# Pair production - conservation of momentum VS conservation of energy

1. Jan 13, 2013

### 71GA

Allover the web i am only seeing a statement similar to this:

"Pair production is not possible in vaccum, 3rd particle is needed so
that conservation of momentum holds."

Well noone out of many writers shows, how to prove this matematically. So this is what interests me here.

First i wanted to know if pair production really cannot happen in vacuum, so i drew a picture and used equations for conservation of energy and conservation of momentum to calculate energy of a photon $(h \nu)$ needed for pair production.

It turns out $h\nu$ is different if i calculate it out of conservation of energy or conservation of momentum. And even more! It can never be the same because equallity would mean parts $v_1 \cos \alpha$ and $v_2 \cos \beta$ should equall speed of light $c$. Well that cannot happen.

Below is my derivation.

CONSERVATION OF ENERGY:
$$\scriptsize \begin{split} W_{1} &= W_{2}\\ W_f &= W_{e^-} + W_{e^+}\\ h\nu &= W_{ke^-} + W_{0e^-} + W_{ke^+} + W_{0e^+}\\ h\nu &=\left[m_ec^2 \gamma(v_1) - m_ec^2\right] + m_ec^2 + \left[m_ec^2 \gamma(v_2) - m_ec^2\right] + m_ec^2\\ h\nu &=m_ec^2 \gamma(v_1) + m_ec^2 \gamma(v_2)\\ h\nu &=m_ec^2 \left[\gamma(v_1) + \gamma(v_2) \right]\\ \end{split}$$

CONSERVATION OF MOMENTUM:

$y$ direction:
$$\scriptsize \begin{split} p_{1} &= p_{2}\\ 0 &= p_{e^-} \sin \alpha - p_{e^+} \sin \beta \\ 0 &= m_e v_{1} \gamma(v_{1}) \sin \alpha - m_e v_{2} \gamma(v_{2}) \sin \beta\\ &\text{If \boxed{\alpha = \beta} \Longrightarrow \boxed{v_1 = v_2} and:}\\ 0 &= 0 \end{split}$$

$x$ direction:

$$\scriptsize \begin{split} p_{1} &= p_{2}\\ \frac{h}{\lambda} &= p_{e^-} \cos \alpha + p_{e^+} \cos \beta \\ \frac{h \nu}{c} &= m_e v_{1} \gamma(v_{1}) \cos \alpha + m_e v_{2} \gamma(v_{2}) \cos \beta\\ h \nu &= m_e c \Big[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Big] \end{split}$$

Alltogether:

Because momentum in $y$ direction equals 0 (holds for some combinations of $\alpha, \beta, v_1, v_2$) whole momentum equals just the momentum in $x$ direction. So if i add them i get:

$$\scriptsize h \nu = m_e c \Big[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Big]$$

From this i can conclude only that i cannot sucessfully apply conservation of energy and conservation of momentum at the same time and therefore pair production in vacuum cannot happen.

QUESTION1:
Why do writers state that 3rd particle is needed so that conservation of momentum holds? What if conservation of momentum holds and conservation of energy doesn't? How can we say which one holds and which one doesnt?

QUESTION2:
Do writters actually mean that if a 3rd particle is included we can achieve $h \nu$ to match in both cases?

QUESTION3:
Can someone show me mathematically how this is done? I mean it should right?

2. Jan 13, 2013

### Staff: Mentor

1) We have never, not once in the history of humanity, ever observed a situation in which energy and momentum are not both conserved. Every time we think we've found such a situation, it has always turned out that we've overlooked or misunderstood something. So it's a pretty safe bet that both always hold.

2) yes.

3. Jan 13, 2013

### vanhees71

First of all you have to define which process you really mean. I don't understand you picture. What's "razpad"?

Usually under pair production one understands the process
$$\gamma + \text{nucleus} \rightarrow e^+ + e^- +\text{nucleus}.$$

Of course both energy and momentum are conserved in this reaction. What you may have read is the correct statement that a single-photon decay in the vacuum is impossible due to energy-momentum conservation.

4. Jan 13, 2013

### Staff: Mentor

"Razpad" is the Slovenian word for "decay". I assumed from the context that OP was using it in the sense of particle decay, although someone fluent in both physics and English might prefer something that translated into "interaction".

