On Feb 6, 11:54�am, Randy Poe wrote:
On Feb 6, 12:39 pm, Albertito wrote:





On 6 feb, 16:48, "Androcles" wrote:

wrote in message

....
On 6 f�v, 02:34, Pentcho Valev wrote:

http://philipball.blogspot.com/2007/...-was-innocent-....
"With the technology then available, measuring the bending of
starlight was very challenging. And contrary to popular belief,
Newtonian physics did not predict that light would remain undeflected
- Einstein himself pointed out in 1911 that Newtonian gravity should
cause some deviation too. So the matter was not that of an all-or-
nothing shift in stars' positions, but hinged on the exact numbers."

| Right.

| Classical Newtonian mechanics predicts that light will bend with
| twice the angle observed.

How?

I'll tell you how. In order to see how newtonian gravity can
bend the trajectory of a photon, relativists fake it. They
magically tranform a photon with energy E into a particle
with mass m = E/2c^2, it is saying �half the mass in E = mc^2.
Then, relativists assume that particle passes by the massive
body from infinity travelling locally at c, and then they can
apply newtonian gravity to see how its trajectory is deflected
into a hyperbolic one. As a result there is a deflection angle
twice the observed one. A particle with mass m = E/c^2,
travelling locally at c from infinity, would be deflected in
the correct angle, under newtonian gravity.

Why don't you do the Newtonian calculation before
making this claim?

� � � � � - Randy- Hide quoted text -

- Show quoted text -

Well, going back to the astronaut with his slower clock, it appears to
me that he would come up with a different gravitational constant than
the observer on earth because the velocity of the satellite relative
to earth would be faster according to his clock.
Robert B. Winn

wrote in news:3abf9bab-c868-4a05-816c-20784a206ac3
@m34g2000hsf.googlegroups.com:

t will confuse nobody except nitpickers. You forget
that this is a free wheeling site for discussion, not
a panel of expert meant to approve papers.

This is NOT a 'site for discussion'.

You are posting to Usenet. Your posts get sent to tens of thousands of
computers around the world that are running NNTP protocol.
Your posts are stored on some of those NNTP servers 'forever'.

The fact that you are using Google's interface to post to Usenet does NOT
make 'this' a 'site'.

Usenet predates 'the web' and Google by quite some time.

Do some research and learn just what you are participating in.

This particular Usenet or 'News Group' has its own rules and FAQ's.
You should always read the FAQs for a group before you post to that group.





--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

remove ch100-5 to avoid spam trap

On 9 fév, 05:54, bz wrote:
wrote in news:3abf9bab-c868-4a05-816c-20784a206ac3
@m34g2000hsf.googlegroups.com:

t will confuse nobody except nitpickers. You forget
that this is a free wheeling site for discussion, not
a panel of expert meant to approve papers.

This is NOT a 'site for discussion'.

You are posting to Usenet. Your posts get sent to tens of thousands of
computers around the world that are running NNTP protocol.
Your posts are stored on some of those NNTP servers 'forever'.

The fact that you are using Google's interface to post to Usenet does NOT
make 'this' a 'site'.

Usenet predates 'the web' and Google by quite some time.

Do some research and learn just what you are participating in.

This particular Usenet or 'News Group' has its own rules and FAQ's.
You should always read the FAQs for a group before you post to that group.

--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

Quite obvious.

André Michaud

remove ch100-5 to avoid spam trap

On Feb 8, 8:34 pm, Eric Gisse wrote:
On Feb 8, 8:53 am, Randy Poe wrote:
[...]

"Exponential growth" means, among other things,
"growing faster than ANY polynomial".

I always interpreted exponential to be a literal meaning. I can write
a^x as exp(ln(a) * x) , so anything of that form remains exponential.

Well, yes. I shouldn't have said that's a meaning, but that's a
property of exponential growth. Algorithms which scale exponentially
with problem size are considered "hard" in complexity theory,
not tractable. The idea of finding a polynomial-time algorithm for
such a problem, ANY polynomial, would be considered a revolutionary
breakthrough.

