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Posted

We constantly see things like atomic nuclei and protons being smashed against each other from opposite directions to see what results but what happens if two electrons are smashed together at near light speed? 

Posted

A lot of the same reactions as with other particles. There is an advantage to doing so, as they are fundamental particles, so they don't make as many extraneous products as with protons (with all of their quarks)

The problem is that for the same energy, electrons have to move faster than heavier particles, which means more energy loss as they accelerate. This ultimately limits the collision energy. You can't easily get them to the TeV energies we see in the LHC, for example. The device that was in place before the LHC was LEP, an electron-positron collider. 

Posted
  On 12/21/2017 at 10:27 PM, J.C.MacSwell said:

Besides the speeds and energies involved, any remarks on the difficulty of getting point like masses to collide?

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They have a non-zero interaction cross section. (Basically because the electromagnetic field does not require a direct hit.)

Posted (edited)

Good answer, couldn't have said it better myself. Unfortunately the only easy non math way to describe a cross section is to recall particles are field excitations, whose wavefunctions will give a probability function that will correspond the probability of scattering. This is also related to the particles mean liftetime.

Under QFT it is part of the Feyman path integrals.

Rutherford scattering under QM

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

The way QM and QFT will mathematically describe this will be different as the two use different operators and QFT will use Klien Gordon as opposed to Schrodinger but that takes a considerable time to describe the distictions. Its easiest to understand under QM first. (QM will teach the required fundamentals)

A very easy way to identify a QM vs a QFT treatment if your self studying outside a textbook or formalized training.

QM operators are position and momentum (priori is the particle)

QFT operators are creation and annihilation ( priori is the field not the particle)

Edited by Mordred
Posted (edited)

Well, Electron on positron collisions create Gamma radiation, Proton on Anti proton create gamma rays and High Energy Proton- anti proton many particles, High Energy Proton on Proton many particles from their energy mass, So High energy Electron on Electron should create many particles from their energy mass like proton on proton through Tachyon Condensation of the Higgs Field.

Low Energy

 

positron1.jpg

Edit: ya, see I thought that looked wrong, I thought it was 6 photons not 10 photons for low energy Proton-Antiproton, but I knew the electron-positron was right so I trusted it. Refer to sensei's Post after this one for low energy Proton-Antiproton.

 

High Energy Matter/Antimatter collider

 

grimm_collision_150c.gif

 

 

High Energy Matter/Matter collider

 

proton-proton-collisions.gif

cern_10.jpg

download.png.6622300964a4879f871cd7658ba30570.png

 

So, basically at high energies collisons cause most of the Energy to be dumped into Rest mass.

Higgs-Field.jpg

+fermion+mass.jpg

 

So Muons or Tauons as a result even in electron on electron depending on what energy is actually required to break the symmetry of electrons higher would be Tauons, lower Muons.

particles.png

Edited by Vmedvil
Posted (edited)

The image above is incorrect. Proton-antiproton annihilation does not produce two gamma photons.

The most typical result of annihilation of proton-antiproton is:

p^+ + p^- \rightarrow \pi^+ + \pi^- + \pi^+ + \pi^- + \pi^0

other paths:

p^+ + p^- \rightarrow \pi^+ + \pi^- + \pi^0 + \pi^0 + \pi^0

p^+ + p^- \rightarrow \pi^0 + \pi^0 + \pi^0 + \pi^0 + \pi^0

(and dozen more)

 

Later:

\pi^0 \rightarrow \gamma + \gamma

\pi^0 \rightarrow \gamma + e^+ + e^-

\pi^+ \rightarrow \mu^+ + v_{\mu}

\pi^- \rightarrow \mu^- + \bar{v}_{\mu}

Edited by Sensei
Posted (edited)
  On 12/23/2017 at 7:42 AM, Sensei said:

The image above is incorrect. Proton-antiproton annihilation does not produce two gamma photons.

The most typical result of annihilation of proton-antiproton is:

p++pπ++π+π++π+π0

other paths:

p++pπ++π+π0+π0+π0

p++pπ0+π0+π0+π0+π0

(and dozen more)

 

Later:

π0γ+γ

π0γ+e++e

π+μ++vμ

πμ+v¯μ

Expand  

Correct form of low energy Proton-Antiproton.