5. Jan 13, 2013

### 71GA

Oh i am sorry it means "splits". The picture shows an impossible situation which i intentionally predicted. But still. How can i show mathematically that nucleus is even needed? Where exactly does it come in so that it fixes my equations? By equations i mean these two:

$$\begin{split} h\nu &=m_ec^2 \left[\gamma(v_1) + \gamma(v_2) \right]\\ h \nu &= m_e c \Big[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Big] \end{split}$$

6. Jan 13, 2013

### 71GA

What i need now is only the missing 3). Could anyone provide equations?

Last edited: Jan 13, 2013
7. Jan 13, 2013

### Staff: Mentor

Your inequality shows that the process is not possible without a nucleus.
You can look at the time-reversed process, too: With electron+positron, there is an inertial system with a fixed center of mass. This is not true for a single photon.

With a nucleus, it is a bit lengthy to calculate it in an analytic way. The general idea is that a nucleus is heavy compared to the electrons - it can gain significant momentum with little energy ($E \approx \frac{p^2}{2m}$ (edit: fixed prefactor)). Therefore, you are "allowed" to ignore momentum conservation if you consider photons and electron/positron only: the nucleus will get the momentum difference.

Last edited: Jan 13, 2013
8. Jan 13, 2013

### 71GA

I am more of a newbie and interpreted it more like a: "Oh inequality! So it can't be done." From my perspective there is no proton. I yet have to proove to myself that i really need it. Thats where i saw the other part of your post which i like:
Where does equation $E \approx \frac{p^2}{m}$ come from? Can you explain more in detail why i can ignore momentum conservation? I dont understand this yet.
If i use center of mass system with nucleus before collision (picture below) center of mass is positioned in the proton itself as photon has no mass. Is the basic idea with using the rest mass system in the fact that after collision $p = 0$ and equations are simplified?

If we use center of mass system we get $p_{1} = p_{2} = 0$ out of conservation of momentum. I can't help myself with this but could i help myself with a conservation of energy?

http://shrani.si/f/1j/3E/20P1xg9s/screenshot-from-2013-01-.png [Broken]

Last edited by a moderator: May 6, 2017
9. Jan 13, 2013

### Staff: Mentor

That is correct.
Without proton (=as you calculated), it cannot be done.

Classical mechanics, but I forgot a 2 in the denominator. $p=mv$, $E=\frac{1}{2}mv^2$, solve first equation for v and plug it into the second.

Any "deviation" from momentum conservation is the momentum the nucleus gets.

There is a similar effect in classical collisions: For elastic collisions, you have both energy and momentum conservation. For inelastic collisions, energy is still conserved (of course, it is a very fundamental law), but a part of the energy can go to heat or other forms of energy you don't care about. Therefore, for (perfectly) inelastic collisions, energy conservation is ignored, and momentum conservation only is considered.

Center of mass = center of energy, and the photon has energy. But that part was related to the (hypothetical) process $e^- + e^+ \to \gamma$.

Right. For electron+positron, there is an inertial system where their added momentum is zero. Momentum is conserved, so without an additional nucleus (to get momentum), the resulting photon would have zero momentum -> impossible.

?

Let's look at an actual example with a nucleus:

We have an incoming photon with an energy of 1277 keV and a momentum of 1277 keV (using c=1), approacing a 208Pb nucleus (m=194 GeV) and performing pair production there. Assume that the angle between electron and positron is 0 (easier to calculate):

Electron+positron fly in x-direction with v=0.6c (giving gamma=1.25). Their summed energy is 2.5*511keV = 1277 keV, and their summed momentum is 1277 keV * 0.6 = 766.5 keV (with c=1).
The 208Pb nucleus (m=194 GeV) moves in x-direction with a momentum of 510.5 keV and a kinetic energy of 0.00067 keV. As you can see, its energy is neglibile - you could dump basically every amount of momentum you like into it, and it would not make a difference. Therefore, energy conservation is relevant, momentum conservation is not (unless you want to get the velocity of the nucleus after the collision).

10. Jan 13, 2013

### 71GA

Thank you soooo much!

There is only one more thing.

Does anyone know of a good center of mass = center of energy system animation or. good picture for pair production.