I Electron-Positron Annihilation, low energy vs high energy mode- do both end in 511 keV Annihilation?

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High school physics demonstrations teach and show us that a positron and an electron must be at rest with their environment before they can join together and annihilate. The process makes them disappear and they are replaced by two 511 keV rays. It is also said ( but no way to demonstrate) that under high energy conditions (like a collider) the two particles can be forced to annihilate with a more powerful end result. I've searched for days to find a reference that shows this one way or another. I do understand that in a collider, a great deal of energy is added, then stopped(?) in a collision, which if the energy is high enough, even new physical particles can be created, plus all sorts of other things related to kinetic energy, not to annihilation, but what about the rest energy of the electron and positron? It seems that they would still be 511 keV each, and the two rays produced would still be 511 keV.
A reference along with the answer would be great.

George

[Reference Equations:

1B+ + 1B- = 2 X 511 keV

511=511]
 
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BvU

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Not really.


e+e- collisions is the simplest of search terms !
$
Not really.
we are taught that this is the low energy model, and all the measurements in the experiment bear that out. Please do you have a reference when this is different?
Thank you.
George
 

BvU

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'At rest with their environment' is in contradiction with principles of relativity.
Don't have a reference (not expert enough :frown:) . @mfb ?

Wiki ?
 
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High school physics demonstrations teach and show us that a positron and an electron must be at rest with their environment before they can join together and annihilate.
That is wrong and if some teacher says that they are probably a bad teacher.
It is also said ( but no way to demonstrate) that under high energy conditions (like a collider) the two particles can be forced to annihilate with a more powerful end result.
"It is said" is a weird phrase for something that is routinely done (and has been done for decades).

Yes, a great deal of energy is added to the particles. Often thousands of times their rest energy. The rest energy doesn't change (as they are still electrons and positrons) but the total energy increases.
a great deal of energy is added, then stopped(?) in a collision
There is nothing "stopped" in a collision.
and the two rays produced would still be 511 keV.
Where did you get that idea from? Just check conservation of energy: Can that happen? You have an electron, e.g. at 5 GeV, colliding with a positron at 5 GeV. Total energy: 10 GeV. Energy is conserved, so what do you expect the final state to be? The energy has to add up to 10 GeV.
 
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MFB said:
"Where did you get that idea from? Just check conservation of energy: Can that happen? You have an electron, e.g. at 5 GeV, colliding with a positron at 5 GeV. Total energy: 10 GeV. Energy is conserved, so what do you expect the final state to be? The energy has to add up to 10 GeV."
'At rest with their environment' is in contradiction with principles of relativity.
Don't have a reference (not expert enough :frown:) . @mfb ?

Wiki ?
You are wrong. Do you understand the (atomic) physics term "thermalized"? I will be glad to explain but that's not what I'm here for.

This is getting tedious, these constant attacks don't include definitions or citations.

George Dowell
 
That is wrong and if some teacher says that they are probably a bad teacher."It is said" is a weird phrase for something that is routinely done (and has been done for decades).
MFB You are wrong. If you would like I can correct you, with citations form MIT or Princeton (NOT WIKI!!!), one point at a time.. If you don't want that, explain to me with your citations to prove these false statements.

George Dowell
 
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MFB Challenger #1:

OP Question #1 and MFB Answer #1

Citation:

Nature of the Experiment
In our experiment, positrons from a 22Na source are allowed to strike an aluminum sheet. The kinetic
energy of these positrons, initially in the range of 0–0.544 MeV, is rapidly lost to electrostatic interactions
as the positrons move about in the solid aluminum. During this slowing down process, which is similar
to what would be experienced by any charged particle, annihilation of the positron is unlikely. When
a positron has become “thermalized” at the thermal energy level of the lattice (about 0.025 eV at room
temperature), the probability of its annihilation becomes large.

Source: https://www.princeton.edu/~romalis/PHYS312/Positron.pdf


George Dowell
 
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"It is said" is a weird phrase for something that is routinely done (and has been done for decades).

