# No pair production with light alone

#### DaTario

Hi,

I have heard someplace in my graduation course that two gamma photons are not allowed to yield a pair of particle anti-particle unless there is some material agent nearby. I remember also that this result was told to be very easy to demonstrate. Unfortunatelly there is no room for that in this page (recalling Fermat..).

Is there anyone who knows this proof ?

Thankx

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#### Berislav

Try writing a S-matrix and drawing a Feynman diagram for such a process.

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#### DaTario

But as I remember one can derive this result from conservation laws, using Lorentz invariants or related things.

#### ZapperZ

Staff Emeritus
2018 Award
DaTario said:
Hi,

I have heard someplace in my graduation course that two gamma photons are not allowed to yield a pair of particle anti-particle unless there is some material agent nearby. I remember also that this result was told to be very easy to demonstrate. Unfortunatelly there is no room for that in this page (recalling Fermat..).

Is there anyone who knows this proof ?

Thankx
This is a very naive explanation, but it'll do for now.

Take a photon with energy exactly equal to the rest mass energy of an electron-positron pair. It has energy JUST enough to create that pair. So conservation energy is fine.

However, the photon had a MOMENTUM before. If it has exactly just enough energy to create an electron-positron pair, there's nothing leftover to be converted into a net momentum for the electron-positron. So there's violation of conservation of momentum.

It requires a coupling to another "massive" entity to provide the necessary recoil momentum.

Zz.

#### marlon

DaTario said:
Hi,

I have heard someplace in my graduation course that two gamma photons are not allowed to yield a pair of particle anti-particle unless there is some material agent nearby. I remember also that this result was told to be very easy to demonstrate. Unfortunatelly there is no room for that in this page (recalling Fermat..).

Is there anyone who knows this proof ?

Thankx
you will find everything in here
http://laser.phys.ualberta.ca/~egerton/pair-p&a.htm [Broken]

marlon

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#### Berislav

Well, I didn't really understand which process you were describing. Is it two photons decaying into two electrons and two positrons, or something else?
I thought that maybe the S-matrix would show that there is a violation of 4-momentum. Or that a Feynman diagram would reveal an open internal vertix.

EDIT: Oops. I'm too late.

#### EL

The question was about two photons producing a electron-positron pair.
If the photons collide heads on, momentum can be conserved...why do we need some other material then?

#### DaTario

ZapperZ said:
This is a very naive explanation, but it'll do for now.

Take a photon with energy exactly equal to the rest mass energy of an electron-positron pair. It has energy JUST enough to create that pair. So conservation energy is fine.

However, the photon had a MOMENTUM before. If it has exactly just enough energy to create an electron-positron pair, there's nothing leftover to be converted into a net momentum for the electron-positron. So there's violation of conservation of momentum.

It requires a coupling to another "massive" entity to provide the necessary recoil momentum.

Zz.

It is quite like that, I can remember the outlines. But if the incoming photon had more energy? Wouldn't it be sufficient to guarantee the extra kinetic energy which would be noticeable also in the form of momentum (recoil momentum) ?

Best Regards

DaTario

#### ZapperZ

Staff Emeritus
2018 Award
EL said:
The question was about two photons producing a electron-positron pair.
If the photons collide heads on, momentum can be conserved...why do we need some other material then?
You may or may not. But when was the last time we had a "photon collider", and how often does this happen? Yet, we HAVE generated antimatter via pair production.

This means that there is an inherent and more universal conservation consideration that should work for one, two, three, four, etc... photon conversion into matter. ALL electron-positron pair production, for example, are produce via gamma rays going through a crystal, where the crystal is the medium that supply the necessary momentum conservation. And unless I missed something, the feynman diagram for matter-antimatter pair production in general always seem to have an extra coupling to a nearby object [Someone doing high energy physics can correct me on this].

Zz.

#### EL

ZapperZ said:
You may or may not. But when was the last time we had a "photon collider", and how often does this happen? Yet, we HAVE generated antimatter via pair production.

This means that there is an inherent and more universal conservation consideration that should work for one, two, three, four, etc... photon conversion into matter. ALL electron-positron pair production, for example, are produce via gamma rays going through a crystal, where the crystal is the medium that supply the necessary momentum conservation. And unless I missed something, the feynman diagram for matter-antimatter pair production in general always seem to have an extra coupling to a nearby object [Someone doing high energy physics can correct me on this].

Zz.
Sure, but in principle it is possible to create electron-positrons pairs out of two colliding photons. (In fact the collision need not be heads on, but that's the case when it's most effective.)
However, we do not have sufficently good lasers at the moment.
But at least it may soon be possible to detect photon-photon scattering by colliding laser beams.

#### ZapperZ

Staff Emeritus
2018 Award
EL said:
Sure, but in principle it is possible to create electron-positrons pairs out of two colliding photons. (In fact the collision need not be heads on, but that's the case when it's most effective.)
However, we do not have sufficently good lasers at the moment.
But at least it may soon be possible to detect photon-photon scattering by colliding laser beams.
I think that for any appreciable cross-section, gamma-gamma is the only choice currently. If that is the case, then I know for sure we don't have gamma lasers as of yet. :)

I believe this is one type of collision that is being considered for LHC as one of its many functions.

Zz.

#### DaTario

marlon said:
you will find everything in here
http://laser.phys.ualberta.ca/~egerton/pair-p&a.htm [Broken]

marlon

Thank you for the reference.

