# Annihilation: calculation of photon energies

• Frostman
In summary, the conversation discusses the conservation of quadri-momenta in the problem of two particles annihilating and producing two photons. The speaker considers the two photons as a single particle with energy equal to twice the energy of a single photon. However, the other person points out that this assumption is invalid as the two photons must have different energies and be moving in opposite directions. The conversation then moves on to discuss finding the energies in the reference system integral with the incident particle, and the speaker suggests using Lorentz transformations to do so.
Frostman
Homework Statement
An antiparticle of mass ##m## and energy ##E## hits a stationary particle in the laboratory and the system annihilates itself producing two photons, which are directed along the direction of the motion of the incident antiparticle.
1 - Calculate the energies of the two photons in the laboratory's reference system.
2 - Calculate the energies in the reference system integral with the incident particle, explicitly carrying out the Lorentz transformation. Discuss the result obtained.
Relevant Equations
I set up this problem this way:

##p_a^{\mu}=(E, \sqrt{E^2-m^2}, 0, 0)##
##p_b^{\mu}=(m, 0, 0, 0)##
##p_c^{\mu}=(2E_\gamma, 2E_\gamma, 0, 0)##

I have chosen to consider the two photons as a single particle of energy equal to ##2E_\gamma##. At this point I applied conservation of the quadri-momenta:

##E+m=2E_\gamma##
##\sqrt{E^2-m^2}=2E_\gamma##

From the first equation I find the first point: ##E_\gamma=\frac {E+m}2##. But if I put this in the second equation I don't have the equality. Why?
I'm supposing that the particle ##a## is a ##e^+## and particle ##b## is an ##e^-##, because the statement said that the system annihilates itself.

Frostman said:
I have chosen to consider the two photons as a single particle of energy equal to ##2E_\gamma##.
How can this possibly be justified?

PeroK said:
How can this possibly be justified?
Because they are the same particle. I could write:
##p_{\gamma 1}^\mu=(E_\gamma, E_\gamma, 0,0)##
##p_{\gamma 2}^\mu=(E_\gamma, E_\gamma, 0,0)##

PeroK
Frostman said:
Because they are the same particle. I could write:
##p_{\gamma 1}^\mu=(E_\gamma, E_\gamma, 0,0)##
##p_{\gamma 2}^\mu=(E_\gamma, E_\gamma, 0,0)##
This is nonsense. 1) They must be moving in opposite directions and 2) They must have different energies.

PeroK said:
This is nonsense. 1) They must be moving in opposite directions and 2) They must have different energies.
For
...which are directed along the direction of the motion of the incident antiparticle...
I was supposing this situation in LAB-frame:

So then, they have the same direction, but opposite verse and different module?

Frostman said:
For

I was supposing this situation in LAB-frame:
View attachment 276393
So then, they have the same direction, but opposite verse and different module?
They must be in opposite directions in all IRF's, as they must be in opposite directions in the CM frame.

Frostman
I understand. I had to start thinking from the CM how the situation should be and then think about how the situation changed in the different IRF's.
In this case so I should start from:

##p_a^{\mu}=(E, \sqrt{E^2-m^2}, 0, 0)##
##p_b^{\mu}=(m, 0, 0, 0)##
##p_c^{\mu}=(E_{\gamma 1}, E_{\gamma 1}, 0, 0)##
##p_d^{\mu}=(E_{\gamma 2}, - E_{\gamma 2}, 0, 0)##

Considering the conservation of the quadrimoment, moving the term ##p_c^\mu## to the first member and squaring the members I obtain:

##p_a^{\mu}+p_b^{\mu}=p_c^{\mu}+p_d^{\mu}##
##(p_a^{\mu}+p_b^{\mu}-p_c^{\mu})^2=(p_d^{\mu})^2##
##p_a^2+p_b^2+p_c^2+2p_a^\mu p_{\mu b}-2p_a^\mu p_{\mu c}-2p_b^\mu p_{\mu c}=0##
##m^2+m^2+2Em-2(EE_{\gamma 1}-E_{\gamma1}\sqrt{E^2-m^2})-2mE_{\gamma 1}##

##E_{\gamma 1}=\frac{m(E+m)}{E+m-\sqrt{E^2-m^2}}=\frac{m}{1-\sqrt{\frac{E-m}{E+m}}}##

Similarly for ##E_{\gamma 2}##.

