Ekpyrotic Posted June 24, 2007 Posted June 24, 2007 Warning: I'm kinda' crossing physics and philosophy here. I've constructed a logically proof for the inexistence of a Higgs particle, and I need some physicists to point out the logical fallacy in the argument because I can't see it myself, although I can see potential flaws. (1) Massless particles can't interact. (this is a major flaw, but I can't seem to find information for this either way) (2) In order for a massless particle to attain mass it must interact in a Higgs' field. Therefore the Higgs particle requires a mass. (3) For a Higgs particle to aquire mass it must interact with another Higgs that already has mass. (4) Step (3) repeat until we work back to the first Higgs particle. (5) The first Higgs couldn't have come into existence because it would have been massless, which makes it useless. I'm sure it will be shot down on the first point. Thanks for reading. Jordan (Ekpyrotic)
insane_alien Posted June 24, 2007 Posted June 24, 2007 (1) Massless particles can't interact. (this is a major flaw, but I can't seem to find information for this either way) (2) In order for a massless particle to attain mass it must interact in a Higgs' field. Therefore the Higgs particle requires a mass. (3) For a Higgs particle to aquire mass it must interact with another Higgs that already has mass. (4) Step (3) repeat until we work back to the first Higgs particle. (5) The first Higgs couldn't have come into existence because it would have been massless, which makes it useless. 1/ photons have no rest mass, they interact. IIRC higgs boson has a mass bigger than anything wehave seen so far(for an elementary particle) which is why we need to build the most massive accelerator yet to find it) 5/ being massless doesn't mean useless.
Ekpyrotic Posted June 24, 2007 Author Posted June 24, 2007 Thanks for the fast response insane, I'd like to take the time to discuss your points, please excuse my ignorance. 1, photons have no rest mass, they interact. Can you quote a source that I can read up on. I thought this was the case but I couldn't find any description online for this behavior. All I could think of was Young's double slit, which would mean the particles are interacting with a 'Higgs wave', is this the case? 5, being massless doesn't mean useless. This refute is dependent on the integrity of your argument against the first point. Thanks, Jordan (Ekpyrotic)
insane_alien Posted June 24, 2007 Posted June 24, 2007 http://en.wikipedia.org/wiki/Photons massless just means that it always zooms about at the speed of light. thats all. can't go slower, can't go faster.
Ekpyrotic Posted June 24, 2007 Author Posted June 24, 2007 I read the Wikipedia entry through, and I couldn't find any implicit mention of massless particle interaction. The closest I get was the following: as a particle, it can only interact with matter by transferring the amount of energy I know I'm being demanding, but I like to be throughout. Thanks, Jordan (Ekpyrotic)
BenTheMan Posted June 24, 2007 Posted June 24, 2007 Yeah Ekpyrotic--- Depending on how technical you want to get, a mass term in the lagrangian is an interraction between two fields and a dimensionful coupling constant. So mass IS an interraction (you're right about that). But we can also have an interraction between three fields or four fields, with a coupling constant that is dimensionless (i.e. has no units if hbar = c = 1). In fact, when everything is massless, the mass terms for fermions are interractions between two fermions and a higgs, with a dimensionless coupling constant. It is only when the higgs gains a vev that the effective coupling constant becomes dimensionful, and the terms become mass terms.
insane_alien Posted June 24, 2007 Posted June 24, 2007 In modern physics the photon is the elementary particle responsible for electromagnetic phenomena. It is the carrier of electromagnetic radiation of all wavelengths, including gamma rays, X-rays, ultraviolet light, visible light, infrared light, microwaves, and radio waves. The photon differs from many other elementary particles, such as the electron and the quark, in that it has zero mass[3]; therefore, it travels (in vacuum) at the speed of light, c. ^ from the first paragraph of the wiki. its not an implicit statement but an explicit one. The photon is massless[3], has no electric charge[12] and does not decay spontaneously in empty space. oh, there we go again. first 4 words of the physical properties section. did you read the article?
