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Natural Science Forum / Physics / Relativity / December 2004Relativists cannot count to three.
If I stand before a vertical surface and throw a ball 'o' with speed 'u', catching it as it returns, I need a minumum of two coordinates to describe the events of emission, reflection and reception. Ao----------(+u - >)------------* emission *--------------------------------oB reflection Ao----------( -u <- )------------* reception In agreement with experience it is clear that the time for the ball to reach the wall and back again is given by 2AB/(t'A-tA) = u, the velocity of the ball. If I do this whilst riding a moving platform travelling at velocity v, the wall also moving, I need a minimum of three coordinates for a stationary observer to evaluate the events. Ao----------(v+u)-----------* emission *---------------------------oB reflection Co---------(v-u)------------* reception In the frame of the observer, the ball travels from A to B' at v+u and returns from B' to C at v-u. Einstein only uses two cordinates, A = C = (0,0,0) and B = (x',0,0). Proof: "From the origin of system k let a ray be emitted at the time tau0 along the X-axis to x', and at the time tau1 be reflected thence to the origin of the co-ordinates, arriving there at the time tau2" ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) Reference : http://www.fourmilab.ch/etexts/einstein/specrel/www/ Hence Einstein and his relativist disciples cannot count to three, QED. "the velocity of light in our theory plays the part, physically, of an infinitely great velocity." -Einstein. Androcles. > If I stand before a vertical surface and throw a ball 'o' with speed > 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 19 lines] > > Einstein only uses two cordinates, A = C = (0,0,0) and B = (x',0,0). *You* *do* *not* *understand* *the* *concept* *of* *coordinates*. You don't understand vectors: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/VectorLength.html You don't understand limits: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegativeInfinity.html You don't understand equations: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Pythagoras2.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve2.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Persuasive.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/AndroDistri.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Pythagoras.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ToothlessBite.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Competent.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/UseTrans.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Sheesh.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/DivZero.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Think.html You don't understand square roots: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/STILL.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/CanSpecify.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Nearly.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Quadratic.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GrowUp.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Tautology.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Material.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/GIVEN.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PythagoRescue.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtRev.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/NegSqrt.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SqrtAnswers.html You don't understand exclusive ors: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Gibberish.html You don't understand partial differential equations: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff2.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/PartialDiff3.html Dirk Vdm >> If I stand before a vertical surface and throw a ball 'o' with speed >> 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 28 lines] > *of* > *coordinates*. Is it not true that SR only requires the origin of the moving observer's coordinate system and the stationary observer's coordinate system to be coincident at a single instant? After that both observers see the other's origin shifting position. In any case this model is useless for evaluating SR. The velocities are miniscule compared to the speed of light, and the velocity composition is linear here, unlike the requirement of SR. If you used the relativistic velocity composition formula and then examined the results as one of the velocities approached c, then it might be meaningful. >>> If I stand before a vertical surface and throw a ball 'o' with >>> speed [quoted text clipped - 37 lines] > single instant? After that both observers see the other's origin > shifting position. Correct. Dinky the spermless can't count to three either. > In any case this model is useless for evaluating SR. The velocities > are miniscule > compared to the speed of light, and the velocity composition is linear > here, > unlike the requirement of SR. This is not about "evaluating" SR, v can have any value. If you used the relativistic velocity > composition > formula and then examined the results as one of the velocities > approached c, > then it might be meaningful. Sure thing. c = (c+v)/(1+v/c), giving ½[tau(0,0,0,t)+tau(0,0,0,t+x'/c+x'/c)] = tau(x',0,0,t+x'/c) instead of ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) so now derive the Lorentz Transforms from that and then derive the composition of velocities from the Lorentz Transforms. Reference : http://www.fourmilab.ch/etexts/einstein/specrel/www/ Androcles. [snip] > Sure thing. > [quoted text clipped - 6 lines] > so now derive the Lorentz Transforms from that and then derive the > composition of velocities from the Lorentz Transforms. And then try to invert the transformation http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/SetSolve2.html Dirk Vdm > [snip] > [quoted text clipped - 14 lines] > > Dirk Vdm Sorry to disappoint you Dirk, but your reference site does not prove anything. If you had taken the time to actually solve the equations you posted you would have found that they show that the inverse is actually achieved by changing the sign of the relative velocity and exchanging the primed values with the unprimed ones. The simple reason the equations don't look identical is that the second set has a mixture of primed and unprimed coordinates on the RHS. If you simply collect all the primed terms on one side and all the unprimed terms on the other, the symmetry becomes obvious. If you don't believe me, try it yourself. Rearrange the second equation in the inverse set to solve for x, and then equate the result with the first equation, and collect terms so that primed coordinates are on one side and unprimed are on the other. You get an equation for t(x', t') instead of t(x, t'). Substituting the result back into the first equation of the second set gives you similar results - an equation for x(x', t') instead of x(x', t). Having read Einstein's derivation, I am unclear as to why he gets away with assuming the stationary observer sees a light ray travel at c-v, then c+v. > > [snip] > > [quoted text clipped - 17 lines] > Sorry to disappoint you Dirk, but your reference site does not prove > anything. It proves that Androcles is an idiot. > If you had taken the time to actually solve the equations you posted I did not post equations. Androcles did. > you > would have found that they show that the inverse is actually achieved by > changing the sign of the relative velocity and exchanging the primed values > with the unprimed ones. Don't tell me. Try to explain it to the idiot. > The simple reason the equations don't look identical > is that the second set has a mixture of primed and unprimed coordinates on > the RHS. Yes, I know. Explain it to the idiot. > If you simply collect all the primed terms on one side and all the > unprimed terms on the other, the symmetry becomes obvious. If you don't [quoted text clipped - 8 lines] > an > equation for x(x', t') instead of x(x', t). I'm not sure what you are after and why you are telling *me* this. For the record and to be sure, when you solve the set { x' = g(x-vt) { t' = g(t-vx/c^2) with g = 1/sqrt(1-v^2/c^2) for x and t, you get { x = g(x'+vt') { t = g(t'+vx'/c^2) If *you* don't believe it, try it, otherwise, no problem. If *you* can explain it to the idiot, try that as well. > Having read Einstein's derivation, I am unclear as to why he gets away with > assuming the stationary observer sees a light ray travel at c-v, then c+v. He most certainly does not. Have a long and close look at http://groups-beta.google.com/group/sci.physics/msg/4ac73e8ad8da3906 or http://groups.google.co.uk/groups?as_umsgid=Oymwd.2255$OF6.228383@phobos.telenet -ops.be Follow the pointers. Dirk Vdm >> > [snip] >> > [quoted text clipped - 80 lines] > > Dirk Vdm Sorry for the attribution error. I misunderstood the source of the Fumble. I guess my comments should be directed to Androcles: you appear to be able to understand simple algebra - finish the job using the sequence I outlined and your argument vanishes. My other comment was directed at you and still stands. Einstein