Yes, a great deal of energy is added to the particles. Often thousands of times their rest energy. The rest energy doesn't change (as they are still electrons and positrons) but the total energy increases.There is nothing "stopped" in a collision.Where did you get that idea from? Just check conservation of energy: Can that happen? You have an electron, e.g. at 5 GeV, colliding with a positron at 5 GeV. Total energy: 10 GeV. Energy is conserved, so what do you expect the final state to be? The energy has to add up to 10 GeV.
""It is said" is a weird phrase for something that is routinely done (and has been done for decades)."
Not in high schools. We can't demonstrate the high energy mode in schools any more. Lat time I tried was 1965 and was shut down.

"Yes, a great deal of energy is added to the particles. Often thousands of times their rest energy. The rest energy doesn't change (as they are still electrons and positrons) but the total energy increases.There is nothing "stopped" in a collision.Where did you get that idea from? Just check conservation of energy: Can that happen? You have an electron, e.g. at 5 GeV, colliding with a positron at 5 GeV. Total energy: 10 GeV. Energy is conserved, so what do you expect the final state to be? The energy has to add up to 10 GeV."

We are not far apart. My question was specifically about the electron-proton annihilation event. Originally you told me that GeV were added to the Annihilation Radiation, I need proof. Facts are the GeV of radiation are SHED by the positron (some can be rays, some can be new physical particles of matter). No one disputes that. As a matter of fact, it bears out my claim that X-Rays energy has no upper limit.

"The rest energy doesn't change (as they are still electrons and positrons) "

OK I agree 100%, so how does the Annihilation Radiation energy increase beyond the native rest energy equivalence 511 keV X c^2 of the two particles? Unless they are forced to annihilate by some additional kinetic energy.....which I doubt but am not sure of. If you know (or anyone) please tell me. Perhaps after whipping me at the post for a week, you now understand the original question asked is a perfectly valid one.

George Dowell
 
Then why does the thread title say electron-positron?
Obviously I a made an error while trying to respond to multiple question posts. Positrons it is. Can't edit it now, too late, sorry.

Rewording my question might help : In the high energy e-positron collider, do the newly formed particles come about by the kinetic energy collisions of two equal mass/ equal high kinetic energy but oppositely charged particles, due to energy imparted by the collider first and then the particle/antiparticle undergo thermalized e-p annihilation? Or does the e-positron pair undergo some sort of super annihilation event, then the energy from those new much higher energy Annihilation Radiation photons cause the new particles to appear? Normally in the low energy context, the only one I am familiar with, dictates for electron and positron to be a rest (thermalized), and two 511 keV Annihilation Photons replace them, being a product of their rest mass of 511 keV X c^2 each.

Sorry if my original question was unclear, and my one reply just confused things more, but I'm coming at this from a different frame of reference than many here and am somewhat confused myself.

George Dowell
 
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Moderator's note: Thread moved to Quantum Physics as it is not homework and does not belong in the homework forum. Thread level marked as "I".
 

BvU

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You are wrong.
I don't consider being told 'you are wrong' as an attack. It has happened (quite) often that I could be proven wrong and that constituted a learning opportunity.

What also happened frequently is that in fact there was some miscommunication.
(in this case: what does it mean to you that 'e+ and e- are at rest with respect to their environment'? )

So: can you prove me wrong or clarify your claims ?
 
I don't consider being told 'you are wrong' as an attack. It has happened (quite) often that I could be proven wrong and that constituted a learning opportunity.

What also happened frequently is that in fact there was some miscommunication.
(in this case: what does it mean to you that 'e+ and e- are at rest with respect to their environment'? )

So: can you prove me wrong or clarify your claims ?
Sure, I'll be back in 10 or so minutes with links, citations, and exceprts. Meanwhile, ponder on why positronium can exist.Hint: it's to allow excess kinetic enegy to be shed by interaction with other particles, to prepare for e-p annihilation. This is known as the low energy mode and is the only mode used in indusrtry and medicine.

You asked:"(in this case: what does it mean to you that 'e+ and e- are at rest with respect to their environment'? ) "
My Answer: It's called "thermalized" a very common physics term used in many different nuclear interactions, including neutron absorption. It means to be at rest in the medium, you could also say "with no momentum within the medium" etc.
We all no nothing anywhere can be truly at rest, because the earth, the Solar system, the galaxy and the universe , time-space inflation- all impart vectors on every experiment .

