On May 13, 8:57 pm, WIlly wrote:
On May 12, 3:01 pm, Albertito wrote:



On May 12, 1:13 pm, Curt Jurgens wrote:

On May 12, 1:56 pm, Albertito wrote:

On May 11, 10:07 pm, PD wrote:

On May 9, 6:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

f' = Exp(-r) f ,
where r = arctanh(v/c),

can't still be experimentally distinguished from this one

f' = Exp(-v/c) f

There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.

PD

Interpretation under my model and that provided
by SR can't still be experimentally distinguished.

Consider this sum of betas,

v/c = v_p/c + v_mu'/c ,
with
v_p, pion speed at the moment of its decay wrt muon
detector.
v_mu', the relative speed of a muon with respect
to the point where the pion has decayed

Consider this sum of rapidities,

arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)

I spot a big mistake here, is atanh distributive?

tanh seems to be not that

tanh(x+y) = {tanh x + tanh y} / {1 + tanh x tanh y}

arctanh(v_p/c) = x - v_p/c = tanh(x),
arctanh(v_mu'/c) = y - v_mu'/c = tanh(y),
arctanh(w/c) = x + y,
w/c = tanh(x + y) = (tanh x + tanh y)/(1 + tanh x tanh y)

where is the great mistake?

You said this ****

v/c = v_p/c + v_mu'/c

Then you take arctanh of both sides

arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)

Which you may not, but

arctanh(w/c) = arctanh(v_p/c + v_mu'/c)

Which gives something completely else

No, that's wrong. v and w are different variables.
Of course, we get
arctanh(v/c) = arctanh(v_p/c + v_mu'/c), and
arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c),
and both expressions are correct, because w and v
are not the same variable.

On May 13, 10:17 pm, Albertito wrote:
On May 13, 8:57 pm, WIlly wrote:



On May 12, 3:01 pm, Albertito wrote:

On May 12, 1:13 pm, Curt Jurgens wrote:

On May 12, 1:56 pm, Albertito wrote:

On May 11, 10:07 pm, PD wrote:

On May 9, 6:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

f' = Exp(-r) f ,
where r = arctanh(v/c),

can't still be experimentally distinguished from this one

f' = Exp(-v/c) f

There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.

PD

Interpretation under my model and that provided
by SR can't still be experimentally distinguished.

Consider this sum of betas,

v/c = v_p/c + v_mu'/c ,
with
v_p, pion speed at the moment of its decay wrt muon
detector.
v_mu', the relative speed of a muon with respect
to the point where the pion has decayed

Consider this sum of rapidities,

arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)

I spot a big mistake here, is atanh distributive?

tanh seems to be not that

tanh(x+y) = {tanh x + tanh y} / {1 + tanh x tanh y}

arctanh(v_p/c) = x - v_p/c = tanh(x),
arctanh(v_mu'/c) = y - v_mu'/c = tanh(y),
arctanh(w/c) = x + y,
w/c = tanh(x + y) = (tanh x + tanh y)/(1 + tanh x tanh y)

where is the great mistake?

You said this ****

v/c = v_p/c + v_mu'/c

Then you take arctanh of both sides

arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)

Which you may not, but

arctanh(w/c) = arctanh(v_p/c + v_mu'/c)

Which gives something completely else

No, that's wrong. v and w are different variables.
Of course, we get
arctanh(v/c) = arctanh(v_p/c + v_mu'/c), and
arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c),
and both expressions are correct, because w and v
are not the same variable.

Yes, sorry

The difference seems to be exactly the error I just made.

BTW, what is the difference?

On May 13, 1:07Â*pm, Eric Gisse wrote:
On May 13, 11:34Â*am, Albertito wrote:



out �those three generations of charged leptons, are just exact
ratios derived from multiples of c.

I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist.- Hide quoted text -

- Show quoted text -

Ptolemy was an actual published astronomer when Galileo decided that
the earth was rotating instead of the sun orbiting the earth.
Special relativity has just as many epicycles as Ptolemaic astronomy..
Robert B. Winn

Where are the epicycles, Robert?

Dark matter and dark energy are epicycles.

Bullet cluster? Casimir effect?

