Electron’s puzzles.
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From: glird on 15 May 2010 13:29 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"? > phi(v) = 1 means phi(v), or _a_ , or a(v) (all the same thing) is a > constant function such as: (0 * V) + 1 > no matter what value I put in for v it always returns 1 > > you said above: > > > 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. > > of course everyone realized that, that's the behavior of a constant function If, as your comment implies, there is no change in the value of -a_ = ? as v changes, then how can _a_ be a function of v? (If it isn't, then eta = ay -> phi(v)y was defective to start with. > > Btw, E's "proof" that "phi(v) = 1" was defective anyhow. > > no, only your assumptions about it are defective. Do you know what my assumptions were? glird
From: glird on 15 May 2010 15:13 On May 14, 10:39 am, waldofj <wald...(a)verizon.net> wrote: > >< As I said above it turns out that _a_ doesn't denote anything so there's nothing to define. btw we have both been making a mistake when we write a = phi(v). that implies that a is a variable that is being set to the value returned by phi(v). That's not what it is. We should be using the symbol for "is the same as" (I can't reproduce it here with ascii text) instead of equals. In other words, _a_ is just a typographical substitution for phi(v), easier to type _a_ than phi(v). > That's an excellent point! (I think you may be right. But why phi("v")? Why not phi(q)? (Or, as Anrdocles would automatically say, Why not phu(q)? ><As to what _a_ is (here we go again) _a_ is a mathematical artifact that arises from the method used to derive the equations, nothing more, nothing less. It requires a subsequent analysis of the problem to determine if it denotes anything or not. As it turns out, _a_ doesn't denote anything.> Even so, if - as E later said, a = phu(q) = 1, then in wadicall Eq 3, it disappears, leaving us with tau = t - v(x-vt)/(c^2-v*2). That is NOT what the LTE demand! To get the LTE's equation from eq 3, _a_ (thus phi(v) MUST equal 1/ beta! glird
From: glird on 15 May 2010 18:28 On May 14, 10:39 am, waldofj <wald...(a)verizon.net> wrote: > > If you look at the equations on page 45 in the Dover publication and > then look at the equations at the top of page 46 you will see that > beta has been divided out of all four equations. Look again, and then -- after you see that beta remains present in the first two, ask yourself "How does beta get into the first two and disappear from the next two, even though it is present in the last two on pg 45 and not in the tau or zi equations to there?" > This step has been > skipped and no explanation is given for it. That's why I say a page is > missing from the Dover publication. There are a lot of pages "missing" from E's published paper; but that isn't one of them. > The short answer: this is done to > make the equations compatible with the principle of relativity while > maintaining compatibility with the constancy of the speed of light. The short answer is: That was done in order to let E arrive at Poincare's LTE. The trouble is that his inserted (not derived) eq at the top og pg 45 fits the LTE only if a = q. For all other values, they fit he eqs on top of pg 46; which i call Einstein's General transformation Equations th GTE), because they permit and explain the results of the M&M experiments. glird
From: waldofj on 16 May 2010 01:12 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. for example, if I want to say y is a function of x and that function is x^2 I'll write something like this: y = f(x) = x^2 nothing unique here. > > > phi(v) = 1 means phi(v), or _a_ , or a(v) (all the same thing) is a > > constant function such as: (0 * V) + 1 > > no matter what value I put in for v it always returns 1 > > > you said above: > > > > 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. > > > of course everyone realized that, that's the behavior of a constant function > > If, as your comment implies, there is no change in the value of -a_ > = ? as v changes, then how can _a_ be a function of v? (If it isn't, > then > eta = ay -> phi(v)y > was defective to start with. this question came up in another thread some time ago. First look at it from the standpoint of pure math. What