As you say, anything of the form a^x, or actually a^x + b^x +
polynomials, is exponential and would be called O(exp(x)).

And there are probably worse things than exponential.

But I do think it's funny how Andre continually makes up his own
meanings for words [relativistic, exponential, etc] and gets upset
when folks who are educated get confused and request a translation.

And then insisting he is using standard meanings. It's
strange.

If your whole point of being on a newsgroup is to present your
"original thinking", why pretend it's not original?

I thought the French were _for_ language preservation? Or does that
only apply to the French language?

Ah yes, m'sieu, but the language Andre is abusing is
anglais. In particular, anglais scientifique.

- Randy

On Feb 9, 5:29 am, rbwinn wrote:
On Feb 6, 12:15�pm, Randy Poe wrote:



On Feb 6, 2:00 pm, Albertito wrote:

On 6 feb, 18:54, Randy Poe wrote:

On Feb 6, 12:39 pm, Albertito wrote:

On 6 feb, 16:48, "Androcles" wrote:

wrote in message

...
On 6 f�v, 02:34, Pentcho Valev wrote:

http://philipball.blogspot.com/2007/...-was-innocent-...
"With the technology then available, measuring the bending of
starlight was very challenging. And contrary to popular belief,
Newtonian physics did not predict that light would remain undeflected
- Einstein himself pointed out in 1911 that Newtonian gravity should
cause some deviation too. So the matter was not that of an all-or-
nothing shift in stars' positions, but hinged on the exact numbers."

| Right.

| Classical Newtonian mechanics predicts that light will bend with
| twice the angle observed.

How?

I'll tell you how. In order to see how newtonian gravity can
bend the trajectory of a photon, relativists fake it. They
magically tranform a photon with energy E into a particle
with mass m = E/2c^2, it is saying �half the mass in E = mc^2.
Then, relativists assume that particle passes by the massive
body from infinity travelling locally at c, and then they can
apply newtonian gravity to see how its trajectory is deflected
into a hyperbolic one. As a result there is a deflection angle
twice the observed one. A particle with mass m = E/c^2,
travelling locally at c from infinity, would be deflected in
the correct angle, under newtonian gravity.

Why don't you do the Newtonian calculation before
making this claim?

� � � � � - Randy

The calculations have been already done by you relativists,http://www.theory.caltech.edu/people.../obsertop.html

Correct.

I'm only witness of that, and I see that if we double the mass
assumed for a photon we attain the correct answer.

There are a great many assumptions in those
equations which you are not aware of. You are trying
to change the meaning of the symbols midstream.

Here's what those orbital mechanics equations look like for
planetary orbits:http://www.go.ednet.ns.ca/~larry/orbits/kepler.html
Note that E is the sum of kinetic energy and potential
energy. So what are you going to use for this E? Do
you really want to use mc^2 and claim it's a Newtonian
model?

If you were doing Newtonian gravitation the way Newton
did it to calculate the planetary orbits, you'd say
the kinetic energy is (1/2)*mc^2, and the potential
energy is -GMm/r. Then you'd find out that the mass m
drops out, that is the parameter E/m = [c^2/2 - GM/r]
does not depend on m.

Let's start from the basics: What is the initial velocity,
what is the acceleration, what is the result?

Start out with a velocity c, mass m, distance R
as it passes near mass M.

The acceleration experienced by the mass m is
GM/R^2. That is what you would apply to the velocity
vector to get the change in velocity. It does not
depend on m.

Trajectories of small bodies through solar systems do
not depend on mass.

� � � � � � - Randy- Hide quoted text -

- Show quoted text -

Well, you are also using the concept of absolute time if you say
this.

If I'm doing a Newtonian calculation, I do it with Newtonian
physics.

But according to scientists, they have proven that a clock in
S' will run slower than a clock in S, so my question is, What does
that do to Newton's equations?

Whatever it does (and I have no interest in your nonsense),
the result is not Newtonian physics. Call it something else.
If you want to ask, "what does pure Newtonian physics predict"
then the answer is done with Newtonian physics, not some
hybrid.

