# Matter and antimatter must interchangible

1. Jan 4, 2005

### Gamish

If I have 1 KG of matter, and 1 KG of antimatter, and I combine the 2, then I get 2c^2 = 1.79751036 × 10^17 joules, then in turn, I convert the 1.79751036 × 10^17 joules back to 2 KG of mass. So, if antimatter is so anti, why can they be converted? :uhh:

2. Jan 4, 2005

### dextercioby

If u knew a little bit more physics,u'd know that antiparticles don't differ from particles by anything,except electric charge.They have the same mass,the same spin,the same energy.Since mass is equivalent to energy,then pure energy (the 2 gamma photons) can be converted to mass (the particle and its antiparticle) and viceversa,mass (particles with rest mass) go into pure energy (gamma photons).
Actually the problem is a bit more tricky or subtle.The two photons should be one photon and one antiphoton,but since photons are chargeless (by their nature),the photon and the antiphoton are identical.

That's one great thing about relativity.Prediction of antiparticles.

Daniel.

3. Jan 4, 2005

### chroot

Staff Emeritus
When you convert the radiation back to matter, you will not produce just matter. You will, in fact, produce exactly one particle of antimatter for each particle of matter. There is no other way to conserve, for example, electric charge. Photons have no charge, but electrons do; you therefore cannot convert a photon into an electron without breaking the known laws of physics. You can, however, convert a photon into a electron and a positron, conserving the net zero charge.

- Warren

4. Jan 4, 2005

### Gamish

Well, I'm sorry that I am not the best at physics, I have not even taken a class on physics, but I try to learn as much as posible.

And yes, I do know that the positron and the electron (same for all particles) have the exact same properties. But we still refer to them as anti-particles. If we combine matter with antimatter, why do they anniahlate, if they are the exact same thing?

5. Jan 4, 2005

### marlon

The "annihilate" because the interaction of a positron with some electron needs to respect certain conservation laws like charge, just like chroot mentionned. The reason why this is the case can be proven by QED...An electron and a positron have OPPOSITE charge so the net charge must be ZERO. For example a photon has zero charge and that is why an electron and a positron will be "transformed" into a photon. The energy of this photon is determined by other conservation laws like mass and energy...
This photon is the radiation in which these two particles are converted...

regards
marlon

6. Jan 4, 2005

### DB

A couple of problems with what your doing and asking. Your are right to use E=mc^2 though I think you are multiplying Kg times m/s, instead of Kg tmes km/s. Also doing what you are doing will keep your running circles, which is pretty much the basis of anti-matter. As dextercioby said, anti-matter is just difference in charge of a particle; an electron being
$$e-$$,
its anti-matter particle is shown $$\bar{e}$$ a positron.
Same with any particle:

$$\bar{p},\bar\mu,\bar\tau$$

Particles and their anti particles combined to form pure energy, which happens often at microscopic levels. This can happen in just empty space with an energy fluctuation, over and over again. So your question is somewhat meaningless.

Last edited: Jan 4, 2005
7. Jan 4, 2005

### DB

P.S, sorry if I seem repetitive, by the time I had posted there were 3 more replies...!

8. Jan 4, 2005

### dextercioby

:surprised :surprised :surprised :surprised :surprised :surprised
I'm sorry,guys,BUT WRONG,WRONG!!!!!!!!!!!!!!!!!!!!!!

I'm surprised at Marlon,i didn't expect that answer,especially having mentioned QED.

The Feynman diagram in QED is in the second order of pertubation theory and is topologically equivalent to the one describing the Compton effect ("switch propagators and change the direction of time).It has one electron,one positron and 2 PHOTONS,NOT ONE.I don't need QFT to prove to you guys that for free scattering electron and positron,u must have either the second photon (as i said,it should be an antiphoton,but massless+gauge invariance dictate zero electric charge for the photon,therefore photon=antiphoton),or another particle (boson for spin conservation) for having conserving 4 momentum in the scattering.This is 'classical' stuff,how could u f*** it up...???? :surprised :surprised :yuck:

Daniel.