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#### EL

ZapperZ said:
I think that for any appreciable cross-section, gamma-gamma is the only choice currently. If that is the case, then I know for sure we don't have gamma lasers as of yet. :)

I believe this is one type of collision that is being considered for LHC as one of its many functions.

Zz.
Appreciable cross-section for what: Pair production or photon-photon scattering?

#### ZapperZ

Staff Emeritus
2018 Award
EL said:
Appreciable cross-section for what: Pair production or photon-photon scattering?
Photon-photon collision cross-section.

Zz.

#### BenLillie

ZapperZ said:
I think that for any appreciable cross-section, gamma-gamma is the only choice currently. If that is the case, then I know for sure we don't have gamma lasers as of yet. :)

I believe this is one type of collision that is being considered for LHC as one of its many functions.

Zz.
I don't know that the LHC can work in gamma-gamma mode. However, one of the operating modes of the proposed International Linear Collider (which accelerates electrons and positrons) is as a photon collider. Since this machine works in the TeV region there is certainly enough energy to produce all kinds of interesting particles, and the cross-sections are comparable to the cross-section for e+ e- annihilation at those energies.

#### BenLillie

ZapperZ said:
And unless I missed something, the feynman diagram for matter-antimatter pair production in general always seem to have an extra coupling to a nearby object [Someone doing high energy physics can correct me on this].

Zz.
You can show using Lorentz invariance that a massless particle can not decay into any two particles. There is no such restriction on two massless particles colliding. So, as long as there is enough center-of-mass energy, two photons can collide to produce an electron-positron pair.

If you like, you can think of pair production by a single photon in a medium as a two photon collision, where the second photon is provided by an electrically charged particle in the medium. After that, the physics is exactly the same as in any other gamma-gamma collision.

#### EL

ZapperZ said:
Photon-photon collision cross-section.

Zz.
Well that depends on what comes out after the collision. For pair production the photons surely must have a large frequency. (A minimum frequency is easy to work out, just equal to the electron mass, but I don't have the numbers right now. Anyway to get any cross-section to talk about we would probably need more photon energy than that).
However, in photon-photon scattering you need not have that large photon energy since the fermions are only virtual.

In fact by using three incoming laser beams, it will probably soon be possible to generate scattered photons, even with wavelengths as large as in the visible spectrum. We just need lasers with somewhat better power than the existing ones. Well in fact it could be enough if we just had three of todays most powerful lasers at the same place...

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#### reilly

One photon can convert to a pair only if there is some stuff around, with which to interact. That's a result of standard energy and momentum conservation laws.

But, these conservation laws do not preclude a direct conversion of 2 gammas into a pair, or vice versa. What's to prevent the process?

Note: Rotate Compton scattering diagrams (by 90 degrees with the usual convention that "going up" means increasing time.) and, lo and behold, you have the diagrams for 2 gamma -> one pair.

Regards,
Reilly Atkinson

#### EL

reilly said:
But, these conservation laws do not preclude a direct conversion of 2 gammas into a pair, or vice versa. What's to prevent the process?
Nothing, but as ZZ noted, there exist no lasers with high enough frequency today.

#### DaTario

A sub - question on this subject: does photon - photon scattering contradict superpostition principle and / or independent luminal path principle ?

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#### Meir Achuz

Homework Helper
Gold Member
DaTario said:
Hi,

I have heard someplace in my graduation course that two gamma photons are not allowed to yield a pair of particle anti-particle unless there is some material agent nearby. I remember also that this result was told to be very easy to demonstrate. Unfortunatelly there is no room for that in this page (recalling Fermat..).

Is there anyone who knows this proof ?

Thankx
I think you just remember the statement incorrectly.
Two photons do produce particles, as several posts have mentioned.
It is a single photon that cannot produce a pair without additional matter. This follows from conservation of 4-momentum.
The photon has p^2=0, while any particle pair needs p^2>0.

#### DaTario

Meir Achuz said:
I think you just remember the statement incorrectly.
Two photons do produce particles, as several posts have mentioned.
It is a single photon that cannot produce a pair without additional matter. This follows from conservation of 4-momentum.
The photon has p^2=0, while any particle pair needs p^2>0.
Yes, you are correct, I have already noted that, by the already sent posts.

Thank you.

Best Wishes,

DaTario

#### James Jackson

Meir Achuz said:
The photon has p^2=0, while any particle pair needs p^2>0.
Do you mean a photon has $m^2=0$ while any particle pair needs $m^2 > 0$ (in natural units)? Recalling that $E^2=m^2 + p^2$, with photons having no mass and all that jazz...

#### Berislav

Do you mean a photon has LaTeX graphic is being generated. Reload this page in a moment. while any particle pair needs LaTeX graphic is being generated. Reload this page in a moment. (in natural units)? Recalling that LaTeX graphic is being generated. Reload this page in a moment., with photons having no mass and all that jazz...
It's 4-momentum actually.

#### Meir Achuz

Homework Helper
Gold Member
James Jackson said:
Do you mean a photon has $m^2=0$ while any particle pair needs $m^2 > 0$ (in natural units)? Recalling that $E^2=m^2 + p^2$, with photons having no mass and all that jazz...
Yes and, as Berislav implies, the equation m^2=E^2-p^2 just represents the length squared of the momentum 4-vector.

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