# Anti-matter question

1. Sep 30, 2004

### Slacker

Hello, I'm new here and I have a question about anti-matter. Does annihilation occur when a particle and it's anti-particle touch, or does any combination of particles and anti-particles annihilate? Can someone help explain this to me?

2. Sep 30, 2004

### humanino

Welcome Slacker !

You may like to view antimatter as Feynman understood it : antimatter is ordinary matter traveling backwards in time.

Hard to swallow. Well, it is very efficient to picture it this way, and besides, nobody could tell you "you are wrong" with this conception. Imagine a one + one dimensional world : there is one time dimension, and only one space dimension. It is very convenient, because spacetime is reduced to a plane, and we can actually visualize it. Now, there is only one particle around. What do we see : a single curve. The particle moving in spacetime draws a curve. (This is the same in actual 3+1 diemsional spacetime, we just cannot visualize spacetime, that would require to go in a 5 dimensional space and contemplate the 4-dim hypersurface with a curve in it. See, let us stick to the 1+1 world for now.)

OK. In order to make the discussion easier, the time axis is vertically oriented, future goes up. The unique space dimension is horizontal. Now our fuzzy particle decides to change direction in time. Let us say, it should emit a photon in order to conserve energy and momentum, but let us not care about this outgoing photon for now. It goes backwards in time for a while (down), then meets another photon, and reverses once again to go back forwards in time (up). Let us also not care for now about the ingoing photon required to make the second flow reversal. We can consider 3 regions : in the far past, there is only our single and sad particle going towards the future. Likewise, in the far future, there is the same sad single curve. But in between somewhere, we see two curves ! As it appears to us, they look like different particles. Besides, the curve corresponding to our buddy travelling backwards in time has all the properties of an antiparticle : if for instance our particle carries electrical charge, this charge will seem to us to be opposite when the flow is towards the past !

Now let us go back to our photons : there are two photons in the plane. As it seems to us, one photon in the lower part is creating a particle-antiparticle pair, the particle escapes to infinity, and the antiparticle eventually meets an identical particle and this meeting produces a decay in the outgoing photon.

The photon carry no quantum number such as electrical charge, or other kinds of charges (except for angular momentum, but this is also linked to spatial movement, so it does not need to be conserved at the level of internal charges). You see that you need exactly the opposite quantum numbers to produce annihilation. And from the Feynman-backwards-in-time-flow view point, this is obvious.

3. Sep 30, 2004

### Slacker

OK, I think I understand. So, let's say there was an isotope of hydrogen with one neutron. If an atom composed of one anti-proton, one neutron, and one positron were to touch it, they would annihilate. Does the number of neutrons affect whether or not they annihilate?

4. Sep 30, 2004

### humanino

I could not tell for sure, but in my opinion if such a reaction was performed, the first meeting would be the electron/positron annihilation, and this would already produce so much energy (twice 511 keV) that both nuclei could blow up before they even meet. So I cannot even tell if the proton/antiproton annihilation would actually occur

Besides, is there really a bound state antiproton/neutron ? The annihilation would have already occured at the quark level before one can even produce the second "atom".

5. Oct 2, 2004

### Mk

Thanks humanino, I didn't know that, but when the anti-particle and particle annihiate, is it as if they were never there? Forward in time and backward in time cancel each other out? If so, where does the information go? According to quantum theory information can't be destroyed, but then again it isn't, 'cause it was never there... but it was there before it was never there... my head hurts.

6. Oct 2, 2004

### zefram_c

When they annihilate, their energy, momentum and information content is carried by the decay products: usually photons for charged particles, but other end products are also possible.

Personally, I find that thinking of antiparticles as particles going back in time is just a useful mathematical trick - while some physics about the particle/antiparticle symmetry emerges, we always observe the antiparticle as moving forward through time and we can't observe stuff moving backwards through time anyway. So I try not to read too much into this - someone here can correct me if I'm missing something.

7. Oct 2, 2004

### Mk

Thank's zefram_c.

8. Oct 23, 2004

### NEOclassic

Couldn't the neutron be ambidextrous?

On page 1122 of my 5th Ed. of Halliday-R-W's Fundamentals of Physics extended, there is a photograph of the annihilation of an anti-proton that was propelled into a real proton loaded target that produced 4 pi+ mesons and 4 pi- mesons. 7 of these were in flight at the time of the photograph but the first +/_ pair produced were directed in the direction opposed to that of incident anti-proton which is hard to understand for momentum conservation purposes. There was a fork in the image of the first Pi+ (perhaps the magnetic field of the bubble chamber was responsible for the absence of the anticipated fork in the track of the Pi- meson.) which fork represented the decay to the mu+ meson that deays, 100 %, to a positron that annihilates with any nearby electron. The upshot is that the anti-proton annihilates in pieces with each annihilation producing 1.022 MeV of total photon energy(with no photon more energetic than 0.511 MeV).

When it is remembered that the spin axis of the anti-neutron is merely the opposite end of the spin axis of the neutron, it merely flips 180-degrees. The neutron binds equally well to either real- or anti- protons. Cheers, Jim

9. Oct 27, 2004

### humanino

I am not saying $$p^+/p^-$$ annihilation can not occur. I was questionning whether in the case of hydrogen/antihydrogen the energy released by $$e^+/e^-$$ (which occurs first) could prevent the protons from meeting at all. It might depend on the experimental conditions really.

10. Nov 2, 2004

### NEOclassic

Hi Humanino,
In actual experiments, anti protons are fired into a hydride target with the result that 4 plus pions and 4 minus pions were pictured in a bubble chamber. If the exposure had been longer all mesons would have decayed to like charged muons. One positive pion that moved more slowly than the others did show a kink in its trajectory when the muon appeared. Cheers, Jim

PS. Lest your forget; opposite charges attract each other so that nothing can happen before annihilation.

Last edited: Nov 2, 2004
11. Nov 3, 2004

### humanino

In actual experiments 20 years ago ? Nobody uses bubble chambers anymore !
Right. What is the point ?
I do not hear this argument :uhh: Let us look at energies : the binding energy of two unit charges at a distance of the order of the atomic scale is around 13.6 eV, whereas the annihilation of the electron/positron pair yields twice 511 keV. So once the $$e^+/e^-$$ pair has annihilated, what is happening to the $$p^+$$ remaining ? Eventually, it will annihilate for sure, ok. The question was whether this would occur with the nucleus of the original hydrogen nuclei, or with any other random proton somewhere else.

12. Nov 6, 2004

### Antimatter

Yup, indeed, big problem p~=(u~u~d~), n= (udd) [~stands for anti], so we are looking at 2x2 quark annihilations (u~u)+(d~d) (therefore quite unstable) if you ask me.