# Electron vs. positron? / spin?

1. Feb 18, 2005

### 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

2. Feb 18, 2005

### 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.

3. Feb 18, 2005

### 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?

4. Feb 18, 2005

### 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!

5. Feb 18, 2005

### dextercioby

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

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

No,there's no connection.

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

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

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

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.

6. Feb 19, 2005

### Physicist

Thanks all.

What does leptonic charge mean?

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

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

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 alot

7. Feb 19, 2005

### 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.

8. Feb 19, 2005

### 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 im just as confused as the layman on what it really *means* at an intuitive level, and ive been practising physics for over ten years.

9. Feb 19, 2005

### dextercioby

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...

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"...

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

Daniel.

10. Feb 20, 2005

### Physicist

OK.. things are getting clearer now.

Thanks all !

11. Feb 23, 2005

### 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)

12. Feb 23, 2005

### dextercioby

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

Daniel.

13. Mar 2, 2005

### da_willem

Last edited by a moderator: Apr 21, 2017
14. Mar 2, 2005

### 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.

15. Apr 3, 2005

### 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 ?

16. Apr 4, 2005

### 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.

17. Apr 4, 2005

### humanino

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.

18. Apr 5, 2005

### Creator

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.--

Last edited: Apr 5, 2005
19. Apr 14, 2005

### Sterj

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

20. Apr 14, 2005

### Astronuc

Staff Emeritus
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.