only needed to count to 2 because he was talking about a stationary point and the origin and an observer in the moving coordinate system. Your point that an observer in the stationary coordinate system would need 3 points is correct as well, but in no way contradicts Einstein's argument. [snip] > >> Having read Einstein's derivation, I am unclear as to why he gets away > >> with [quoted text clipped - 13 lines] > guess my comments should be directed to Androcles: you appear to be > able to understand simple algebra He does not understand simple algebra. > - finish the job using the sequence I > outlined and your argument vanishes. My other comment was directed at > you and still stands. So do the explanations at http://groups-beta.google.com/group/sci.physics/msg/4ac73e8ad8da3906 or http://groups.google.co.uk/groups?as_umsgid=Oymwd.2255$OF6.228383@phobos.telenet -ops.be Follow the pointers. > Einstein only needed to count to 2 because he was > talking about a stationary point and the origin and an observer in the > moving coordinate system. Your point that an observer in the stationary > coordinate system would need 3 points is correct as well, but in no way > contradicts Einstein's argument. I have no idea what you are talking about. Dirk Vdm > [snip] > [quoted text clipped - 40 lines] > > Dirk Vdm I was referring to comments in my previous post that were directed to Androcles, and still are. They refer to the premise of this thread which he started, that relativists can't count to three and Androcles can. I will try to restrict my comments to one party per posting in the future. [snip] > I was referring to comments in my previous post that were directed to > Androcles, and still are. They refer to the premise of this thread which > he started, that relativists can't count to three and Androcles can. That might be so, but it is not obvious that he can do anything more than that. > I will try to restrict my comments to one party per posting in the future. That is essential if you want your readers to know what is going on. As a matter of fact, there is enough of interest in your mangled replies to warrant considering whether you should not reboot and repeat the conversations in the manner you said, and with the correct attributions. Franz > Sorry for the attribution error. I misunderstood the source of the > Fumble. I [quoted text clipped - 6 lines] > talking about a stationary point and the origin and an observer in the > moving coordinate system. Einstein couldn't count to three anymore than Dinky Van de Torquemada can. There is no observer in the moving coordinate system. ALL the parameters to the function tau() are stationary frame parameters. In fact he specifically says "If we place x'=x-vt, it is clear that a point at rest in the system k must have a system of values x', y, z, independent of time." x' is NOT at rest in the system K, though. |__0 ____________x __| t0 |__0'____________x'__| |__0 ____________x __| t1 |__0'____________x'__| |__0 ____________x __| t2 |__0'____________x'__| See, x' remains a fixed distance from 0' and is at rest in system k, independent of time. It is at a moving distance from 0, though,by x' = x-vt which is not independent of time. What Einstein's equation says is |__0 ____________x __| t2 |__0'____________x'__| Because 0' still coincides with 0 at t2. > Your point that an observer in the stationary > coordinate system would need 3 points is correct as well, Exactly. > but in no way > contradicts Einstein's argument. Well, it works when v = 0 or c = infinity. Otherwise it is nonsense. "the velocity of light in our theory plays the part, physically, of an infinitely great velocity." - Einstein. ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ Androcles. >> Sorry for the attribution error. I misunderstood the source of the >> Fumble. I [quoted text clipped - 10 lines] > can. > There is no observer in the moving coordinate system. ALL the Wrong. This is what Einstein wrote: "From the origin of system k let a ray be emitted at the time along the X-axis to x', and at the time be reflected thence to the origin of the co-ordinates, arriving there at the time ; ..." The system k is the moving coordinate system. > parameters to the function tau() are stationary frame parameters. Because tau() is a transform equation to relate a moving system coordinate to the stationary coordinates. In the moving coordinate system, tau is not dependent on any other coordinates. > In fact he specifically says > "If we place x'=x-vt, it is clear that a point at rest in the system k > must > have a system of values x', y, z, independent of time." > x' is NOT at rest in the system K, though. Who cares? > |__0 ____________x __| t0 > |__0'____________x'__| [quoted text clipped - 14 lines] > > Because 0' still coincides with 0 at t2. Because Einstein is describing the system k which is moving. Of course its origin moves relative to the stationary frame K, but, again, who cares? >> Your point that an observer in the stationary >> coordinate system would need 3 points is correct as well, [quoted text clipped - 5 lines] > > Well, it works when v = 0 or c = infinity. Otherwise it is nonsense. You have failed to show this. > "the velocity of light in our theory plays the part, physically, of an > infinitely great velocity." - Einstein. > ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ This statement proves nothing. It is the second part of an analogy to Newton, and describes the asymptotic behavior of velocities in the two different systems. > Androcles. Tom Capizzi >>> Sorry for the attribution error. I misunderstood the source of the >>> Fumble. I [quoted text clipped - 21 lines] > co-ordinates, arriving there at the time ; ..." > The system k is the moving coordinate system. Yeah, so? I don't see 'observer' mentioned in that. Learn to read. You obviously can't count to three either. >> parameters to the function tau() are stationary frame parameters. > [quoted text clipped - 4 lines] > dependent on > any other coordinates. So what? Nobody disagrees with that. You obviously can't count to three either. >> In fact he specifically says >> "If we place x'=x-vt, it is clear that a point at rest in the system [quoted text clipped - 28 lines] > origin moves relative to the stationary frame K, but, again, who > cares? Sure, so what your point? >>> Your point that an observer in the stationary >>> coordinate system would need 3 points is correct as well, [quoted text clipped - 7 lines] > > You have failed to show this. Kinda hard for those that cannot count to three, I know. >> "the velocity of light in our theory plays the part, physically, of >> an >> infinitely great velocity." - Einstein. >> ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ > > This statement proves nothing. Strange that he had to say it then. > It is the second part of an analogy to > Newton, and describes the asymptotic behavior of velocities in the > two different systems. In your dreams. >> Androcles. > > Tom Capizzi >>>> Sorry for the attribution error. I misunderstood the source of the >>>> Fumble. I [quoted text clipped - 22 lines] > > Yeah, so? I don't see 'observer' mentioned in that. Actually, there is no observer mentioned anywhere in Section 3 of Einstein's paper. However, he clearly locates the ray of light in the moving frame. It returns to the origin of the moving frame. In the moving frame, the origin does not move, whether there is an observer or not. That's 2 points, not 3. Maybe you should try to understand what you read instead of insulting everyone's intelligence. > Learn to read. You obviously can't count to three either. > [quoted text clipped - 37 lines] >>> >>> Because 0' still coincides with 0 at t2. Your cutesy diagrams are meaningless and incorrect. For the sake of argument, let the symbols K and k represent the origins of their respective systems. If x' = x - v t, when x' = 0, x = v t. For a stationary point in k, x' is a constant, and x = v t + constant. The light ray is shown from the perspective of the stationary frame K. From the moving frame k the two paths appear to be identical in length Then your drawings should look like this: |__K___________x __|t = 0 |__k ___________x'__|tau = 0 0 |__K_____________x __|t = t0 v t0|__k ___________x'__|light ray leaves k at tau0 | |--------------------------------->| |__K________________x __|t = t1 v t1|__k ____________x'__|light ray bounces off x' at tau1 | |<-------------------------- | |__K___________________x __|t = t2 v t2|__k _____________x'__|light ray returns to k at tau2 | The origins k and K only coincide once, when t = tau = 0, so your last drawing is just plain wrong, as is your conclusion that v can only equal 0 or infinity. The ray begins at some arbitrary time tau0, when t = t0. It ends at tau2, when t = t2. As long as tau0 >= 0, the origin of k never again coincides with the origin of K. >> Because Einstein is describing the system k which is moving. Of course >> its [quoted text clipped - 31 lines] > > In your dreams. What Einstein said, exactly, is this: " For velocities greater than that of light our deliberations become meaningless; we shall, however, find in what follows, that the velocity of light in our theory plays the part, physically, of an infinitely great