George Dowell

OK 60 seconds and here's the first quote. Its from Princeton U. Nuclear course, I've posted this on PF before.

Excerpt, full text attached as a file:
"Nature of the Experiment
In our experiment, positrons from a 22Na source are allowed to strike an aluminum sheet. The kinetic
energy of these positrons, initially in the range of 0–0.544 MeV, is rapidly lost to electrostatic interactions
as the positrons move about in the solid aluminum. During this slowing down process, which is similar
to what would be experienced by any charged particle, annihilation of the positron is unlikely. When
a positron has become “thermalized” at the thermal energy level of the lattice (about 0.025 eV at room
temperature), the probability of its annihilation becomes large.
The most likely annihilation process involves conversion of the total energy (mostly the rest mass
energies) of a positron and an electron into two gamma-ray photons. "

Also attached is the textbook we in the nuclear metrology science go by. I can leave it for a short time out of respect for Dr. Gilmore's patent (this is an old edition, please buy his latest edition anywhere- online or in libraries>)

George Dowell

edit #2. Ironically while gathering the references for you, I found the answer to the question that I came here to ask.
from the man himself:
Source - http://serious-science.org/videos/447 MIT Senior Research Scientist Frank Taylor on a boson of finite width, quarks, muons, and the Standard Model

If you want to continue this conversation, I'll be happy to take the time to provide further references. For students that might be reading now or someday, I am always available to answer questions on the subject of detecting nuclear and atomic radiation.


George Dowell
email
GEOelectronics@rallstech.net
 

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BvU

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we in the nuclear metrology science
To me that explains a lot of misunderstandings.

'We' in high-energy physics don't understand an expression like 'at rest with their environment'. Lab, center of mass, a missile on the way to Alpha Centauri -- whatever frame of reference: it has to yield the same physics. An e+ and an e- moving in the same direction with .99 c wrt the lab annihilate happily; the photons will not be back to back in the lab and have a combined energy of 7.24 keV .

And yes, this is all low energy stuff. Just not at rest.

And yes, energy conservation dictates that the center-of mass energy is conserved. A bunch of quantum numbers are conserved as well.

I don't see the relevance of Taylor's video for this issue.

Now to the high-energy case

Facts are the GeV of radiation are SHED by the positron (some can be rays, some can be new physical particles of matter). No one disputes that.
In #14 you describe a particular scenario, where a 'high'-energy positron is directed at a static target and ends up in an annihilation event. Since the center-of mass energy isn't enough to produce other stuff with the right quantum numbers, that's the most likely outcome. That changes only a little if you go to e.g. the linac at SLAC, but a beam dump experiment is messy.

In high-energy physics 'we' mainly think of e+ e- colliders, where all the energy (in the tens of GeV range) is in the center of mass. Much cleaner. Shedding radiation and thermalizing are hardly in 'our' vocabulary. The concentration of energy in one very small place is what allows the creation of new stuff from the original particles. There is no stopping.
We are not far apart
our mental pictures are
after whipping me at the post for a week
For a thread you started two days ago ? Or is there some leftover luggage from earlier threads ?
 
I don't think it's logically possible to post that you've been banned. 😉
Not banned from the site yet, but from the threads where I posted my questions and views, and yes my definitions.



George Dowell
 
you said:
"An e+ and an e- moving in the same direction with .99 c wrt the lab annihilate happily; the photons will not be back to back in the lab and have a combined energy of 7.24 keV . "

Please rephrase this quote and correct if if you see fit. As it is written I can't understand it.
Oh and please add if you work at an accelerator?
Thank you.
George Dowell

OK nothing heard so far, so I'll rephrase it as I THINK you mean:
Scenario: The lab is moving, and the sample is moving, (which they are, multiple vectors as I mentioned).
"An e+ and an e- moving in the same direction with .99 c wrt WITH the lab annihilate happily WHEN THERMALIZED ; the photons will not WILL be back to back in the lab and have a combined energy of 7.24 keV 2 X 511 keV = 1022 keV. "

Or in other words, just as the Princeton University nuclear course excerpt I posted says.

