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From: glird on 4 Apr 2010 16:30 Several years ago, after I realized that Einstein was doing algebra, not calculus, in his 4rth, 5th, 6th and 7nth equations and thereafter, I posted that on sci.physics.relativity. After several people said I was wrong, I said, If you think he got to eq 7 by calculus, please show how he did it. When several replied, You use the chain rule; or, You use this and that, I said, Don't tell me what to do, just DO IT for us. Nobody did. A few weeks ago, I posted this challenge: < Using calculus, show how he got from eq 3 to eq 7, "tau = a(t-vx'/ {c^2-v^2})". If you can't do it by calculus, try to show it by any method at all. (Having done it years ago, I am fairly sure that you can't!) > Although several ninnies again replied, only ONE person tried to show us how to do it. He said, "ok, I accept the challenge and of course you do it with calculus!" Here are some of the discussions that ensued over the next few weeks: me: The General Transformation Equations ("GTE") ARE compatible with the first postulate (principle of relativity) he: well, as it turns out they aren't. When I say they satisfy the Por what I mean is the forward and reverse transforms should have the same form and if you apply the reverse transform to the forward transform you should get what you started with, i.e. x = x and t = t. When I did that that's just what I got. Then I realized I did something I do so automatically that I don't even think about it. I set phi(v) = 1. If phi(v) is set to anything else these equations (top of page 46) don't satisfy the Por. And with phi(v) = 1 they are then the LTE. Although he is wrong about that because, as I said, "They DO fit the PoR for ANY value of phi(v), but other than 1, they do NOT fit the LTE!!" me: Ask yourself, therefore, Is THAT why E tried to prove that phi(v) = 1? (The answer is YES!!) he: of course yes! me: Ask yourself, therefore, What was the value of _a_ B4 he tried that failed proof? he: what failed proof? He showed correctly that a must equal 1 for the equations to work. And it never had a previous value. All prior references say a is a function of v yet to be determined. I will interrupt today's thesis by correcting another of his misconceptions. Although the prior reference did say that "_a_ is a function phi (v) at present unknown", and although eq 7 is the first place it appears, its value to there had been the same as that of (delta tau/delta t); which was algebraically equal to q. What Einstein unsuccessfully tried to show is that *phi(v)* = 1 (NOT that _a_ "must equal 1"). Indeed, if _a_ = 1 (as it DID in P1), that rules out the LTE; in which _a_ = beta = gamma = 1/q. me: The answer, taken from the above ALGEBRA, was: a = dtau/dt = q = 1/beta =/= the LTE!! he: I think it has been pretty well established that if a doesn't equal 1 then no LTE. See! (Paraphrased, he said, "if @tau/@t doesn't equal 1 then no LTE", which is of course ridiculous.) Back to today's thesis: he: All four equations have a factor of gamma divided out. That step isn't shown nor is any explanation given for it. A whole page is missing! Although i didn't realize it when I went to bed about a week ago; I WON !!!!! The next few exchanges show why I hadn't realized it immediately. me: Look again! YOU missed page 47, on the way to the eqs at the top of page 48! he: 47 comes after 46, not before he: From my own work with the LTE I know why it's done. The equations on page 45 me [PAGE 46!!] he (!): no, at this point I was still trying to get from page 45 to the top of page 46. Ahhhhh so. THAT's why I failed to know instantly that I'd won. (I thought he was trying to get from the GTE at the top of pg 46 to the LTE at the top of pg 48!) As