5614 Posted March 16, 2005 Share Posted March 16, 2005 An inertial observer can never be in a frame where the photon is at rest... so you could imagine one, you couldn't be in that frame in any way. If the speed of a light is c in any inertial frame, then it cannot be 0 in any frame. momentum = plank's constant / wavelength ok, thanks. p=mv cannot work for photons, we both proved it! (making the safe assumption that a photon's mass is 0) Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 An inertial observer can never be in a frame where the photon is at rest This is untrue. Suppose that S is an inertial reference frame. Define the frame S` to be a coordinate system with a photon at the origin, and let the axes not be rotating in S. S` is also an inertial frame, and the photon is at rest in it. Link to comment Share on other sites More sharing options...
5614 Posted March 16, 2005 Share Posted March 16, 2005 2nd postulate of special relativety.... photons dont have a rest frame, they are never at rest. OK, so you can imagine this, I can imagine time being frozen and seeing a still photon... get back to reality! OK, that just sounds rude, take it light heartedly and as a joke, no offence meant, but I think you get which road I'm going down too? Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 2nd postulate of special relativety.... photons dont have a rest frame' date=' they are never at rest. OK, so you can imagine this, I can imagine time being frozen and seeing a still photon... get back to reality! OK, that just sounds rude, take it light heartedly and as a joke, no offence meant, but I think you get which road I'm going down too?[/quote'] I know alot about photon rest frames, be nice. Let me ask you something... how do you state the second postulate. How do YOU formulate it in words, or symbols. I just want to check something. Link to comment Share on other sites More sharing options...
5614 Posted March 16, 2005 Share Posted March 16, 2005 The second postulate of special relativity is that c is the same for all inertial observers or the speed of light is measured as constant in all frames of reference. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 The second postulate of special relativity is that c is the same for all inertial observers or the speed of light is measured as constant in all frames of reference. Ok we are talking you here, not Einstein... Do you literally mean all frames of reference, or do you mean all inertial reference frames, or do you mean all reference frames in which Maxwell's equations are true? And for that matter, how are you defining inertial reference frame? Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Author Share Posted March 16, 2005 Where does the hf = 2(mc²)² come from? EDIT: And please' date=' for the future: Add the steps that led you to a conclusion. You cannot expect any better answer than "you are wrong" when you don´t.[/quote'] i assumed nothing abouit mass. the energy equation for a particle at c simplifies to e^2=2m^2c^4. i put h^2f^2 in for e^2 because e=hf. i to the square root of both sides and got [math]hf=mc^2\sqrt{2}[/math] then solve for m: [math]\frac{hf}{c^2\sqrt{2}}[/math] when simplified is [math]\frac{hf\sqrt{2}}{2c^2}[/math] Link to comment Share on other sites More sharing options...
5614 Posted March 16, 2005 Share Posted March 16, 2005 Ok we are talking you here' date=' not Einstein... Do you literally mean all frames of reference, or do you mean all inertial reference frames, or do you mean all reference frames in which Maxwell's equations are true? And for that matter, how are you defining inertial reference frame?[/quote'] Ohhh, I like Einstein, he's my buddy!!!! Well swansont said "all inertial observers" in another thread, I'd have said "all reference frames" but as you didn't pick me up on saying two different things what's the difference? Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 Well swansont said "all inertial observers" in another thread' date=' I'd have said "all reference frames" but as you didn't pick me up on saying two different things what's the difference?[/quote'] That's not answering the question. It is impossible for the speed of light to be 299792458 meters per second in all frames of reference, and this is rapidly provable, so what on earth do you mean? State the second postulate of the theory of special relativity. Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Author Share Posted March 16, 2005 this doesn't have anything to do with this thread, but i want you to prove c CAN"T be constant in all reference frames. Link to comment Share on other sites More sharing options...
swansont Posted March 16, 2005 Share Posted March 16, 2005 This is untrue. Suppose that S is an inertial reference frame. Define the frame S` to be a coordinate system with a photon at the origin, and let the axes not be rotating in S. S` is also an inertial frame, and the photon is at rest in it. Go ahead and define it. What is the transformation that gets you from S to S'? Link to comment Share on other sites More sharing options...
5614 Posted March 16, 2005 Share Posted March 16, 2005 but i want you to prove c CAN"T be constant in all reference frames. Because if I am moving at 1mph then c is c relative to me, if I am moving at 100,000mph c is still c relative to me. c, from my point of view is always c relative to me therefore it cannot be the same if I have a varying speed. State the second postulate of the theory of special relativity. So how about: c is the same for all inertial observers. Link to comment Share on other sites More sharing options...
