How do you create protons, neutrons, and electrons from energy?

trewsx7
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Quick question, regarding mass-energy equivalence (e=mc2) and matter creation.

Knowing that there are many concete visual examples of matter turning into energy using Einstein's equation, how do you use energy to create matter? If I'm not mistaken, I believe electron-positron pairs are created using photons, but how does one use energy to create protons and neutrons? What are the "energy ingredients" in this process? Thanks!
 
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Gluons, and quarks.

For instance in an electron-positron collider, you can create a quark-antiquark pair that undergoes hadronization and gluon emission -> in those processes protons and neutrons (and other hadrons) can be created.

Also in a proton-antiproton collider and in a proton-proton collider, you will get free quarks and gluons that will undergo hadronization.

In particle physics, a photon is a particle.
 


Protons and neutron are created by a process called Pair production.
Pair production refers to the creation of an elementary particle and its antiparticle via Quantum Chromodynamics (QCD).

Pair production can only occur if the photon has an energy exceeding the twice the rest mass(me) of an electron (1.022 MeV); the same applies for the generation of other higher energy leptons such as the muon and tau.

m_e = 0.00051099891844 \; \text{GeV}
m_p = 0.9382720298 \; \text{GeV}
m_n = 0.9395655681 \; \text{GeV}

E_{\gamma} = 2 m_e = 0.00102199783688 \; \text{GeV}
E_{\gamma} = 2 m_p = 1.8765440596 \; \text{GeV}
E_{\gamma} = 2 m_n = 1.8791311362 \; \text{GeV}

\gamma( E_{\gamma} ) \rightarrow \; e^+ + e^- - Beta Pair production
\gamma( E_{\gamma} ) \rightarrow \; p^+ + p^- - Proton Pair production
\gamma( E_{\gamma} ) \rightarrow \; n^+ + n^- - Neutron Pair Production

In semiclassical general relativity, pair production is also invoked to explain the Hawking radiation effect.

Pair production is also the hypothesized mechanism behind the Pair instability supernova type of stellar explosions, where pair production suddenly lowers pressure inside a supergiant star, leading to a partial implosion, and then explosive thermonuclear burning.
[/Color]
Reference:
http://en.wikipedia.org/wiki/Pair_production"
http://en.wikipedia.org/wiki/Quantum_chromodynamics"
 
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Two comments:

- In particle physics energy is not an entity by itself but a property of particles. You do not turn energy into particles but you turn one collection of particles into another collection of particles where energy and momentum conservation dictate that both collections of particles have the same energy and the same momentum. The example of two photons (or one photon and some atom) turning into an electron-positron pair is not an example of energy turning into electron and positron but an example of a collection of particles (photon+X) turning into a different collection of particles (e+ + e-). Not related to what I wanted to say but for completeness: In the photon+atom reaction, the result is e+ + e- + atom; the atom remains in the final collection and merely absorbs some energy and momentum.

- In theory there is (at least usually) not THE process to create some target collection. For example you can create an electron-positron pair not only with photons but also with different reactions, the simplemost being a Z-boson decaying into electron and positron. You don't find many (for any practical solution you could even say "you don't find any") Z-bosons naturally existing but there's quite some (in theory an infinite amount of) possible configurations that could create the Z for you, e.g. shooting a suitable lepton-antilepton pair or a suitable quark-antiquark pair onto each other. So if you happen to read collection X is created from collection Y, then it should in this context not be read as Y being the only collection that can create X but read as Y being the best-achievable, the most common (in nature or applications), the easiest to understand or perhaps the only experimentally used collection to create X.
 
trewsx7,
As has been said here, our most common way of "creating matter" from energy via E =mc^2 is a particle collider. For instance you can have e + e and lots of kinetic energy go to e + e + p + n and lots of other particles. You can think of this as converting kinetic energy to rest mass, or "creating matter from energy".
Jim Graber
 
Matter creation from photons only occurs in close proximity to atomic nuclei. It is said that a sufficiently energy photon (>1.022 MeV) can cause the production of an electron and a proton just as is stated in the Wiki article on pair production.

