how can a photon be massless if it decays into a proton, nutron, and electron.?
92282
The photon has energy. And energy can be converted into mass. Remember \n\nE^2=(m^2)( c^4) + (p^2) (c^2)\n\n----------\n\nI would agree with cedykeman1 that it does seem somewhat controversial to suggest that the photon is unstable. However, its energy can still probably be converted to matter, although the chance of getting those 3 exact particles seems a bit far-fetched (probably their antiparticles have to be produced as well, and various conservation laws followed carefully).\n\n--------\nI do not know what to say about goring, except to smile.
458495
The theoretical physics equation E = mc2 states a relationship between energy (E), in whatever form, and mass (m). In this formula, c², the square of the speed of light in vacuum, is the conversion factor required to formally convert from units of mass to units of energy, i.e. the energy per unit mass. In unit specific terms, E (Joules) = M (kilograms) · (299792458 m/s) 2.\n\nThis formula, which was discovered by Albert Einstein, proposes that when a body has a mass (measured at rest), it has a certain (a very large) amount of energy associated with this mass. This is opposed to the Newtonian mechanics, in which a massive body at rest has no kinetic energy, and may or may not have other (relatively small) amounts of internal stored energy (such as chemical or thermal energy), in addition to any potential energy it may have from its position in a field of force. That is why a body's mass, in Einstein's theory, is often called the rest energy of the body. The E of the formula can be seen as the total energy of the body, which is proportional to the mass when the body is at rest.\n\nConversely, a single photon travelling in empty space cannot be considered to have an effective mass, m, according to the above equation. The reason is that such a photon cannot be measured in any way to be at "rest" and the formula above applies only to single particles when they are at rest. Photons are generally considered to be "massless," (i.e., they have no rest mass or invariant mass) even though they have varying amounts of energy.\n\n\nENERGY=MASS x Speed square \n\n\nbtw  check out my website \n\nhttp://oldman.netfirms.com/
745459
light has no mass, therefore no weight.\n\nHowever, if all the energy of the light were somehow converted to mass, its mass would be m=E/c^2.
182197
Disregarding the pseudoscientific ramblings above, use Einstein's energy equation which states that the energy of a particle is the magnitude of its momentum 4-vector, E^2 = (mc^2)^2 - (pc)^2. Since the photon is massless, this becomes E^2 = (pc)^2, and we get the momentum of a photon as directly proportional to its energy p = E/c. \nDon't be reluctant to use the equations, and don't try to shoehorn Newtonian mechanics where it does not belong (speeds at or near the speed of light). In this case, you are using the fallacy that classical momentum mv applies to a relativistic object. For more information about similar fallacies, see the second source below.
554379
light.\n\nlet me refer you to e=mc2...light has mass energy and inertia.\na light is a wave and a particle.\nIt is in motion, and objects in motion have inertia and thusly power.\n\nThe true absence of light would exist in a vacuum. There is no pwer in a true vacuum. Even if the absence of light was "massive", that word you chose is a relative term, which is to say that massive CAN NOT imply mass, inertia, etc. The absence of light is NOT measurable because there is nothing there to exist.
185055
The equations imply that it is impossible for any object with >0 rest mass to accelerate from below the speed of light up to exactly the speed of light. This is because, as has been mentioned, that would take an infinite amount of energy.\n\nThe equations don't prohibit an object traveling faster than the speed of light from the outset though. Photons have zero rest mass, and the instant they are created they are traveling at exactly c. If it were possible for a particle to exist that was traveling faster than c from the instant it was created, this would not be a problem for relativity. I'm not sure what it would imply about it's mass though. The name for such a particle is a tachyon (although there is no evidence these actually exist as far as I'm aware).
300067
A photon, which is a quanta of evergy below which you cannot go-this is the smallest, and only, particle which carries radiant energy.  The photon is the energetic equivalent of the quark in matter.
184874
All the fundamental particles are thought of as having no size.  They are all point particles as far as we know.  In reality they probably are all made up of something else...but we haven't been able to probe deep enough to know what that would be.\n\nSo, they are are smallest: quarks, leptons (electrons, neutrinos), and bosons (photon, gluons, W/Z, graviton), and more?