velocity." Just as it is meaningless to talk about velocities greater than infinity in Newton's physics, it is meaningless to talk about velocities greater than the speed of light in Einstein's. >>> Androcles. >> >> Tom Capizzi I no longer care if you understand what I have tried to explain. I post for the benefit of less experienced readers who may have been confused by your bs. I refuse to engage in a battle of wits with someone who is unarmed. Grow up. >>>>> Sorry for the attribution error. I misunderstood the source of the >>>>> Fumble. I [quoted text clipped - 29 lines] > Einstein's paper. However, he clearly locates the ray of light in the > moving frame. Yes, That's right. > It returns to the origin of the moving frame. That's right. Those that can count to two will quickly realize two points are needed in the moving frame. > In the > moving frame, the origin does not move, whether there is an observer > or not. That's 2 points, not 3. Well done. > Maybe you should try to understand > what you read instead of insulting everyone's intelligence. Yes, well, I'm not trying to insult you intelligence, it is what it is. >> Learn to read. You obviously can't count to three either. Oh, but I can. Three points are needed in the stationary frame, and two are needed in the moving frame. Lacking sufficient intelligence to count to three, I don't expect you to understand that. >>>> parameters to the function tau() are stationary frame parameters. >>> [quoted text clipped - 40 lines] > > Your cutesy diagrams are meaningless and incorrect. Yes, well, we know you can't count to three. > For the sake of > argument, let the symbols K and k represent the origins of their > respective systems. Learn to read. The symbols K and k represent frames, not origins of frames. > If x' = x - v t, when x' = 0, x = v t. Well done, you've just found the third point in K, 0-vt2. The other two are 0-vt0 and x' - vt1. You'll be counting to three anytime soon. > For a stationary > point in k, x' is a constant, and x = v t + constant. The light ray is [quoted text clipped - 16 lines] > The origins k and K only coincide once, when t = tau = 0, so your last > drawing is just plain wrong, Well, of course it is wrong! It is, however, what Einstein's equation says, so I agree that his equation is wrong. Two points in the k-frame and three in the K-frame. Learn to count to three. > as is your conclusion that v can only equal 0 > or infinity. The ray begins at some arbitrary time tau0, when t = t0. [quoted text clipped - 3 lines] > again > coincides with the origin of K. Well of course, so why does Einstein's equation say it does? Learn to count to three. >>> Because Einstein is describing the system k which is moving. Of >>> course [quoted text clipped - 43 lines] > infinity in Newton's physics, it is meaningless to talk about > velocities greater than the speed of light in Einstein's. Sure, since it is already infinite. Androcles. >>>> Androcles. >>> [quoted text clipped - 4 lines] > confused by your bs. I refuse to engage in a battle of wits with > someone who is unarmed. Grow up. <snip> >>>>> "the velocity of light in our theory plays the part, physically, of >>>>> an [quoted text clipped - 20 lines] >> infinity in Newton's physics, it is meaningless to talk about >> velocities greater than the speed of light in Einstein's. >Sure, since it is already infinite. And Androcles shows once again what a complete and utter moral vacuum he is. He has been informed, several times, that what Einstein meant by that statement is that, just as infinite speed is the unattainable supremum for the attainable speed of material bodies in classical mechanics, the speed of light is the unattainable supremum for the attainable speed of material bodies in special relativity. He has been informed several times that Einstein NEVER claimed that the speed of light is infinite. Of course, he is so dishonest that he will continue to claim what he has already been told is false, and he will refuse to even acknowledge being told otherwise. How could anybody discuss anything with a person so devoid of personal integrity as Androcles? David ----- > >>> If I stand before a vertical surface and throw a ball 'o' with > >>> speed [quoted text clipped - 68 lines] > > Androcles. Hey idiot Androcles, <http://www.google.com/search?q=Androcles+fumble+site%3Ausers.pandora.be> reposting the same idiot drool that has been so thoroughly, utterly publicly discredited by those who can do math (e.g., Randy Poe, in disgustingly punctilious counterpoint) merely demonstrates what an intractible idiot you are. Empirical physical reality casts the only votes that count. Your idiot spew is falsified by trivial empirical observation. You are a psychotic ineducable idiot. Where are your citations, idiot Androcles? Where are your literature references, idiot Androcles? Where is your empirical observational support, idiot Androcles? You drown in explicit empirical falsfification, idiot Androcles. Your ignorance, incompetence, and psychosis are not of interest to the world at large. Quite the contrary. You are not even an interesting laughingstock. <http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html> Hafele-Keating experiment. You are f.cked, idiot Androcles. Nature 425 374 (2003) <http://relativity.livingreviews.org/Articles/lrr-2003-1/> http://www.eftaylor.com/pub/projecta.pdf <http://www.public.asu.edu/~rjjacob/Lecture16.pdf> Relativity in the GPS system. You are f.cked, idiot Androcles. <http://relativity.livingreviews.org/Articles/lrr-2003-1/> http://arXiv.org/abs/gr-qc/0311039 <http://www.weburbia.demon.co.uk/physics/experiments.html> Experimental constraints on General Relativity. You are f.cked, idiot Androcles. Science 303(5661) 1143;1153 (2004) http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries. You are f.cked, idiot Androcles. Physics Today 57(7) 40 (2004) http://physicstoday.org/vol-57/iss-7/p40.shtml No aether. You are f.cked, idiot Androcles. http://fsweb.berry.edu/academic/mans/clane/ http://physicsweb.org/articles/world/17/3/7 No Lorentz violation. You are f.cked, idiot Androcles. <http://math.ucr.edu/home/baez/RelWWW/tests.html> Mathematics of gravitation. You are f.cked, idiot Androcles. http://www.hep.upenn.edu/~max/toe.html You are f.cked, idiot Androcles. http://www.iancgbell.clara.net/maths/spctime.htm You are f.cked, idiot Androcles. http://insti.physics.sunysb.edu/~siegel/Fields2.pdf You are f.cked, idiot Androcles. Idiot Androcles is a eunuch in a brothel, a capon in a henhouse, a steer amidst cows; a stot, a gelding, a gelt, a havier, a gib, a lapin, a seg, a hog, a wether... a butt-f.cked psychotic idiot spewing in a science newsgroup.  SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf >> > "Dirk Van de moortel" >> > <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> [quoted text clipped - 79 lines] >> >> Androcles. Hey idiot Schwartz, you are one of those that cannot count to three. Androcles > If I stand before a vertical surface and throw a ball 'o' with speed > 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 35 lines] > > Androcles. Hey idiot Androcles, <http://www.google.com/search?q=Androcles+fumble+site%3Ausers.pandora.be> reposting the same idiot drool that has been so thoroughly, utterly publicly discredited by those who can do math (e.g., Randy Poe, in disgustingly punctilious counterpoint) merely demonstrates what an intractible idiot you are. Empirical physical reality casts the only votes that count. Your idiot spew is falsified by trivial empirical observation. You are a psychotic ineducable idiot. Where are your citations, idiot Androcles? Where are your literature references, idiot Androcles? Where is your empirical observational support, idiot Androcles? You drown in explicit empirical falsfification, idiot Androcles. Your ignorance, incompetence, and psychosis are not of interest to the world at large. Quite the contrary. You are not even an interesting laughingstock. <http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/airtim.html> Hafele-Keating experiment. You are f.cked, idiot Androcles. Nature 425 374 (2003) <http://relativity.livingreviews.org/Articles/lrr-2003-1/> http://www.eftaylor.com/pub/projecta.pdf <http://www.public.asu.edu/~rjjacob/Lecture16.pdf> Relativity in the GPS system. You are f.cked, idiot Androcles. <http://relativity.livingreviews.org/Articles/lrr-2003-1/> http://arXiv.org/abs/gr-qc/0311039 <http://www.weburbia.demon.co.uk/physics/experiments.html> Experimental constraints on General Relativity. You are f.cked, idiot Androcles. Science 303(5661) 1143;1153 (2004) http://arXiv.org/abs/astro-ph/0401086 http://arxiv.org/abs/astro-ph/0312071 Deeply relativistic neutron star binaries. You are f.cked, idiot Androcles. Physics Today 57(7) 40 (2004) http://physicstoday.org/vol-57/iss-7/p40.shtml No aether. You are f.cked, idiot Androcles. http://fsweb.berry.edu/academic/mans/clane/ http://physicsweb.org/articles/world/17/3/7 No Lorentz violation. You are f.cked, idiot Androcles. <http://math.ucr.edu/home/baez/RelWWW/tests.html> Mathematics of gravitation. You are f.cked, idiot Androcles. http://www.hep.upenn.edu/~max/toe.html You are f.cked, idiot Androcles. http://www.iancgbell.clara.net/maths/spctime.htm You are f.cked, idiot