The question I have about matter creation in this process is: How do we know/decide/measure if one of the neutrons within the core has or has not been "annihilated" or "converted" into the "so-called" created matter (electron and proton) from the interaction of the gamma ray with the atom?
 
Timo said:
Two comments:

- In particle physics energy is not an entity by itself but a property of particles. You do not turn energy into particles but you turn one collection of particles into another collection of particles where energy and momentum conservation dictate that both collections of particles have the same energy and the same momentum. The example of two photons (or one photon and some atom) turning into an electron-positron pair is not an example of energy turning into electron and positron but an example of a collection of particles (photon+X) turning into a different collection of particles (e+ + e-). Not related to what I wanted to say but for completeness: In the photon+atom reaction, the result is e+ + e- + atom; the atom remains in the final collection and merely absorbs some energy and momentum.

- In theory there is (at least usually) not THE process to create some target collection. For example you can create an electron-positron pair not only with photons but also with different reactions, the simplemost being a Z-boson decaying into electron and positron. You don't find many (for any practical solution you could even say "you don't find any") Z-bosons naturally existing but there's quite some (in theory an infinite amount of) possible configurations that could create the Z for you, e.g. shooting a suitable lepton-antilepton pair or a suitable quark-antiquark pair onto each other. So if you happen to read collection X is created from collection Y, then it should in this context not be read as Y being the only collection that can create X but read as Y being the best-achievable, the most common (in nature or applications), the easiest to understand or perhaps the only experimentally used collection to create X.



Nice explanation of particles possessing energy.

My question on this is: If energy is not an entity on its own, and light is massless, then what should I think light is, if it is not an entity on its own?
 
Energy is a property of a system, not an entity on its own. A car which runs 100 miles/h have energy, due to E^2 = p^2 + m^2.

A photon have energy according to E^2 = p^2 + m^2

A photon is in particle physics, a massless particle. A photon in classic electrodynamics is a wave.

I have to quote my professor in Quantum mechanics: "An electron is an electron. It have wave properties, and particle properties. But it is neither of those. An electron is an electron".

A photon/light is an entity on its own, and has energy. But there is no equivalence between light and energy. Light is (a form of) energy, but energy is not light.

Also, that was a quite off-topic question of yours Bucky, I think we have some old threads about this question "what is energy?". Perhaps have a look at those, or make a new thread in the appropriate subforum.
 
Orion1 said:
...
\gamma( E_{\gamma} ) \rightarrow \; n^+ + n^- - Neutron Pair Production
...

That's a process I'm not familiar with...

Protons, neutrons and other hadrons are complicated composite particles. It's best to think of pair production of quarks and antiquarks, followed by a process of hadronization (see en.wikipedia.org/wiki/Hadronization for a start), as described by malawi_glenn.
 
  • #10
daschaich said:
That's a process I'm not familiar with...

Protons, neutrons and other hadrons are complicated composite particles. It's best to think of pair production of quarks and antiquarks, followed by a process of hadronization (see en.wikipedia.org/wiki/Hadronization for a start), as described by malawi_glenn.

The photon might be able to pair produce a quark - anitquark pair, that will undergo hadronization. But neutrons are composite objects in the standard model, so you're right.

Orion1: Can you try drawing the Feynman diagram for the process:
\gamma( E_{\gamma} ) \rightarrow \; n^+ + n^-

Also, what is a positive and negative neutron?..
 
  • #11
malawi_glenn said:
...Also, what is a positive and negative neutron?..

That's all I was saying :wink:. \gamma \to n + \overline{n} through pair production and hadronization is fine by me.
 
  • #12
daschaich said:
That's all I was saying :wink:. \gamma \to n + \overline{n} through pair production and hadronization is fine by me.

what are the odds that only n + \overline{n} are produced? There are thousands of possible hadronization channels.