is a function? Isn't it an expression that I plug in a value and it returns another? Of course it can get more complicated than that but lets keep it simple. Lets say the function is x^2. I plug in 1 it returns 1. I plug in 2 it returns 4. I plug in 3 it returns 9. And so on. Now, if I want to make a graph of this function, what do I do? First I get some graph paper, label the axes on the paper, horizontal axis x, vertical axis y. Now I use my function to generate some xy points and come up with a list like: (-3, 9) (-2, 4) (-1, 1) (0, 0) (1,1) (2, 4) (3, 9) and so on. now I plot these points and connect them with a smooth line and PRESTO! I have a parabola. So now my question is, what is the function for a horizontal line, say, one unit above the x axis? on this line there will be points like: (0, 1) (1, 1) (2, 1) (3, 1) What function generates a set of points like this? A constant function! I could write it as y = f(x) = 1 but I can hear it now. How is this a function of x when x doesn't appear anywhere in the equation? It's a function of x simply because I labeled the horizontal axes on my graph x. It's as simple as that. If it helps I can write the function like this: y = f(x) = (0 * x) + 1 now x appears in the function and as you can see, for any value I plug in for x it returns 1. And so it is for phi(v) (or a or a(v)) it is a constant function that returns 1 for any value of v. If you graph it, it's the graph of a horizontal line. Why is that so hard to understand? As I said before, it arises from the method used to derive the equations, but as it turns out it is just a trivial constant function that always returns 1. period. You keep trying to read more into it then there is. Let it go. > > > Btw, E's "proof" that "phi(v) = 1" was defective anyhow. > > > no, only your assumptions about it are defective. > > Do you know what my assumptions were? Only the basic assumption that you keep trying to treat this an algebra problem when it's not. > > glird
From: waldofj on 16 May 2010 01:35
On May 15, 3:13 pm, glird <gl...(a)aol.com> wrote: > On May 14, 10:39 am, waldofj <wald...(a)verizon.net> wrote: > > >< As I said above it turns out that _a_ doesn't denote > > anything so there's nothing to define. > btw we have both been making a mistake when we write a = phi(v).. > that implies that a is a variable that is being set to the value > returned by phi(v). > That's not what it is. > We should be using the symbol for "is the same > as" (I can't reproduce it here with ascii text) instead of equals. In > other words, _a_ is just a typographical substitution for phi(v), > easier to type _a_ than phi(v). > > > That's an excellent point! (I think you may be right. > But why phi("v")? Why not phi(q)? (Or, as Anrdocles would > automatically say, Why not phu(q)? You can call it anything you want but, well, it's a point I keep harping on, I don't mean to sound like a broken record and I don't mean to sound like I'm talking down my nose at you but (here we go again), without calculus you will NOT understand the long answer. The short answer is: it is a function of v because of the the way it arises from the derivations. It has to be a function of v, period. > > ><As to what _a_ is (here we go again) > > _a_ is a mathematical artifact that arises from the method used to > derive the equations, nothing more, nothing less. It requires a > subsequent analysis of the problem to determine if it denotes > anything > or not. As it turns out, _a_ doesn't denote anything.> > > Even so, if - as E later said, a = phu(q) = 1, then in wadicall Eq > 3, > it disappears, leaving us with > tau = t - v(x-vt)/(c^2-v*2). so: tau = t - (vx - tv^2)/(c^2 - v^2) tau = (tc^2 - tv^2 - vx + tv^2)/(c^2 - v^2) tau = (tc^2 - vx)/ (c^2 - v^2) tau = (t - vx/c^2)(c^2/(c^2 - v^2) tau (t - vx/c^2) (1 / (1 - v^2/c^2) tau = (t - vx/c^2) beta^2 > That is NOT what the LTE demand! correct! > To get the LTE's equation from eq 3, _a_ (thus phi(v) MUST equal 1/ > beta! well, you finally noticed that a factor of beta is missing from those equations. I say it's been divided out, you say multiplied by 1/beta. Same thing. But if you're thinking that 1/beta has somehow been absorbed into _a_. No. It's a missing step. (one of the assumptions you're getting wrong) _a_ is accounted for later. > > glird |