Androcles. http://insti.physics.sunysb.edu/~siegel/Fields2.pdf You are f.cked, idiot Androcles. Idiot Androcles is a eunuch in a brothel, a capon in a henhouse, a steer amidst cows; a stot, a gelding, a gelt, a havier, a gib, a lapin, a seg, a hog, a wether... a butt-f.cked psychotic idiot spewing in a science newsgroup. SignatureUncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) http://www.mazepath.com/uncleal/qz.pdf >> If I stand before a vertical surface and throw a ball 'o' with speed >> 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 37 lines] >> > Hey idiot Schwartzfification, we know you can't count to three. Androcles. > >> If I stand before a vertical surface and throw a ball 'o' with speed > >> 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 39 lines] > > Androcles. And Androcles can't use attribution marks. Franz <snip> >> >> "the velocity of light in our theory plays the part, physically, >> >> of an infinitely great velocity." -Einstein. Androcles insists, even in the face of overwhelming refutation, that what Einstein meant by this sentence was that the speed of light is infinite. He refuses to even acknowledge the possibility that what Einstein actually meant was that, just as infinite speed in the unattainable supremum for the possible speeds of material bodies in Newtonian mechanics, the speed of light is the unattainable supremum for the possible speeds of material bodies in special relativity. In other words, the relation of infinite speed to Newtonian mechanics and the relation of the speed of light to special relativity are exactly the same: each speed is the unattainable supremum of the possible speeds of material bodies in the corresponding theory. David ----- Showing Androclese he's wrong is a fool's errand. He LIKES being pubicly humiliated because it's the only attention he can get. This started in 1995 when he became sexually aroused whenever his mommy spanked him. He then repeatedly misbehaved, just to get the spanking. I know this because his mother told me last night in bed. So I told her "Shut up bitch, and get back to work! That mouth isn't for talking with!" I didn't pay her the two dollars, either. =[ d > If I stand before a vertical surface and throw a ball 'o' with speed > 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 6 lines] > to reach the wall and back again is given by 2AB/(t'A-tA) = u, > the velocity of the ball. Wrong. Velocity is not time. The velocity is u. The time taken is distance/velocity = 2AB/u (assuming perfect, instantaneous bounce) They said you were useless at math and they weren't wrong. > > If I stand before a vertical surface and throw a ball 'o' with speed > > 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 13 lines] > > They said you were useless at math and they weren't wrong. I think that sentence was a clumsy shortcut for: | " ... it is clear that the time for the ball | to reach the wall and back again is given by t'A-tA in | the equation 2AB/(t'A-tA) = u, where u is the velocity | of the ball." Rigour is not Androcles' forte ;-) Dirk Vdm > If I stand before a vertical surface and throw a ball 'o' with speed > 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 35 lines] > > Androcles. To understand Einstein's formula you have to realize and keep in mind that Einstein is defining tau as a function of (x',y,z,t) rather than (x,y,z,t). x' is defined by the equation x'=x-vt tau is the time of an event as measured in Einstein's 'moving' frame K. (x,y,z,t) are the coordinates of the event as measured in the 'stationary' frame k, and v is the speed of frame k relative to frame K. Einstein introduced x' merely for mathematical convenience. Note that x' does not represent the x-coordinate of the event in frame k (unless t happens to be zero) nor does it represent the x-coordinate of the event as measured in the 'moving' frame K. The physical interpretation of x' is that it represents the x-distance between the location of the event and the origin of the 'moving' frame K as measured in the 'stationary' frame k. (Failure to understand the meaning of x' is a primary stumbling block in following Einstein's derivation.) With this in mind and letting A, B, and C denote the events of emission, reflection, and reception, respectively, of the light signal, then we can write Einstein's formula as (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) This equation merely expresses Einstein's clock synchronization in frame K. Now x'_A represents the x-distance (as measured in frame k) between the point of emission of the light signal and origin of frame K. This is clearly zero since the signal was emitted from the origin of K. Likewise, x'_C represents the x-distance (as measured in k) between the point of reception of the light signal and the origin of frame K. Again this is clearly zero since the signal was received at the origin of K. Thus, x'_A = x'_C = 0. Then the formula may be written (1/2)[tau(0,0,0,t_A)+tau(0,0,0,t_C)] = tau(x'_B,0,0,t_B) Next, note that t_B is the time of reflection of the signal at B as measured by k. Since light is assumed to propagate at speed c in frame k (as well as frame K), we have t_B = t_A + (x_B - x_A)/c Using x'_A = x_A - vt_A and x'_B = x_B - vt_B we have t_B = t_A + (x'_B + vt_B - x'_A - vt_A)/c This is easily solved for t_B to get (remembering x'_A = 0) t_B = t_A + x'_B/(c-v) Likewise, t_C = t_B + (x_B - x_C)/c and a similar calculation (remembering x'_C = 0) shows t_C = t_B + x'_B/(c+v) and then substituting t_B = t_A + x'_B/(c-v) yields t_C = t_A + x'_B/(c-v) + x'_B/(c+v) Now Einstein simply denotes t_A by t and x'_B by x'. So, the equation (1/2)[tau(0,0,0,t_A)+tau(0,0,0,t_C)] = tau(x'_B,0,0,t_B) becomes Einstein's (1/2)[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) Todd >> If I stand before a vertical surface and throw a ball 'o' with speed >> 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 42 lines] > (x,y,z,t). x' is defined by the equation > x'=x-vt Of course. I works for a ball bouncing, too. There is no magic to it. > tau is the time of an event as measured in Einstein's 'moving' frame > K. Yep. > (x,y,z,t) are the coordinates of the event as measured in the > 'stationary' frame k, and v is the speed of frame k relative to frame > K. Yep. Simple, isn't it? > Einstein introduced x' merely for mathematical convenience. Sure. Writing x-vt every time get old. > Note that x' does not represent the x-coordinate of the event in frame > k (unless t happens to be zero) nor does it represent the x-coordinate > of the event as measured in the 'moving' frame K. 'Moving' frame K? What's that then? K is the 'stationary' frame. You are getting mixed up. > The physical interpretation of x' is that it represents the x-distance > between the location of the event and the origin of the 'moving' frame > K as measured in the 'stationary' frame k. You've got your frames crossed. Read the paper again. K is the stationary frame, k is the 'moving frame'. > (Failure to understand the meaning of x' is a primary stumbling block > in following Einstein's derivation.) Not to me.. It might be harder for someone that confuses K-frame with 'moving' frame though. > With this in mind and letting A, B, and C denote the events of > emission, reflection, and reception, respectively, of the light > signal, then we can write Einstein's formula as > > (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) That's right. Well done. > This equation merely expresses Einstein's clock synchronization in > frame K. Well, not really. That half at the beginning is nonsense. It certainly wouldn't be true if I threw a ball against the wind and i never caught up with the wall. > Now x'_A represents the x-distance (as measured in frame k) between > the point of emission of the light signal and origin of frame K. Yep, that's right. This is > clearly zero since the signal was emitted from the origin of K. No no, you are still getting your frames crossed over. Quote: From the origin of system k let a ray be emitted... Unquote. Reference http://www.fourmilab.ch/etexts/einstein/specrel/www/ It is zero in the stationary frame K, though, so you are almost right. > Likewise, x'_C represents the x-distance (as measured in k) between > the point of reception of the light signal and the origin of frame K. > Again this is clearly zero since the signal was received at the origin > of K. I can't really answer that since you have your frames crossed over. I did provide a diagram, though. Ao----------(v+u)-----------* emission *---------------------------oB reflection <==vt==>Co---------(v-u)------------* reception As you can see, C is displaced from A by a distance vt. Putting that into your equation (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) we have (1/2)[tau(x'_A,0,0,t_A)+tau(x'_A-vt,0,0, t_C)] = tau(x'_B,0,0,t_B) but obviously Einstein doesn't have your equation, does he? > Thus, x'_A = x'_C = 0. No no, that is why I use a ball instead of light and said that relativists can't count to three. x'_C =0-vt , x'_A = 0. Snip remainder, you've blundered. Androcles. >>> If I stand before a vertical surface and throw a ball 'o' with speed >>> 'u', catching it as it returns, I need a minumum of two coordinates [quoted text clipped - 62 lines] > 