The photon must produce quark-antiquark pairs.
 
  • #13
malawi_glenn said:
what are the odds that only n + \overline{n} are produced? There are thousands of possible hadronization channels.

The photon must produce quark-antiquark pairs.

True, but at least I know what those symbols mean!
 
  • #14

Correction, the reaction listed on Post #3 should be:
[/Color]
\gamma( E_{\gamma} ) + N \rightarrow \; \overline{e}^+ + e^- + N - Positron + Electron Pair production
\gamma( E_{\gamma} ) + N \rightarrow \; p^+ + \overline{p}^- + N Proton + Anti-proton Pair production
\gamma( E_{\gamma} ) + N \rightarrow \; n^0 + \overline{n}^0 + N Neutron + Anti-Neutron Pair production

Above their respective threshold energies, these are the most probable pair production reaction types because of conservation of mass, charge and colour.

My esteemed colleagues, being experts at conservation laws, will now demonstrate in the next posts what conservation laws this reaction violates:

Universe Pair production:
E_{\gamma} = 2 m_U - Universe rest mass
\gamma( E_{\gamma} ) + N \rightarrow \; U + \overline{U} + N - Universe + Anti-Universe Pair production
[/Color]
Reference:
http://en.wikipedia.org/wiki/Universe"
 
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  • #15
Orion1 said:
Correction, the reaction listed on Post #3 should be:
[/Color]
\gamma( E_{\gamma} ) + N \rightarrow \; \overline{e}^+ + e^- + N - Positron + Electron Pair production
\gamma( E_{\gamma} ) + N \rightarrow \; p^+ + \overline{p}^- + N Proton + Anti-proton Pair production
\gamma( E_{\gamma} ) + N \rightarrow \; n^0 + \overline{n}^0 + N Neutron + Anti-Neutron Pair production

Above their respective threshold energies, these are the most probable pair production reaction types because of conservation of mass, charge and colour.

My esteemed colleagues, being experts at conservation laws, will now demonstrate in the next posts what conservation laws this reaction violates:

Universe Pair production:
\gamma( E_{\gamma} ) + N \rightarrow \; U + \overline{U} + N - Universe + Anti-Universe Pair production
[/Color]
Reference:
http://en.wikipedia.org/wiki/Universe"



What??

Can you draw a feynman diagram for:
\gamma( E_{\gamma} ) + N \rightarrow \; p^+ + \overline{p}^- + N

And a reference to an article about neutron-antineutron photon pair production.

-From your wiki reference on Pair production:
"Pair production refers to the creation of an elementary particle and its antiparticle, usually from a photon (or another neutral boson"

Now a neutron is not an elementary particle, neither the proton.

It is just not a matter of conservation laws, it is how particle interacts. Not everything that are ok with some conservation laws are physical.
 
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  • #16
Orion1 said:
...

Above their respective threshold energies, these are the most probable pair production reaction types because of conservation of mass, charge and colour.

...

No, pion production and other hadronic states involving light mesons are far more probable than the latter two. As malawi_glenn points out, there are thousands of possible final states that can result from the hadronization of quark-antiquark pair production (the more fundamental process), all of which conserve energy, charge and colour. As a rough rule of thumb, processes with larger available phase spaces (lighter particles, generally) will be more likely.
 
  • #17
try to get only two pions from quark-antiquark pair production... (A)
 
  • #18
malawi_glenn said:
try to get only two pions from quark-antiquark pair production... (A)

I never said anything about only two pions, though I did gloss over the complications that phase spaces increase both with the number of particles and with their momenta, while increasing the number of particles reduces the momentum available to each. I think we're in agreement.

PS. Is there any way to turn off the auto-linking to "momentum"?
 
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  • #19
Yes I know that we agree, just wanted to point that out for Orion1 :-)
 
  • #20

The reaction is two-Photon annihilation into baryon + anti-baryon pairs and it is the the most probable PAIR PRODUCTION reaction types because of conservation of mass, charge and colour.

