Why don't Proton and Anti-proton collisions produce 3 Neutral Pions.

[Vmedvil]

 

Original Thread being hijacked, why isn't the color field conserved in QCD on P+ , P- collisions, which I brushed on in this topic of why it should be 3 neutral pions. that decay into 6 Photons.

 

The Color field has symmetry so why isn't this used at low energies, p+ + p− → π₀ + π₀ + π₀  Then π₀ → γ + γ

https://webific.ific.uv.es/web/en/content/lattice-qcd-numerical-approach-strong-force

 

https://courses.lumenlearning.com/suny-physics/chapter/33-5-quarks-is-that-all-there-is/

 

[Strange]

4 hours ago, Vmedvil said:

why isn't the color field conserved in QCD on P+ , P- collisions

Both protons and pions are colour-neutral, so it is conserved, isn't it?

(p.s. thanks for providing links; very helpful.)

[Vmedvil]

1 hour ago, Strange said:

Both protons and pions are colour-neutral, so it is conserved, isn't it?

(p.s. thanks for providing links; very helpful.)

Well, my Issue is with (-/+) Charged pions being (DU , UD) The Proton is UUD while the Antiproton is UUD  when they bind it should not have more than three pairs of pions and they should all be neutral for everything to be conserved as neutral pions are (DD . UU) making   

                                                                                                                                                                                    Quarks (Flavor field)

                                                                                                                                                                                   UUD +  UUD ------> UU + UU + DD

                                                                                                                                                                                         Color Field

                                                                                                                                                                                    RBG + RBG -------> RR +BB +GG

                                                                                                                                                                                             Charge Field 

                                                                                                                                                                                                  p+ + p− → π₀ + π₀ + π₀

but for some reason Sensei said it was otherwise.

Edited December 24, 2017 by Vmedvil

[John Cuthber]

Would the masses balance?

[swansont]

A proton or antiproton contains gluons and virtual quarks as well. The assumption that the quarks will automatically match up with their counterpart in the other particle is too simplistic. Additional quark-antiquark pairs can be created, by conversion of kinetic energy into mass energy.

[Vmedvil]

13 hours ago, swansont said:

A proton or antiproton contains gluons and virtual quarks as well. The assumption that the quarks will automatically match up with their counterpart in the other particle is too simplistic. Additional quark-antiquark pairs can be created, by conversion of kinetic energy into mass energy.

Well, this is based on the idea that they are at low energies where kinetic energy is near zero @ higher kinetic energy that would be true.

if  (P+,P-) + KE  ≥  (P+,P-) + UD + DU

Which doesn't explain why there is missing Green Color Charge symmetry in.

RBG + RBG + KE =  RR + BB + GG + RR + BB

Or in 

(P+,P-) + KE  ≥ DU + UD +UD + DU + DD

Which is missing  Green in sensei's model.

RBG + RBG  + KE = RR + BB + GG + RR + BB

 

All the ones Sensei posted is missing a Green Component being 5 and not 3,6,9,12,15 ..... N + 3)

 

The Strong nuclear force disagrees with that statement and tells you it will not bind this without a green component. 

Like (P+,P-) + KE --> π₀ + π₀ + π-  + π₊  + π-  + π₊  would be fine 

(P+,P-) + KE ----->  DD + UU + DU + UD + DU + UD

Then  RBG + RBG  + KE = RR +BB + GG + RR + BB + GG

Which is not missing a Green component. 

Edited December 25, 2017 by Vmedvil

[Sensei]

Check table that I gave in other thread..

There can be created 3 pions as well.

[math]p^+ + p^- \rightarrow \pi^+ + \pi^- + \pi^0[/math] This branch has ratio 6.9%

[math]p^+ + p^- \rightarrow \pi^+ + \pi^- + \pi^+ + \pi^- + \pi^0[/math] This branch has ratio 19.6%

[math]p^+ + p^- \rightarrow \pi^+ + \pi^- + n \pi^0[/math] This branch has ratio 35.8%, n>1 (so at least two neutral pions)

 

[Vmedvil]

8 minutes ago, Sensei said:

Check table that I gave in other thread..

There can be created 3 pions as well.

p++p−→π++π−+π0 This branch has ratio 6.9%

p++p−→π++π−+π++π−+π0 This branch has ratio 19.6%

p++p−→π++π−+nπ0 This branch has ratio 35.8%, n>1 (so at least two neutral pions)

 

The first is fine but the middle one is impossible via color charge symmetry, that last one may or may not be depends on whether you mean a n₀ or π₀ , if it is both then that is impossible.

Edited December 25, 2017 by Vmedvil

[Sensei]

3 minutes ago, Vmedvil said:

The first is fine but the middle one is impossible via color charge symmetry, that last one may or may not be depends on whether you mean a n0 or π0 , if it is both then that is impossible.

n is number of pions produced..

[math]p^+ + p^- \rightarrow \pi^+ + \pi^-[/math] This branch has ratio 0.37%

 

[Vmedvil]

14 minutes ago, Sensei said:

n is number of pions produced..

p++p−→π++π− This branch has ratio 0.37%

 

That is color charge impossible missing green again.

