What is the energy of photons in pion decay at rest?

In summary, the conversation discusses the decay of a pion at rest and the calculation of the energy and momentum of the resulting photons using conservation of energy and momentum. It is noted that the photon has zero rest mass and the energy equation reduces to E^2 = p^2c^2. The question of how to calculate gamma without given velocities is raised, and it is clarified that the initial velocity of the pion is zero. The momentum of the pion is also discussed and it is concluded that the initial momentum is zero but not the initial energy. The conversation ends with a thank you and the acknowledgement that the pointers were helpful.
  • #1
rwooduk
762
59
Apologies, this is going to be a bit disjointed, I don't want to write the full question down as I don't want anyone to give me a solution as its an assignment question.

1. am i correct in assuming when a pion decays at rest its energy will be given by:

E^2 = p^2 c^2 + m^2 c^4

which goes to:

E^2 = m^2 c^4

because I get something ridiculous like 1394J. I've canceled the c^2 of the pion mass with the c^4 in the energy eqn.

2. to find the energy of the photons i use conservation of energy:

E(pion) = E(photon1) + E(photon2)

E(photon1) = GAMMA * m c^2

i) what do i use for the mass of a photon?
ii) how do i calculate gamma when no velocity of the photons (or the pion) is given?

thanks in advance for any suggestions.
 
Physics news on Phys.org
  • #2
The photon has a rest mass of zero, so the energy equation reduces to:

E^2 = p^2 c^2, or E = pc.

The equation E = gamma m c^2 applies only for a massive particle.

So take the equation E = pc, then conserve energy and momentum to calculate the energy of the two photons.
 
  • #3
phyzguy said:
The photon has a rest mass of zero, so the energy equation reduces to:

E^2 = p^2 c^2, or E = pc.

The equation E = gamma m c^2 applies only for a massive particle.

So take the equation E = pc, then conserve energy and momentum to calculate the energy of the two photons.

thanks for the reply!

but using conservation of energy E(pion) = E(photon1) + E(photon2) doesn't seem to help because although we now have E(pion) the energies of each photon may not be equal. also how would you calculate momentum without a velocity?

thanks again.

btw the question says to assume relativistic, so I am back to the problem of how to find gamma with no velocities given.
 
  • #4
What is the momentum of the initial pion at rest? What does this tell you about the momentum of the final system?

Also on your question about momentum, for the photon E = pc. What is p?
 
Last edited:
  • #5
phyzguy said:
What is the momentum of the initial pion at rest? What does this tell you about the momentum of the final system?

Also on your question about momentum, for the photon E = pc. What is p?

i) the momentum of the pion is zero at rest, so it's energy would be zero, so you would have to assume that it had a momentum before it came to rest and that it is the dissipation of this momentum that is 'given' to the photons?

how else could it impart energy if it had none at the point of decay?

unless it has an intrinsic energy inherent to the particle itself, would that be in the form of mass?

so E = GAMMA * mc^2

so the momentum of the final system would be only the energy its mass imparts

so E= GAMMA * mc^2 = pc

but again I am stuck with GAMMA when i have no given velocities.
 
  • #6
OK, think I am there,

E(pion) = GAMMA * mc^2

COE:

E(photon) = 1/2 E(pion)

p(photon) = E(photon) / c

Only thing is how to find GAMMA with no velocities given? should i just write the answer in terms of GAMMA?

I see that K = (GAMMA - 1) mc^2 is the kinetic energy, can you assume this is equal to its rest mass, hence find GAMMA this way, or does it have to be moving for this to be true?

Thanks for the direction!
 
Last edited:
  • #7
rwooduk said:
i) the momentum of the pion is zero at rest, so it's energy would be zero, so you would have to assume that it had a momentum before it came to rest and that it is the dissipation of this momentum that is 'given' to the photons?

No. The pion has a mass m. E^2 = p^2c^2 + m^2c^4. If p = 0, then E = mc^2.

Also, as you wrote, E = gamma mc^2. If p=0, v=0, so gamma = 1, giving again E = mc^2.

So the initial momentum is zero, but not the initial energy.
 
  • Like
Likes 1 person
  • #8
rwooduk said:
Only thing is how to find GAMMA with no velocities given? should i just write the answer in terms of GAMMA?

Thanks for the direction!

You are given the initial velocity. It says at rest.
 
  • Like
Likes 1 person
  • #9
ahhh i see and you can calculate the momentum of the photon from COE because we know the energy of the pion.

thanks for all the pointers, really appreciated!
 

1. What is pion decay of two photons?

Pion decay of two photons is a process in which a neutral pion particle decays into two photons (particles of light).

2. How does pion decay of two photons occur?

Pion decay of two photons occurs through the strong nuclear force, which is responsible for holding protons and neutrons together in an atom's nucleus. The neutral pion particle is made up of a quark and an antiquark, and when these two particles interact with each other, they produce two photons.

3. What is the significance of pion decay of two photons in particle physics?

Pion decay of two photons is important in particle physics because it provides evidence for the existence of the strong nuclear force, which is one of the four fundamental forces in nature. It also helps scientists understand the structure of matter and the interactions between subatomic particles.

4. Can pion decay of two photons be observed in nature?

Yes, pion decay of two photons has been observed in nature through experiments conducted at particle accelerators. These experiments involve colliding particles at high energies and then studying the particles that are produced as a result.

5. How is pion decay of two photons relevant to everyday life?

Pion decay of two photons may not have direct relevance to everyday life, but it is an important process to study for understanding the fundamental laws of the universe. The technology and knowledge gained from studying pion decay of two photons can also have practical applications in fields such as medicine and energy production.

Similar threads

  • Advanced Physics Homework Help
Replies
4
Views
449
  • Advanced Physics Homework Help
Replies
6
Views
3K
Replies
10
Views
3K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
667
  • Advanced Physics Homework Help
Replies
6
Views
7K
  • Advanced Physics Homework Help
Replies
6
Views
2K
Back
Top