The fundamental reaction is:
\gamma + \gamma \rightarrow q + \overline{q}

The Baryon + anti-baryon reaction is:
\gamma + \gamma \rightarrow B + \overline{B}

The feynman diagrams are listed in reference 1.

Universe Pair production:
E_{\gamma} = 2 m_U
\gamma( E_{\gamma} ) + \gamma( E_{\gamma} ) \rightarrow \; U + \overline{U} - Universe + Anti-Universe Pair production
[/Color]
Reference:
http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Ahep-ph%2F0206288"
 
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  • #21
Orion1 said:
The reaction is two-Photon annihilation into Baryon + anti-baryon pairs.
[/Color]
This is OK in principle. But now come over and measure it :smile:
See, the process is extremely clean once you have extracted it from your background, and that is the all difficulty. But believe me, the authors of the paper are perfectly aware of this difficulty.
edit
and BTW, the process is much easier to investigate in the cross diagram photon/proton -> photon/proton in the deep regime. Actually, since we have been measuring this process already, the all point of the above paper is to present the cross channel in case anybody would come with a bright experimental proposal.
 
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  • #22
I was just about to write what Humanino wrote ;)

"The quark hadronizes into B and the antiquark into B, with any number of soft partons connecting the two parton-hadron vertices."

So it is basically what we have discussed earlier, that what is produced is the "fundamental" reaction. Then there is of course a probability that the hadronization process leads to baryon-antibaryon.
 
  • #23

What is a more probable reaction than a two-photon annihilation into a quark + anti-quark pair production?
\gamma + \gamma \rightarrow q + \overline{q}
[/Color]
 
  • #24
electrons, muons?

also you can not create free quarks, they must have enough energy to hadronize, otherwise colour confinement is not obtained.
 
  • #25

Why would not this reaction be the most probable?
\gamma + \gamma \rightarrow \gamma + \gamma
[/Color]
 
  • #26
Orion1 said:
Why would not this reaction be the most probable?
\gamma + \gamma \rightarrow \gamma + \gamma
[/Color]
Because there is no first order for this reaction (photons are chargeless).
 
  • #27

The reaction of two-photon annihilation into lepton + anti-lepton pairs is the most probable fundamental reaction type.
\gamma + \gamma \rightarrow L + \overline{L}
[/Color]
Reference:
http://en.wikipedia.org/wiki/Lepton"
 
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  • #28
Orion_1, who do we have two photons in initial state?

I thought there was only one photon in pair production?

How is the feynman diagram for two photon annihilation? Can anyone show me?
 
  • #29
malawi_glenn said:
How is the feynman diagram for two photon annihilation? Can anyone show me?
I found this web page (with unfortunately bad resolution plots)
Light interacting with Light[/color]
Below you'll find a little introduction to two-photon physics - porbabely more than you ever wanted to know about it.
:rolleyes:
Also
http://durpdg.dur.ac.uk/spires/hepdata/online/2gamma/opal.html
 
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  • #30
malawi_glenn said:
Orion_1, why do we have two photons in initial state?

I thought there was only one photon in pair production?

Pair production cannot occur in empty space out of a single photon; the nucleus or another photon is required to conserve both momentu_m and energy.

\gamma + N \rightarrow \; e^- + \overline{e}^+ + N
\gamma + \gamma \rightarrow L + \overline{L}

Universe Pair production:
E_{\gamma} = m_U c^2 - Universe mass

\gamma( E_{\gamma} ) + \gamma( E_{\gamma} ) \rightarrow \; U + \overline{U} - Universe + Anti-Universe pair production
[/Color]
Reference:
http://en.wikipedia.org/wiki/Pair_production"
http://en.wikipedia.org/wiki/Universe"
 
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  • #31
haha Darth Vader! =D

I must show this to my professor in elementary particle physics tomorrow =D
 

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