Sayonara Posted October 16, 2013 Share Posted October 16, 2013 He didn't make a conclusion, he said the results suggested a motive. Imprecise wording on my part, mea culpa. He says the results suggest a specific motive, and I characterise this specificity as the 'conclusion'. Perhaps I should have said that he could not possibly justify selecting one of many possible motives as being one that is suggested by the available evidence, given that there is no such evidence accompanying the reputation total on each post. If you are going to attack his logic you should make sure yours is impeccable. Ironically, this was my point to him. Link to comment Share on other sites More sharing options...
Ophiolite Posted October 16, 2013 Share Posted October 16, 2013 Ironically, this was my point to him. Yes, I thought you would appreciate the irony. Link to comment Share on other sites More sharing options...
md65536 Posted October 17, 2013 Share Posted October 17, 2013 (edited) Imprecise wording on my part, mea culpa. He says the results suggest a specific motive, and I characterise this specificity as the 'conclusion'. Perhaps I should have said that he could not possibly justify selecting one of many possible motives as being one that is suggested by the available evidence, given that there is no such evidence accompanying the reputation total on each post.Upvotes are encouraging. I didn't say that was the motive, let alone only motive, for the upvote. By the way, ad hominem is an informal logical fallacy. I think I used it correctly, arguably. I suggest you drop this now, as you encouraged me to do. I will drop it too, but it's unfair to tell me to drop it and then continue arguing against me. I have never seen such strong defence of making fun of someone. I think it's sad, but as others disagree with me I'll leave it alone. Edit: Sorry for once again being a dick. I've gotten frustrated by not getting my point across and strayed off course. My point wasn't that there was any malice here (I don't think there was), just that I think that the types of arguments that wouldn't be permitted against mainstream science (like appeal to ridicule or authority) should not be used against speculations. I think that getting used to using bad arguments in defence of science weakens our ability to do science properly. Edited October 17, 2013 by md65536 2 Link to comment Share on other sites More sharing options...
DimaMazin Posted October 17, 2013 Share Posted October 17, 2013 The only way to refute relativity is with experimental data. No.Mathematically wrong thing can't be experimentally right in the same case. -2 Link to comment Share on other sites More sharing options...
Strange Posted October 17, 2013 Share Posted October 17, 2013 No.Mathematically wrong thing can't be experimentally right in the same case. I'm not quite sure what that means, but it appears to be wrong. No, it appears to be right... I think it means, "something which is mathematically wrong, will not give correct the results in experiments". Which is correct. Experimental data is used to falsify incorrect models all the time; that is how science works. For example, experimental data would show that the OP's incorrect mathematics gave the wrong results. Experiment confirms the correct results of relativity (and other models). Link to comment Share on other sites More sharing options...
swansont Posted October 17, 2013 Share Posted October 17, 2013 No.Mathematically wrong thing can't be experimentally right in the same case. Relativity isn't mathematically wrong. The work in the OP is. Link to comment Share on other sites More sharing options...
Bart Posted October 17, 2013 Author Share Posted October 17, 2013 He does so, because he thinks he can simply take the distances traveled reative to the ship and divide them by the speed of sound. But this isn't what SR say you have to do. SR says you have to apply the relativistic formula for the addition of velocities to get the speed at which the sound travels with respect to the ship in each direction according to A and B. Thus, in forward direction the sound travels at: [math] \frac{0.866c+1.11e-6c}{1+\frac{0.886c(1.11e-6c}{c^2}}- 0.866c = ~2.78e-6c = 0.0832646 km/s. [/math] If L' = 0.08325 km, then the trip time for the forward leg is ~1 sec. For the return trip, we get a speed of [math] \frac{0.866c-1.11e-6c}{1-\frac{0.886c(1.11e-6c}{c^2}}- 0.866c = 0.0832647 km/s. [/math] Which gives a return trip time of ~1 sec The total round trip take 2 sec. Which matches the result of the light clock. In the case of the tranverse trip, you have to apply the formula for orthogonal velocities. [math]S= \sqrt{v^2+u^2-v^2u^2}[/math] to work out the speed at which the sound travels with respect to the ship according to A and B as predicted by SR. And if you do so you get the same agreement with the light clock. Your explanation for forvard and return is not clear to me. Could you give the full physical formula for your expression please. What is the parameter e? For transverse direction: Can you prove it by calculation by above formula and show us the value of the sound speed and the ticks period for transverse trip? For radio clock: Can you show your math formula and calculation of the ticks period for radio clock seen by observer A and by observer B?. Link to comment Share on other sites More sharing options...