  On 12/23/2017 at 7:42 AM, Sensei said:

The image above is incorrect. Proton-antiproton annihilation does not produce two gamma photons.

The most typical result of annihilation of proton-antiproton is:

p++pπ++π+π++π+π0

other paths:

p++pπ++π+π0+π0+π0

p++pπ0+π0+π0+π0+π0

(and dozen more)

 

Later:

π0γ+γ

π0γ+e++e

π+μ++vμ

πμ+v¯μ

Expand  

 Sensei, why is it 5 (0,-,+) pions and not 3 or 6?

250px-Quark_structure_proton.svg.png151401856841679.png.451dcaff82f6cc0bd306e9b96a47f515.png

Edited by Vmedvil
Posted (edited)
  On 12/23/2017 at 7:53 AM, Vmedvil said:

 Sensei, why is it 5 (0,-,+) pions and not 3 or 6?

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I just gave the most typical annihilation modes (the highest probability of happening).

Theoretically there is even 13 pions plausible (above it, is not possible, because there is not enough mass-energy. 2 m_p / m_{\pi} )

 

Edited by Sensei
Posted (edited)
  On 12/23/2017 at 8:44 AM, Sensei said:

I just gave the most typical decay modes.

Theoretically there is even 13 pions plausible (above it, is not possible, because there is not enough mass-energy. 2mp/mπ )

 

Expand  

well, ya I had always thought it was pp− → π0 π0 π Then π→ γ γ

Edited by Vmedvil
Posted (edited)
  On 12/23/2017 at 8:48 AM, Sensei said:

Here you have table:

 

Expand  

Okay, but I don't get why it doesn't hold to color charge and Anti color charge symmetry rules being mesons generated in QCD.

400px-Qcd_fields_field_(physics)_svg.png.cbaa8ce1a25757702f99c0e5782388c1.png

Untitled.thumb.png.3dd096d15d22c81ea6a3846dca026883.png

Edited by Vmedvil
Posted
  On 12/23/2017 at 8:50 AM, Vmedvil said:

Okay, but I don't get why it doesn't hold to color charge and Anti color charge symmetry rules being mesons generated in QCD.

400px-Qcd_fields_field_(physics)_svg.png.cbaa8ce1a25757702f99c0e5782388c1.png

Untitled.thumb.png.3dd096d15d22c81ea6a3846dca026883.png

Expand  

Please stop posting images with no reference or credit.

Posted (edited)

What Strange wanted to say, is that you have to include link to the reference material in the post where you're using it.

 

Edited by Sensei
Posted
  On 12/23/2017 at 9:12 AM, Vmedvil said:

Google QCD first link

Expand  
!

Moderator Note

From Rule 2.2 (click on guidelines near top of page) "Plagiarism/copyright violation is unacceptable."

That applies to text and images.

 
Posted (edited)

Well, it is not like they didn't steal it someone who did the actual work like maybe a scientist. isn't that a form of plagiarism. I don't see any credit to the creators of QCD or the Standard model on their websites either.

Edited by Vmedvil
Posted
  On 12/23/2017 at 10:09 PM, Vmedvil said:

Well, it is not like they didn't steal it someone who did the actual work like maybe a scientist. isn't that a form of plagiarism. I don't see any credit to the creators of QCD or the Standard model on their websites either.

Expand  

As you don't provide a link to your sources, one can't verify this. But if you are referring to wikipedia, for example, they have very strict rules about providing references to support the statements made. You would do well to follow their lead.

And you might also want to get yourself a decent dictionary and look up the word "plagiarism".

Posted (edited)

 

According to the Merriam-Webster online dictionary, to "plagiarize" means:

  • to steal and pass off (the ideas or words of another) as one's own
  • to use (another's production) without crediting the source
  • to commit literary theft
  • to present as new and original an idea or product derived from an existing source.

 

well, I guess I broke the second one, but it wasn't theirs to begin with those sites. I usually credit the actual person that discovered it. combinations of multiple ideas do not fall under 4 so, I have never broken that as long as they are 50% to 33% different. Ask a patent lawyer about that one.

Edited by Vmedvil

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