# Electron-positron annihilation, photon emission angle

• Frostman
In summary, the conversation discusses the use of conservation of four-momentum in a laboratory system to find the energy and angle of two emitted photons in a positron-electron annihilation scenario. The initial four-momentums of the positron, electron, and two photons are given, and conservation of energy and momentum equations are set up and solved to find the energy of the second photon and the angle between the two photons.f

#### Frostman

Homework Statement
In the annihilation process ##e^+e^- \rightarrow \gamma \gamma##, seen in the reference system of the laboratory, in which the electron is stationary and the positron has energy E, it may happen that one of the two photons is emitted at right angles to the direction of incidence of the positron. Calculate, in this case, the energy and the scattering angle of the other photon.
Relevant Equations
Conservation of the four-momentum
I consider the laboratory system. The four momentums in this reference system are respectively:
##p^\mu = \big(\sqrt{|p|^2+m^2}, 0, 0, |p| \big)##
##p'^\mu= \big(m, 0, 0, 0 \big)##
##k^\mu = E\big(1, 0, 1, 0\big)##
##k'^\mu = E'\big(1, 0, -\sin \varphi, \cos \varphi \big)##

I used conservation of four-momentum:
##(k^\mu)^2=(p^\mu+p'^\mu-k'^\mu)^2##

And I find out that:

##E'=\frac{m^2+m\sqrt{|p|^2+m^2}}{m+\sqrt{|p|^2+m^2}-|p|\cos\varphi}##

But I still have two unknown variables...

I used conservation of four-momentum
Did you?
I'm not well versed in relativity, but I would have thought conservation of four-momentum was a vectorial equation, just as it is for 3D momentum. Looks to me like you only considered the norm conserved.

vela
Did you?
I'm not well versed in relativity, but I would have thought conservation of four-momentum was a vectorial equation, just as it is for 3D momentum. Looks to me like you only considered the norm conserved.
Yes, I used the fact that the quantity ##p^\mu+p'^\mu = k^\mu+k'^\mu## must be equal both in the initial state and in the final state after that I elevated and got that relation.

Yes, I used the fact that the quantity ##p^\mu+p'^\mu = k^\mu+k'^\mu## must be equal both in the initial state and in the final state after that I elevated and got that relation.
But surely you would have got more information than is represented in a single scalar equation? I'm probably missing something, but you seem to have, in scalar terms, four equations and three unknowns.

I'm also a bit confused by your four expressions for the four momenta. I assume they are intended to be, in order, positron, electron, gamma of known direction, gamma of unknown direction. But although E is given as the energy of the positron, E appears in your expression for a gamma, not in the expression for the positron.

As @haruspex says, you are missing too many equations here. You should have three equations: conservation of energy and momentum in the ##y## and ##z## directions, if that's what you are using.

First, you need to eliminate the angle and get an equation for the energy of the second photon, in terms of ##E## and ##m##.

As @haruspex says, you are missing too many equations here. You should have three equations: conservation of energy and momentum in the ##y## and ##z## directions, if that's what you are using.

First, you need to eliminate the angle and get an equation for the energy of the second photon, in terms of ##E## and ##m##.

##p^\mu_{e^+} = \big(E, 0, 0, \sqrt{E^2-m^2} \big)##
##p^\mu_{e^-}= \big(m, 0, 0, 0 \big)##
##k^\mu_{\gamma_1} = E_1\big(1, 0, 1, 0\big)##
##k^\mu_{\gamma_2} = E_2\big(1, 0, -\sin \varphi, \cos \varphi \big)##

In the lab-frame, ##e^+## comes from left to right along z-axe.
In the lab-frame, ##e^-## is at rest.
In the lab-frame, ##\gamma_1## goes along y-axe, because, by hypothesis, it's emitted orthogonally to the ##e^+## direction.

I need to find ##E_2## and ##\varphi## for ##\gamma_2##.