xyzt Posted October 17, 2013 Share Posted October 17, 2013 Your explanation for forvard and return is not clear to me. Could you give the full physical formula for your expression please. What is the parameter e? For transverse direction: Can you prove it by calculation by above formula and show us the value of the sound speed and the ticks period for transverse trip? For radio clock: Can you show your math formula and calculation of the ticks period for radio clock seen by observer A and by observer B?. Let me re-write Janus LatEx formulas for you: [math] \frac{0.866c+1.11e^{-6}c}{1+\frac{0.886c(1.11e^{-6}c)}{c^2}}- 0.866c = ~2.78e^{-6}c = 0.0832646 km/s. [/math] If L' = 0.08325 km, then the trip time for the forward leg is ~1 sec. For the return trip, we get a speed of [math] \frac{0.866c-1.11e^{-6}c}{1-\frac{0.886c(1.11e^{-6}c)}{c^2}}- 0.866c = 0.0832647 km/s. [/math] Which gives a return trip time of ~1 sec He'll explain the formulas for you (they are simply the relativistic speed composition formulas). 1 Link to comment Share on other sites More sharing options...
Janus Posted October 18, 2013 Share Posted October 18, 2013 Your explanation for forvard and return is not clear to me. Could you give the full physical formula for your expression please. What is the parameter e? It's shorthand for exponential notation. Thus 1.11e-6c = 1.11x10^-6c = 0.00000111c, which is the speed of sound as expressed as a fraction of the speed of light. The general formula is [math] S = \frac{u+v}{1+\frac{uv}{c^2}}[/math] In this case u is the speed of the ship relative to A and B, and u is the speed of the sound clock pulse as measured in the ship S is the speed of the same pulse relative to A and B as measured by A and B. For transverse direction: Can you prove it by calculation by above formula and show us the value of the sound speed and the ticks period for transverse trip? You can do yourself. Take the equation I gave for the orthogonal velocity addition. Plug in 0.866 for v and 0.00000111 for u (warning: you will need to use something that will carry out to quite a few decimal points) Once you get an answer, then plug S into: [math] \sqrt{S^2-0.866^2}[/math] Which gives you the tranverse component of the sound pulse crossing the ship according to A and B. You will get an answer of 0.000000555, or half the speed as measured by someone in the ship. Since the tranverse width of the ship is the same in both frames, the pulse will take twice as long to make the round trip according to A and B, than it does for someone in the ship. For radio clock: Can you show your math formula and calculation of the ticks period for radio clock seen by observer A and by observer B?. I have already spent more time on this than it deserves. I will simply repeat what I know that I have told you before and Swansont has already alluded to in this thread. SR is mathematically and logically consistant. If it wasn't, it would have been tossed in the trash bin a long time ago. It is impossible to come up with any situation, scenario or thought experiment that, if SR is properly applied, will result in a contradiction or flaw in SR. The only way to expose a problem with SR is to provide real experimental results from an actually physically performed experiment that gives results that disagree with what SR predicts or provide a real-life observation of a natural event that disagrees with SR. Otherwise, you are just wasting your and everyone else's time. 3 Link to comment Share on other sites More sharing options...
