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Lorentz transforms physical incoherence
_______________________________________ (Based on an example presented by Daryl McCullough) We assume that a rocket is moving at speed v = 0.866 c relative to the Earth, and we want to know what is the relationship between the coordinates of an event, in the frame of the Earth, and the coordinates of the same event, in the frame of the rocket. We assume that the rocket passes the Earth at an event with Earth-coordinates tR=0, xR=0 and with rocket-coordinates tR'=0, xR'=0. (The primed variables are observed in the rocket frame). Let's pick an event E, in this case an explosion, that takes place on a planet, at rest relatively to the Earth. The planet is situated at a distance xE form the Earth, and the explosion is observed after a time interval tE, hence the Earth coordinates of the event E are (xE, tE), and the corresponding rocket coordinates are (xE', tE'). For simplicity, let's assume that the Earth, the planet, and the rocket are all lined up, so that we only need to consider the x-axis. In this scenario, the distances are expressed in light-years (ly), and the times in years. After a time interval T, the rocket will be at a distance v*T from the Earth, hence the distance between the rocket and the event E is D = xE-vT. In the rocket frame, the corresponding distance is D' = D/gamma (as v = 0.866 c, gamma = 1/sqrt(1-v^2/c^2) = 2). Let xE = 3 ly and tE = 5 years. Examples: 1) After a time interval T = xE/v, D = 0 and D' = 0. Let's notice that T' = T/gamma. This is always the case, as the ratio T/T' depends on the velocity of the rocket, not on its position. 2) Let's take an arbitrary value T, for instance T = 10 years. From the formula D = xE -vT, we get D = 3-0.866*10 = -5.66 ly, and D' = D/gamma = -2.83 ly. T' = T/gamma = 5 years. 3) Let's take T = 9.608 years. Then D = 3-0.866*9.608 = -5.32 ly D' = D/gamma = -5.32/2 = -2.66 ly Of course, T' = 9.608/2 = 4.804 years. Let's notice that those results have been obtained by applying "Lorentzian" formulae. Now let's use the Lorentz (or rather Einstein) transformations: t' = gamma(t-vx/c^2) and x' = gamma(x-vt) Using the above notation, we get tE' = gamma(tE-v*xE/c^2), thus = 2(5-0.866*3) = 4.804 years D' = gamma(xE-v*tE) = gamma(3-0.866*5) = -2.66 ly, which is exactly the value obtained in example 3, by using T = tE'*gamma = 9.608 years. Thus, gamma(xE-v*tE) = (xE-vT)/gamma. But, in fact, *there is no physical relation between time and position*. When the rocket travels an arbitrary distance v*T wrt the Earth, T' is always given by T/gamma, in other words, T' is independant from v*T. If T = tE = 5 years, T' = 5/2 = 2.5 years, not 4.804 years, and of course, D = xE-v*tE = 3-0.866*5 = -1.33 ly, and D'= D/gamma = -0.665 ly, not -2.66 ly. (notice "en passant" that -0.665*gamma^2 = -2.66) Let's now illustrate the physical incoherence of the Lorentz time transformation tE' = gamma(tE-v*xE/c^2). We have mathematically seen above that gamma (xE-v*tE) = (xE-v*T)/gamma, or gamma^2 (v*tE - xE) = v*T - xE (1) Let's tE = xE/v + delta t, thus delta t = tE - xE/v v*delta t = v*tE - xE Replacing (v*tE - xE) by v*delta t in relation (1), one gets gamma^2 * v*delta t = v*T - xE v*T = xE + gamma^2 * v*delta t T = xE/v + gamma^2 * delta t, T = xE/v + gamma^2 * (tE - xE/v) (2) And indeed, with xE=3, tE=5 and gamma=2, equation (2) leads to T = 3/0.866 + 4(5 - 3/0.866) = 3.464 + 4(5 - 3.464) = 9.608 years Equation (2) shows that the t' in the Lorentz time transform t' = gamma(t-vx/c^2), expressed in the Earth frame by T = t'*gamma, corresponds to the time tau needed by the rocket to travel the distance xE at the velocity v, increased by the difference between the time coordinate tE of the event E and the time tau, multiplied by the factor gamma^2. If the Lorentz time transformation were coherent, the correction by gamma^2 would also apply to tau, not only to the difference (tE-tau). Indeed, there is no physical justification for not treating homogenously the time coordinate tE. Btw, such fancy correction by gamma^2 leads to absurdities. For instance, when v is close to the speed of light, the time corresponding to the product gamma^2(tE-tau) can exceed the age of the Universe! The incoherence of the Lorenz time transform is very obvious when the event coordinates are (xE=0, tE=5). Then the transform gives T = gamma^2 * tE = 4 * 5 = 20 years, instead of T = 5 years! Marcel Luttgens |
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![]() wrote in message ups.com... Lorentz transforms physical incoherence _______________________________________ (Based on an example presented by Daryl McCullough) Marcel Luttgens based on Marcel Luttgens: Marcel Luttgens' fumbles: http://www.google.com/search?q=luttg...ers.pandora.be Marcel Luttgens' home page: http://perso.wanadoo.fr/mluttgens/ March 2003 "The Lorentz transformation (LT) are false": http://perso.wanadoo.fr/mluttgens/LTfalse.htm shot down in http://groups.google.com/groups?&thr... g.google.com May 2003 "There is no length contraction" http://perso.wanadoo.fr/mluttgens/mmx.htm shot down in http://groups.google.com/groups?&thr... ng.google.com Sep 2001 "The Twin paradox falsifies SR" http://perso.wanadoo.fr/mluttgens/twinpdx1.htm shot down in http://groups.google.com/groups?&thr...nso-cl.aol.com "Newton vs Einstein" http://redshift.vif.com/JournalFiles...F/V05n1lut.pdf Dirk Vdm |
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Moronic crackpot!
