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From: Androcles on 16 May 2010 01:31 "waldofj" <waldofj(a)verizon.net> wrote in message news:adefbf7f-6782-4699-ba07-6b9a33c30259(a)q8g2000vbm.googlegroups.com... On May 15, 1:29 pm, glird <gl...(a)aol.com> wrote: > On May 14, 10:39 am, waldofj <wald...(a)verizon.net> wrote:> On May 12, 3:11 > pm, glird <gl...(a)aol.com> wrote: > > > << Since eta and zeta are independent of v or any change in it, neither > > he nor anyone since seems to have realized that the latter requires that > > da/dv = 0; i.e. that there is NO change in the value of a as v changes. > > The point is this: If a = phi(v) means that a is a function of v, > > > > > it does > > > > and if a(v) means the same thing, > > > it does > > > > and if da/dv is another way to express that relation; > > > stop right there, it most definitely is not > > i meant to ask on these newsgroups, > What is the calculus expression for > "_a_ is a function phi of the velocity v"? calculus uses the same expressions for functions that any other branch of mathematics does. ======================================== I know what the derivative of a function is, but what's the derivative of tau(0,0,0,0), a coordinate?
From: waldofj on 16 May 2010 02:09 On May 16, 1:31 am, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > "waldofj" <wald...(a)verizon.net> wrote in message > > news:adefbf7f-6782-4699-ba07-6b9a33c30259(a)q8g2000vbm.googlegroups.com... > On May 15, 1:29 pm, glird <gl...(a)aol.com> wrote: > > > > > On May 14, 10:39 am, waldofj <wald...(a)verizon.net> wrote:> On May 12, 3:11 > > pm, glird <gl...(a)aol.com> wrote: > > > > << Since eta and zeta are independent of v or any change in it, neither > > > he nor anyone since seems to have realized that the latter requires that > > > da/dv = 0; i.e. that there is NO change in the value of a as v changes. > > > The point is this: If a = phi(v) means that a is a function of v, > > > > it does > > > > > and if a(v) means the same thing, > > > > it does > > > > > and if da/dv is another way to express that relation; > > > > stop right there, it most definitely is not > > > i meant to ask on these newsgroups, > > What is the calculus expression for > > "_a_ is a function phi of the velocity v"? > > calculus uses the same expressions for functions that any other branch > of mathematics does. > ======================================== > I know what the derivative of a function is, I doubt it. > but what's > the derivative of tau(0,0,0,0), a coordinate? If you knew what a derivative is you wouldn't ask such a ridicules question.
From: glird on 16 May 2010 11:00 On May 16, 2:09 am, waldofj <wald...(a)verizon.net> wrote: > > stop right there Thank you, Waldofj! Now that John Walker "Androcles" is replying to you we know he is paying attention to the line of thought beneath this discussion. Since that's what I wanted to know; there is little reason to continue this thread. Take a look at the thread callled "OOPS, says "Androcles"", which I am now finished with also. Now that Anyone has replied to that thread it is time to begin presenting the line of thought itself. Its purpose is to teach J P Walker, and through him everyone else, the meaning of every symbol and equation along Einstein's way to his Sept, 1905 "derivation" of Poincare's LTE, published in June of that same year. girdle :-)
From: glird on 16 May 2010 11:54
On May 16, 2:26 am, "Androcles" <Headmas...(a)Hogwarts.physics_a^z> wrote: > > I know what the derivative of a function is, > but what's the derivative of tau(0,0,0,0), a coordinate? In the expression, "tau is a function of 0,0,0,0" tau is the time of a given clock of the moving system k and 0,0,0,0 are where the clock is on system K at t = 0. The first three numbers denote x=y=z=0,0,0, which means that the moving clock is at the origin of K at the K time t, denoted by the fourth zero. > Let me put it this way. What is the rate of change of a > stationary frame coordinate? In Einstein's paper, a "co-ordinate" is a point on one of the three spatial axes of a given co-ordinate system. (See the title of his segment 3.) Although the rate of change of a co-ordinate of any system is zero, that has nothing to do with the rate if change of the time on a given clock. As of there, neither does the expression "tau is a function of 0,0,0,0". Indeed, in the equation tau = a(t - vx'/Q) if a = 1, the rate of change of the time of a moving clock is identical to the rate of change of the time of a stationary clock, so (delta tau)/(delta t) = dtau/dt = 1. THAT is one of the things to be proved by "you", in "your" discussion of Einstein's segment 3's equations. [Btw, Johny, in the LTE, dtau/dt = 1/q =/= 1!] glird |