Frostman said:
I understand. I had to start thinking from the CM how the situation should be and then think about how the situation changed in the different IRF's.
In this case so I should start from:

##p_a^{\mu}=(E, \sqrt{E^2-m^2}, 0, 0)##
##p_b^{\mu}=(m, 0, 0, 0)##
##p_c^{\mu}=(E_{\gamma 1}, E_{\gamma 1}, 0, 0)##
##p_d^{\mu}=(E_{\gamma 2}, - E_{\gamma 2}, 0, 0)##

Considering the conservation of the quadrimoment, moving the term ##p_c^\mu## to the first member and squaring the members I obtain:

##p_a^{\mu}+p_b^{\mu}=p_c^{\mu}+p_d^{\mu}##
##(p_a^{\mu}+p_b^{\mu}-p_c^{\mu})^2=(p_d^{\mu})^2##
##p_a^2+p_b^2+p_c^2+2p_a^\mu p_{\mu b}-2p_a^\mu p_{\mu c}-2p_b^\mu p_{\mu c}=0##
##m^2+m^2+2Em-2(EE_{\gamma 1}-E_{\gamma1}\sqrt{E^2-m^2})-2mE_{\gamma 1}##

##E_{\gamma 1}=\frac{m(E+m)}{E+m+\sqrt{E^2-m^2}}=\frac{m}{1+\sqrt{\frac{E-m}{E+m}}}##

Similarly for ##E_{\gamma 2}##.
I can't follow that at all, but it's definitely wrong.

Why not use conservation of energy and conservation of momentum?

PeroK said:
I can't follow that at all, but it's definitely wrong.

Why not use conservation of energy and conservation of momentum?
Yes, I understand that I have to relax on the weekend or I shoot nonsense

##E_{\gamma 1} = \frac{E+m+\sqrt{E^2-m^2}}{2}##
##E_{\gamma 2} = \frac{E+m-\sqrt{E^2-m^2}}{2}##

PeroK
2 - Calculate the energies in the reference system integral with the incident particle, explicitly carrying out the Lorentz transformation. Discuss the result obtained.
I suppose now to start from ##p_c^\mu## and ##p_d^\mu## and find ##p_c^{\mu'}## and ##p_d^{\mu'}## using Lorentz transformation for energy and momentum with ##\gamma=\frac E m## and ##v=\frac{\sqrt{E^2-m^2}}{E}##, or I have to find a different way?

Frostman said:
I suppose now to start from ##p_c^\mu## and ##p_d^\mu## and find ##p_c^{\mu'}## and ##p_d^{\mu'}## using Lorentz transformation for energy and momentum with ##\gamma=\frac E m## and ##v=\frac{\sqrt{E^2-m^2}}{E}##, or I have to find a different way?
The question asks you to do the LT explicitly. So, I guess you have to do it that way.

Frostman said:
Homework Statement:: An antiparticle of mass ##m## and energy ##E## hits a stationary particle in the laboratory and the system annihilates itself producing two photons
The stationary particle also has rest mass m (e.g. if we are dealing with a positron and an electron). So in the Lab' frame, you will want to ensure the sum of the two photon energies is E+2m, not E+m ((taking c=1).

PeroK
Steve4Physics said:
The stationary particle also has rest mass m (e.g. if we are dealing with a positron and an electron). So in the Lab' frame, you will want to ensure the sum of the two photon energies is E+2m, not E+m ((taking c=1).
##E## is energy, not Kinetic Energy.

Steve4Physics
Thanks. I misinterpreted that.

## 1. What is annihilation and how is it related to the calculation of photon energies?

Annihilation is a process in which a particle and its corresponding antiparticle collide and are both converted into energy. This energy is released in the form of photons, which have specific energies that can be calculated using mathematical equations.

## 2. What is the formula for calculating the energy of a photon in an annihilation process?

The formula for calculating the energy of a photon in an annihilation process is E = m0c^2, where E is the energy of the photon, m0 is the rest mass of the particle and antiparticle, and c is the speed of light.

## 3. How is the energy of a photon in an annihilation process related to the mass of the particles involved?

The energy of a photon is directly proportional to the mass of the particles involved in the annihilation process. This means that the more massive the particles are, the higher the energy of the resulting photon will be.

## 4. Can the energy of a photon in an annihilation process be calculated without knowing the mass of the particles?

No, the energy of a photon in an annihilation process cannot be calculated without knowing the mass of the particles involved. This is because the mass of the particles is a crucial factor in the calculation of photon energies.

## 5. Are there any other factors that can affect the energy of a photon in an annihilation process?

Yes, there are other factors that can affect the energy of a photon in an annihilation process, such as the velocity of the particles and the direction of the resulting photon. These factors can be taken into account in more complex equations for calculating photon energies.

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