9. Jan 4, 2005

### chroot

Staff Emeritus
I'm aware that one photon cannot by itself turn into a pair as that reaction cannot conserve momentum; however, it wasn't really relevant to the discussion. In the presence of matter, however -- the usual case -- a single photon can turn into a pair.

- Warren

10. Jan 4, 2005

### Gamish

1.I did use m/s in my calculations. I think you should re-check you math.

Let me as my question in a different way.

If I had some antimatter (-e) , can I convert it to energy, the convert it to positive energy, the convert that energy to positive matter (+e)?

11. Jan 4, 2005

### chroot

Staff Emeritus
You cannot convert antimatter by itself directly into energy (radiation) because doing so would violate the conservation of electric charge. You cannot directly turn energy (radiation) into only normal matter, again, because that would violate the conservation of electric charge. The only kinds of reactions that are permissible are those that destroy both a particle of matter and a particle of antimatter at the same time, or those that create a particle of matter and a particle of antimatter at the same time. What you propose is not possible, and has already been explained by several people in this thread. Please try to read and fully understand the responses you have been given before continuing to argue.

- Warren

12. Jan 5, 2005

### Astronuc

Staff Emeritus
What form of matter/anti-matter would be important, e.g. molecules of hydrogen (e-, p) and anti-hydrogen (e+, p-). The presence of anti-neutrons would be interesting.

The positron/electron pairs would annhiliate yielding 2 gamma-rays of approximately energy of 0.511 MeV or the rest mass of the electron/positron). The gamma-rays would scatter off elctrons (Compton scattering) and loose energy, and would not be able to reform electons or positrons. Electrons and positrons are formed in pairs (pair production) by the interaction of gamma-rays with a nucleus of an atom, but this requires a threshold of ~1.022 MeV or twice the electron rest mass.

The proton-antiproton annihilation would result in a whole lot of mesons, then these would decay into muons, electrons, neutrinos and more gammas, and I imagine that some the pions would interact with protons form K-mesons and resonance particles.

I just don't want to be nearby when someone annihilates 1 kg of antimatter.

Also, it is not said how the matter and anti-matter are brought together - instanteously, at rest or collided, solid, liquid or gaseous.

Last edited: Jan 5, 2005
13. Jan 5, 2005

### DB

I know you used m/s in your calculations, I said that you would have had to convert into km/s because your anti/matter was being measured in kg. Though it's irrelevent now.

14. Jan 5, 2005

### DB

Either way, Astronuc explains it all very well.

15. Jan 5, 2005

### chroot

Staff Emeritus
Actually, the fundamental SI unit of mass is the kilogram, not the gram.

- Warren

16. Jan 5, 2005

### marlon

Don't get so agitated man...What you are saying is true and both Chroot and me are well aware of that...check out previous posts if you will ...We were just making a sketch of the situation at hand by giving examples of why conservation laws are used. What you are saying isn't relevant to this discussion because we were not describing a complete annihilation reaction of a positron and an electron. Again we were exemplifying...

regards
marlon

17. Jan 5, 2005

### Gamish

OK, so, let me get this straight. If I annihilate 1 positron, and 1 electron, it will yield 1 anti-gamma-ray, and 1 gamma-ray, then, if we were to somehow convert this energy into matter again, it would yield 1 positron, and 1 electron?

18. Jan 5, 2005

Staff Emeritus
The photon is its own anti-particle, so you only produce the one gamma ray. This ray will carry all the energy and momentum of the original particle and anti-particle.

19. Jan 5, 2005

### Gamish

So, is it posible to convert this ray of energy back to either/and/or matter/antimatter?

20. Jan 5, 2005

### dextercioby

At elementary quantum level,the laws of physics are CPT invariant,and,in this case,and if i'm not mistaking in all cases,T invariant.So the answer is 'yes'.

Daniel.

PS.Sidenotes:
a)T-invariant:means the equations and all physically measurable quantities have the same form if:
$$t\rightarrow -t$$
b)CPT invariant:the....remain invariant if SIMULTANEOUSLY the following transformations are perfomed:
$$q\rightarrow -q;\vec{r}\rightarrow -\vec{r};t\rightarrow -t$$