Electron vs. positron? / spin?

[Physicist]

Hi all,

(1) What is the difference between electron & positron?

I know that they have opposite charges, but what does that deeply mean?

Someone told me that the different sign of the charge is related to the spin:
Electron is said to have spin up if it is spinning CCW, but the positron is said to have spin up if it is spinning CW.

Is that right? If so, can someone explain it more?

& If that's true what will be the difference between an electron with spin up & a positron with spin down?

(2) What does different values of spin physically mean? Is it related to the speed of spinning? What is the difference physically between integer spin & half integer spin?

Thanks

 

[arivero]

No, both electron and positron have the same possible values of spin.

In some sense, charge is due to the complex character of the wave function.

 

[arivero]

In insteresting, related question, is how do we distinghish between atraction and repulsion. It is even better if one things about two different particles. ¿How, or why, should an electron be attracted to an +2/3 quark but repelled from a -1/3 one?

 

[HallsofIvy]

It was Feynman who pointed out that we can think of a positron as an electron that is "moving backward in time". Draw a graph with the vertical axis representing position on a line and the horizontal axis representing time. Draw a broken line zig-zagging across the graph. If we place a "mask" with a narrow open vertical line, representing a specific time, over the graph we can observe each section of the line as separate particle. Sections of the line with positive slope represent electrons, negative slope positrons. Moving the mask slowly to the left (time passing) we see the "particles" moving toward or away from each other. When we pass a bend in the line, if it jogs back to the left, we see positron and electron annihilating each other; if it jogs to the right, we see an electron-positron pair being created.

That explains why all electrons are identical- there is really only one electron bouncing back and forth in time!

 

[dextercioby]

Hi all,

(1) What is the difference between electron & positron?

Taken as particles from a table of particles,simply the electric and leptonic charges are different.

I know that they have opposite charges, but what does that deeply mean?

That,one being the antiparticle of the other,under certain conditions,they would annihilate each other into one or 2 gamma photons...

Someone told me that the different sign of the charge is related to the spin

No,there's no connection.

Electron is said to have spin up if it is spinning CCW, but the positron is said to have spin up if it is spinning CW.

Elementary particles do NOT spin.That's a false hence incorrect image.

(2) What does different values of spin physically mean?

On the surface,different behavior of the classical fields under the Poincaré group.

Is it related to the speed of spinning?

Already told u,nothing is spinning...in the real world.

What is the difference physically between integer spin & half integer spin?

That's not so easy to explain:it's better if you get a good reading on Sakurai to understand the concept of "spin"...

Daniel.

 

[Physicist]

Thanks all.

Taken as particles from a table of particles,simply the electric and leptonic charges are different.

What does leptonic charge mean?

Elementary particles do NOT spin.That's a false hence incorrect image.

This is really strange! all the time we were studying about spin I thought it means rotating around itself.

nothing is spinning...in the real world.

What do you mean "real world"? Did physicist assume the spin theorotically only?

On the surface,different behavior of the classical fields under the Poincaré group.

Could you explain in simpler way?

I think I need to understand more about physical meaning of the spin.

Sakurai

Is that the author of a book? Can you give full name plaese?

Thanks a lot

 

[misogynisticfeminist]

the electron or any particle does not spin in the classical sense, they usually present spin as something spinning clockwise or counterclockwise in lower level classes (especially in chemistry). If I'm not wrong, spin is purely mathematical and cannot be visualized.

 

[Haelfix]

The term spin is more of a historical word in the context of quantum mechanics, in the early days people thought it might be isomorphic in some strange context to classical spinning tops. And in one kinda naive way it is, but in general it is not.

Its best to think of spin like all quantum numbers, just some *thing* that satisfies various mathematical properties. Later in the life of quantum mechanics, Gell-Mann and others invented nifty, but meaningless words to describe other quantum numbers. Flavor, color, etc etc

Personally I am just as confused as the layman on what it really *means* at an intuitive level, and I've been practising physics for over ten years.

 

[dextercioby]

What does leptonic charge mean?

It's a quantum numbaer especially for the leptons and it is very useful in studying decays...For example,this number tells that in the beta decay,one must have an antineutrino alongside the other lepton,the electron...

What do you mean "real world"? Did physicist assume the spin theorotically only?

Those rotations assumed by the definition of "spin" are in an abstract way,basically not particles (or any thing with physical relevance),but mathematical objects "rotate"...