it is, he was still trying to get from eqs 3, 4, 5 on page 45 to EQUATION SEVEN, which IS at "the top of page 46". Here's what I won: After weeks of trying, he was (and is and will be) unable to show, by calculus or by any other means, "how E got from eq 3 to eq 7". Neither he or any other person has EVER shown how to get from eq 5 to eq 7! In my next posting I will tell you why "I WON!!" seems so important to me. he: Even you started to point out that to get there you have to set a = 1/gamma. But E doesn't do that, he leaves it there and later shows it has to be set to 1. So if I want to get from page 45 to the top of page 46 only using E's methodology I have to do an extra step. Divide out a factor of gamma. It's not enough to simply follow the algebra, you have to understand what he's trying to do in the first place. Clearly you are very confused. me: I is clear that although you may be an expert at calculus, you are blind to the meanings of your own equations; and of the simple algebra equations E was doing. One more thing. You use delta tau / delta t to refer to the symbols in Eq 5. When I see delta used like this I think of an interval like x2 - x1 = delta x. If that's how you intend to use it here you're making a big mistake. The symbols used in Eq 5 (for which I substitute the @ symbol) are differential operators, not ratio of variables or intervals. For example, referring to the LTE @tau/@t = gamma while delta tau/delta t = 1/gamma. Via algebra, it is easy to understand all of E's equations on the way to the LTE. When you do, it turns out that the symbols used in Eqs 4 thru 7 ARE ratios of variables, NOT differential operators, For example, dtau = tau2 tau1 = 1 second is the "proper time" of system k and dt = 1 second is the proper time of system K, so dtau/dt is the ratio of proper times of the two systems as viewed by K. (In physics, the "proper time" is the time of a clock of a given system as measured by THAT system.) In the LTE, @tau/@t = gamma = 1/q, but in E's published paper (delta tau)/(delta t) equaled 1/gamma = q UNTIL he inserted equation 7; which fits ANY value of _a_. As I long ago discovered, "Calculus is a blindfold." (It blocks its users from understanding the physical meanings of the symbols in the equations.) So it was here. glird
From: Dirk Van de moortel on 4 Apr 2010 18:30 glird <glird(a)aol.com> wrote in message b069b431-9c5e-472c-b0c6-48b059b211fd(a)q15g2000yqj.googlegroups.com > Several years ago, after I realized that Einstein was doing algebra, > not calculus, in his 4rth, 5th, 6th and 7nth equations and thereafter, > I posted that on sci.physics.relativity. After several people said I > was wrong, I said, If you think he got to eq 7 by calculus, please > show how he did it. When several replied, You use the chain rule; or, > You use this and that, I said, Don't tell me what to do, just DO IT > for us. Nobody did. > A few weeks ago, I posted this challenge: > < Using calculus, show how he got from eq 3 to eq 7, "tau = a(t-vx'/ > {c^2-v^2})". If you can't do it by calculus, try to show it by any > method at all. (Having done it years ago, I am fairly sure that you > can't!) > Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/ Here is how he goes from eq 3 to eq 5: Put: A = 1/(c-v), B = 1/(c+v), and forget about y and z then eq 3 1/2 [ T(0,t) + T(0,t + (A+B) x' ] - T(x', t + A x') = 0 gives, using the Taylor expansions T( x, t + A x' ) = T(0,t) + x' @T/@x'(0,t) + A x' @T/@t(0,t) + Order(x'^2) T( 0, t + A x' ) = T(0,t) + A x' @T/@t(0,t) + Order(x'^2) and throwing away Order(x'^2), 1/2 [ T(0,t) + T(0,t) + (A+B) x' @T/@t(0,t) ] - T(0,t) - x' @T/@x'(0,t) - A x' @T/@t(0,t) = 0 or, rearranging and leaving out the arguments (0,t): 1/2 (A+B) x' @T/@t - x' @T/@x' - A x' @T/@t = 