Conservation Laws and Symmetries Particle Physics

[highwayman1739]

2) A π0 of kinetic energy 350 MeV decays in flight into 2 γ rays of equal energies. Determine the angles of the γ rays from the incident π0 direction.

Not sure where I am going wrong but my answer is not correct.

Energy of π0 meson
E = 350 MeV

Rest mass Energy of π0 meson
E0 = 135 MeV

Initial momentum of π0 meson
P = √E^2 – Eo^2 / c
P = √(350 MeV)^2 – (135 MeV)^2 / c
P = 322.9163978 MeV/c

Law of conservation of Energy
E = Egamma + Egamma
Egamma = E/2
Egamma = 350MeV/2
Egamma = 175 MeV

Angle of gamma ray
θ = cos-1(Pc / 2E)
θ = cos-1(322.9163978 MeV/c) c / (2)(175 MeV)
θ = cos-1(0.922618279)
θ = 22.688 degrees

 

[collinsmark]

2) A π0 of kinetic energy 350 MeV decays in flight into 2 γ rays of equal energies. Determine the angles of the γ rays from the incident π0 direction.

Not sure where I am going wrong but my answer is not correct.

Energy of π0 meson
E = 350 MeV

Rest mass Energy of π0 meson
E0 = 135 MeV

According to the problem statement, the kinetic energy is 350 MeV. The total energy, E, is the kinetic energy plus the rest mass energy.

Initial momentum of π0 meson
P = √E^2 – Eo^2 / c

It wouldn't hurt to use parentheses or brackets in your notation.

p = √(E^2 – Eo^2)/ c

or

p = √(E² – E_o²)/ c

or

[itex] p = \frac{\sqrt{E^2 - E_0^2}}{c} [/itex]

P = √(350 MeV)^2 – (135 MeV)^2 / c

Here you need to add the rest mass energy to the kinetic energy to find the total energy, E.

 

[highwayman1739]

If I add kinetic energy and rest energy (350+135 = 485 MeV) and use this value for E my momentum seems to be to large (p = 465.83). Unless I am making another mistake in the part below.
θ = cos-1(Pc / 2E)
θ = cos-1(465.83 MeV/c) c / (2)(175 MeV)
θ = cos-1(1.33)(which cannot be )

 

[collinsmark]

If I add kinetic energy and rest energy (350+135 = 485 MeV) and use this value for E my momentum seems to be to large (p = 465.83).

So far that looks right to me. Specifically, I got 465.83 MeV/c.

Unless I am making another mistake in the part below.
θ = cos-1(Pc / 2E)

That's almost right, but you're missing an additional factor of 2 in there somewhere. I see you've put one '2' in there already, presumably because each photon's contributes half of the total momentum in the x direction. But also remember that each photon contributes half the total energy too. I only see one of the factor of '2's in there.

----

There is another way to do this, which is the long way. Use the formula E = hc/λ, and solve the the wavelength of each photon (remember that each photon only contributes half the total energy. Then realize |p| = h/λ, and that the x-component of each photon's momentum, p_x = pcosθ = hcosθ/λ, contributes to half the total momentum of the original pion. (Don't worry about the momentums in the y-direction, they are equal and opposite, so they cancel.)

You'll end up with a cosθ = ratio, where the "ratio" is guaranteed to be be less than 1 (assuming the original pion had non-zero rest mass, which of course is true).

[Edit: corrected a mistake.]

 

[Nickie]

The angles of the γ rays from the incident π0 direction are 22.688 degrees and 157.312 degrees. This is because the decay of the π0 meson into two γ rays is a symmetric process, where the two γ rays have equal energies and are emitted in opposite directions from the initial direction of the π0 meson. This conservation of energy and symmetry can be explained by the laws of conservation of energy and momentum, which are fundamental principles in particle physics. The angle of 22.688 degrees is the angle between one of the γ rays and the initial direction of the π0 meson, and the angle of 157.312 degrees is the angle between the other γ ray and the initial direction. This result is in accordance with the laws of conservation of energy and momentum, and demonstrates the importance of these laws in understanding the behavior of particles in particle physics experiments.