- Randy

On Feb 9, 6:37Â*am, Randy Poe wrote:
On Feb 9, 5:29 am, rbwinn wrote:





On Feb 6, 12:15�pm, Randy Poe wrote:

On Feb 6, 2:00 pm, Albertito wrote:

On 6 feb, 18:54, Randy Poe wrote:

On Feb 6, 12:39 pm, Albertito wrote:

On 6 feb, 16:48, "Androcles" wrote:

wrote in message

...
On 6 f�v, 02:34, Pentcho Valev wrote:

http://philipball.blogspot.com/2007/...-was-innocent-...
"With the technology then available, measuring the bending of
starlight was very challenging. And contrary to popular belief,
Newtonian physics did not predict that light would remain undeflected
- Einstein himself pointed out in 1911 that Newtonian gravity should
cause some deviation too. So the matter was not that of an all-or-
nothing shift in stars' positions, but hinged on the exact numbers."

| Right.

| Classical Newtonian mechanics predicts that light will bend with
| twice the angle observed.

How?

I'll tell you how. In order to see how newtonian gravity can
bend the trajectory of a photon, relativists fake it. They
magically tranform a photon with energy E into a particle
with mass m = E/2c^2, it is saying �half the mass in E = mc^2.
Then, relativists assume that particle passes by the massive
body from infinity travelling locally at c, and then they can
apply newtonian gravity to see how its trajectory is deflected
into a hyperbolic one. As a result there is a deflection angle
twice the observed one. A particle with mass m = E/c^2,
travelling locally at c from infinity, would be deflected in
the correct angle, under newtonian gravity.

Why don't you do the Newtonian calculation before
making this claim?

� � � � � - Randy

The calculations have been already done by you relativists,http://www.theory.caltech.edu/people.../obsertop.html

Correct.

I'm only witness of that, and I see that if we double the mass
assumed for a photon we attain the correct answer.

There are a great many assumptions in those
equations which you are not aware of. You are trying
to change the meaning of the symbols midstream.

Here's what those orbital mechanics equations look like for
planetary orbits:http://www.go.ednet.ns.ca/~larry/orbits/kepler.html
Note that E is the sum of kinetic energy and potential
energy. So what are you going to use for this E? Do
you really want to use mc^2 and claim it's a Newtonian
model?

If you were doing Newtonian gravitation the way Newton
did it to calculate the planetary orbits, you'd say
the kinetic energy is (1/2)*mc^2, and the potential
energy is -GMm/r. Then you'd find out that the mass m
drops out, that is the parameter E/m = [c^2/2 - GM/r]
does not depend on m.

Let's start from the basics: What is the initial velocity,
what is the acceleration, what is the result?

Start out with a velocity c, mass m, distance R
as it passes near mass M.

The acceleration experienced by the mass m is
GM/R^2. That is what you would apply to the velocity
vector to get the change in velocity. It does not
depend on m.

Trajectories of small bodies through solar systems do
not depend on mass.

� � � � � � - Randy- Hide quoted text -

- Show quoted text -

Well, you are also using the concept of absolute time if you say
this.

If I'm doing a Newtonian calculation, I do it with Newtonian
physics.

But according to scientists, they have proven that a clock in
S' will run slower than a clock in S, so my question is, What does
that do to Newton's equations?

Whatever it does (and I have no interest in your nonsense),
the result is not Newtonian physics. Call it something else.
If you want to ask, "what does pure Newtonian physics predict"
then the answer is done with Newtonian physics, not some
hybrid.

