DimaMazin Posted November 28, 2013 Posted November 28, 2013 Yes, it does. Google "photon orbits". Photon energy-momentum is just relative thing.Doesn't it make any paradox?
xyzt Posted November 28, 2013 Posted November 28, 2013 Photon energy-momentum is just relative thing.Doesn't it make any paradox? No. That's an old interrogation I have, still unclear to me... You guessed, I hoped to explain cosmic rays as many photons of lesser energy that stick together somehow. But: two photons travelling together would be stationary to an other, hence have no energy to an other. Or? This is not correct, you cannot attach frames of reference to photons. Therefore you cannot say "two photons travelling together would be stationary to an other".
DimaMazin Posted November 28, 2013 Author Posted November 28, 2013 No. This is not correct, you cannot attach frames of reference to photons. Therefore you cannot say "two photons travelling together would be stationary to an other". Fast traveler with the same direction of the photons sees less energy of the photons,and fast traveler with return direction sees more energy of the photons.Then they should see different distance of sticking together of the rays.But if each traveler have the same module of speed relative to the owner of the lasers,they shouldn't see different distance of sticking together of the rays according to length contraction of relativity.
xyzt Posted November 28, 2013 Posted November 28, 2013 (edited) Fast traveler with the same direction of the photons sees less energy of the photons,and fast traveler with return direction sees more energy of the photons. False, the photons travel at c wrt to ANY observer. Then they should see different distance of sticking together of the rays. But if each traveler have the same module of speed relative to the owner of the lasers,they shouldn't see different distance of sticking together of the rays according to length contraction of relativity. Try again. In English , please. Edited November 28, 2013 by xyzt -1
DimaMazin Posted November 28, 2013 Author Posted November 28, 2013 False, the photons travel at c wrt to ANY observer. Try again. In English , please. Fast traveler1 has direction of moving with moving of the photons(relative to lasers).Fast traveler2 has return direction.They see different energy of the photons,therefore they should see different length of wedge of the rays till sticking together.They have identical speeds,only speed of traveler1 is positive and speed of traveler2 is negative,therefore they should see identical length of wedge of the rays till sticking together.This is paradox.
xyzt Posted November 28, 2013 Posted November 28, 2013 (edited) Fast traveler1 has direction of moving with moving of the photons(relative to lasers).Fast traveler2 has return direction.They see different energy of the photons,therefore they should see different length of wedge of the rays till sticking together. Nope. They have identical speeds,only speed of traveler1 is positive and speed of traveler2 is negative,therefore they should see identical length of wedge of the rays till sticking together.This is paradox Also nope. Edited November 28, 2013 by xyzt
DimaMazin Posted November 29, 2013 Author Posted November 29, 2013 Nope. Also nope. Rays of blue photons have more energy than rays of red photons have energy(when quantities of photons are identical in the pairs of the rays), therefore rays of blue photons create more gravitation.Therefore blue photons should have less length( in observer frame) of wedge of their rays till sticking together.
ajb Posted November 29, 2013 Posted November 29, 2013 Rays of blue photons have more energy than rays of red photons have energy(when quantities of photons are identical in the pairs of the rays), therefore rays of blue photons create more gravitation.Therefore blue photons should have less length( in observer frame) of wedge of their rays till sticking together. Now think about an observer moving relative to the source! It is the energy-momentum tensor that is important here rather than the energy carried by the EM wave as measured by some observer.
DimaMazin Posted November 30, 2013 Author Posted November 30, 2013 Now think about an observer moving relative to the source! It is the energy-momentum tensor that is important here rather than the energy carried by the EM wave as measured by some observer. Physics laws identically work in any frame. In observer frame the photons are free from the lasers .
Sensei Posted November 30, 2013 Posted November 30, 2013 (edited) Fast traveler with the same direction of the photons sees less energy of the photons,and fast traveler with return direction sees more energy of the photons. False, the photons travel at c wrt to ANY observer. Do not you see he is speaking about Relativistic Doppler effect? It has nothing to do with speed of light. http://en.wikipedia.org/wiki/Relativistic_Doppler_effect Edited November 30, 2013 by Sensei
DimaMazin Posted December 1, 2013 Author Posted December 1, 2013 Do not you see he is speaking about Relativistic Doppler effect? It has nothing to do with speed of light. http://en.wikipedia.org/wiki/Relativistic_Doppler_effect Firstly xyzt written about this, but edited after.
decraig Posted December 1, 2013 Posted December 1, 2013 (edited) No. This is not correct, you cannot attach frames of reference to photons. Therefore you cannot say "two photons travelling together would be stationary to an other". By "photon" do you mean a tiny-little point having electromagnetic characterics? If we try really hard, we can we can compare two frames of reference, each with velocity c---almost. In relativity we can compare more than inertial frames. When you do this, with an "almost c" velocity, a velocity of c+h where abs(h) is the less than any number you can think of, but great than zero, you would obtain some interesting results. In these frames we have a coordinate system collapse. Not surprising, considering we dealing in numbers not finite. Each frame in normal spacetime has 2 finite dimensions, one transfinite, and one that is infinitessimal. Each frame is independent, having no finite-time communication with the other. It is as if these hypothetical photons each had their own universe. What where you saying about photons? Edited December 1, 2013 by decraig
pritikamehra Posted December 3, 2013 Posted December 3, 2013 Hey Enthalpy, I wasn't exactly thinking of cosmic rays, honestly it was just a completely random thought/scenario I made in my head If we're looking into the photon motion in that much detail, then what causes the photons of one ray of light to coherently travel together? Einstein said that every observer will measure light to be travelling at the speed c. Perhaps, the photons of a particular ray of light also perceive the other photons of that same ray to be travelling at the speed c (they are not stationary relative to one another). This energy-momentum of each individual photon could be the source of gravity that allows these photons from the same ray of light to 'cluster/group' together; hence they move together in the same path? What I'm saying could be completely wrong as I'm just hypothesizing here, but it's just a thought.
swansont Posted December 3, 2013 Posted December 3, 2013 Coherence comes from the production of the photon — in stimulated emission the light will be in phase.
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