"Darwin123" escreveu na mensagem
...
On May 8, 6:21 am, "El Enrrabadore-mor"
wrote:

I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).
This is not true. The precession of spinning objects is fully
explained both in Newtonian and relativistic physics. It is you who
don't know the necessary physics involved. It seems to me that you
haven't even tried.

Do you think you know top/gyroscopes?
I've already faced tens of sharks around here with that
challenge and I'm fishing right now (why you think I've
changed my name - I'm fishing).

Do you want to try some basic general questions yourself?

Wait, you know something about precession.
(And only precession of an horizontal gyroscope for sure).
What about a tilted gyroscope, 360 degrees around?
What about nutation?
What about nutation under precession?
What about precession under nutation?
What about energy conservation?
What about gyroscope harmonic motion (clearly seen
combined with external mass inertias - or the usual
top harmonic oscillations)?
What about gyroscope impedance?
Just to mention a few...

I've already spent more then 4,000 hours in the past 3 Years,
looking for gyroscope equations that could predict
basic experimental results.
And I'm not doing this as a hobby. I really need then.
I got:
- Newton method done (T=dL/dt)
- Lagrangian method done
- Vectorial method done
All them full solutions done by myself, zero simplification.
(I did that because couldn't find a book, link, whatever).
What I need is a genius that could explain me
how to get energy conservation out of it.

Euler's equations of motion don't qualify.
Just try them for I1 = I2 = I3 (sphere).

"Greg Neill" escreveu na mensagem
...
"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...


\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

If that simple, then the proper time of the moving object
will be (according to Tom Roberts):
\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

Or else (solving the integral):
\tau = 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)
http://www.vertex42.com/edu/Files/IntegralSummary.pdf

Now, v(t) = r w(t) = constant

The original problem claimed that v = 0.999c = c
so that (for v = c):
\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?
I've started with something constant and I guess I've
end up with a constant time dilatation, independent
of the radius.

El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
El Enrrabadore-mor wrote:
"Tom Roberts" escreveu na mensagem
...
[...]
Do you have an equation, dimensionally consistent, where one
could see what you are talking about?
Consider an object moving relative to an inertial frame with a velocity
v(t) (= dr/dt where r is its position 3-vector relative to that frame, and
t is the time coordinate of the frame); v(t) can be an arbitrary function
of time. Its elapsed proper time between t=T1 and t=T2 is:

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

T1 is the lower limit and T2 is the upper limit of the \integral.

Around here, "_" introduces a subscript and "^" introduces
a superscript (which can be a power).


Meanwhile:
integral_sqrt(1-v(t)^2/c^2) dt =
= 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)

That's hopeless and wrong -- you cannot begin to do the integral until
v(t) is specified. For circular motion, v(t) is constant, the integral
is trivial:

\tau = sqrt(1-|v|^2/c^2) (T2 - T1)

Note that there is no term related to acceleration, and no term related
to "centrifugal force"; all that matters is SPEED (|v|) relative to the
inertial frame.


Tom Roberts

"El Enrrabadore-mor" wrote in message

"Greg Neill" escreveu na mensagem
...
"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...


\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

If that simple, then the proper time of the moving object
will be (according to Tom Roberts):
\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

Right, where v(t) is some function of time.


Or else (solving the integral):
\tau = 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)
http://www.vertex42.com/edu/Files/IntegralSummary.pdf

I don't think you can get away with this unless v(t) is
a linear function of t or constant.


Now, v(t) = r w(t) = constant

The original problem claimed that v = 0.999c = c
so that (for v = c):

v cannot equal c, at least not in our universe.

If you're going to assume a constant v(t) then sqrt(1 - (v/c)^2)
is also a constant. So let k = sqrt(1 - (v/c)^2) and you've
got:

\tau = \integral_T1^T2 k dt

\tau = _T1^T2 k*t

\tau = (T2 - T1)*k = DT*k where k = sqrt(1 - (v/c)^2)

which is the usual Lorentz equation for time dilation
(as expected for a body moving with constant velocity
with respect to an inertial frame).