Could you explain in simpler way?
I think I need to understand more about physical meaning of the spin.
Is that the author of a book? Can you give full name plaese?
Thanks a lot

From what i know of,J.J.Sakurai ("Modern Quantum Mechanics") gives the best account on spin and group theory...

Daniel.

 

[Physicist]

OK.. things are getting clearer now.

Thanks all !

 

[bbtuna]

Boy, my first two posts are nitpicks. Not that I am always this way, but when I see something wrong, I'll try and fix it. That, and the fact that this subject relates what I do at work.

dextercioby wrote:
"they would annihilate each other into one or 2 gamma photons..."

Positrons cannot annihilate by only one photon. That would violate conservation of momentum. In my lab, we create positronium. Looking at this atom in the center of mass, the net momentum is zero. When an annihilation occurs, there must be at least two photons. How can there be more than two? In some cases (the annihilation of orthopositronium, o-Ps, a bound state of an electron and positron where the spins of the particles are parallel, there needs to be at least 3 photons to conserve momentum and spin), there can be more than 3 photons, but more than 3 is a rare case. Parapositronium (p-Ps) has the spins of the particles anitparallel, so there must be 2 photons emitted. In these cases, not only is momentum conserved, but spin is also (photons have a spin of 1)

As to what spin really is, like others have said, it's just a thing made up to explain observations and have the theories make sense. One could just as easily say the same thing about mass. What is it really other than something made up to explain observations?

(sorry for the nitpick)

 

[dextercioby]

Boy, my first two posts are nitpicks. Not that I am always this way, but when I see something wrong, I'll try and fix it. That, and the fact that this subject relates what I do at work.

dextercioby wrote:
"they would annihilate each other into one or 2 gamma photons..."

Positrons cannot annihilate by only one photon. That would violate conservation of momentum.

(sorry for the nitpick)

If i thought this nitpick would be correct,then i would not have posted this one... Theory (QED,to be exact) can very much prove the validity of my assertion:

they (an electron & a positron (added now for clarity))would annihilate into one or 2 gamma photons..."

Of course,the 3 photon case was,up until your post,unknown to me.

Daniel.

P.S.Looking forward to an answer.

 

[da_willem]

Personally I am just as confused as the layman on what it really *means* at an intuitive level, and I've been practising physics for over ten years.

For an alternative interpretation see: AJP -- June 1986 -- Volume 54, Issue 6, pp. 500-505

Abstract& Discussion: https://www.physicsforums.com/showthread.php?t=60845, http://www.lns.cornell.edu/spr/1999-11/msg0019568.html

Btw: Dextercioby, have you already read Belinfantes original work? What do you think?

 

[dextercioby]

Nope.I told you then what my feelings were.Discussing Dirac field in that ancient notation will require too much energy and patience from my side...

Daniel.

 

[tritonphysics]

Speaking of annihilation ...

It's pretty clear that an electron and positron coming together
result in two photons created and moving off in opposite directions.
This forms the basis for PET imaging.

However, pair production (creation of electron and positron)
occurs from only a single photon. Isn't this odd ?
Why would annihilation produce two photons, but only one
photon causes pair production ?

Any thoughts ?

 

[dextercioby]

If the 2 photons are real,then u'd invoke conservation of energy-momentum.For the second case,the same conservation of energy momentum tells u that at least one of the 3 particles in a QED vertex must be virtual.So it's a virtual photon------>a real pair,or a a virtual photon emitted/absorbed by a real electron/positron,or a real photon ---------->a virtual pair...

Daniel.

 

[humanino]

Positrons cannot annihilate by only one photon. That would violate conservation of momentum.

If i thought this nitpick would be correct,then i would not have posted this one... Theory (QED,to be exact) can very much prove the validity of my assertion

I'd like to make a clarification. I think dextercioby missed bbtuna's point. An electron and a positron can indeed annihilate into only one virtual photon. But by no means this photon is directly measurable. The point is the following : real photons are massless.

I think it is very simple to see why $$e^+e^-\rightarrow\gamma^*$$ requires the photon to be virtual. It is merely geometrical. Since the electron and the positron are massive, they cannot be on the light cone. Real photons on the contrary are massless, so they are on the light cone. Instead of doing calculations of energy-momentum conservation to show that the virtual photon $$\gamma^*$$ resulting from $$e^+e^-$$ annihilation cannot be on the light cone, argue the following way : consider light cone coordinates. The total momentum of the system $$e^+e^-$$ must have a component orthogonal to the light cone, because both particles are inside the light cone. That's it. A real photon would not be able to carry this component.