346637
Photon has mass so far it has energy but it depend on the nature of particle of ligth. You can apply this formula to get the mass of the photon; m = hf/c^2. (Where h=Planck's constant, f= frequency, c= speed of ligth)
522802
Of course energy can be converted into mass. Have you heard about the process called 'pair-production'?\n\nPair production refers to a process of creation of an elementary particle and its antiparticle, usually from a photon or neutral bosonic fields (which are just energies). This is allowed if the following conditions are satisfied:\n1. There is enough energy available to create the pair: at least the total rest mass energy of the two particles. \n2. Linear momentum, angular momentum, electric charge, etc. must be conserved during the process. Therefore, the created particles must have those quantities to be exactly opposite (say, if one particle has spin +1/2 then another one must have spin −1/2).\n\nPair production from an energetic photon:\nFor example, pair-production occurs when a 1.022 MeV gamma-photon interacts with a heavy atomic nucleus, and produces an electron and a positron without violating conservation of momentum. Pair production cannot occur in empty space out of a single photon because the momentum of the initial photon must be absorbed by something, and the heavy nucleus is thus required to conserve both momentum and energy.\n\nThe energy of the photon 1.022 MeV, as described above,is the total energy of twice the electron mass (according to the mass-energy relationship, as given in the question!). If we have much more energetic photons, heavier particles may also be produced. These interactions were first observed in Patrick Blackett's counter-controlled bubble chamber, leading to the 1948 Nobel Prize in Physics.\n\nPair-production in a neutral bosonic field:\nIn semiclassical general relativity, pair production is also invoked to explain the Hawking radiation effect. According to quantum mechanics, at short scales short-lived particle-pairs are constantly appearing and disappearing in a region of strong gravitational tidal forces, the two particles in a pair may sometimes be wrenched apart before they have a chance to mutually annihilate. When this happens in the region around a black hole, one particle may escape, with its antiparticle being captured by the hole.
995142
You are surrounded by zero rest mass particles. They are called photons and they transmit the light from the screen to your eye. They transmit the electrostatic force that holds the atoms in your body together. They carry momentum but have no mass if at rest - in fact, they cannot exist at rest (though their phase velociy can be slowed to be near at rest).\n\nAny suitably modern and suitably advanced book on physics in any store in any country will give you the details, and the experimental verification that led to the conclusions. Try Eisberg and Resnick or Resnick and Halliday.\n\nZero rest mass particles are also affected by gravity, and this has also been measured using the mossbauer effect.\n\nThe zero rest mass particles in nature are those that carry the interactions between quarks and leptons.\n\nContrary to what is said elsewhere here, a charged particle cannot have zero mass because charge is a property of quarks, all of which have mass.\n\nSee ref below for more.
376534
Mass is defined via acting a force on an object produces an acceleration.  \nalso there is a universal kilogram in (i think) france that is the deffinition of a kilogram.\n\nNow how can a particle be massless... \nThis is defined as its rest mass, so if a particle does not exist at rest then it is going to be massless.  \n\nexample:\nA photon can have energy, and is moving.  so it can be said to have a momentum, and can then be said to have a "mass" but if a photon has zero energy its momentum is zero and its "mass" is zero.  \n\nThe mass of a particle is deffined when it is at rest, so some particles will have a zero mass, especial theoretical particals.
707260
The virtual particle that mediates electromagnetism is the photon, which is mass-less. therefore, it will not curve the spacetime it lives in.
816053
looking at particle nature "mass of photon" don't have a simple general meaning as the "mass of a cricket ball".it is some property of photon which comes from quantum mechanics and theory of relativity  but not "mass" exactly.That is why we say that photon has ZERO rest mass.
405474
This is the rest energy of a particle. Energy equals mass times the speed of light in a vacuum squared. This expression became famous for its simplicity, even though it's not really all that practical. I read once that even at absolute zero particles will still have some motion. I'm not sure how one would go about proving that since it's impossible to get anything to a temperature of absolute zero. Anyway since there will always be some movement, the more practical version is:\n\nE = mc^2 /(sqrt(1 - v^2/c^2)), where v is the speed of the particle. (sqrt() is meant to mean the square root of the argument inside the parenthesis)\n\nIt's important to note that this is only true for a particle since you have to consider binding energies if you want to look at anything more complex.
870020
While photons don't have any force-mitigated interactions, there can be interference.  Matter (fermions) has the Pauli Exclusion Priniciple preventing, say, two electrons from being in the same spin-state in the same sub-orbital...there's a destructive interference term in the probability wave equation that comes from their half-spin states.  \n\nSimilarly, there's an inverse sort of "Pauli Inclusion Principle" for bosons (which have integral spin states and includes photons) to constructively interfere in the wave equation such that there's an increased probability for them to be in the same place at the same time.  \n\nOne might be able to, as you say, 'channel' this and force them to work together by creating an energetic population inversion of electrons ready to emit, then sending through a photon of the proper wavelength to produce a cascade of stimulated emission...Oh wait, that's a laser.
14731
Photons, gluons, gravitons and Z are neutral and elementary so they are their own anti-particles.\n\nThe W+ and W-, the weak force carriers, are charged and are anti-particles.  W+ and W- can annihilate each other.