Klaynos Posted May 20, 2006 Share Posted May 20, 2006 I think people are confused. Photons, gravitons, regardless, are not their own anti particles. It is just that it is impossible to tell them APART from their anti particles. If they were their own anti particles, photons in laser would destroy each other, which they don't. However, if we had a method of making anti-photons, they would. Just because we can tell them apart, doesn't mean their all the same. Could you please define what properties the anti-photon would take? Link to comment Share on other sites More sharing options...
GutZ Posted May 20, 2006 Share Posted May 20, 2006 So what determines their charge? Link to comment Share on other sites More sharing options...
abskebabs Posted May 20, 2006 Share Posted May 20, 2006 I think people are confused. Photons, gravitons, regardless, are not their own anti particles. It is just that it is impossible to tell them APART from their anti particles. If they were their own anti particles, photons in laser would destroy each other, which they don't. However, if we had a method of making anti-photons, they would. Just because we can tell them apart, doesn't mean their all the same. I understand that when antiparticles such as a positron and an electron annihilate, high energy gamma photons are produced. So what would be produced if a photon and an antiphoton(theoretically) annihilate? Link to comment Share on other sites More sharing options...
Ragib Posted May 21, 2006 Share Posted May 21, 2006 Klaynos- Like i said, they would be completely identical to photons. GutZ- photons dont have charges. If u refer to anti particles in general, it is as if they are going back in time, making them appear to have opposite charge etc. abskebabs- i am not 100% sure, as this has never had any practical purpose to any physicists i can think of, but i would imagine that they don't annihilate at all, but just keep going, as they are already pure energy. You never know, maybe it becomes matter. i dont know. Link to comment Share on other sites More sharing options...
Klaynos Posted May 21, 2006 Share Posted May 21, 2006 Klaynos- Like i said' date=' they would be completely identical to photons.[/quote'] So they are infact photons? Two things with identicle properties are infact the same... Link to comment Share on other sites More sharing options...
Severian Posted May 21, 2006 Share Posted May 21, 2006 If they were their own anti particles, photons in laser would destroy each other, which they don't. You are misunderstanding what happens when matter and antimatter come into contact. They 'destroy themselves' only in the sense that they are converted into something else. And this something else is usually photons. So if you hit an electron with a positron you get a photon (you could get a Z-boson as well, but that is a needless complication). Similarly two photons can 'annihilate' into one photon in much the same way. In fact, this does happen in a laser. Link to comment Share on other sites More sharing options...
abskebabs Posted May 23, 2006 Share Posted May 23, 2006 Similarly two photons can 'annihilate' into one photon in much the same way. In fact, this does happen in a laser. Hmm.... I thought this was because they share the property of all bosons, that allows them to exist in exactly the same quantum state. In fact I think there is a high lilkelihood of this when the scenario presents it as a possibility(like in a laser). Link to comment Share on other sites More sharing options...
Meir Achuz Posted May 23, 2006 Share Posted May 23, 2006 So if you hit an electron with a positron you get a photon (you could get a Z-boson as well' date=' but that is a needless complication). Similarly two photons can 'annihilate' into one photon in much the same way. In fact, this does happen in a laser.[/quote'] 1. el-pos can only annihilate into two or three photons. One photon is not allowed by E p conservation. 2. Two photons can't annihilate into one photon for the same reason. They can't even do it virtually, due to C invariance. Link to comment Share on other sites More sharing options...
ydoaPs Posted May 23, 2006 Share Posted May 23, 2006 /me compares Severian's user title to Meir Achuz's Link to comment Share on other sites More sharing options...
swansont Posted May 23, 2006 Share Posted May 23, 2006 MA is right; if you analyze the e-p annihilation you can't get a single photon without violation either energy or momentum conservation. Left as an exercise for the diligent student. One photon from two, or two from one, wouldn't violate this, if they were co-propagating, since the energies and momenta would add, but photon-photon interactions in free space would be unusual, IIRC (possible in QED but very small cross-section). The interactions that typically combine photons occur in nonlinear crystals, as part of four-wave mixing. There you can frequency double (or get frequency additions or differences), under the right conditions, 'till the cows come home. Link to comment Share on other sites More sharing options...
Severian Posted May 23, 2006 Share Posted May 23, 2006 1. el-pos can only annihilate into two or three photons.One photon is not allowed by E p conservation. 2. Two photons can't annihilate into one photon for the same reason. They can't even do it virtually' date=' due to C invariance.[/quote'] Yes, it has to be (at least) two, but that is not the issue. For e+e-, they can annihilate to a single virtual photon which quickly decays. Two photons on the other hand can scatter into two photons via a fermion loop, or even turn into an e+e- pair or a quark pair etc Link to comment Share on other sites More sharing options...
Meir Achuz Posted May 30, 2006 Share Posted May 30, 2006 /me compares Severian's user title to Meir Achuz's What does that mean? Link to comment Share on other sites More sharing options...
timo Posted May 30, 2006 Share Posted May 30, 2006 It probalby means that he doesn´t know what a baryon is Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now