\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?
I've started with something constant and I guess I've
end up with a constant time dilatation, independent
of the radius.

1. Dubious evaluation of an integral
2. Assuming a velocity equal to the speed of light

On May 8, 9:47 am, "Greg Neill" wrote:
"El Enrrabadore-mor" wrote in message



"Tom Roberts" escreveu na mensagem
. ..

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

Standard ASCII, that's an oxymoron.
Ken

"Greg Neill" escreveu na mensagem
...
"El Enrrabadore-mor" wrote in message

"Greg Neill" escreveu na mensagem
...
"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
...


\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

If that simple, then the proper time of the moving object
will be (according to Tom Roberts):
\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

Right, where v(t) is some function of time.


Or else (solving the integral):
\tau = 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)
http://www.vertex42.com/edu/Files/IntegralSummary.pdf

I don't think you can get away with this unless v(t) is
a linear function of t or constant.

The only thing I can guarantee is that the derivative
of the above solution of the integral, gives the function
used by Tom Roberts in first place.
I've been doing those kind of derivatives a few time
ago and I know that *arcsin(u)" and *sqrt(1-u^2)* cancel
each other out easy in the process.

Wait, now that I'm thinking on the derivatives, I notice
that I've screwed the integral solution.
I have some square terms which shouldn't be squared.
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)


Now, v(t) = r w(t) = constant

The original problem claimed that v = 0.999c = c
so that (for v = c):

v cannot equal c, at least not in our universe.

I was just testing the limits of the solution/example.
In practice v can be only a very, very, small fraction of c.


If you're going to assume a constant v(t) then sqrt(1 - (v/c)^2)
is also a constant. So let k = sqrt(1 - (v/c)^2) and you've
got:

\tau = \integral_T1^T2 k dt

\tau = _T1^T2 k*t

\tau = (T2 - T1)*k = DT*k where k = sqrt(1 - (v/c)^2)

which is the usual Lorentz equation for time dilation
(as expected for a body moving with constant velocity
with respect to an inertial frame).

Even so, my experience tell me that one shouldn't
ignore the full solution of the problem.
One must not assume simplifications to avoid
surprises.
There's an *arcsin(v/c)* function, a *v/c* function
and the whole solution is times *1/2*.

In general yes. The solution is of the form of
the usual Lorentz equation.
Lorentz equation is trivial for circular motion,
gyroscopes, and whenever Pitagoras Theorem
is a good working tool - electricity, and so on.
Gamma and Lorentz equation are trivial for
everything that cycles, circles, and so on.


\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?
I've started with something constant and I guess I've
end up with a constant time dilatation, independent
of the radius.

1. Dubious evaluation of an integral
2. Assuming a velocity equal to the speed of light

The derivative of *arcsin(u)* is *1/sqrt(1-u^2)*du/dt
and the integral is perfect.

On May 8, 1:34 pm, "El Enrrabadore-mor"
wrote:
"Darwin123" escreveu na ...

On May 8, 6:21 am, "El Enrrabadore-mor"
wrote:
I've been claiming for Years that nobody ever fully explained
the top/gyroscope by means of Classic Mechanics, including
in this thread. (relativity not even has a clue).
This is not true. The precession of spinning objects is fully
explained both in Newtonian and relativistic physics. It is you who
don't know the necessary physics involved. It seems to me that you
haven't even tried.

Do you think you know top/gyroscopes?
I've already faced tens of sharks around here with that
challenge and I'm fishing right now (why you think I've
changed my name - I'm fishing).

Do you want to try some basic general questions yourself?

