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From: waldofj on 14 May 2010 10:39
On May 12, 3:11 pm, glird <gl...(a)aol.com> wrote:
> On May 9, 8:24 pm, waldofj <wald...(a)verizon.net> wrote:
>
> glird: >< In his 1905 SR paper Einstein wrote an equation,
>          tau  a(t - vx'/(c^2 - v^2)
> in which he said that a is a function of v; which he
> expressed as phi(v). A bit later he said that phi(v) = 1. He never did
> say what "a" dnotes all by itself. >
>
> W: He never did because he showed it doesn't denote anything. It's
> just a constant function of v which always equals one.
>
> >< If we let a denote acceleration, >
>
>  You can't just decide to give an arbitrary meaning and then start
> drawing conclusions. Well, you can but you'll just end up with
> nonsense. Garbage in garbage out.
>
> >< then a=phi(v)=1 denotes acceleration = d(dx/dt) = 1. >
>
>   Actually acceleration = (d/dt)(dx/dt)
> or (d^2/dt^2)x or, since we are talking physics here,
>  dx/dt = v so acceleration = dv/dt
> of course I realize none of this means anything to you since you
> refuse to learn calculus.
>
> >< That means that for any given change in the value of v, the value of _a_ changes by an identical amount. >
>
> now you don't know basic physics and you're contradicting yourself.
> You have set a = phi(v) = 1. One is a constant, constants don't change
> (didn't you learn that in algebra class?) Constant acceleration (one
> is a constant, right?) means velocity is always increasing at a
> constant rate. However that last point is moot since _a_ is not
> acceleration.
>
>   True. As I said, Einstein never did define what _a_ denotes.
> Neither did you.

that's correct. 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). 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.

>
> > However, after writing  eta=phi(v)y  and finding that phi(v) = 1, instead of realizing that this means that
>
> the value of phi(v)y, thus of eta, CHANGES as v does; >
>
> How can you say phi(v) is constant (one is a constant, right?) and
> then say it changes?
>
>  ALMOST the same way you did when you wrote, "Constant acceleration
> (one is a constant, right?) means velocity is always increasing at a
> constant rate."

that doesn't make any sense.

>  Btw, how do you think that E got from
>   "eta = a{c/sqrt(c^2-v^2)}y"  to  "eta = phi(v)y"?
> To be explicit, even if a = phi(v) allowed him to replace a with
> phi(v), how come c/sqrt(c^2-v^2) disappeared?

good question. I have commented on this before but I don't mind
repeating myself.
"I ain't proud, or tired"
first of all c/sqrt(c^2-v^2) is just beta so I'll just write beta if
you don't mind. (actually nowadays it's called gamma but I'll stick to
the old terminology for this discussion).
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. 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. 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. If
you want the long answer, ask me in a different thread.

>
> >< he DELETED the symbol from his equations. >
>
> W: What's the point of writing one times something, you do realize
> multiplying by one doesn't change the expression, right?
>
>  Do you realize that the only way for c/sqrt(c^2-v^2) to equal 1 is
> that v = 0?  If so, then why bother with transformation equations at
> all?

who says beta should equal one?

>
> >< Although that did leave him with the LTE, in which 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   phi(v) = 0; i.e. that there is NO change in the value of a = phi(v) as v changes. >
>
> W: Well, zero is also a constant but phi(v) is being MULTIPLIED in
> each expression. If you multiply by zero you'll just set everything to
> zero, not what you want.
> Or is it you don't know what phi(v) = 0 means?
>
>   Thank you for taking the time to answer me.  (I knew that my wording
> was misleading, and hoped that someone would correct it.) Perhaps it
> would have been clearer if I'd written it this way:
>   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 definatly is not

> then if phi(v) = da/dv = 1, and dv = 4 or 3 or
> 6 then da = 4 or 3 or 6 also. But if there is no change in the
> relation between eta and y as v changes, as is the case in the LTE,
> then that relation is NOT a function of 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

>
>  Btw, E's "proof" that "phi(v) = 1" was defective anyhow.

no, only your assumptions about it are defective.

>
> glird

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

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