Therefore, there are diagrams in QED such as $$e^+e^-\rightarrow\gamma^*$$ but those describe only one single term in the expansion of a partial amplitude. The total amplitude for a real process require both

• all terms of the expansion (radiative corrections and so on) This might be irrelevant if one is satisfied why the approximation given by the first term.
• the remaining part of the process, where at least the virtual photon has to decay into something else. This cannot be irrelevant, and is bbtuna's point.

 

[Creator]

It was Feynman who pointed out that we can think of a positron as an electron that is "moving backward in time". ...
That explains why all electrons are identical- there is really only one electron bouncing back and forth in time!

Well, would you please tell him to explain why so many more electrons are bouncing forward in time than backward. :tongue2:

Creator

--Time is God's way of preventing everything from happening all at once.--

 

[Sterj]

Why can't an electron and a positron "decay" into one photon?

 

[Astronuc]

However, pair production (creation of electron and positron)
occurs from only a single photon. Isn't this odd ?
Why would annihilation produce two photons, but only one
photon causes pair production ?

Pair production requires that a photon (usually a gamma ray) of at least 1.022 Mev (2 * electron rest mass) interact with a particle (usually atomic nucleus). The nucleus shares the energy/momentum with the electron-positron pair, and so mass-energy and momentum are conserved.

 

[Sterj]

Why has a nucleus (or such a thing) to be there? I mean almost everything is conserved (energy, charge) if there is an electron and positron created from a photon. So why these strong electric field? Why is it acquired?

 

[Meir Achuz]

"Almost" is not enough.
A photon cannot decay to just a positon and electron, because this cannot conserve both energy and momentum. A simple way to see this is to form the invariant
E^2-p^2 (letting c=1). This must be postive for the electron and positron state, but zero for a photon. The nucleus must be there to take up some of the momentum,
just as A said.

 

[juvenal]

"Almost" is not enough.
A photon cannot decay to just a positon and electron, because this cannot conserve both energy and momentum. A simple way to see this is to form the invariant
E^2-p^2 (letting c=1). This must be postive for the electron and positron state, but zero for a photon. The nucleus must be there to take up some of the momentum,
just as A said.

A person who doesn't understand the concept of momentum conservation is probably not going to understand the concept of a Lorentz invariant.

 

[Sterj]

I know what he means:
I think he is speaking of this formula: E^2=p^2+m^2 (c=1). And a photon has no restmass. thus: E^2-p^2=0 and a positron and electron have restmass:
E^2-p^2=2*m^2.

 

[juvenal]

Do you understand why conservation of momentum and energy mean that the photon cannot decay into an electron and positron in the vacuum? Or reverse? Because both have been mentioned multiple times in this thread before you asked your questions.

Draw a "T" shape. The vertical represents the path of the photon, and the horizontal represents the electron in one direction and the positron in the other direction. It pretty clear that momentum will not be conserved in the vertical direction. How about a "Y" shape? Well, I can always do a Lorentz transformation to get a "T" shape, and the laws of physics still must hold. Also, due to conservation of energy, the photon momentum (= energy) must be at least twice the rest mass of the electron, so one cannot get the photon momentum even infinitesmally close to zero via Lorentz transformation.

 

[tritonphysics]

Thank you Astrounc for clarifying that pair production requires
interaction with a nucleus. This answers the question as to why only
a single photon can result in the production of a pair of particles.

But, why a pair ? WHen the gamma ray photon (as produced by
pi-0 decay e.g.) encounters a nucleon, you see the result as a pair
of e-/p+ ... why not just an e- sometimes ?

 

[marlon]

I think it is very simple to see why $$e^+e^-\rightarrow\gamma^*$$ requires the photon to be virtual. It is merely geometrical. Since the electron and the positron are massive, they cannot be on the light cone. Real photons on the contrary are massless, so they are on the light cone. Instead of doing calculations of energy-momentum conservation to show that the virtual photon $$\gamma^*$$ resulting from $$e^+e^-$$ annihilation cannot be on the light cone, argue the following way : consider light cone coordinates. The total momentum of the system $$e^+e^-$$ must have a component orthogonal to the light cone, because both particles are inside the light cone. That's it. A real photon would not be able to carry this component.

That is a great way of explaining this, Humanino...This is one to remember. I have encountered this approach once in my astrophysics course at college but i totally forgot about it...Nice work


regards
marlon