'Moving' frame K? What's that then? K is the 'stationary' frame. > You are getting mixed up. You're right! Sorry for mixing up the notation. >> The physical interpretation of x' is that it represents the x-distance >> between the location of the event and the origin of the 'moving' frame K >> as measured in the 'stationary' frame k. > > You've got your frames crossed. Read the paper again. > K is the stationary frame, k is the 'moving frame'. Yes, as noted above. >> (Failure to understand the meaning of x' is a primary stumbling block in >> following Einstein's derivation.) [quoted text clipped - 15 lines] > wouldn't be true if I threw a ball against the wind and i never caught up > with the wall. But Einstein was considering a light signal where the equation above does indeed express that the clocks in the moving frame are Einstein-synchronized. >> Now x'_A represents the x-distance (as measured in frame k) between the >> point of emission of the light signal and origin of frame K. [quoted text clipped - 5 lines] > > No no, you are still getting your frames crossed over. Yes, I switched the notations k and K compared to Einstein throughout my post. Again, sorry for the inconvenience. If you like, I can repost it with the notation fixed. > Quote: > From the origin of system k let a ray be emitted... [quoted text clipped - 14 lines] > > As you can see, C is displaced from A by a distance vt. OK, but in looking at Einstein's equation we need to keep in mind that he uses the symbol t to stand for the time in the stationary frame that the light signal was emitted. So, to be more precise, it might be better to say that C is displaced from A according to the stationary frame by the distance x_C - x_A = v*(t_C - t_A). > Putting that into your equation > (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) > we have > (1/2)[tau(x'_A,0,0,t_A)+tau(x'_A-vt,0,0, t_C)] = tau(x'_B,0,0,t_B) No, by definition we have x'_C = x_C - v*t_C and x'_A = x_A - v*t_A Using these relations in x_C - x_A = v*(t_C - t_A) merely yields x'_A - x'_C = 0 or x'_A = x'_C. But we already know this since, in fact, x'_A and x'_B are both zero. But you mangled my equation by substituting x'_C = x'_A-vt which is clearly incorrect. > but obviously Einstein doesn't have your equation, does he? Well, he naturally doesn't have your incorrect modification of my equation. >> Thus, x'_A = x'_C = 0. > No no, that is why I use a ball instead of light and said that relativists > can't count to three. x'_C =0-vt , x'_A = 0. Even with the light replaced by a ball, we would still have x'_A=x'_C=0. Remember, Einstein uses x' to denote the x-distance (as measured in the stationary frame) between an event and the origin of the moving frame. Since A and C for the ball both occur at the origin of the moving frame, we must have x'_A=x'_C=0. Todd > Snip remainder, you've blundered. > > Androcles. >>>> If I stand before a vertical surface and throw a ball 'o' with >>>> speed [quoted text clipped - 104 lines] > does indeed express that the clocks in the moving frame are > Einstein-synchronized. I don't care what he was considering, he doesn't get to say the journey time out equals the journey time back when two different speeds are used. If I fly to New York and get a rowboat back to London I don't estimate 7 hours + 193 hours divided by 2 = 100 hours each way. The half is pulled out of a hat and no self respecting mathematician or physicist would fall for such a silly story. I'm certainly not going to. He can keep his idiotic method of sychronization. >>> Now x'_A represents the x-distance (as measured in frame k) between >>> the point of emission of the light signal and origin of frame K. [quoted text clipped - 38 lines] > > x_C - x_A = v*(t_C - t_A). That's (0,0,0) - (0,0,0) = v * (anything at all) According to those that cannot count to three, C = A. Hence v = 0 and trivial. >> Putting that into your equation >> (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) [quoted text clipped - 10 lines] > > x'_A - x'_C = 0 or x'_A = x'_C. That's right, I fully agree. The origin of the k-frame doesn't move from the origin of the K-frame while the tip of the ray is in flight, so either a) v = 0 or b) the speed of light is infinite. or c) both of the above. Relativists cannot count to three, as you have now demonstrated. > But we already know this since, in fact, x'_A and x'_B are both zero. Not wise. The idiot wants to differentiate his equation, so he says "Hence, if x' be chosen infinitesimally small" f(x+h) - f(x) f'(x) = ---------------------- h doesn't work if h is already zero. > But you mangled my equation by substituting x'_C = x'_A-vt which is > clearly incorrect. Only if you cannot count to three. >> but obviously Einstein doesn't have your equation, does he? > > Well, he naturally doesn't have your incorrect modification of my > equation. That's why I said relativists cannot count to three. The picture clearly shows a separation between A and C which you've said above is zero. Quote: x'_A - x'_C = 0 or x'_A = x'_C. Unquote. >>> Thus, x'_A = x'_C = 0. >> No no, that is why I use a ball instead of light and said that [quoted text clipped - 3 lines] > Even with the light replaced by a ball, we would still have > x'_A=x'_C=0. Fine. tau = (t - vx/(ball speed)^2) / sqrt ( 1 - v^2 / (ball speed)^2) I don't like to be deliberately obtuse, but I'm not the crackpot, I can count to three. > Remember, Einstein uses x' to denote the x-distance (as measured in > the stationary frame) between an event and the origin of the moving > frame. Since A and C for the ball both occur at the origin of the > moving frame, we must have x'_A=x'_C=0. Really? We are trying to find tau in the moving frame as seem from the stationary frame, so the parameters of the function tau() are all coordinates in K. We don't get to use moving frame parameters in the calculation, which is what you are doing. |__A ____________x'__| k-frame tau = 0 |__C ____________x __| K-frame t = 0 |__A____________x'__| k-frame tau = 1 <-vt1->|__C ____________x __| K-frame t = 1 |__A____________x'__| k-frame tau = 2 <-----vt2------>|__C ____________x __| K-frame t = 2 Placing A = C. |__A____________x'__| k-frame tau = 2 |__C____________x __| K-frame t = 2 v = 0. You've also said x'_B = 0 as well, and called it a fact. Quote: "But we already know this since, in fact, x'_A and x'_B are both zero. "-Todd. Unquote: I gotta be honest with ya. You are one of those that cannot count to three. Androcles. [snip] > > But we already know this since, in fact, x'_A and x'_B are both zero. > [quoted text clipped - 5 lines] > h > doesn't work if h is already zero. Todd, that is what you get when you try to explain derivatives to an ape: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/Sheesh.html Dirk Vdm ----- Original Message ----- From: "Androcles" <dummy@dummy.net> Newsgroups: sci.physics.relativity,sci.physics Sent: Tuesday, December 28, 2004 3:07 AM Subject: Re: Relativists cannot count to three. [snip some of the earlier parts] >>>> With this in mind and letting A, B, and C denote the events of >>>> emission, reflection, and reception, respectively, of the light signal, [quoted text clipped - 22 lines] > idiotic method > of sychronization. That's not what Einstein does. I've already shown how the (c-v) and (c+v) quantities occur in Einstein's expressions. These quantities do no represent the speed of light relative to either frame. They simply represent the rate at which the distance between the light signal and the origin of the moving frames is changing as observed in the stationary frame. You have to understand the meaning of the x-primes. >>>> Now x'_A represents the x-distance (as measured in frame k) between the >>>> point of emission of the light signal and origin of frame K. [quoted text clipped - 40 lines] > According to those that cannot count to three, C = A. Hence v = 0 > and trivial. No. x_C does not equal x_A. But the corresponding primed quantities are equal (each is equal to zero). Again, you must distinguish the meaning of the primed and the unprimed x's. >>> Putting that into your equation >>> (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) [quoted text clipped - 20 lines] > c) both of the above. > Relativists cannot count to three, as you have now demonstrated. The origin of the k-frame certainly does move relative to K while the ray is in flight. It seems to me that you are not understanding the meaning of the primed x's. >> But we already know this since, in fact, x'_A and x'_B are both zero. Whoops - another unfortunate typo on my part. I meant x'_C rather than x'_B. The value of x'_B is definitely NOT zero as it measures the distance of separation between the origin of the moving frame and the point of reflection of the light as measured in the stationary frame. > Not wise. The idiot wants to differentiate his equation, so he says > "Hence, if x' be chosen infinitesimally small" [quoted text clipped - 3 lines] > h > doesn't work if h is already zero. To say x' is infinitesimal is not the same as saying it is zero. >> But you mangled my equation by substituting x'_C = x'_A-vt which is >> clearly incorrect. [quoted text clipped - 12 lines] > x'_A - x'_C = 0 or x'_A = x'_C. > Unquote. Again, you are not understanding the