Here's an excerpt from a 3rd year physics course at MIT, and the rest of the document explains the other aspects of my sub-discipline, as they call it X-Ray Physics:
"
Annihilation of electron-positron pairs: Some unstable
nuclei (e.g. 22Na) undergo an inverse betadecay
process in which a proton in the nucleus is
transformed into a neutron with the emission of a
positron (anti-electron) and an electron neutrino.
The ejected positron eventually interacts with an
electron in the surrounding material, annihilating
into photons. Such annihilations usually yield two
photons which, in the center of mass frame, travel
in exactly opposite directions and with each carrying
an energy of precisely mec2 in accordance with
the conservation of momentum and energy."

source attached: MIT-JLExp31.pdf

This is pretty much what I teach to gifted 14 year olds and up.

There are some notes on the official course description and pretty much sums up the attitude of many folks who wander into this field:
  1. Estimated Effort
    8.13 requires at least 18 hours per week.
    By nature, laboratory work has no definite point of completion, so 8.13 students sometimes continue working well beyond the required level. While there have been cases of students spending excessive amounts of time on 8.13, this should be viewed as neither required nor recommended, despite rumors to the contrary. Urban legends regarding 8.13 should be treated with a healthy skepticism, lest they become a self-fulfilling prophecy.
  2. Requirements
    Students enrolling in 8.13 are expected to have recently completed 8.04. If you have not passed 8.04, please see the head Junior Lab instructor.
haha, it's an understatement, I'll be in my lab 18 hours on a good DAY.



George Dowell
 

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I do understand that in a collider, a great deal of energy is added, then stopped(?) in a collision, which if the energy is high enough, even new physical particles can be created, plus all sorts of other things related to kinetic energy, not to annihilation, but what about the rest energy of the electron and positron? It seems that they would still be 511 keV each, and the two rays produced would still be 511 keV.
Your logic in the last sentence here is incorrect.

In a collider experiment, such as the Large Electron Positron Collider, the total energies of the electron and positron in the center of mass frame, which is also the collider's rest frame, are much greater than 511 keV. In the (unlikely) event that the energy of the e-p annihilation when they collide was entirely contained in two gamma rays when detected, the energies of those gamma rays would also be much greater than 511 keV each. The rays don't just contain the rest energies of the electron and positron; they contain the total energies. They have to, otherwise conservation of energy is violated. So your statement...

the two rays produced would still be 511 keV.
...is incorrect.

In actual experiments like the LEP, the energy of the e-p annihilation event is not contained entirely in two gamma rays at the detector. In fact there might not be any gamma rays detected at all. The energy comes out as various heavy particles like W or Z bosons, which is the point of having an experiment like the LEP in the first place: to produce such particles in experiments and measure their properties. But conservation of total energy still has to hold: the total of the energies of all the particles detected at the detector has to equal the total energy of the e+p pair before they annihilate. There is no case where only the rest energies of the electron and positron are contained in the decay products.
 
Your logic in the last sentence here is incorrect.

In a collider experiment, such as the Large Electron Positron Collider, the total energies of the electron and positron in the center of mass frame, which is also the collider's rest frame, are much greater than 511 keV. In the (unlikely) event that the energy of the e-p annihilation when they collide was entirely contained in two gamma rays when detected, the energies of those gamma rays would also be much greater than 511 keV each. The rays don't just contain the rest energies of the electron and positron; they contain the total energies. They have to, otherwise conservation of energy is violated. So your statement...



...is incorrect.

In actual experiments like the LEP, the energy of the e-p annihilation event is not contained entirely in two gamma rays at the detector. In fact there might not be any gamma rays detected at all. The energy comes out as various heavy particles like W or Z bosons, which is the point of having an experiment like the LEP in the first place: to produce such particles in experiments and measure their properties. But conservation of total energy still has to hold: the total of the energies of all the particles detected at the detector has to equal the total energy of the e+p pair before they annihilate. There is no case where only the rest energies of the electron and positron are contained in the decay products.
Thank you for your kind reply. So you are saying the added kinetic energy somehow goes through the annihilation? Would not two negatrons (negative electrons) with the same kinetic energy also shed that energy in a collision and produce similar results without mentioning the concept of antimatter annihilation? Maybe the experiment's don't even look for it specifically because they think everyone knows this already?

