = 1130625000 = 1131 MeV

Thanks for your help again!