timo Posted March 16, 2005 Share Posted March 16, 2005 i assumed nothing abouit mass. the energy equation for a particle at c simplifies to e^2=2m^2c^4. [...'] Didin´t follow the rest of the post (mainly due to the broken TeX). My main point is the following anyways: E² = 2(mc²)² is not the correct expression for a particle´s energy. How did you come up with this? Did you assume pc = mc²? If so, why so? Well, I do have a guess but I´d like to hear how you got to E² = 2(mc²)². Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Author Share Posted March 16, 2005 no assumption about mass. e^2=(mc^2)^2+(pc)^2. p=mv. v=c. e^2=(mc^2)^2+(mcc)^2. e^2=(mc^2)^2+(mc^2)^2. e^2=2(mc^2)^2. i didn't think you would need me to do every step for you. i did that much, i might ass well finish. e=hf. e^2=h^2f^2. h^2f^2=2(mc^2)^2. hf=(m)(c^2)(sqrt(2)). (hf)/((c^2)(sqrt(2)))=m. (hf(sqrt(2)))/((2)(c^2))=m. [math]m=\frac{hf\sqrt{2}}{2c^2}[/math]. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 16, 2005 Share Posted March 16, 2005 this doesn't have anything to do with this thread, but i want you to prove c CAN"T be constant in all reference frames. This isn't a proof but I would suggest trying the "Earth Frame" . (0,0,0) is the center of the Earth and frame and you and I are at "rest" on the surface. Are you suggesting the speed of light to be constant in this frame? Obviously most of the mass of the Universe exceeds 300,000 km/s in this frame. Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Author Share Posted March 16, 2005 there are galaxies that are moving ftl relative to us, but not spacetime. that has nothing to do with c being constant in all frames. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 this doesn't have anything to do with this thread, but i want you to prove c CAN"T be constant in all reference frames. You can do physics in photon rest frames, in which the speed of a particular photon is zero. QED A longer proof would be to compare the speed of light in one frame, to the speed of the same photon in a reference frame which is accelerating away from the photon, and a reference frame which is accelerating towards the photon. In the frame which is accelerating away from the photon, the photon will have a measured speed which is slower than in the frame which is accelerating towards the photon. QED Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 So how about: c is the same for all inertial observers. That's a bit better, what is an inertial observer? Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Author Share Posted March 16, 2005 give links. Link to comment Share on other sites More sharing options...
timo Posted March 16, 2005 Share Posted March 16, 2005 EDIT: Wow, so much action in a rather long-dead thread. Seems that I should add that the following message is in reply to mydadonapogo: Momentum isn´t p=mv, that´s only an approximation for small velocities. If it was, any mass could easily reach and succeed lightspeed for example, as the nessecary energy to reach lightspeed would be given by your E² = 2(mc²)². For small particles these energies can be achieved relatively easily (energy would be 1 MeV/c² for electrons - modern acellerators operate in the range of 1000 MeV/c², I think). Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 Originally Posted by Johnny5This is untrue. Suppose that S is an inertial reference frame. Define the frame S` to be a coordinate system with a photon at the origin' date=' and let the axes not be rotating in S. S` is also an inertial frame, and the photon is at rest in it.[/quote'] Go ahead and define it. What is the transformation that gets you from S to S'? I already did define it. Any three non-collinear points in S are moving in a straight line at a constant speed in reference frame S. So if there was a solid body in frame S`, and the three points were on the surface of the body, then the body isn't rotating in reference frame S`. And if that body was being viewed in reference frame S, then those three points on the surface of that body would be moving in straight lines at constant speeds in S, and the three points would be at rest in S`. Was that clear enough? Link to comment Share on other sites More sharing options...
ydoaPs Posted March 16, 2005 Author Share Posted March 16, 2005 johnny, after VERY LITTLE searching, i am finding that c is constant for all frames. http://www.glenbrook.k12.il.us/gbssci/phys/Class/relativity/relpost2.html http://en.wikipedia.org/wiki/Relativity_physics Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 16, 2005 Share Posted March 16, 2005 there are galaxies that are moving ftl relative to us, but not spacetime. that has nothing to do with c being constant in all frames. Allow me to bring you "closer to home". I'll pick a frame (Let's call it "Spinny")centered at the same point (Earth center) and rotating at 300,000 rps wrt the Earth. Now you and I are well above 300,000 km/s in that frame. I'm sure you can prove to yourself that the speed of light is not constant in "Spinny". Link to comment Share on other sites More sharing options...
timo Posted March 16, 2005 Share Posted March 16, 2005 johnny, after VERY LITTLE searching, i am finding that c is constant for all frames. You´ll find that everywhere. What he´s trying to say is that he knows better than our current theories. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 16, 2005 Share Posted March 16, 2005 johnny' date=' after VERY LITTLE searching, i am finding that c is constant for all frames.http://www.glenbrook.k12.il.us/gbssci/phys/Class/relativity/relpost2.html http://en.wikipedia.org/wiki/Relativity_physics[/quote'] it's impossible for the speed of any particular object to be the same in all frames, I don't have to bother reading anything to understand that. Here is the quote from wikipedia: The speed of light in vacuum is a constant and is equal to 299' date='792,458 metres per second. [/quote'] That statement right there is false. Link to comment Share on other sites More sharing options...
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