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From: Dirk Van de moortel on 6 Apr 2010 11:45 glird wrote: > 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. This was a *temporary* change of notation in *that* particular post where "x'=x-vt" was not used. I had put it there for readability. > Since E stipulated that at t = 0, x = xi = T = 0, goodbye to the rest > of this 2003 "calculus". I knew that you wouldn't be able to understand that, so I reprocduced the calculation without that substitution here - see above. Perhaps I shouldn't have put the pointer to the 2003 post here. I'm sorry if this caused confusion with you. Anyway, the above is typical first year calculus, combined with what we call "substitution values of variables in equations". > 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). It looks like you have SEVERE problem with basic algebra. It's no surprise that you never managed to aquaint yourself with basic calculus. Hm... this really explains quite a lot. > 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. Hm...it also looks that you have a severe problem reading written text. That explains quite some more. > 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. We have covered this. It is your problem with basic algebra. Nothing new here. > 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? It is what we call a "temporary substitution" again. We talk about --and manipulate-- an equation @T/@x' + v/(c^2-v^2) @T/@t = 0 which is eq 3, by temporarily defining K as K = -v/(c^2-v^2) and then writing the equation as @T/@x' = K @T/@t so it becomes easier to work. Most kids learn this technique around the age of 14. > (Tomorrow I will think > about whether or not thatis the right value of K. Yes, do think about it, but try not to hurt yourself in the process. > 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? Well, never mind. > > >> 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. I'm sorry. I shouldn't have said it, given the fact that you can't possibly understand anything of this. I should have known better. Perhaps it was useful for someone with basic insight in dealing with variables and equations. > 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. Forget it. You really don't have what it takes to go along with this. I'm really sorry for having wasted your valuable time. > >>> 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. Did I write that? Silly me. That definitely should have been: "You LOSE" - with one "O". Sorry. Dirk Vdm
From: Dirk Van de moortel on 6 Apr 2010 11:46 glird <glird(a)aol.com> wrote in message aad6c90d-a68d-449a-aa44-a6bbbedcf1f1(a)g30g2000yqc.googlegroups.com > 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 *facepalm* Dirk Vdm
From: glird on 6 Apr 2010 21:27 On Apr 6, 9:40 am, glird <gl...(a)aol.com> wrote: > >< 3. While trying to find out whether these values do or don't fit Dirk's 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 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. > Thanks for interrupting me there, Nat. (Gave me time to think about what I wrote in step 4; which I did while scrambling two eggs.) Here, then, is the corrected version of step 4 (Wait while I now write it out. {It is now 2:12 p.m. so let's see how much time it will take me to do it right}.) The subject equation is: @T/@x' [ = (@T/@t)t vx'/c^2)]. Voight's 1887 equation, which Lorentz called the "local time" one, is t = t - vx/c^2, in which the first t is the time of clock B, the second t is the time of clock A at that instant, x is the distance from clock 1 to clock 2; and v is the velocity of the given system through either Lorentz's universally stationary ether or Einstein's similarly stationary empty space. Note that ALL THREE co-ordinates t, t, and x are those of the same system AS PLOTTED BY ITSELF; and v is the velocity of THAT system. Now I will reconstruct item b in step 4, which was, "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 function of the the missing symbol, dxi/dx." Here's what that meant: In my "local time" equation, @T/@x' = (@T/@t)t vx'/c^2), in the left-side expression, T = t is the time of clock B of a system moving at v, and @x' = dx = x is the distance between clocks A and B of that system as measured by itself. Hence, @T = dtau = dt is the difference in simultaneous clock settings (the "offset") of the two clocks that are @x' = dx' = dx apart in their own esynched system. In the right-side expression, @T/@t = dtau/dt