BenTheMan Posted June 24, 2007 Posted June 24, 2007 I wrote something on another forum about the higgs, here: http://www.sciforums.com/showthread.php?t=68189 It was subsequently attacked by crackpots. Ekpyrotic--- Plus, if the photon were massive, it would make EM a short range force, with a range proportional to the Compton wavelength of the photon.
BenTheMan Posted June 24, 2007 Posted June 24, 2007 Also, I wrote this about the mass of a photon. You can do some back of the envelope type calculations to see how incredibly small it has to be. http://www.sciforums.com/showpost.php?p=1391396&postcount=48 PS: Maybe this forum doesn't have the same problem, but SciForums seems to be polluted with crackpots. Be careful of what you believe.
Ekpyrotic Posted June 24, 2007 Author Posted June 24, 2007 I'll respond to each commentor in turn. insane_alien: Thanks again for your response. did you read the article? I very much did, and it took a good 15 minutes too. The quoted references in your response seem rather vague, let us remember we want evidence that photons interact with each other. I can't see that anywhere in these quotes: In modern physics the photon is the elementary particle responsible for electromagnetic phenomena. It is the carrier of electromagnetic radiation of all wavelengths, including gamma rays, X-rays, ultraviolet light, visible light, infrared light, microwaves, and radio waves. The photon differs from many other elementary particles, such as the electron and the quark, in that it has zero mass[3]; therefore, it travels (in vacuum) at the speed of light, c... ...The photon is massless[3], has no electric charge[12] and does not decay spontaneously in empty space. This tells me the following: photons have 0 mass, they are the perpetrator of the electromagnetic force, they have no charge, and travel at c in a vacuum. Not that they interact with each other - which I don't doubt - I just like sturdy evidence. BenTheMan: You really know the art of writing well informed, insightful posts - if you don't mind the flattery. This particularly took my interest: But how does the higgs give things mass? Well, imagine every point in space as a small pendulum. You start the pendulum rocking, and it rocks back and forth and back again. But where is the average position of the pendulum? Its average is at the minimum---zero. This is how the higgs acts---except for one important caveat. The exception is that the higgs doesn't oscillate around zero, it oscillates around another place. The fact that it doesn't oscillate around zero MEANS that all of the particles that it couples to now have mass. (This statement can be mathematically motivated, but I don't know how useful a bunch of maths would be!) And however geeky this might seem I'd love to see the maths behind it, it always seems clearer that way. In fact, when everything is massless, the mass terms for fermions are interractions between two fermions and a higgs, with a dimensionless coupling constant. It is only when the higgs gains a vev that the effective coupling constant becomes dimensionful, and the terms become mass terms. This makes it a lot clearer, I'll need to spent some time reading up on the subject as of yet - but thanks again. Thanks, Jordan (Ekpyrotic) hehe, I got moved to the pseudoscience.
insane_alien Posted June 24, 2007 Posted June 24, 2007 http://en.wikipedia.org/wiki/Two-photon_physics there you go, photons interacting with each other.
Ekpyrotic Posted June 24, 2007 Author Posted June 24, 2007 http://en.wikipedia.org/wiki/Two-photon_physics there you go, photons interacting with each other. Thanks insane_alien. It's actually really interesting: A photon can, within the bounds of the uncertainty principle, fluctuate into a charged fermion/ anti-fermion pair, to either of which the other photon can couple. This fermion pair can be leptons or quarks. In the latter case, we distinguish several cases: * Direct or pointlike: The photon couples directly to a quark inside the target photon. At this small scale, alpha_s is small and this process can be calculated in perturbative QCD. * Single resolved: The quark pair of the target photon form a vector meson. The probing photon couples to a constituent of this meson. * Double resolbed: Both targed and probe photon have formed a vector meson. This results in an interactions between two hadrons. For the latter two cases alpha_s is large, and this Vector Meson Dominance (VDM) process has to be modelled in non-perturbative QCD. Time to research. Thanks, Jordan (Ekpyrotic)
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