PS Yes everyone knows that the energy added to any particle in any collider can result in massive particles, rays etc. being formed. What has this to do with e-p annihilation? The reason they used e-p was not for the annihilation, it was for the two equal mass but opposite polarity particles which would naturally spin in opposite directions in a circular magnet.

Every e-p annihilation in the low energy model is created with excess energy and has momentum, which in every case is shed by normal electron mechanisms, radiative and collision alike. At the end of that is left a thermalized positron. This is not a personal theory, please look at the Princeton, MIT, NRC and other references I have provided in this and other threads.
Thank you for your reply.

PS I've just sent same question to SLAC. There are 17 DOE National Labs in USA, one of them will eventually answer me with a report or some other citation.

George Dowell
 
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So you are saying the added kinetic energy somehow goes through the annihilation?
Of course it does; it has to, by conservation of energy. It can't be carried by the electron and positron because they no longer exist, so it has to be carried by whatever does exist after the annihilation.

Would not two negatrons (negative electrons) with the same kinetic energy also shed that energy in a collision
No, because they don't annihilate each other; they're still there after the collision. Total energy has to be conserved.

If the two electrons (e-) collided with enough energy to produce other particles, then some of the total energy could be carried out by those other particles as kinetic (as well as rest) energy. But the total energy still has to come out somewhere; it can't just disappear.

The simple answer is that total energy is always conserved; total energy includes rest energy and kinetic energy. This is such a basic principle that I am extremely surprised to see you talking as if it doesn't have to hold.

Yes everyone knows that the energy added to any particle in any collider can result in massive particles, rays etc. being formed. What has this to do with e-p annihilation?
It doesn't have anything to do with e-p annihilation specifically; the principle holds for any collision with sufficient energy, not just an annihilation. You are the one that started talking on e-p annihilation and making incorrect claims about it, not me.

Every e-p annihilation in the low energy model is created with excess energy and has momentum, which in every case if shed by normal electron mechanisms, radiative and collision alike. At at the end of that is left a thermalized positron.
Sure, for this particular type of experiment. But that does not mean that in other experiments like the LEP, where the energies are much higher, that e-p annihilation can't occur, or that somehow thermalization has to take place before it can occur, or that somehow the much higher kinetic energies of the electron and positron just disappear instead of appearing in the collision products. In other words, you are incorrectly extending a property that applies to one particular class of experiments involving e-p annihilation, to any e-p annihilation whatsoever.

I've just sent same question to SLAC. There are 17 DOE National Labs in USA, one of them will eventually answer me with a report or some other citation.
Nobody here needs a citation to document how low energy e-p annihilation experiments of the type you describe work. That's not where the problem lies with the claims you are making. See above.
 
or that somehow the much higher kinetic energies of the electron and positron just disappear instead of appearing in the collision products. In other words, you are incorrectly extending a property that applies to one particular class of experiments involving e-p annihilation, to any e-p annihilation whatsoever.
No sir, I am not meaning to do that. I am asking the question, and the real answer has ramifications at every level of science, particularly astro-physics where these high energy phenomenon are common place.

Obviously we agree that the e and p are carriers of induced kinetic energy. Their rest mass is inconsequential in terms of the collision, as are their ultimate fate. Please show me two opposed Annihilation Photons of any energy that are 180 degree opposition of any energy if you will. That would be proof of excess energy annihilation in the matter-antimatter sense.
George Dowell
 
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Please show me two opposed Annihilation Photons of any energy that are 180 degree opposition of any energy if you will. That would be proof of excess energy annihilation in the matter-antimatter sense.
You are basically asking for proof that conservation of energy holds. Every high energy physics experiment ever conducted shows that.

Thread closed.
 
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[Moderator's note: Added for the sake of completion.]

Summary: Electron-Positron Annihilation, low energy vs high energy mode- do both end in 511 keV Annihilation? ANSWERED!

From the friendly folks at Berkeley/ DOE National Lab:
www.particleadventure.org/eedd.html


1566164248973.png


George Dowell
 
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