0 or, if x' # 0: 1/2 (A+B) @T/@t - @T/@x' - A @T/@t = 0 or 1/2 (A+B) @T/@t = @T/@x' + A @T/@t which is eq 4, or @T/@x' + 1/2 (A-B) @T/@t or, using the defs of A and B @T/@x' + v/(c^2-v^2) @T/@t which is eq 5. (see also my post in 2003: http://groups.google.be/group/sci.math/msg/45e61b375de93eb7 It is called Taylor expansion - see first year calculus.) Here is how he goes from eq 5 to eq 7: So we have a linear function T(x,y) with. @T/@x' = K @T/@t (eq A) with K = -v/(c^2-v^2) Since T is linear in x' and t, it can b written as a function of p x' + t for some p, so T = a( p x' + t ) We calculate the partial derivatives @T/@x' = p a'( p x' + t ) @T/@t = a'( p x + t ) and we demand eq A, giving p a'( p x' + t ) = K a'( p x' + t ) which must be true for all x' and t, so p = K so T = a( K x' + t ) = a( t - x' v/(C^2-v^2) ) > Although several ninnies again replied, only ONE person tried to > show us how to do it. He said, "ok, I accept the challenge and of > course you do it with calculus!" This ninnie showed. You loose. Dirk Vdm
From: glird on 5 Apr 2010 22:27 On Apr 4, 6:30 pm, "Dirk Van de moortel" <dirkvandemoor...(a)nospAm.hotmail.com> wrote: > glird <gl...(a)aol.com> wrote > > > A few weeks ago, I posted this challenge: > > < Using calculus, show how he got from eq 3 to eq 7, "tau = a(t-vx'/ > > {c^2-v^2})". If you can't do it by calculus, try to show it by any > > method at all. (Having done it years ago, I am fairly sure that you > > can't!) > > >< Here is how he goes from eq 3 to eq 5: Put: A = 1/(c-v), B = 1/(c+v), and forget about y and z then eq 3 1/2 [ T(0,t) + T(0,t + (A+B) x' ] - T(x', t + A x') = 0 gives, using the Taylor expansions T( x, t + A x' ) = T(0,t) + x' @T/@x'(0,t) + A x' @T/@t(0,t) + Order(x'^2) T( 0, t + A x' ) = T(0,t) + A x' @T/@t(0,t) + Order(x'^2) and throwing away Order(x'^2), 1/2 [ T(0,t) + T(0,t) + (A+B) x' @T/@t(0,t) ] - T(0,t) - x' @T/@x'(0,t) - A x' @T/@t(0,t) = 0 or, rearranging and leaving out the arguments (0,t): 1/2 (A+B) x' @T/@t - x' @T/@x' - A x' @T/@t = 0 or, if x' # 0: 1/2 (A+B) @T/@t - @T/@x' - A @T/@t = 0 or 1/2 (A+B) @T/@t = @T/@x' + A @T/@t which is eq 4, or @T/@x' + 1/2 (A-B) @T/@t or, using the defs of A and B @T/@x' + v/(c^2-v^2) @T/@t which is eq 5. (see also my post in 2003: http://groups.google.be/group/sci.math/msg/45e61b375de93eb7 It is called Taylor expansion - see first year calculus.) I looked. It includes "Put x = x'"; which is valid only if t = 0. Since E stipulated that at t = 0, x = xi = T = 0, goodbye to the rest of this 2003 "calculus". Coming back to today's math, you claim to have gone by a Taylor expansion from eq 3 to 1/2 [ T(0,t) + T(0,t) + (A+B) x' @T/@t(0,t) ] - T(0,t) - x' @T/@x'(0,t) - A x' @T/@t(0,t) = 0 However, eq 3's 1/2 [ T(0,t) + T(0,t + (A+B) x' ] can become your 1/2 [ T(0,t) + T(0,t) + (A+B) x' @T/@t(0,t) ] only if [@T/@t(0,t) ] = 1; and even if it does, your equation is correct only if - T(x', t + A x') = - T(0,t) - x' @T/@x'(0,t) - A x' @T/@t(0,t). Algebraically speaking, though, since x' is unequal to 0 UNLESS t = 0 in the left-side expression's x',t (which is actually the right- side expression in eq 3); your right-side expression reduces to - T(a function of x = x' = t = 0) - 0 @T/@x'(0,0) - (0/(c-v) @T/ @0(0,0) = zero. Given THAT, then your entire calculus demo says that 0 - 0 = 0; which though correct is NOT what eq 5 means, even to freshmen in college > Here is how he goes from eq 5 to eq 7: >So we have a linear function T(x,y) with T isn't a function of x and y; it is a function of x and t. So the rest of your demo isn't worth reviewing, though I will... > @T/@x' = K @T/@t (eq A) > with K = -v/(c^2-v^2) If @xi/@x = 1, then @T/@t = Q; and if x' = x vt = 1 at t = 1, then the value of @T/@x' IS a function of v. But WHERE did