Bart Posted October 20, 2013 Author Share Posted October 20, 2013 (edited) It's shorthand for exponential notation. Thus 1.11e-6c = 1.11x10^-6c = 0.00000111c, which is the speed of sound as expressed as a fraction of the speed of light. The general formula is [math] S = \frac{u+v}{1+\frac{uv}{c^2}}[/math] In this case u is the speed of the ship relative to A and B, and u is the speed of the sound clock pulse as measured in the ship S is the speed of the same pulse relative to A and B as measured by A and B. You can do yourself. Take the equation I gave for the orthogonal velocity addition. Plug in 0.866 for v and 0.00000111 for u (warning: you will need to use something that will carry out to quite a few decimal points) Once you get an answer, then plug S into: [math] \sqrt{S^2-0.866^2}[/math] Which gives you the tranverse component of the sound pulse crossing the ship according to A and B. You will get an answer of 0.000000555, or half the speed as measured by someone in the ship. Since the tranverse width of the ship is the same in both frames, the pulse will take twice as long to make the round trip according to A and B, than it does for someone in the ship. I have already spent more time on this than it deserves. I will simply repeat what I know that I have told you before and Swansont has already alluded to in this thread. SR is mathematically and logically consistant. If it wasn't, it would have been tossed in the trash bin a long time ago. It is impossible to come up with any situation, scenario or thought experiment that, if SR is properly applied, will result in a contradiction or flaw in SR. The only way to expose a problem with SR is to provide real experimental results from an actually physically performed experiment that gives results that disagree with what SR predicts or provide a real-life observation of a natural event that disagrees with SR. Otherwise, you are just wasting your and everyone else's time. Thank you very much for your reply, but to dispel my doubts, please confirm what ticks of sound clock in seconds will see observer A, according to the theory of relativity at speed of the ship 260 000 km / s? And what ticks period should see the observer B? I'm sorry, but I must repeat yet my second request for the professional explanations, if it is possible, what ticks of the radio clock (pulsar) should see passenger in the ship and each of stationary observers A and B. I believe that the time spent on the search for truth is not a waste of time! Edited October 20, 2013 by Bart Link to comment Share on other sites More sharing options...
Bart Posted October 21, 2013 Author Share Posted October 21, 2013 (edited) Version of the article in my opennig post was a bit of my provocation to ignite a professional discussion on the presented example. Unfortunately provocation was unsuccessful. In fact, only Mr. Janus tried to explaina little but not quite. So now on the link as in post #1: http://dl.dropbox.com/u/26262175/ClocksNotConfirmRelativity.pdf is placed the right article on the subject, at 1 and 1/2 pages plus tables and figures. Total 4 pages. Extended article presents a further example of unreliability of SR theory. Once some of the famous physics said somthing like this: "If you can not explain to the bartender a given theory, it means that you yourself do not undertstand the theory, or the theory is just a fairy tale." I admit that I still do not understand the theory of relativity. Bart (ender) Edited October 21, 2013 by Bart -1 Link to comment Share on other sites More sharing options...
swansont Posted October 21, 2013 Share Posted October 21, 2013 Once again: if you use relativity to predict the behavior of the clocks, you cannot get a contradiction with the theory of relativity if you have done your math correctly. Relativity is self-consistent; it's just algebra. I admit that I still do not understand the theory of relativity. And yet you argue that it's wrong… Link to comment Share on other sites More sharing options...
timo Posted October 21, 2013 Share Posted October 21, 2013 And yet you argue that [relativity] is wrong… That's because bartenders don't understand it. On the other hand: Who really goes to a bar to chat about relativity with the bartender? Perhaps they do know relativity. Link to comment Share on other sites More sharing options...
John Cuthber Posted October 21, 2013 Share Posted October 21, 2013 There are two universities in my town. Plenty of the bar staff are students. I'm pretty sure I could find a bar tender who could explain relativity to the customers. 1 Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted October 21, 2013 Share Posted October 21, 2013 Version of the article in my opennig post was a bit of my provocation to ignite a professional discussion on the presented example. Unfortunately provocation was unsuccessful. In fact, only Mr. Janus tried to explaina little but not quite. So now on the link as in post #1: http://dl.dropbox.com/u/26262175/ClocksNotConfirmRelativity.pdf is placed the right article on the subject, at 1 and 1/2 pages plus tables and figures. Total 4 pages. Extended article presents a further example of unreliability of SR theory. Once some of the famous physics said somthing like this: "If you can not explain to the bartender a given theory, it means that you yourself do not undertstand the theory, or the theory is just a fairy tale." I admit that I still do not understand the theory of relativity. Bart (ender) ! Moderator Note If you don't understand it, then how is it that you think you are able to disprove it? We don't tolerate trolling here. I am closing this, pending full staff review. Link to comment Share on other sites More sharing options...
imatfaal Posted October 22, 2013 Share Posted October 22, 2013 ! Moderator Note After staff review it was decided that the thread should remain locked. Do not open a new thread on the same topic. Link to comment Share on other sites More sharing options...
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