J.C.MacSwell Posted March 2, 2005 Share Posted March 2, 2005 Where on earth did you come up with that??? In a photonic frame' date=' a photon's speed is zero by fiat. But, where did you get the idea that photons travel at c in all frames? That's just wrong. Regards[/quote'] From your"photonic frame" perspective are all massive objects at "c" or some other constant? Link to comment Share on other sites More sharing options...
Tom Mattson Posted March 2, 2005 Share Posted March 2, 2005 You have it backwards. The barn and ladder paradox was designed to show that simultaneity is absolute' date=' not relative. It is mean't to show that simultaneity is NOT frame dependent.[/quote'] No, it's you who has it backwards. The barn and ladder paradox is resolved if you abandon absolute simultaneity. It is only a paradox if you insist on that concept. This is true of many of the so-called paradoxes of SR. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 2, 2005 Share Posted March 2, 2005 You have it backwards. The barn and ladder paradox was designed to show that simultaneity is absolute' date=' not relative. It is mean't to show that simultaneity is NOT frame dependent. Regards[/quote'] I see the 40 foot ladder (at speed lorentz contracted) in my 20 foot barn. You see the front end of the ladder exit the barn before the aft end enters. We cannot agree on simultaneity because it is relative. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 2, 2005 Author Share Posted March 2, 2005 No, it's you who has it backwards. The barn and ladder paradox is resolved[/i'] if you abandon absolute simultaneity. It is only a paradox if you insist on that concept. This is true of many of the so-called paradoxes of SR. I don't want to get into this debate with you. Regards Link to comment Share on other sites More sharing options...
Tom Mattson Posted March 2, 2005 Share Posted March 2, 2005 I don't want to get into this debate with you. Fine with me, because there's no debate to be had. Section 3.3 of the following document explains quite clearly why you are wrong. http://www.physics.nyu.edu/hogg/sr/sr.pdf Link to comment Share on other sites More sharing options...
Johnny5 Posted March 3, 2005 Author Share Posted March 3, 2005 Fine with me' date=' because there's no debate to be had. Section 3.3 of the following document explains quite clearly why you are wrong. http://www.physics.nyu.edu/hogg/sr/sr.pdf[/quote'] I did look over section 3.3, but it had no effect. You can't fit a twenty foot ladder inside a ten foot barn no matter how fast you run. I would think something is wrong with the math, before I would think something is wrong with reality. Regards Link to comment Share on other sites More sharing options...
Tom Mattson Posted March 3, 2005 Share Posted March 3, 2005 I did look over section 3.3' date=' but it had no effect.[/quote'] Didn't read it, eh? You can't fit a twenty foot ladder inside a ten foot barn no matter how fast you run. I would think something is wrong with the math, before I would think something is wrong with reality. Unfortunately for you, the universe is not subject to your a priori assumptions about it. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 3, 2005 Author Share Posted March 3, 2005 Didn't read it' date=' eh?[/quote'] I started to. In article 3.5 he is already discussing boosted frames, but I have a problem with his worldline argument in 3.3. Regards Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2005 Share Posted March 3, 2005 I did look over section 3.3' date=' but it had no effect. You can't fit a twenty foot ladder inside a ten foot barn no matter how fast you run. I would think something is wrong with the math, before I would think something is wrong with reality. Regards[/quote'] If you agree with relativity then you know it's not a 20 foot ladder in the barn's frame. If you disagree with that, then I think you just don't understand relativity. Link to comment Share on other sites More sharing options...
5614 Posted March 3, 2005 Share Posted March 3, 2005 ... or he is disagreeing with it, in which case I suppose rather than saying "i disagree" you should explain why. Has the ladder/barn experiment ever been conducted or is it purely mathematical? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted March 3, 2005 Share Posted March 3, 2005 ... or he is disagreeing with it' date=' in which case I suppose rather than saying "i disagree" you should explain why. Has the ladder/barn experiment ever been conducted or is it purely mathematical?[/quote'] The ladder disintegrated and the barn burned down. Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2005 Share Posted March 3, 2005 Has the ladder/barn experiment ever been conducted or is it purely mathematical? What do you think a reasonable answer to this is? Link to comment Share on other sites More sharing options...
5614 Posted March 3, 2005 Share Posted March 3, 2005 well travelling at 0.8666C (from Tom Mattson link) and on a much smaller scale and considering particle accelerators I thought that possibly on an atomic scale something similar could theoretically be recreated... just wondered if it was physically possible, im guessing not now.... how fast to atoms go (in multiples of c, like 0.8666C) in particle accelerators? Link to comment Share on other sites More sharing options...