meaning of the primed x's. The relation x'_A = x'_C does not mean that there is no spatial separation between the events A and C in the stationary frame. x'_A = x'_C does not imply x_A = x_C. >>>> Thus, x'_A = x'_C = 0. >>> No no, that is why I use a ball instead of light and said that [quoted text clipped - 5 lines] > Fine. > tau = (t - vx/(ball speed)^2) / sqrt ( 1 - v^2 / (ball speed)^2) Sorry, that went right past me. > I don't like to be deliberately obtuse, but I'm not the crackpot, I can > count to three. [quoted text clipped - 7 lines] > We are trying to find tau in the moving frame as seem from the stationary > frame, so the parameters of the function tau() are all coordinates in K. Einstein clearly states that tau is a function of x' rather than x. You have to understand the meaning of x'. x' is not the x-coordinate in K. > We don't get to use moving frame parameters in the calculation, which is > what you are doing. [quoted text clipped - 12 lines] > > v = 0. Nobody is saying that A and C are spatially coincident in K. x_A does not equal x_C even though x'_A = x'_C. It is important to understand the meaning of x'=x-vt. > You've also said x'_B = 0 as well, and called it a fact. > Quote: > "But we already know this since, in fact, x'_A and x'_B are both zero. As I noted above, I meant to say x'_A and x'_C are both zero. x'_B is not zero. Sorry for the confusion. Todd > Unquote: > I gotta be honest with ya. You are one of those that cannot count to > three. > Androcles. > ----- Original Message ----- > [quoted text clipped - 94 lines] > > No. x_C does not equal x_A. Well done. They are not equal. > But the corresponding primed quantities are equal (each is equal to > zero). > Again, you must distinguish the meaning of the primed and the unprimed > x's. We already know the meaning. x' = x-vt. The origins of the two systems are displaced by vt. YOU must distinguish which frame you are using to calculate time in the other frame. >>>> Putting that into your equation >>>> (1/2)[tau(x'_A,0,0,t_A)+tau(x'_C,0,0, t_C)] = tau(x'_B,0,0,t_B) [quoted text clipped - 25 lines] > It seems to me that you are not understanding the meaning of the > primed x's. It seems to me that I understand what x' = x-vt very well, and you seem to think x' = 0. quote: But the corresponding primed quantities are equal (each is equal to zero). unquote. >>> But we already know this since, in fact, x'_A and x'_B are both >>> zero. [quoted text clipped - 4 lines] > the point of reflection of the light as measured in the stationary > frame. Oh good. At least that is cleared up. Now, it is true that x'_C = x'_A, and x_C differs from x_A. It is NOT true that the speed of light bouncing between x'_A, x'_B and x'_C is anything other than c. It IS true that the speed of light bouncing between x_A, x_B' and x_C is c-v, c+v. It is also true that Einstein is using x'_A and x'_B with the wrong speed of light. >> Not wise. The idiot wants to differentiate his equation, so he says >> "Hence, if x' be chosen infinitesimally small" [quoted text clipped - 5 lines] > > To say x' is infinitesimal is not the same as saying it is zero. Then I advise you against it. quote: But the corresponding primed quantities are equal (each is equal to zero). unquote. >>> But you mangled my equation by substituting x'_C = x'_A-vt which is >>> clearly incorrect. [quoted text clipped - 14 lines] > > Again, you are not understanding the meaning of the primed x's. x' = x-vt. How difficult is that? The > relation x'_A = x'_C does not mean that there is no spatial separation > between the events A and C in the stationary frame. x'_A = x'_C does > not imply x_A = x_C. x'_C = x'_A implies x_C = x_A >>>>> Thus, x'_A = x'_C = 0. >>>> No no, that is why I use a ball instead of light and said that [quoted text clipped - 8 lines] > > Sorry, that went right past me. Ok, I'll elaborate. Einstein's equations apply as much to the speed of a ball as they do to the speed of light. Hence if they were correct (which they are not, but it is your claim that they are) the equation above is then valid. >> I don't like to be deliberately obtuse, but I'm not the crackpot, I >> can count to three. [quoted text clipped - 10 lines] > > Einstein clearly states that tau is a function of x' rather than x. That's right, and x' = x-vt. You > have to understand the meaning of x'. x' is not the x-coordinate in > K. It is a moving coordinate in K, x' = x-vt. It isn't a fixed coordinate in k either, since xi = x' * beta. The parameters to the function tau are ALL K- frame parameters. You have to understand the meaning of x', it is given as x' = x-vt and x, v and t are all K-frame parameters. >> We don't get to use moving frame parameters in the calculation, which >> is what you are doing. [quoted text clipped - 14 lines] > > Nobody is saying that A and C are spatially coincident in K. Einstein is. He uses the same K-frame position for the return of the light and the source of the light. > x_A does not equal x_C even though x'_A = x'_C. Nonsense. x'_A = x_A-vt and x'_C = x_C-vt. Hence if x'_A = x'_C then x_A = x_C. > It is important to understand the meaning of x'=x-vt. Then try it. >> You've also said x'_B = 0 as well, and called it a fact. >> Quote: [quoted text clipped - 3 lines] > As I noted above, I meant to say x'_A and x'_C are both zero. x'_B is > not zero. Sorry for the confusion. That's ok, we are past that. x'_A = x_A-vt and x'_C = x_C-vt. Hence if x'_A = x'_C then x_A = x_C. Androcles. > Todd > >> Unquote: >> I gotta be honest with ya. You are one of those that cannot count to >> three. >> Androcles. ----- Original Message ----- From: "Androcles" <dummy@dummy.net> Newsgroups: sci.physics.relativity,sci.physics Sent: Tuesday, December 28, 2004 11:39 AM Subject: Re: Relativists cannot count to three. > "Todd" <nope@nospam.com> wrote in message >> x_A does not equal x_C even though x'_A = x'_C. > > Nonsense. x'_A = x_A-vt and x'_C = x_C-vt. > Hence if x'_A = x'_C then x_A = x_C. This seems to be a central point of misunderstanding on your part. You have inadvertently assumed that t has the same value for A as it does for C. Please note that when the equation x'=x-vt is applied to the event A, then it should be interpreted as x'_A = x_A - vt_A where t_A is the time in the K frame at which event A occurs. Likewise x'_C = x_C - vt_C where t_C is the time in the K frame at which event C occurs. Clearly t_C does not equal t_A. Hence, x_C does not equal x_A even though x'_C does equal x'_A. The correct relation for the unprimed x's is x_C = x_A + v(t_C - t_A). Todd > ----- Original Message ----- > [quoted text clipped - 16 lines] > You have inadvertently assumed that t has the same value for A as it > does for C. Not at all. x' = x-vt. If x = A then x' = C. > Please note that when the equation x'=x-vt is applied to the event A, > then it should be interpreted as > > x'_A = x_A - vt_A Yeah well, t_A < t_C but x_A= x_C, so I think you are clutching at straws now. This is a central point of misunderstanding on your part. In Einstein's equation x_A = x_C or x'_A = x'_C, which amounts to the same thing. > where t_A is the time in the K frame at which event A occurs. and t_C is the time in the K frame at which event C occurs, and t_C > t_A but x_C = x_A according to Einstein so there is no motion and relativists cannot count to three. Likewise > x'_C = x_C - vt_C > > where t_C is the time in the K frame at which event C occurs. and t_A is the time in the K frame at which event A occurs, and t_C > t_A but x_C = x_A according to Einstein so there is no motion and relativists cannot count to three. > Clearly t_C does not equal t_A. Clearly x_A DOES equal x_C in Einstein's equation so the moving k-frame isn't moving and relativists cannot count to three. Hence, x_C does not equal x_A even though > x'_C does equal x'_A. And relativist cannot count to three. The correct relation for the unprimed x's is > x_C = x_A + v(t_C - t_A), hence v = 0 and relativists cannot count to three. Androcles > > ----- Original Message ----- > > [quoted text clipped - 19 lines] > Not at all. x' = x-vt. > If x = A then x' = C. How to Explain the Difference Between a Coordinate and an Event to an Unwilling Ape. Dirk Vdm >> >>> x_A does not equal x_C even though x'_A = x'_C. >> >> >> >> Nonsense. x'_A = x_A-vt and x'_C = x_C-vt. >> >> Hence if x'_A = x'_C then x_A = x_C. Isn't it interesting that Androcles does not realize that the value of t at the event A and the value of t at the event C are not equal. This is the same sort of attitude as Eleaticus who refuses to acknowledge that t can be traeted as a coordinate. >> > This seems to be a central point of misunderstanding on your part. >> > You have inadvertently assumed that t has the same value for A as it >> > does for C. >> >> Not at all. x' = x-vt. >> If x = A then x' = C. >How to Explain the Difference Between >a Coordinate and an Event to an Unwilling Ape. Has anybody ever tried to explain Einstein's derivation in his 1905 paper in minute detail (including the significance of x' = x-vt) in one go, so that a person of average intelligence would be able to see everything in context? (Not that I am suggesting here that Androcles is even of average intelligence.) David ----- > Has anybody ever tried to explain