is the ratio of clock-rates of two differently moving systems as measured by the one taken as stationary; and (@T/@t)t is the time, tau > t, of clock 1 of the moving system. (If t is NOT the time of the moving system, then one has to multiply t by 1/(the ratio of rates) in order to find the value of -> tau.) That was the easy part. Now comes the troublesome bit. On the right side of my local time eq I wrote (@T/@t)t - vx'/c^2) and 4 b said, "the value of its xi -> x > x' is a function of the missing symbol, dxi/dx." Though Ok, that might mislead some people. Here's why: Let my local time equation be written as tau = (@T/@t)t - vx'/c^2), in which tau is the time on clock B when a ray arrives (see E's 1905 paper for what that means) and t is the time of the viewing system at that instant. If x' = x - vt is the distance between co- moving clocks A and B, then the value of xi at point x (which is where clock B is at t of the viewing system) is a function of the ratio of lengths, @xi/@x = dxi/dx, as measured by the stationary system.7 Accordingly, if we let tau and xi be co-ordinates of the moving system k, and t and x be those of the viewing system K, then the equation should have been written thus: tau_B = (dtau/dt)t - v(dxi/dx)x'/c^2), in which x' = x - vt is the distance between clocks A and B as measured by K. Notice that there is no symbol_a_ there. Where, then, did Einstein get it and what did it denote? Since his eq 7 has no visible symbol for the dtau/dt in my similar equation I thought his _a_ denoted it. WOW!" i said to myself, "THAT was brilliant!! All he did to reach eq 7 was to rewrite my local time equation thus: tau = (dtau/dt)(t v(dxi/dx)x'/c^2 and then replace dtau/dt with the letter "a" and substitute the inverse of his prior value of dxi/dx = Q = c^2 - v^2 (because a Q-shorter unit would fit 1/q more times into given length dx = x') to arrive at tau = (at vx'/(c^2 -v^2)), in which a = dtau/dt = 1 had been his prior value of the ratio of rates; and then in a flash of brilliance take the letter a out of the parenthetical clause to arrive at tau = a(t - vx'/(c^2 - v^2)) Eq 7!" But I now think I was wrong!! Explore it with me. This time, math experts, hold your instant objections until after we finish the next few steps. Tomorrow. Been a long and busy day, and I'm tired. G'nite now. glird
From: glird on 7 Apr 2010 10:10 On Apr 6, 9:27 pm, glird <gl...(a)aol.com> wrote: > >< On the right side of my local time eq I wrote (@T/@t)t - vx'/c^2) and 4 b said, "the value of its xi -> x > x' is a function of the missing symbol, dxi/dx." Though Ok, that might mislead some people. Here's why: Let my local time equation be written as tau = (@T/@t)t - vx'/c^2), in which tau is the time on clock B when a ray arrives (see E's 1905 paper for what that means) and t is the time of the viewing system at that instant. If x' = x - vt is the distance between co- moving clocks A and B, then the value of xi at point x (which is where clock B is at t of the viewing system) is a function of the ratio of lengths, @xi/@x = dxi/dx, as measured by the stationary system. Accordingly, if we let tau and xi be co-ordinates of the moving system k, and t and x be those of the viewing system K, then the equation should have been written thus: tau_B = (dtau/dt)t - v(dxi/dx)x'/c^2), in which x' = x - vt is the distance between clocks A and B as measured by K. Notice that there is no symbol_a_ there. Where, then, did Einstein get it and what did it denote? Since his eq 7 has no visible symbol for the dtau/dt in my similar equation I thought his _a_ denoted it. WOW!" i said to myself, "THAT was brilliant!! All he did to reach eq 7 was to rewrite my local time equation thus: tau = (dtau/dt)(t v(dxi/dx)x'/c^2 and then replace dtau/dt with the letter "a" and substitute the inverse of his prior value of dxi/dx = Q = c^2 - v^2 (because a Q- shorter unit would fit 1/q more times into given length dx = x') to arrive at tau = (at vx'/(c^2 -v^2)), in which a = dtau/dt = 1 had been his prior value of the ratio of rates; and then in a flash of brilliance take the letter a out of the parenthetical clause to arrive at tau = a(t - vx'/(c^2 - v^2)) Eq 7!" But I now think I was wrong!! Explore it with me. This time, math experts, hold your instant objections until after we finish the next few steps. > Woke this morning at 6:15, went to the john and came back to bed. But instead of going to sleep I began chuckling to myself as the final step clarified in my mind. So I jumped