your "calculus" reach that value of K? (Tomorrow I will think about whether or not thatis the right value of K. Meanwhile, I wonder how it can be, since @tau/@x' is the offset between two esynched clocks of the moving system k, that are x' apart as measured by the stationary cs K and @tau/@t is the ratio of clock rates; which is equal to beta = 1/q in the LTE.) >< Since T is linear in x' and t, it can be written as a function of p x' + t [ = 2p; since x' = t = 1] for some p, so T = a( p x' + t ) [ = a(2p) = 2ap] > Btw; WHERE did this "a" come from? >< We calculate the partial derivatives @T/@x' [ = (@T/@t)t vx'/c^2)] = p a'( p x' + t ) @T/@t = a'( p x + t ) and we demand eq A, giving p a'( p x' + t ) = K a'( p x' + t ) which must be true for all x' and t, so p = K so T = a( K x' + t ) = a( t - x' v/(C^2-v^2) ) > So you say. Now SHOW how CALCULUS arrives at your final equation. Along the way, show us where calculus got your undefined "a"; and what it denotes; and then if you can tell us its mathematical value before, after, and in eq 7. > > Although several ninnies again replied, only ONE person tried to > > show us how to do it. He said, "ok, I accept the challenge and of > > course you do it with calculus!" > > This ninnie showed [that he too is a nincompoop]. > You loose. Speaking loosely, YOU lose. Speaking directly, Dirk Vdm, Thank you for trying! glird
From: glird on 6 Apr 2010 09:40 On Apr 5, 10:27 pm, glird <gl...(a)aol.com> wrote: > On Apr 4, 6:30 pm, "Dirk Van de moortel"<dirkvandemoor...(a)nospAm.hotmail.com> wrote: > > > @T/@x' = K @T/@t (eq A) > > with K = -v/(c^2-v^2) > > If @xi/@x = 1, then @T/@t = Q; and if x' = x vt = 1 at t = 1, > then the value of @T/@x' IS a function of v. But WHERE did > your "calculus" reach that value of K? (Tomorrow I will think > about whether or not that is the right value of K. Meanwhile, > I wonder how it can be, since @tau/@x' is the offset between > two esynched clocks of the moving system k, that are x' apart > as measured by the stationary cs K and @tau/@t is the ratio > of clock rates; which is equal to beta = 1/q in the LTE.) (snip) > Speaking directly, Dirk Vdm, Thank you for trying! Woke up this morning thinking about this and suddenly knew why i was right in thanking Dirk for trying, His value of K was the final link in the mathematical chain from P1 to P2, Here's how: 1. He wrote: K = -v/(c^2-v^2). 2. If we set v = .8c, that says K = -.8/.64 = 1.25, If we set v = .5 it says K = -2. 3. While trying to find out whether these values do or don't fit his prior equations, i remembered a momentary disquiet while I was typing my insert into his equation that followed his statement, "We calculate the partial derivatives". His equation was @T/@x' = p a'( p x' + t ) @T/@t = a'( p x + t ) My insert was @T/@x' [ = (@T/@t)t vx'/c^2)] 4. The disquiet, though momentary, had two causes: a. There, all by itself, was the essence of eq 7! b. BUT!! Although the tau -> t in this local time equation is a function of the ratio of rates of the two systems, the value of its xi -> x > x' is a is a function of the the missing symbol, dxi/dx. 5. Knowing that the amount of offset between two co-moving clocks is independent of the values of dtau/dt or dxi/dx, I wondered if it could depend on the value of Dirk's K. It took all of the next two minutes to figure out that the answer is No. Since my wife is impatiently calling me to come have breakfast, I will finish typing this message this afternoon. (It is 9:40 a.m. now, and I have an appointment with a YMCA class at 11 a.m.; so I won't be back here until after noon. Bye now. glird
From: Dono. on 6 Apr 2010 09:47
On Apr 6, 6:40 am, glird <gl...(a)aol.com> wrote: > > Woke up this morning You woke up the same idiot you went to bed, Lebau |