swansont Posted February 12, 2013 Posted February 12, 2013 O.K. How then do you change the energy of a particular photon? In essence you don't. You destroy it and create a new one. Even in Compton scattering it is modeled this way.
derek w Posted February 12, 2013 Posted February 12, 2013 The wave length of photons,can expand in a expanding space or shrink in a shrinking space.
swansont Posted February 13, 2013 Posted February 13, 2013 ! Moderator Note Tangential discussion on whether photons are multiple particles split http://www.scienceforums.net/topic/72831-photon-mass/#entry728833
michel123456 Posted February 13, 2013 Posted February 13, 2013 (edited) I don't understand (once again). a photon has no rest mass, O.K. But the photons that we catch (and the others that we don't catch) are not in state of rest, thus they have mass, be it relativistic. Those photons that travel at C exert and are sensible to gravity, as much as I know. Correct? Edited February 13, 2013 by michel123456
ajb Posted February 13, 2013 Posted February 13, 2013 But the photons that we catch (and the others that we don't catch) are not in state of rest, thus they have mass, be it relativistic.By catch you mean they are destroyed and their energy is absorbed by some medium?
swansont Posted February 13, 2013 Posted February 13, 2013 I don't understand (once again). a photon has no rest mass, O.K. But the photons that we catch (and the others that we don't catch) are not in state of rest, thus they have mass, be it relativistic. No. Because in OUR frame, i.e. where we are doing the measurement, E = pc. There is nothing left over. Thus, m = 0. This is true regardless of what specific name we give that term.
michel123456 Posted February 13, 2013 Posted February 13, 2013 By catch you mean they are destroyed and their energy is absorbed by some medium? Oh sorry, no, I ment "observe". The Universe is full of photons that travel in all directions and do not reach the Earth & our eyes. No. Because in OUR frame, i.e. where we are doing the measurement, E = pc. There is nothing left over. Thus, m = 0. This is true regardless of what specific name we give that term. A photon has not even relativistic mass?
Przemyslaw.Gruchala Posted February 13, 2013 Posted February 13, 2013 (edited) Oh sorry, no, I ment "observe". The Universe is full of photons that travel in all directions and do not reach the Earth & our eyes. You can't observe photon that wasn't absorbed by your eye. By observation you're absorbing photons from world. Photon detector in double slit experiment is so good in absorption that whole effect is disappearing. Edited February 13, 2013 by Przemyslaw.Gruchala
D H Posted February 13, 2013 Posted February 13, 2013 I don't understand (once again). a photon has no rest mass, O.K. But the photons that we catch (and the others that we don't catch) are not in state of rest, thus they have mass, be it relativistic. Those photons that travel at C exert and are sensible to gravity, as much as I know. Correct? Stop calling it "rest mass". It's just a name, and you are getting far to hung up on that name. Use the term "invariant mass" instead, where "invariant" means "the same in all inertial frames of reference."
swansont Posted February 13, 2013 Posted February 13, 2013 A photon has not even relativistic mass? Relativistic mass is an ad-hoc, extraneous term that just means energy. It adds nothing to the discussion but confusion.
michel123456 Posted February 13, 2013 Posted February 13, 2013 Correcting post #54 I don't understand (still). A photon has no invariant mass, O.K. But the photons that we observe (and the others that we don't)are not in state of rest. They have energy and momentum. Thosephotons that travel at C in all directions exert and are sensible to gravity, as much as Iknow. Correct?
Przemyslaw.Gruchala Posted February 13, 2013 Posted February 13, 2013 Everything is "sensible" to gravity, because it's bending of space..
michel123456 Posted February 13, 2013 Posted February 13, 2013 So what's the question? i would be very amazed if astronomers did not count for those gravitational effects in their calculations.