Â* Â* Â* Â* Â* Â* Â* - Randy- Hide quoted text -

- Show quoted text -

Well, OK, Randy, let's use pure Newtonian physics. Newton is sitting
under a tree trying to figure out the orbit of the moon, and an apple
falls and hits him on the head. So Newton remembers Galileo's
experiment with the two different size rocks on the leaning tower or
Pisa and decides that the moon is falling toward the earth at a rate
that keeps it always at the same altitude. So he uses Kepler's
equation for the radius of an orbit and figures the force required.
All I asked was how a slower clock on the moon would apply to the
equations. You people were the ones who said that a clock on the moon
would be running slower. Now you have all of these GPS satellites,
etc. How do the times on the clocks in those satellites relate to
their orbits?
You keep claiming that the equations that Newton used are
scripture of some kind. Kepler's number was just an approximation.
It did not exactly match the orbit of anything from the start. It was
just an average taken from the orbits that could be observed.
Maybe if clock time was used instead of absolute time, Kepler's
calculation of the orbits would be closer, maybe it would be less
accurate. I have no way of determining any of this. I have no time
to run an experiment of that kind. I have to work every day welding.
So I ask a scientist. No, Newton's equations are scripture.
You cannot use them any way except the way Newton did. OK, I do not
want to get into an argument with you over your religion. Just go
ahead and use Newton's equations the same way he did. He used
absolute time. According to him, a clock in a satellite would read
the same as a clock on earth.
I don't want to get you all irrational over something like this.
Robert B. Winn

On Feb 9, 10:38 am, rbwinn wrote:
On Feb 9, 6:37 am, Randy Poe wrote:



On Feb 9, 5:29 am, rbwinn wrote:

On Feb 6, 12:15�pm, Randy Poe wrote:

On Feb 6, 2:00 pm, Albertito wrote:

On 6 feb, 18:54, Randy Poe wrote:

On Feb 6, 12:39 pm, Albertito wrote:

On 6 feb, 16:48, "Androcles" wrote:

wrote in message

...
On 6 f�v, 02:34, Pentcho Valev wrote:

http://philipball.blogspot.com/2007/...-was-innocent-...
"With the technology then available, measuring the bending of
starlight was very challenging. And contrary to popular belief,
Newtonian physics did not predict that light would remain undeflected
- Einstein himself pointed out in 1911 that Newtonian gravity should
cause some deviation too. So the matter was not that of an all-or-
nothing shift in stars' positions, but hinged on the exact numbers."

| Right.

| Classical Newtonian mechanics predicts that light will bend with
| twice the angle observed.

How?

I'll tell you how. In order to see how newtonian gravity can
bend the trajectory of a photon, relativists fake it. They
magically tranform a photon with energy E into a particle
with mass m = E/2c^2, it is saying �half the mass in E = mc^2.
Then, relativists assume that particle passes by the massive
body from infinity travelling locally at c, and then they can
apply newtonian gravity to see how its trajectory is deflected
into a hyperbolic one. As a result there is a deflection angle
twice the observed one. A particle with mass m = E/c^2,
travelling locally at c from infinity, would be deflected in
the correct angle, under newtonian gravity.

Why don't you do the Newtonian calculation before
making this claim?

� � � � � - Randy

The calculations have been already done by you relativists,http://www.theory.caltech.edu/people.../obsertop.html

Correct.

I'm only witness of that, and I see that if we double the mass
assumed for a photon we attain the correct answer.

There are a great many assumptions in those
equations which you are not aware of. You are trying
to change the meaning of the symbols midstream.

Here's what those orbital mechanics equations look like for
planetary orbits:http://www.go.ednet.ns.ca/~larry/orbits/kepler.html
Note that E is the sum of kinetic energy and potential
energy. So what are you going to use for this E? Do
you really want to use mc^2 and claim it's a Newtonian
model?

If you were doing Newtonian gravitation the way Newton
did it to calculate the planetary orbits, you'd say
the kinetic energy is (1/2)*mc^2, and the potential
energy is -GMm/r. Then you'd find out that the mass m
drops out, that is the parameter E/m = [c^2/2 - GM/r]
does not depend on m.

Let's start from the basics: What is the initial velocity,
what is the acceleration, what is the result?

Start out with a velocity c, mass m, distance R
as it passes near mass M.

The acceleration experienced by the mass m is
GM/R^2. That is what you would apply to the velocity
vector to get the change in velocity. It does not
depend on m.

Trajectories of small bodies through solar systems do
not depend on mass.

� � � � � � - Randy- Hide quoted text -

- Show quoted text -

Well, you are also using the concept of absolute time if you say
this.

If I'm doing a Newtonian calculation, I do it with Newtonian
physics.

But according to scientists, they have proven that a clock in
S' will run slower than a clock in S, so my question is, What does
that do to Newton's equations?