Einstein's derivation in his 1905 paper > in minute detail (including the significance of x' = x-vt) in one go, so [quoted text clipped - 3 lines] > > David There is a June 2004 paper in the American Journal of Physics by Alberto A. Martinez entitled ''Kinematic subtleties in Einstein's first derivation of the Lorentz Transformations''. This is probably not quite the type of paper you had in mind, though. The paper is mostly concerned with Einstein's use of the x' variable. Personally, I feel that Martinez is too harsh on Einstein. He implies that Einstein is inconsistent and/or ambiguous in his use of notation. Also, I think that Martinez overstates the difficulties in following Einstein. For example he says, Einstein presented his derivation [of the Lorentz transformation equations] in about fifty-five equations, but worked out explicitly the derivation involves more than three hundred equations, consisting of roughly five hundred algebraic and differential operations. Nevertheless, the paper is worth reading and it does point out some errors that other authors have made when trying to explain Einstein's derivation. Todd ----- Original Message ----- From: "Androcles" <dummy@dummy.net> Newsgroups: sci.physics.relativity,sci.physics Sent: Tuesday, December 28, 2004 2:00 PM Subject: Re: Relativists cannot count to three. >> ----- Original Message ----- >> [quoted text clipped - 19 lines] > Not at all. x' = x-vt. > If x = A then x' = C. I find your notation confusing. A is the label of an event - namely, the emission of the light signal. This event has an x-coordinate in frame K that I've been denoting as x_A. C is a different event - the reception of the light signal. C has an x-coordinate in the frame K denoted by x_C. Each of these events also has a value of x'. x' may be thought of as the *difference* in two x coordinates. Specifically, x' for an event is the difference in the x-coordinate of the event and the x-coordinate of the origin of the moving frame at a time that is considered simultaneous (according to K) with the event . (Note, all of these x-coordinates are locations as measured along the x-axis of the stationary frame K.) Since events A and C both occur at the origin of the moving frame k, it should be obvious that the value of x' for each of these events is zero. It might help if you could explain what you mean by x = A and x' = C and also explain your reasoning behind saying that if x = A then x' = C. >> Please note that when the equation x'=x-vt is applied to the event A, >> then it should be interpreted as [quoted text clipped - 5 lines] > Einstein's equation x_A = x_C or x'_A = x'_C, which amounts to the same > thing. No, Einstein's equation doesn't even use x_A or x_C because his function tau is defined as a function of x', not x. The equation only uses x'_A and x'_C which are both zero (and also x'_B which is not zero). >> where t_A is the time in the K frame at which event A occurs. > > and t_C is the time in the K frame at which event C occurs, and > t_C > t_A but x_C = x_A according to Einstein so there is no motion > and relativists cannot count to three. How do you get that x_C = x_A according to Einstein? Can you reference a specific place in Einstein's paper to justify your claim? > Likewise >> [quoted text clipped - 10 lines] > Clearly x_A DOES equal x_C in Einstein's equation so the moving > k-frame isn't moving and relativists cannot count to three. Again, on what basis are you concluding that x_A = x_C in Einstein's equation? Todd > Hence, x_C does not equal x_A even though >> x'_C does equal x'_A. [quoted text clipped - 7 lines] > > Androcles Nope. x_C does not equal x_A and Einstein never implied that they are equal. Todd > ----- Original Message ----- Todd, you will get precisely nowhere in pursuing arguments with Androcles. Many have tried and failed. The onlt way to deal with him is to tell him quite bluntly that he is a brainless cretin. Franz >> ----- Original Message ----- > [quoted text clipped - 3 lines] > > Franz Yes, I'm not getting anywhere with this and I'll probably give up soon. Thanks. Todd > >> ----- Original Message ----- > > [quoted text clipped - 3 lines] > > > > Franz > Yes, I'm not getting anywhere with this and I'll probably give up soon. Of course you aren't. You cannot teach a baboon anything other than eating bananas Have you seen the desperate attempts by Randy Poe to make contact with Androcles' solitary neuron? Franz Franz > ----- Original Message ----- > [quoted text clipped - 31 lines] > > I find your notation confusing. Mine? You are the one that started putting x_ and x'_ in front of A and C, so that's your problem. > A is the label of an event - namely, the emission of the light signal. Not the way I used it. It was simply a position which I showed on a diagram, but you thought of it that way so I went along with it. > This event has an x-coordinate in frame K that I've been denoting as > x_A. C is a different event - the reception of the light signal. C > has an x-coordinate in the frame K denoted by x_C. Well ok, then. x_A = x_C in Einstein's equation, and t_A < t_C. Hence the moving frame isn't moving. Why is that so hard to understand? > Each of these events also has a value of x'. x' may be thought of as > the *difference* in two x coordinates. Yes, and it is constant if you write (x-vt) - (0-vt) and not constant if you write (x-vt) - (0). Einstein didn't write (0-vt), he wrote (0). Why is that hard to understand? > Specifically, x' for an event is the difference in the x-coordinate of > the event and the x-coordinate of the origin of the moving frame at a > time that is considered simultaneous (according to K) with the event . K k (0,0,0) (0,0,0) emission (x',0,0) (x,0,0) reflection (0,0,0) (0,0,0) reception Should be (0-vt,0,0) (0,0,0) reception > (Note, all of these x-coordinates are locations as measured along the > x-axis of the stationary frame K.) That's right. (0,0,0) (0,0,0) reception Should be (0-vt,0,0) (0,0,0) reception > Since events A and C both occur at the origin of the moving frame k, > it should be obvious that the value of x' for each of these events is > zero. In k, sure, but we are not measuring in k. (Note, all of these x-coordinates are locations as measured along the x-axis of the stationary frame K.) > It might help if you could explain what you mean by x = A and x' = C > and also explain your reasoning behind saying that if x = A then x' = > C. It's your notation. To me, A and C are positions (events if you like) in K. I say they have different positions and different times. Einstein says they have the same position and different times, which turns out to be the same time, since he says "the velocity of light in our theory plays the part, physically, of an infinitely great velocity." >>> Please note that when the equation x'=x-vt is applied to the event >>> A, then it should be interpreted as [quoted text clipped - 9 lines] > No, Einstein's equation doesn't even use x_A or x_C because his > function tau is defined as a function of x', not x. Ok. I didn't use x_A or x_C either. You did. > The equation only uses x'_A and x'_C which are both zero (and also > x'_B which is not zero). Einstein doesn't use that notation in his equation. He uses x' which is stretching the frame as it moves, with the coincident origins firmly pinned together. But yes, you are quite correct, x'_A and x'_C are both zero. They shouldn't be, at the other end is x'_B which differs from x_B by -vt. The moving frame is a rubber band, only one end of which moves.. >>> where t_A is the time in the K frame at which event A occurs. >> [quoted text clipped - 4 lines] > How do you get that x_C = x_A according to Einstein? Can you > reference a specific place in Einstein's paper to justify your claim? Well, the notation is yours, of course, but yes, I can. It is in section 3 of "On the Electrodynamics of Moving Bodies", Reference : http://www.fourmilab.ch/etexts/einstein/specrel/www/ ½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)) x_A x_C x'_B (Note, all of these x-coordinates are locations as measured along the x-axis of the stationary frame K.) Notice that the position of x_C is the same as the position of x_A, although time x'/(c-v)+x'/(c+v) has passed. Also I have used x'_B to show the stretch. had Einstein used x'_A and x'_C, there wouldn't be a problem, but then he'd have no paper to write either. >> Likewise >>> [quoted text clipped - 13 lines] > Again, on what basis are you concluding that x_A = x_C in Einstein's > equation? See above. > Todd > [quoted text clipped - 12 lines] > Nope. x_C does not equal x_A and Einstein never implied that they are > equal. I agree, he didn't imply it. He said it directly in his equation. Only two positions are given, (0,0,0) and (x',0,0). (Note, all of these x-coordinates are locations as measured along the x-axis of the stationary frame K.) Three are needed. Relativists cannot count to three. Androcles ----- Original Message ----- From: "Androcles" <dummy@dummy.net> Newsgroups: sci.physics.relativity,sci.physics Sent: Tuesday, December 28, 2004 5:32 PM Subject: Re: Relativists cannot count to three. >> ----- Original Message ----- >> [quoted text clipped - 63 lines] > Should be > (0-vt,0,0) (0,0,0) reception For the reception event