D H Posted February 13, 2013 Posted February 13, 2013 So what's the question? i would be very amazed if astronomers did not count for those gravitational effects in their calculations. Why should they? That light is subject to gravitation, sure. Astronomers see this in eclipses and in images such as this: That's the Einstein cross. That light gravitates, why should they? The effect is incredibly small, much smaller than observational errors, and much smaller than numerical errors in the case of a simulation. There's no reason to include this effect.
michel123456 Posted February 13, 2013 Posted February 13, 2013 Nobody cared making an evaluation? That is 13 billion years of burning stars radiation + the radiation of the BB that did not change in massive particles. I am amazed.
swansont Posted February 13, 2013 Posted February 13, 2013 There's also the fact that isotropic radiation, such as the CMWB, does not contribute to gravitational effects outside of the radius in question. Nobody cared making an evaluation? That is 13 billion years of burning stars radiation + the radiation of the BB that did not change in massive particles. I am amazed. At what, exactly?
michel123456 Posted February 14, 2013 Posted February 14, 2013 That nobody cared making an evaluation of the gravitational effect of light.
ajb Posted February 14, 2013 Posted February 14, 2013 Light can act as a source of gravity. I imagine for a focused beam people have calculated the corresponding metric, but I don't know a reference.
elfmotat Posted February 14, 2013 Posted February 14, 2013 Light can act as a source of gravity. I imagine for a focused beam people have calculated the corresponding metric, but I don't know a reference. http://en.wikipedia.org/wiki/Pp-wave_spacetime
ajb Posted February 14, 2013 Posted February 14, 2013 http://en.wikipedia.org/wiki/Pp-wave_spacetimeThanks. I know pp-waves are studied in relation to gravitational waves, I did not know that they also model EM waves.
derek w Posted February 14, 2013 Posted February 14, 2013 quote from Wikipedia :- QUANTUM FOAM. According to Einstein's theory of relativity,energy curves space time.This suggests that-at sufficiently small scales-the energy of these fluctuations would be large enough to cause significant departures from the smooth space time seen at larger scales,giving space time a "foamy" character.
swansont Posted February 14, 2013 Posted February 14, 2013 That nobody cared making an evaluation of the gravitational effect of light. That's not what happened. That's not what a physicist generally means by "ignore" in this context. The sun converts about 4 billion* kg/s into other forms of energy, which is eventually radiated away at c. In the volume inside the earth, that's 500 s of travel, or the equivalent of 2 x 10^12 kg. The mass of the sun is 2 x 10^30 kg. Ignoring the effect of the light gives an error of a part in 10^18 in terms of effect on the earth. At that point you ignore it, unless you are doing an experiment where a part in 10^18 matters. Not including in a calculation is not the same as refusing to consider. *edit to correct value and subsequent numbers 1
michel123456 Posted February 14, 2013 Posted February 14, 2013 (edited) That's not what happened. That's not what a physicist generally means by "ignore" in this context. The sun converts about 4 kg/s into other forms of energy, which is eventually radiated away at c. In the volume inside the earth, that's 500 s of travel, or the equivalent of 2000 kg. The mass of the sun is 2 x 10^30 kg. Ignoring the effect of the light gives an error of a part in 10^27 in terms of effect on the earth. At that point you ignore it, unless you are doing an experiment where a part in 10^27 matters. Not including in a calculation is not the same as refusing to consider. I believe you but I don't understand your calculation. The sun radiates in all directions. What the Earth receives is a tiny part of it. Where do the 500s come from? ------- mm. the 500s is the orbital radius. So you are estimating the volume of the cylinder from Earth to the Sun. ------ Isn't there something missing? Edited February 14, 2013 by michel123456
imatfaal Posted February 14, 2013 Posted February 14, 2013 That's not what happened. That's not what a physicist generally means by "ignore" in this context. The sun converts about 4 kg/s into other forms of energy, which is eventually radiated away at c. In the volume inside the earth, that's 500 s of travel, or the equivalent of 2000 kg. The mass of the sun is 2 x 10^30 kg. Ignoring the effect of the light gives an error of a part in 10^27 in terms of effect on the earth. At that point you ignore it, unless you are doing an experiment where a part in 10^27 matters. Not including in a calculation is not the same as refusing to consider. "The sun converts about 4 kg/s into other forms of energy" - I am not disputing that number but it does blow my mind. I have looked after oil-tankers that burn almost as kilograms of fuel oil per second just to move across the ocean! It just shows the complete difference in scale between chemical energy burning and the energy release in nuclear reactions. All life on earth (well barring hydrothermal vents etc) - that lovely sunshine...waxes lyrical for a while about nature... all from 4kg per second. Amazing
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