Whatever it does (and I have no interest in your nonsense),
the result is not Newtonian physics. Call it something else.
If you want to ask, "what does pure Newtonian physics predict"
then the answer is done with Newtonian physics, not some
hybrid.

- Randy- Hide quoted text -

- Show quoted text -

Well, OK, Randy, let's use pure Newtonian physics. Newton is sitting
under a tree trying to figure out the orbit of the moon, and an apple
falls and hits him on the head. So Newton remembers Galileo's
experiment with the two different size rocks on the leaning tower or
Pisa and decides that the moon is falling toward the earth at a rate
that keeps it always at the same altitude. So he uses Kepler's
equation for the radius of an orbit and figures the force required.

Garbled. But at any rate, in Newtonian physics we can write the
equation
GMm/r^2 = d^2/dt^2 [m*x ]

and calculate the behavior of mass m in the presence of mass M.

All I asked was how a slower clock on the moon would apply to the
equations.

Didn't you start this post with this statement:
"let's use pure Newtonian physics."?

There's no "slower clock on the moon" in pure Newtonian
physics.

You people were the ones who said that a clock on the moon
would be running slower.

Faster, actually. In the theory called GR. Not in
"pure Newtonian physics".

Now you have all of these GPS satellites,
etc. How do the times on the clocks in those satellites relate to
their orbits?

Well, that's a different question. Did you want to ask
a question about satellite orbits in GR or in Newtonian
mechanics? Don't say "Newtonian" and then ask a GR question.

I'm going to try to make an analogy with a subject you
might understand: welding. If I weld the left half of
a Mercedes-Benz to the right half of a Volkswagon Beetle,
and find this creates all sorts of problems, does this
tell me that the Benz is poorly designed?

OK, OK, let's just talk about the pure Benz and how
poorly it handles. I took my welded Benz-Beetle out
for a drive and found... what's that? That's not a pure
Benz anymore? What's wrong with you? I'm just trying
to ask you a question about handling of the Mercedes
Benz!

You keep claiming that the equations that Newton used are
scripture of some kind.

No, I'm claiming that if you want to claim you're
doing Newtonian mechanics, you should be using the
Newtonian physics. If you want to throw something else
in there, you can certainly do it, there's no law against
it. But it's no longer Newtonian physics.

What, is the Mercedes Benz some holy grail that I'm
not allowed to weld onto? Was it created by St. Peter
or something? What's wrong with my welding job? Just
answer the question about why the handling of the
Mercedes is so bad, based on my experiments with
my welded vehicle!

Kepler's number was just an approximation.

What number?

It did not exactly match the orbit of anything from the start.

But Newton's equations matched the behavior astonishingly
well, and could even be extended to model the interaction
of multiple bodies with high accuracy. To the extent that
NASA still uses them, and so does every agency which
tracks the thousands of bits of spacecraft and rubble
traveling around the earth.

It was
just an average taken from the orbits that could be observed.

That's nice, but we were talking about Newton's theory,
not Kepler's. Newton explained Kepler's Laws, and also
in what way they are an approximation.

Maybe if clock time was used instead of absolute time, Kepler's
calculation of the orbits would be closer, maybe it would be less
accurate. I have no way of determining any of this. I have no time
to run an experiment of that kind. I have to work every day welding.
So I ask a scientist. No, Newton's equations are scripture.

If you ask me about Newton's equations, then those are
the equations I'm going to use. You've dragged both
pre-Newton and post-Newton science into this discussion.
Are you asking about Newton's equations or not?

What the hell is wrong with this Mercedes Benz
anyway? It's just a terrible car.

You cannot use them any way except the way Newton did.

I can use them any way I want. And I can even change
them. But then I have to say they aren't Newton's
equations any more.

What's wrong with saying that? Why do you want to
change them and then pretend they're unchanged?
You want to explore the welded theory, let's do it.

- Randy

On 9 fév, 08:34, Randy Poe wrote:
On Feb 8, 8:34 pm, Eric Gisse wrote:

On Feb 8, 8:53 am, Randy Poe wrote:
[...]