out of bed and went to the computer room to close the tiny hole in the middle of the jigsaw puzzle solved by the third and final portion of The Missing Symbol (TML). Took but a few minutes for me to spot "the hole in the middle" in item b of the first paragraph of the introductory remarks: "In this posting we will see what happened after Einstein read Poincare's "Sur la Dynamique des l' Electron" and tried to derive the LTE; and why no mathematician or physicist ever understood the following things: a) In equations containing dxi/dx or dtau/dt, it is implied that these symbols denote the ratio of size of units of measure of the given dimension, x for length and t for time. However, in transformation compared to deformation equations the numerical value per symbol is its own reciprocal; and in inverse cases compared to direct cases the numerical values of the ratio of size of units of length and of time switch places with each other; and so do the symbols themselves! b) Einstein was doing deformation mathematics as per Lorentz, not transformations as per Poincare'. In the deformation equations, dxi/dx = c2 -v2 = Q denotes the ratio of size of units of length of the two systems, in which dxi = 1 is Q shorter than dx = 1. In the inverse deformation equations for this case, dtau/dt = Q is the inverse of the ratio of size of units of time of the two systems, because dtau = 1 is 1/Q larger than dt = 1." For today's discussion, the latter sentence should be revised to: In the inverse deformation equations for this case, dxi/dx = 1/Q is the inverse of the ratio of size of units of length of the two systems, because dxi = 1 is Q shorter than dx = x' = 1. Here's why. Last night, while writing the stuff following "WOW" I wrote, "substitute the inverse of his prior value of dxi/dx = Q = c^2 - v^2 (because a Q-shorter unit would fit 1/q more times into given length dx = x')". I knew that clause was a bimp, because I had initially written it something like this, < replace dtau/dt with the letter "a" and substitute his prior value of dxi/dx = Q = c^2 - v^2 to arrive at etc. > But then I realized that in his eq 7 Einstein didn't write - vQx', he wrote -vx'/Q. So I "corrected" myself, even though I knew that none of you would understand the changed version. The bimp, though, wasn't that YOU wouldn't understand it, it was that Einstein didn't either! So I posted the message and went to bed wondering WHY Einstein wrote x'/Q instead of Qx'. When i realized why, this morning, I said to myself, "The solution to last night's bimp was the word "visible", which I inserted into its sentence a bit later." A moment later, as the ramifications of that word sank in, i softly chuckled, "The proof of the pudding is in the eating". To understand WHY you will have to understand TMS. That's why I stopped what I'd started to do when i ran to the computer over three hours ago, and decided to post this message before I finished updating P1. The second section of TMS introduced P1 thus: "I will now show how Einstein really obtained equations 4 and 5; and then got equation 3 from them; and why he didn't understand what some of their symbols mathematically represent or physically mean. To do that, I will resurrect some of his initially submitted paper ("P1" herein) including the relevant mental experiments he deleted from the published version ("P2"). Along the way I will use d in place of delta [and will add some new things in a different font, [as here.]" Having shown you the solution to last night's bimp (see Posting 3's item b, above, and the latter paragraph from Posting 2) I will now leave you to argue with each other while i go back to fill in the hole in the middle of my reconstruction of P1. Until I return sometime tomorrow or so, you too may again enjoy Posting 2's introduction to P1: "Roused by the pulsating beat and now that they know someone is listening at last, the P1 equations have begun to stir, to rearrange themselves, to talk to me; an excited babble of eager voices anxious to be heard. The living remnants of P1, splattered all over the bloody pages of P2, are ready to come together again. In the following reconstruction of Einstein's initial paper we will watch them be reborn." See you later. glird
From: Dirk Van de moortel on 7 Apr 2010 11:14
glird <glird(a)aol.com> wrote in message 5e3772b4-ea8d-4599-a60b-f818df55bb98(a)11g2000yqr.googlegroups.com > On Apr 6, 9:27 pm, glird <gl...(a)aol.com> wrote: [snip TL;DR] > See you later. > glird Dirk Vdm |