Johnny5 Posted March 3, 2005 Author Share Posted March 3, 2005 From your"photonic frame" perspective are all massive objects at "c" or some other constant? From the perspective of a photon at rest, the speed of other things would not be constant, nor would they all be c, but to be sure, large things would be zipping by. (I am using the Galilean transformations by the way) Regards Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2005 Share Posted March 3, 2005 well travelling at 0.8666C (from Tom Mattson link) and on a much smaller scale and considering particle accelerators I thought that possibly on an atomic scale something similar could theoretically be recreated... just wondered if it was physically possible, im guessing not now.... how fast to atoms go (in multiples of c, like 0.8666C) in particle accelerators? Depends on what's being accelerated. Electrons can go 0.99 (and more digits) times the speed of light because of their small mass. The TRIUMF cyclotron accelerates protons to about 0.75c. Labs that use more massive ions get to lower speeds, generally because you are limited in how much total energy you can give to the particles. Getting the particle moving that fast isn't the problem. It's the barn part. The you have a "barn" that's the same size, roughly, as the particle you are accelerating. Nuclei are about 10-5 the size of atoms. You really have to do a similar experiment with macroscopic objects, and getting a macroscopic object up to a fraction of c is a problem. I'd think you'd have to design some other kind of simultaneity experiment. But you can test relativity in other ways, and just do this as a thought experiment. Link to comment Share on other sites More sharing options...
swansont Posted March 3, 2005 Share Posted March 3, 2005 From the perspective of a photon at rest' date=' the speed of other things would not be constant, nor would they all be c, but to be sure, large things would be zipping by. (I am using the Galilean transformations by the way) Regards[/quote'] That's a problem, because Galilean transformations aren't valid approximations at those speeds. Link to comment Share on other sites More sharing options...
Asimov Pupil Posted March 4, 2005 Share Posted March 4, 2005 sorry to bore you guys with my simplicity but could someone please tell me if i'm getting this. the E^2 = (mc^2)^2 part is derived from the formula E=mc^2 and is thus the energy of the particle if it was converted to energy at the speed of light and the (pv)^2 is the kinetic energy? if so why cant E=hf be the energy of the photon that comes out of the converted particle. or am i just stupid................HIGH SCHOOL IS FUN! Link to comment Share on other sites More sharing options...
Asimov Pupil Posted March 4, 2005 Share Posted March 4, 2005 by the way i love http://www.physics.nyu.edu/hogg/sr/sr.pdf thank you Link to comment Share on other sites More sharing options...
Johnny5 Posted March 4, 2005 Author Share Posted March 4, 2005 sorry to bore you guys with my simplicity but could someone please tell me if i'm getting this. the E^2 = (mc^2)^2 part is derived from the formula E=mc^2 and is thus the energy of the particle if it was converted to energy at the speed of light and the (pv)^2 is the kinetic energy? if so why cant E=hf be the energy of the photon that comes out of the converted particle. or am i just stupid................HIGH SCHOOL IS FUN! No, you are confused about E=Mc^2 in the first place. Let's see your derivation of the formula, if you don't have one, I can provide you with one. Regards Link to comment Share on other sites More sharing options...
Asimov Pupil Posted March 4, 2005 Share Posted March 4, 2005 I posted the question about the total energy of the atom and someone gave me the formula E^2=(mc^2)^2 + (pc)^2 and it got me thinking why did they square everything then i though they must hve done it to e=mc^2 cuz if you didn't and you took E=(m-m)c^2 it came out to a negative and negative energy kinda confused me. then p=mv so pc at the speed of light is mv(v) or mv^2 or kinetic energy if you will. but it just hit me from math class that if you take the square root of something you get a positive and a negative anyway. Link to comment Share on other sites More sharing options...
Johnny5 Posted March 4, 2005 Author Share Posted March 4, 2005 I posted the question about the total energy of the atom and someone gave me the formula E^2=(mc^2)^2 + (pc)^2 and it got me thinking why did they square everything then i though they must hve done it to e=mc^2 cuz if you didn't and you took E=(m-m)c^2 it came out to a negative and negative energy kinda confused me. then p=mv so pc at the speed of light is mv(v) or mv^2 or kinetic energy if you will. but it just hit me from math class that if you take the square root of something you get a positive and a negative anyway. Here is my suggestion to you. Learn how to derive the following formula from scratch: [math] E^2 = (Pc)^2 + (m_0c^2)^2 [/math] Here, I will start you off: [math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math] [math] \vec P = M \vec v [/math] Regards Link to comment Share on other sites More sharing options...
Asimov Pupil Posted March 5, 2005 Share Posted March 5, 2005 Thank you Link to comment Share on other sites More sharing options...
Johnny5 Posted March 5, 2005 Author Share Posted March 5, 2005 Thank you You're welcome. Link to comment Share on other sites More sharing options...
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