"Exponential growth" means, among other things,
"growing faster than ANY polynomial".

I always interpreted exponential to be a literal meaning. I can write
a^x as exp(ln(a) * x) , so anything of that form remains exponential.

Well, yes. I shouldn't have said that's a meaning, but that's a
property of exponential growth. Algorithms which scale exponentially
with problem size are considered "hard" in complexity theory,
not tractable. The idea of finding a polynomial-time algorithm for
such a problem, ANY polynomial, would be considered a revolutionary
breakthrough.

As you say, anything of the form a^x, or actually a^x + b^x +
polynomials, is exponential and would be called O(exp(x)).

And there are probably worse things than exponential.

But I do think it's funny how Andre continually makes up his own
meanings for words [relativistic, exponential, etc] and gets upset
when folks who are educated get confused and request a translation.

And then insisting he is using standard meanings. It's
strange.

If your whole point of being on a newsgroup is to present your
"original thinking", why pretend it's not original?

That's where you have it wrong. I am not here to present
my thinking original or not. Here to offer info to newbees
and occasionally discuss with openminded individuals.

I always mirrorred disingenuity will disingenuity and
close-mindedness with close-mindedness, and it will remain
so for as long as I post here.

You and your bunch will get exactly zip out of me.

André Michaud

On Feb 9, 7:47 am, wrote:
On 9 fév, 08:34, Randy Poe wrote:



On Feb 8, 8:34 pm, Eric Gisse wrote:

On Feb 8, 8:53 am, Randy Poe wrote:
[...]

"Exponential growth" means, among other things,
"growing faster than ANY polynomial".

I always interpreted exponential to be a literal meaning. I can write
a^x as exp(ln(a) * x) , so anything of that form remains exponential.

Well, yes. I shouldn't have said that's a meaning, but that's a
property of exponential growth. Algorithms which scale exponentially
with problem size are considered "hard" in complexity theory,
not tractable. The idea of finding a polynomial-time algorithm for
such a problem, ANY polynomial, would be considered a revolutionary
breakthrough.

As you say, anything of the form a^x, or actually a^x + b^x +
polynomials, is exponential and would be called O(exp(x)).

And there are probably worse things than exponential.

But I do think it's funny how Andre continually makes up his own
meanings for words [relativistic, exponential, etc] and gets upset
when folks who are educated get confused and request a translation.

And then insisting he is using standard meanings. It's
strange.

If your whole point of being on a newsgroup is to present your
"original thinking", why pretend it's not original?

That's where you have it wrong. I am not here to present
my thinking original or not. Here to offer info to newbees
and occasionally discuss with openminded individuals.

....only as long they don't know enough to argue with you, right?


I always mirrorred disingenuity will disingenuity and
close-mindedness with close-mindedness, and it will remain
so for as long as I post here.

You and your bunch will get exactly zip out of me.

André Michaud

On Feb 9, 7:53 pm, rbwinn wrote:

OK, so you do not believe Kepler had a number.

I didn't say whether I believed Kepler had a number of not.
I asked you what number you were referring to.

Kepler thought that he
did. It was R^3/T^2 = 3.353, more or less.

OK, that's a number. What about it?

Newton used Kepler's number in his equation for the centripetal force
on a planet.

No, Newton didn't "use Kepler's number". Newton
proved why R and T have this relation from more fundamental
considerations.

Now I don't know what you are saying about welding.

I'm asking you what theory you want to talk about.

(1) Newton's theory, which would be the theory that, you know,
was Newton's. I will only insist on sticking to Newton physics
if you say that we are having a "pure Newtonian" discussion.
That would be what is implied by the words "Newtonian"
and "pure".

(2) GR, where time is affected by gravitational potential
and orbits are geodesics, or

(3) Some theory where you have welded GR results with
Newtonian equations, which is neither GR nor Newton but
a hybrid monster, and possibly not drivable.

I'm happy to talk about one, but you have to pick. Don't
say you're talking about (1) and then immediately switch
to (3).

You have some question about the GR effect of time
and gravitational fields. So we're in either (2) or
(3). Which is it, and what's your question?

- Randy