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Daryl, you didn't follow the demonstration, especially:
3) Let's take T = 9.608 years. Then D = 3-0.866*9.608 = -5.32 ly D' = D/gamma = -5.32/2 = -2.66 ly Of course, T' = 9.608/2 = 4.804 years. Let's notice that those results have been obtained by applying "Lorentzian" formulae. Now let's use the Lorentz (or rather Einstein) transformations: t' = gamma(t-vx/c^2) and x' = gamma(x-vt) Using the above notation, we get tE' = gamma(tE-v*xE/c^2), thus = 2(5-0.866*3) = 4.804 years D' = gamma(xE-v*tE) = gamma(3-0.866*5) = -2.66 ly, which is exactly the value obtained in example 3, by using T = tE'*gamma = 9.608 years. Take a little more time! Marcel |
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![]() wrote in message ups.com... Daryl, you didn't follow the demonstration, especially: But you were wrong before you even started, silly. You still don't understand coordinates and events. You talk about "the distance between the rocket and event E" but you should say "the distance between the rocket and the planet" You say: "Let's notice that T' = T/gamma. This is always the case, as the ratio T/T' depends on the velocity of the rocket, not on its position." but this is clearly not always the case. It is only the case if T' happens to be a time interval between two *two physical ticks of the rocket clock*, and if T is the same time between those same rocket ticks, but now measured on the Earth clock. But that was not the way you defined T' and T. In stead, you defined T as the time measured on the *Earth clock* between the tick of departure of the rocket, and another tick on the Earth clock, namely the tick that is simultaneous with the event where the rocket is at distance vT as seen in the Earth frame. So you are talking about a proper time on the *Earth clock*, so you should be using the expression T' = T*gamma. Remember, Marcel? Many of us explained that to you just as many times as the number of times you miserably failed to understand it. Difficult, isn't it? Shouldn't you turn your attention to collecting poodle droppings or something? 3) Let's take T = 9.608 years. Then D = 3-0.866*9.608 = -5.32 ly D' = D/gamma = -5.32/2 = -2.66 ly Of course, T' = 9.608/2 = 4.804 years. Let's notice that those results have been obtained by applying "Lorentzian" formulae. But you don't even understand the concept of coordinates. What could you possible do with advanced stuff like that? Now let's use the Lorentz (or rather Einstein) transformations: t' = gamma(t-vx/c^2) and x' = gamma(x-vt) Using the above notation, we get tE' = gamma(tE-v*xE/c^2), thus = 2(5-0.866*3) = 4.804 years D' = gamma(xE-v*tE) = gamma(3-0.866*5) = -2.66 ly, which is exactly the value obtained in example 3, by using T = tE'*gamma = 9.608 years. Take a little more time! Take *much* more time. Dirk Vdm |
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Like always, you are "parroting" what you read in textbooks, you can't
think by yourself. You are a typical example of those religious SR crackpots whose mind closes when they are confronting with a dissident opinion. If you want to be taken seriously, try at least to refute my demonstration of the LT incoherence, or at least to explain why gamma(xE-v*tE) = (xE-vT)/gamma. is wrong, according to your faithful follower thinking. Don't forget that when xE=3, v=0.866, tE=5, T=9.608 (T is obtained from the LT), one gets 2(3-0.866*5) = (3-0.866*9.608)/2 , or -2.66 = -2.66 ! Use your brain, not your prejudices! Of course, abuse (your specialty) is easier. Marcel Luttgens |
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![]() wrote in message oups.com... Like always, you are "parroting" what you read in textbooks, you can't think by yourself. You are a typical example of those religious SR crackpots whose mind closes when they are confronting with a dissident opinion. But you don't even *have* a dissident opinion. A complete misunderstanding is what you have. You find inconsistencies in your own view on a theory by simply not having a clue what it is about and by making one silly mistake after the other. I have just shown you that you that your statement | "Let's notice that T' = T/gamma. This is always the | case, as the ratio T/T' depends on the velocity of | the rocket, not on its position." is completely wrong in the context that *you* provided. And, just like you have been doing for so many years, you build on your mistake to arrive at something silly. Exercice for you: In my previous message there is one word that should not be there. Find it. Then explain what is wrong with what I said. If you want to be taken seriously, try at least to refute my demonstration of the LT incoherence But you have not applied the LT correctly, because you made a mistake with your usage of the time dilation. I just explained it. Don't you understand it? Even if I have explained it to you hundreds of times? Prove to me that you understood at least part of it by doing the exercise I gave you. , or at least to explain why gamma(xE-v*tE) = (xE-vT)/gamma. is wrong, according to your faithful follower thinking. I don't have faith and I don't follow anything. I couldn't care less whether special relativity is right or wrong. But unlike you, I can apply it correctly because I understand what the symbols and variables mean. Don't forget that when xE=3, v=0.866, tE=5, T=9.608 (T is obtained from the LT), one gets 2(3-0.866*5) = (3-0.866*9.608)/2 , or -2.66 = -2.66 ! Use your brain, not your prejudices! Of course, abuse (your specialty) is easier. By now, you should be grown up enough to ignore the abuse. Unless that is what you are *really* after :-) Can you do the little exercise? Or do I conclude -again- that you are cheating, Marcel? Dirk Vdm |
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Daryl, you forgot to send your message.
Did you realize that, according to my demonstration, the Lorentz time transformation t' = gamma(t-xv/c^2) can be written t' = (x/v + gamma^2(t-x/v)) / gamma, from which it is clear that the LT doesn't treat homogenously the time coordinate tE. Try with x=3, t=5 and v=0.866 (gamma=2). Marcel |
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