Wait, you know something about precession.
(And only precession of an horizontal gyroscope for sure).
What about a tilted gyroscope, 360 degrees around?
Mathematically, the periodicity of the trigonometric functions
creates a bit of a problem because the inverse solution of the
equations isn't unique. I can't see what you are doing. However, I'll
bet your problem is with solving the mathematics rather than a real
problem with the physics.
What about nutation?
What about nutation under precession?
What about precession under nutation?
In the case spherical "top" resting on a flat surface, which is
orthogonal to the gravitational field, there is no nutation or
precession. You can't get the axis of rotation to face in a different
direction than the gravitational field, not without breaking the
surface. I am not using mathematics, I am merely resting a ball on a
surface and trying to roll it. If you try to use mathematics to
describe nutation or precession here, you will probably get
pathological solutions.
This is not a problem in physics. You have to pay close attention
to those trigonometric functions. They are tricky. You may be getting
a perfectly good description of a broken surface. Look up "holonomic
trajectories."
I suspect that such behavior is possible on a surface that tilts
with respect to the gravitational field, or with an object that isn't
really rigid. However, that will take real analysis and I don't really
have the time.

What about energy conservation?
What about it? Work-energy theorems show when and where mechanical
energy is preserved. As far as I can tell, there is no fundamental
violation of it. Or did you tilt that surface? You know, when a
spinning object falls, potential energy is changed to kinetic energy.
I've already spent more then 4,000 hours in the past 3 Years,
looking for gyroscope equations that could predict
basic experimental results.
How sure are you of the experimental measurements? I mean, are
you really sure about the moment of inertia of your rotators? How do
you calculate them?
And I'm not doing this as a hobby. I really need then.
I got:
- Newton method done (T=dL/dt)
- Lagrangian method done
Already available in classical mechanics textbooks.
- Vectorial method done
You do know that moment of inertia of a nonspherical body is a
matrix. And that the matrix has to be differentiated?
All them full solutions done by myself, zero simplification.
(I did that because couldn't find a book, link, whatever).
What I need is a genius that could explain me
how to get energy conservation out of it.
Try Goldstein, "Classical Mechanics," or Korben and Stehle,
"Classical Mechanics." A world of wonder awaits you.

Euler's equations of motion don't qualify.
Just try them for I1 = I2 = I3 (sphere).
I learned Newtonian mechanics from
"Classical Mechanics" 2nd ed. by H. C. Corben and Philip Stehle
(Krieger 1960). I looked up nutation in the index. In chapter 9, page
149, it qualitatively describes both precession and nutation of a free
rotator.
On page 150, it shows the Euler equations related to a rotator
(equation 52.6). I can't write the equations in the text format, but
they are well known. When I substitute I1=I2=I3, I get a simple spin
with all three axis uncoupled. Basically, it looks like there is no
rotation or nutation for a spherical object. Why should a spherical
object without friction show precession or nutation?
From your complaints, I have a suspicion that you have a problem
both with the physics and the mathematics. Physically: Nutation and
precession are not required in each and every rigid body. They are the
results of different degrees of motion being coupled together. Think
of coupled harmonic oscillators. For some symmetries, the motions are
decoupled. You can't get beats from two pendulums that aren't
connected in some way. Mathematically: you have to analyze
trigonometric functions very carefully. The inverse of a trigonometric
function isn't unique. For instance, consider differentiation. The
derivative of an inverse doesn't have to be the inverse of the
derivative since the function isn't unique.
I don't have my copy of Goldstein available. Someone borrowed it
and didn't return it. Try "Classical Mechanics" by Goldberg. It may
help you.

El Enrrabadore-mor wrote:
The only thing I can guarantee is that the derivative
of the above solution of the integral, gives the function
used by Tom Roberts in first place.[...]
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)

This cannot possibly be correct. Differentiate your expression by t and
you must get a factor of v'(t) (i.e. dv(t)/dt), but there is no such
factor in the original integral.


Tom Roberts

On May 8, 12:34 pm, "El Enrrabadore-mor"
wrote:
"Greg Neill" escreveu na ting.com..."El Enrrabadore-mor" wrote in message





"Greg Neill" escreveu na mensagem
om...
"El Enrrabadore-mor" wrote in message

"Tom Roberts" escreveu na mensagem
. ..