the value of x' (as defined by Einstein) is zero as I've explained several times. It is not 0-vt. I don't know what you mean by the x in (x,0,0) for the reflection under the heading k. It is certainly not the x that occurs in x'=x-vt. But it doesn't matter since it's not relevant to Einstein's derivation. >> (Note, all of these x-coordinates are locations as measured along the >> x-axis of the stationary frame K.) [quoted text clipped - 3 lines] > Should be > (0-vt,0,0) (0,0,0) reception No. The reception occurs at the origin of the moving frame k. If we let, for the moment, t be the time of the reception (according to clocks in K), then the origin of k is located at x = vt in K. Thus, the x-coordinate of the reception is x = vt. Therefore the value of x' for the reception is x' = x - vt = vt - vt = 0 >> Since events A and C both occur at the origin of the moving frame k, it >> should be obvious that the value of x' for each of these events is zero. > > In k, sure, but we are not measuring in k. > (Note, all of these x-coordinates are > locations as measured along the x-axis of the stationary frame K.) Right, we are not measuring in k. x for event C is measured in K and t for event C is measured in K. Then when you calculate x' = x-vt you get x' = 0 for C. All measurements are in K. >> It might help if you could explain what you mean by x = A and x' = C and >> also explain your reasoning behind saying that if x = A then x' = C. [quoted text clipped - 4 lines] > out to be the same time, since he says "the velocity of light in our > theory plays the part, physically, of an infinitely great velocity." Einstein never says or implies that the events A and C have the same position in frame K. Also, you are quoting Einstein out of context. He made this comment in section 4 of his paper and it has no bearing on his derivation in section 3 that we are discussing. >>>> Please note that when the equation x'=x-vt is applied to the event A, >>>> then it should be interpreted as [quoted text clipped - 21 lines] > is x'_B which differs from x_B by -vt. > The moving frame is a rubber band, only one end of which moves.. ???? >>>> where t_A is the time in the K frame at which event A occurs. >>> [quoted text clipped - 15 lines] > (Note, all of these x-coordinates are > locations as measured along the x-axis of the stationary frame K.) No. You need to understand that there are no x-coordinates in tau. They are values of x', not x. Thus, in the first tau expression we have tau(0,0,0,t). The first zero here refers to the value of x' for event A. You seem to think that this zero refers to the value of x for event A since you typed x_A under this zero. You should have typed x'_A instead. Likewise the first zero in tau(0,0,0,t+x'/(c-v)+x'/(c+v)) corresponds to the value of x'_C, not x_C. Note what Einstein says in the second sentence of the paragraph that precedes the paragraph that contains the equation. I will quote: ''We first define tau as a function of x', y, z, and t '' . Thus, the arguments of tau are x', y, z, and t. That is, we should always think tau(x', y, z, t) and never tau(x, y, z, t). Whenever Einstein is writing the arguments of tau, the first argument is x', not x. > Notice that the position of x_C is the same as the position of x_A, > although time x'/(c-v)+x'/(c+v) has passed. Also I have used x'_B > to show the stretch. had Einstein used x'_A and x'_C, there wouldn't > be a problem, but then he'd have no paper to write either. No, x_C is not the same as x_A. You think that Einstein says these are equal because you are interpreting the first argument of the function tau to be x rather than x'. Todd > ----- Original Message ----- > [quoted text clipped - 82 lines] > For the reception event the value of x' (as defined by Einstein) is > zero as I've explained several times. It is not 0-vt. Yes, I know that is what you've explained. That is the reason I gave the thread the title it has. It is not really intended for relativists, or I would have titled it "Relativists cannot count past two", three being beyond their compehension. There are two frames. One of them uses two points and the other uses two points and one more. The two points are used in the frame called k. They are called called (0,0,0) and (x,0,0), and the speed involved is called c. There is no v in the frame called k. It is the 'moving' frame, but that means nothing unless we say what it is moving with respect to. The light ray moves in the frame called k. It travels a distance (x,0,0)-(0,0,0) from a point called (0,0,0) to a point called (x,0,0) and back again to (0,0,0), a distance of (0,0,0) - (x,0,0). The two points and one more are used in the frame called K. They are called (0-vt0,0,0) and (x-vt1,0,0). The one more is called (0-vt2, 0,0). The light ray also moves in the frame called K. It travels a distance (x-vt1) - (0-vt0) from a point called (0-vt0,0,0) to a point called (x-vt1,0,0) at a speed called c-v. Then it travels back from (x-vt1,0,0) to the one extra point (the "third" point for those that are not relativists) called (0-vt2, 0,0) at a speed of c+v. As you have explained to me, 0-vt0 = 0, which is quite correct, and 0-vt2 = 0 which is only correct if v = 0. because t2 > t0. Now, I do thank you for your explanation, it was kind of you to offer it, but since you are unable to count to three (oops, I shoud have said 'beyond two') I am somewhat dubious as to your effort having any validity. Have a nice day. > I don't know what you mean by the x in (x,0,0) for the reflection > under the heading k. Well, it is simple enough to any non-relativist. If you get out your wooden rule that you were given in first grade, you will see some numbers written on it. At one end is 0, then the next is 1, and after that comes 2. There are some higher numbers that you do not understand yet. Now, the number 1 is called a coordinate and represents a Euclidean metric we call "distance" from the number 0. Likewise, the number 2 represents a distance from the number 0. If you really think hard, you may notice that the distance from 0 to 2 is twice the distance from 0 to 1. Those of us that are a little more practiced in algebra can use 'x' to represent either 1 or 2. We can even use to represent those bigger numbers that go beyond 2 that you do not yet understand. Now, slide your wooden rule across the desk. See if the numbers change. If they do, you can explain to me all about relativity. > It is certainly not the x that occurs in x'=x-vt. Yes, I know this is tough for anyone that cannot count to... (oops, I almost said three) cannot count beyond two, but x' = x-vt, and we add -vt to both sides of the equation we get x = x' + vt. That's a little bit too advanced for you, though. > But it doesn't matter since it's not relevant to Einstein's > derivation. Well, that is true too. We are not interested in the frame called k, we ae interested in the frame called K with the two points and the extra one. >>> (Note, all of these x-coordinates are locations as measured along >>> the x-axis of the stationary frame K.) [quoted text clipped - 5 lines] > > No. The reception occurs at the origin of the moving frame k. Well of course it does. How very perceptive of you. (Note, all of these x-coordinates are locations as measured along the x-axis of the stationary frame K.) So where is the origin of the frame k when MEASURED in the stationary frame K? > If we let, for the moment, t be the time of the reception No, no, no, t2 is the time of reception. > then the origin of k is located at x = vt in K. No, no, no. The origin of k is located at x = vt0 AND at x = vt2 in K because it moves. > Thus, the x-coordinate of the reception is x = vt. No, no, no. Even Einstein knew better than that, and he was pretty dim. Look, he even said (1/2)(tau0 + tau2) = tau1. He never said (1/2)(t + t) = t, although that would be true. > Therefore the value of x' for the reception is > > x' = x - vt = vt - vt = 0 Ah yes, I see. Try getting your first grade wooden rule out again, put a sticky note on it and write x. Now put it on top of a sheet of paper to represent the frame called K. Where the 0 of the rule is, make a mark and write 0 on the sheet. Make another mark beside the x, and write x on the sheet. Now, very carefully, slide the rule along the sheet in the x-direction at a speed v for a time t. Now write x' opposite the x on the rule. Can you stretch you imagination enough to realize that x' is moving? >>> Since events A and C both occur at the origin of the moving frame k, >>> it should be obvious that the value of x' for each of these events [quoted text clipped - 5 lines] > > Right, we are not measuring in k. Good. x for event C is measured in K Good. and t for > event C is measured in K. Err, no, not quite. t2 is is measured in K for event C. > Then when you calculate x' = x-vt you get x' = 0 for C. All > measurements are in K. Yes, I know how difficult this is for you, counting beyond 2 is far too tough for a relativist <shrug>. >>> It might help if you could explain what you mean by x = A and x' = C >>> and also explain your reasoning behind saying that if x = A then x' [quoted text clipped - 9 lines] > Einstein never says or implies that the events A and C |
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