If I used conservation of four-momentum, I would have:

##E: \ \ \ \ \ \ \ \ \ \ E+m=E_1+E_2##
##p_x: \ \ \ \ \ \ \ \ \ \ 0+0=0+0##
##p_y: \ \ \ \ \ \ \ \ \ \ 0+0=E_1-E_2\sin\varphi##
##p_z: \ \ \ \ \ \ \ \ \ \ \sqrt{E^2-m^2}+0=0+E_2\cos \varphi##

Can we agree on this?

##p^\mu_{e^+} = \big(E, 0, 0, \sqrt{E^2-m^2} \big)##
##p^\mu_{e^-}= \big(m, 0, 0, 0 \big)##
##k^\mu_{\gamma_1} = E_1\big(1, 0, 1, 0\big)##
##k^\mu_{\gamma_2} = E_2\big(1, 0, -\sin \varphi, \cos \varphi \big)##

In the lab-frame, ##e^+## comes from left to right along z-axe.
In the lab-frame, ##e^-## is at rest.
In the lab-frame, ##\gamma_1## goes along y-axe, because, by hypothesis, it's emitted orthogonally to the ##e^+## direction.

I need to find ##E_2## and ##\varphi## for ##\gamma_2##.

If I used conservation of four-momentum, I would have:

##E: \ \ \ \ \ \ \ \ \ \ E+m=E_1+E_2##
##p_x: \ \ \ \ \ \ \ \ \ \ 0+0=0+0##
##p_y: \ \ \ \ \ \ \ \ \ \ 0+0=E_1-E_2\sin\varphi##
##p_z: \ \ \ \ \ \ \ \ \ \ \sqrt{E^2-m^2}+0=0+E_2\cos \varphi##

Can we agree on this?
Yes. Now, how do you normally get rid of sines and cosines?

Yes. Now, how do you normally get rid of sines and cosines?
Except, in this case, we need to find the angle too, so it might be simpler to start by collapsing sin and cos into a single trig function and eliminate the unknown energies instead?

Except, in this case, we need to find the angle too, so it might be simpler to start by collapsing sin and cos into a single trig function and eliminate the unknown energies instead?
I calculated the energy first and it all comes out. It might be slightly messier to try to find the angle first.

I can go for:

##E_1^2=E_2^2\sin^2\varphi##
##E^2-m^2=E_2^2\cos^2\varphi##

Adding these two equations and exploiting the energy conservation relation I get:

##E^2-m^2+E_1^2=E_2^2##

Remmebering:
##E_1 = E+m-E_2##
##E_1^2=E^2+m^2+E_2^2+2Em-2EE_2-2E_2m##

Replacing:
##E^2-m^2+E^2+m^2+E_2^2+2Em-2EE_2-2E_2m=E_2^2##

Result:
##E_2=\frac{E^2+Em}{E+m}##

While for angle ##\varphi##

##\cos \varphi = \frac{\sqrt{E^2-m^2}}{E^2+Em}(E+m)=\frac{\sqrt{E^2-m^2}}{E}=\sqrt{1-\frac{m^2}{E^2}}##

Result
##E_2=\frac{E^2+Em}{E+m}##
That simplifies!

##\cos \varphi = \frac{\sqrt{E^2-m^2}}{E^2+Em}(E+m)=\frac{\sqrt{E^2-m^2}}{E}=\sqrt{1-\frac{m^2}{E^2}}##

Why not use ##\sin \varphi##?

That simplifies!

Why not use ##\sin \varphi##?
I forgot to simplify, so:

##E_2=E##

And it's much easier to use ##\sin \varphi##
##\sin \varphi=\frac mE##

So I can notice, in this lab-frame, that ##\gamma_2## gets the same energy of ##e^+##, while ##\gamma_1## has, as energy, the mass of the ##e^-## at rest.

PeroK
Thank you so much for the help!

You can avoid much of the algebra if you consider ##(p_{e^+}-k_1)^2 = (k_2 - p_{e^-})^2## to show that ##E_1 = (m/E) E_2##.

Frostman and vanhees71