The most likely from looking at this form QCD is p++p−→π₊ + π₋ + π₀ 

 (P+,P-) + KE --> π₀ + π₀ + π-  + π₊  + π-  + π₊  would be fine 

Even p+ + p− → π₀ + π₀ + π₀  

Even p+ + p− →π+ + π− + n₀ 

But not anything that isn't (N + 3), (N+1) is missing Blue and Green, (N+2) is missing Green

Edited December 25, 2017 by Vmedvil

[Sensei]

Source: http://geant4.cern.ch/G4UsersDocuments/UsersGuides/PhysicsReferenceManual/html/img2497.gif

cern.ch

 

[Vmedvil]

27 minutes ago, Sensei said:

Source: http://geant4.cern.ch/G4UsersDocuments/UsersGuides/PhysicsReferenceManual/html/img2497.gif

cern.ch

 

I dunno maybe the SNF symmetry is broken too in such exchanges, which is odd because Electric charge has conserved symmetry in them.

 

QCD says all those non 3,6,9,12,15,.... N + 3 are impossible.

 

Edited December 25, 2017 by Vmedvil

[Strange]

8 hours ago, Vmedvil said:

Which doesn't explain why there is missing Green Color Charge symmetry in.

As long as the resulting combination is colour neutral (colour + anti-colour in the case of pions) it doesn't mater. You are also ignoring the role (and colour) of gluons and the fact that quarks can change colour. 

[Vmedvil]

5 minutes ago, Strange said:

As long as the resulting combination is colour neutral (colour + anti-colour in the case of pions) it doesn't mater. You are also ignoring the role (and colour) of gluons and the fact that quarks can change colour. 

It must not Strange for the Symmetry breaks, I realize that now. It must not be conserved on those breaks.

Edited December 25, 2017 by Vmedvil

[Strange]

And yet it is:

Quote

Color, like charge, is a conserved quantity which cannot be created or destroyed. 

http://webhome.phy.duke.edu/~kolena/modern/hansen.html

[Vmedvil]

8 minutes ago, Strange said:

And yet it is:

http://webhome.phy.duke.edu/~kolena/modern/hansen.html

Then that CERN model is wrong then, it is missing Green and AntiGreen.

Edited December 25, 2017 by Vmedvil

[Strange]

Gosh. CERN vs some random guy who posts nonsense on the Internet. Tough call.

After some thought, I think I will go with CERN.

[swansont]

11 hours ago, Vmedvil said:

Well, this is based on the idea that they are at low energies where kinetic energy is near zero @ higher kinetic energy that would be true.

if  (P+,P-) + KE  ≥  (P+,P-) + UD + DU

Which doesn't explain why there is missing Green Color Charge symmetry in.

RBG + RBG + KE =  RR + BB + GG + RR + BB

 

You have a green and antigreen. How is it "missing"?

[Vmedvil]

6 minutes ago, swansont said:

You have a green and antigreen. How is it "missing"?

The Second GG pair is missing. Do you see how there is two pairs of every color but green.

Edited December 25, 2017 by Vmedvil

[Vmedvil]

3 hours ago, Strange said:

Gosh. CERN vs some random guy who posts nonsense on the Internet. Tough call.

After some thought, I think I will go with CERN.

It is really sad Strange that you do not understand the higher level physics I am speaking about, This is about QCD not my personal views. The Gauge of the Strong Nuclear force or Color Charge and its symmetry not my personal views.

 

https://www.physi.uni-heidelberg.de/~uwer/lectures/ParticlePhysics/Vorlesung/Lect-5a.pdf

http://www.t39.ph.tum.de/T39_files/Lectures_files/StrongInteraction2011/QCDkap2.pdf

Edited December 25, 2017 by Vmedvil

[swansont]

2 hours ago, Vmedvil said:

The Second GG pair is missing. Do you see how there is two pairs of every color but green.

Why is that necessary? What conservation law does it violate?

[Vmedvil]

15 minutes ago, swansont said:

Why is that necessary? What conservation law does it violate?

It violates color charge being a triplet and Gluon Octet, along with Singlet. 

Edited December 25, 2017 by Vmedvil

[swansont]

1 hour ago, Vmedvil said:

It violates color charge being a triplet and Gluon Octet, along with Singlet. 

But you're the one who presented the state. Why did you use the singlet state?

[Mordred]

A proton/antiproton annihilation is rather complex. The 3 quark/antiquark constituents only account for 1 % the total mass. Upon annihilation of the two particles all mass of both are released. They can form any number of other particles provided those particles has less than the total energy of the previous two. 

In point of detail the 3 quarks is simply the inbalance of the proton. There is literally thousands of quarks/gluons that comprise the proton

Mostly this will be pions and kaons however the number can theoretically be as high as 13 pions.

https://en.m.wikipedia.org/wiki/Annihilation

 

Edited December 25, 2017 by Mordred

[Vmedvil]

6 hours ago, Mordred said:

A proton/antiproton annihilation is rather complex. The 3 quark/antiquark constituents only account for 1 % the total mass. Upon annihilation of the two particles all mass of both are released. They can form any number of other particles provided those particles has less than the total energy of the previous two. 

In point of detail the 3 quarks is simply the inbalance of the proton. There is literally thousands of quarks/gluons that comprise the proton

Mostly this will be pions and kaons however the number can theoretically be as high as 13 pions.

https://en.m.wikipedia.org/wiki/Annihilation

 

I realize that part, that was not the issue, the question is does it hold to color symmetry producing them in sets of 3,6,9,12 and so on as sets of singlets from the two triplets.

Edited December 26, 2017 by Vmedvil