Relativistic momentum conservation

[Frank Einstein]

Homework Statement

Good morning/ afternoon I have a doubt about relativistic collisions, any help will be welcome.

Consider the process of annihilation electron-positron to two photons.

Calculate in the centre of mass system the wavelength of photons produced using momentum of e+e-.

Particularize for the scenario in which they are at rest (The electron and positron I guess)

Consider a reference frame which is moving at constant speed v to the left, which wavelenght will the observer assign to the photons of annihilation?

Homework Equations

wavelenght=(Plank constant/Momentum).
Energy= sqrt(me^2*C^4+(pC)^2) where p= momentum

3. The Attempt at a Solution
I use the conservation of momentum and energy. If we are at the centre of mass, both particles approach us with the same momentum, so the two photons will leave with the same momentum too, so I will call the momentum for all of the four particles of the system p. The initial energy is Energy=2* sqrt(me^2C^4+(pC)2).

The final energy is 2(p/c) because the photons are massless and always travel at speedlight. If we square both sides, we have

4(e^2 C^4 +(pC)^2)=4* (pC)^2)

So, I am stuck here, because the (pC)^2 simplifies at both sides, so I have me C^2=0

Thanks for reading

 

[Orodruin]

Your assertion that the photons must have the same momentum as the electrons is not correct. All that is required is that the (vector) sum of the incoming momenta must equal that of the outgoing. How can you use this information along with conservation of energy?

 

[Frank Einstein]

When I use this equation, conservation of energy, I write P, not Pe₊, Pe^- or P_photon
Energy= sqrt(me^2*C^4+(pC)^2)

 

[Orodruin]

Yes, but the point was that you cannot assume the photons to have the same momentum as the electron and positron. As you have noticed, this would lead to non conservation of energy.

 

[Frank Einstein]

Well, in the centre of mass, the electron and positron have 0 momentum in the y-axis and-p and p in the x, so the total is 0, that's why I thought that both photons would have the same momentum in modulus so I could use the conservation of energy tho find momentum. If I can't asume that both photons have the same momentum, I will have a three variable equation for conservation of energy, original and both photons momentum.

 

[Orodruin]

You were asked to find the photon wavelength in terms of the momenta of the electron/positron. The photon momentum is directly given by its momentum/energy so if you find it you know the wavelength. Asserting the photons to have the same momentum as the electron/positron would give the photon wavelength directly, but be wrong due to nonconservation of energy.

What do you consider to be your variables that you cannot solve for?

 

[Frank Einstein]

Before anything, this is a translation, so I might have not expressed the definition as good as I wanted.
I want to find the momentum of the photons and put them as a function of the momentum of the electron and positron

 

[Orodruin]

Yes, this should be possible using the conservation laws you stated. You also stated that you had three unknowns but I only see one, the momentum of the photons (which have to be of equal magnitude and opposite direction due to momentum conservation in the CoM frame).

 

[Frank Einstein]

Then, If I understand you correctly, Pe⁺ is equal in modulus and opposed in direction to Pe^-, so the total momentum of the photons is 0. For the conservation laws, I see that the total momentum of the photons is 0 too. Which can only mean that they are equal in modulus but opposite in direction.

EDIT: I understand under the matematical point of wiew that if I went to my equation of conservation of energy and I changed P on the right side for P_photon I might find a suitable annwser, but I don't understand how it doesn't violate the conservation of momentum.

SECOND EDIT: Allright, Now I understand what you meant with the difference in momentum, it happens due to the loss of the original particle and in stead the creation of a new one.

Thanks for your help