Pair production - conservation of momentum VS conservation of energy

In summary: Big]\end{split}$$where ##v_1, v_2## are photons. If we include a nucleus, then we can get an equation that matches conservation of energy and conservation of momentum:$$\begin{split}h\nu &= m_ec^2 \left[\gamma(nucleus) + \text{nucleus} \right]\\h\nu &= m_e c \Big[ \gamma(nucleus) \underbrace{nucleus}_{\neq c} + \text{nucleus}_{\neq c}...\Big]
  • #1
71GA
208
0
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 no one 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.

wGF1v.png


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?
 
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  • #2
71GA said:
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?

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
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
[tex]\gamma + \text{nucleus} \rightarrow e^+ + e^- +\text{nucleus}.[/tex]

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
vanhees71 said:
First of all you have to define which process you really mean. I don't understand you picture. What's "razpad"?

"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
vanhees71 said:
What's "razpad"?.

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
Nugatory said:
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.

What i need now is only the missing 3). Could anyone provide equations?
 
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  • #7
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:
  • #8
mfb said:
Your inequality shows that the process is not possible without a nucleus.
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:
mfb said:
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}{m}##). Therefore, you are "allowed" to ignore momentum conservation if you consider photons and electron/positron only: the nucleus will get the momentum difference.
Where does equation ##E \approx \frac{p^2}{m}## come from? Can you explain more in detail why i can ignore momentum conservation? I don't understand this yet.
mfb said:
With electron+positron, there is an inertial system with a fixed center of mass.
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:
  • #9
71GA said:
I am more of a newbie and interpreted it more like a: "Oh inequality! So it can't be done."
That is correct.
Without proton (=as you calculated), it cannot be done.

Where does equation ##E \approx \frac{p^2}{m}## come from?
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.

Can you explain more in detail why i can ignore momentum conservation? I don't understand this yet.
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.

If i understand this right before collision (picture below) center of mass is positioned in the proton itself as photon has no mass.
Center of mass = center of energy, and the photon has energy. But that part was related to the (hypothetical) process ##e^- + e^+ \to \gamma##.

Is the basic idea with using the rest mass system in the fact that after collision $p = 0$ and equations are simplified?
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.

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?
?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
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.
 

1. How does pair production demonstrate the conservation of momentum?

Pair production is a process in which a high-energy photon is converted into an electron and a positron. This conversion occurs in the presence of a nucleus, which absorbs some of the energy and momentum from the photon. The resulting particles, electron and positron, have opposite momenta but equal in magnitude, thus conserving the total momentum of the system.

2. Does pair production also follow the conservation of energy?

Yes, pair production also follows the conservation of energy. The energy of the original photon is converted into the masses of the electron and positron, as well as their kinetic energies. The total energy of the system before and after the process remains the same, satisfying the law of conservation of energy.

3. What is the relationship between conservation of momentum and conservation of energy in pair production?

The conservation of momentum and the conservation of energy are both fundamental laws of physics that must be satisfied in any physical process. In pair production, these two laws are interconnected and work together to ensure that the total momentum and energy of the system are conserved.

4. How is the conservation of momentum and energy demonstrated in pair production experiments?

In experiments, the conservation of momentum and energy can be demonstrated by measuring the momenta and energies of the original photon, electron, and positron. The total momentum and energy before and after the process should be equal, providing evidence that these laws are conserved in pair production.

5. Are there any exceptions to the conservation of momentum and energy in pair production?

As with any physical process, there may be small deviations from the exact conservation of momentum and energy in pair production. However, these deviations are extremely small and are usually well within the margin of error in experiments. Overall, the conservation of momentum and energy in pair production is a well-established and proven concept in physics.

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