\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

You mean that "T1^T2" is T1 exponential to T2?
Or else, is it "T1 times T2"?

Oy vey. It's "standard" ascii notation for the
definite integral from T1 to T2.

If that simple, then the proper time of the moving object
will be (according to Tom Roberts):
\tau = \integral_T1^T2 sqrt(1-v(t)^2/c^2) dt

Right, where v(t) is some function of time.

Or else (solving the integral):
\tau = 1/2*(v(t)^2/c^2)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)^2/c^2)
http://www.vertex42.com/edu/Files/IntegralSummary.pdf

I don't think you can get away with this unless v(t) is
a linear function of t or constant.

The only thing I can guarantee is that the derivative
of the above solution of the integral, gives the function
used by Tom Roberts in first place.
I've been doing those kind of derivatives a few time
ago and I know that *arcsin(u)" and *sqrt(1-u^2)* cancel
each other out easy in the process.

Wait, now that I'm thinking on the derivatives, I notice
that I've screwed the integral solution.
I have some square terms which shouldn't be squared.
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)

Now, v(t) = r w(t) = constant

The original problem claimed that v = 0.999c = c
so that (for v = c):

v cannot equal c, at least not in our universe.

I was just testing the limits of the solution/example.
In practice v can be only a very, very, small fraction of c.

If you're going to assume a constant v(t) then sqrt(1 - (v/c)^2)
is also a constant. So let k = sqrt(1 - (v/c)^2) and you've
got:

\tau = \integral_T1^T2 k dt

\tau = _T1^T2 k*t

\tau = (T2 - T1)*k = DT*k where k = sqrt(1 - (v/c)^2)

which is the usual Lorentz equation for time dilation
(as expected for a body moving with constant velocity
with respect to an inertial frame).

Even so, my experience tell me that one shouldn't
ignore the full solution of the problem.

For Greg's "v(t)" to be a constant he is using
a rotating spatial CS mixed with a Lorentz
Transform, something that is less than rigorous.
(BTW I use that myself, carefully).

It's a bit more complicated, (as El ref'd),
http://hermes.physics.adelaide.edu.a...vity/SR/rigid_...

One must not assume simplifications to avoid
surprises.
There's an *arcsin(v/c)* function, a *v/c* function
and the whole solution is times *1/2*.

In general yes. The solution is of the form of
the usual Lorentz equation.
Lorentz equation is trivial for circular motion,
gyroscopes, and whenever Pitagoras Theorem
is a good working tool - electricity, and so on.
Gamma and Lorentz equation are trivial for
everything that cycles, circles, and so on.

GR is the working tool,
SR provides an approximation.
I did a GR analysis in thread
"Speed of Gravity revisited" that shows
how to apply GR in rotating CS's.
(I can copy them to this thread if you want).

\tau = 45 degrees * (T2-T1) = 0.785 (T2-T1) =
= 0.785 t_rest frame.

Where did I screw up?
I've started with something constant and I guess I've
end up with a constant time dilatation, independent
of the radius.

1. Dubious evaluation of an integral
2. Assuming a velocity equal to the speed of light

The derivative of *arcsin(u)* is *1/sqrt(1-u^2)*du/dt
and the integral is perfect.

So what? A "perfect" solution needs a
painstaking evaluation of GR's geodesic
equations.
Regards
Ken S. Tucker

"El Enrrabadore-mor" wrote in message


Wait, now that I'm thinking on the derivatives, I notice
that I've screwed the integral solution.
I have some square terms which shouldn't be squared.
The right solution must be:
\tau = 1/2*(v(t)/c)*sqrt(1-v(t)^2/c^2) + 1/2*arcsin(v(t)/c)

[snip]


The derivative of *arcsin(u)* is *1/sqrt(1-u^2)*du/dt
and the integral is perfect.

So where is your dv in the original integral? You've
only got dt.

Integral _T1^T2 sqrt(1 - (v(t)/c)^2) dt

If you use a form such as Integral _a^b u*du then you
better make sure that you have that du in place.