Just because _you do not understand_ does not make them "epicycles".- Hide quoted text -

Well, I for sure do not understand the distance contraction. I never
did believe in that. What Galileo believed became "trendy" after his
death among younger scientists, so the Ptolemaic system of astronomy
was eventually taken out of use as old scientists faithful to
epicycles died off, and younger scientists did not remember the
complicated mathematics of epicycles. Today there is no one left who
is even interested in what Ptolemy said.
I think eventually Einstein will fade away the same way.
Robert B. Winn

On May 13, 9:07 pm, Eric Gisse wrote:
On May 13, 11:34 am, Albertito wrote:



On May 13, 8:23 pm, Eric Gisse wrote:

On May 13, 6:19 am, rbwinn wrote:

On May 13, 3:01�am, Eric Gisse wrote:

On May 13, 1:53�am, Albertito wrote:

On May 13, 10:04 am, PD wrote:

On May 13, 3:31 am, Albertito wrote:

On May 12, 11:21 pm, PD wrote:

On May 12, 6:56 am, Albertito wrote:

On May 11, 10:07 pm, PD wrote:

On May 9, 6:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c �below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

� � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't �still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

� � � � � � � �w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

� � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

� � � � � � � �f' = Exp(-r) f ,
� � � � � � � �where r = arctanh(v/c),

can't still be experimentally distinguished from this one

� � � � � � � �f' = Exp(-v/c) f

There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c..

PD

Interpretation under my model and that provided
by SR can't still be experimentally distinguished.

Consider this sum of betas,

� � v/c = v_p/c + v_mu'/c ,
� � with
� � v_p, pion speed at the moment of its decay wrt muon
� � � � � � detector.
� � v_mu', �the relative speed of a muon with respect
� � � � � � � � �to the point where the pion has decayed

Consider this sum of rapidities,

�arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c)

In both sums, neither v_p nor v_m_u' exceed c.

SR assumes the final speed of the muon in the detector
must be

� � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)),

� � � � but it needs also the assumption there is time dilation
� � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2),
� � � � in order to fit the prediction to the observation.

In my model, the final speed of the muon in the detector
is
� � � � � � �v �= v_p + v_mu',
� � � � � � without the assumption of any time dilation at all.

If it still is not clear to you, I can repeat,

� � � � � � �v/c =(v_p + v_mu')/c

And, if you do the math, you find out the left-hand-side of this
formula should be greater than 1, given the amount of kinetic energy
released in pions decaying to muons. Go ahead, plug in the numbers.

What's the problem with finding out the left-hand-side
of this formula is greater than 1?

Well, for one thing, it doesn't happen. As I mentioned to you, the
time of flight of muons is *measured*. We know what the speed is. It's
not greater than 1.

That's only a problem
if you use the relativistic kinetic energy.

No, sir. It's a problem with measurement.

Of course,
v/c = (v_p + v_mu')/c �is greater than 1, because v_p/c
is very close to 1, and v_mu'/c can be greater than
(1 - v_p/c).

You can't keep track of single muons to measure their
'time of flight'. All you can do is to perform statistics and
extrapolate to single muons. You can't say a muon called
A was emitted by a pion called P at time t=0, and then A
was detected at distance R at time t'. The extrapolation for
a single muon you perfom is pure nonsense. That's the
reason why we find out in particle physics there are three
odd generations of charged leptons, and nobody can guess
why there are actually those three generations and why they
exhibit the rest masses they exhibit. SR is unable to account for
the mass of muons and tau leptons with respect to the mass of
electrons. Should we call that impossibility a fraud?. If you avoid
SR, and try to explain why muon and tau lepton have the rest
masses they have wrt that of an electron, you will be able to find
out �those three generations of charged leptons, are just exact
ratios derived from multiples of c.

I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist.- Hide quoted text -

- Show quoted text -

Ptolemy was an actual published astronomer when Galileo decided that
the earth was rotating instead of the sun orbiting the earth.
Special relativity has just as many epicycles as Ptolemaic astronomy..
Robert B. Winn

Where are the epicycles, Robert?

Dark matter and dark energy are epicycles.

Bullet cluster? Casimir effect?

Just because _you do not understand_ does not make them "epicycles".

Under GR, dark matter is a hypothetical form of matter to
explain the variations in speed and direction of the apparent
motion of stars in spiral galaxies, which yield apparent flat
rotation curves.

Under Ptolemaic system of astronomy, the epicycle is a
geometric model to explain the variations in speed and
direction of the apparent motion of the Moon, Sun, and
planets.

Thus,

GR -- Ptolemaic system of astronomy,
dark matter -- epicycle,
hypothetical form of matter -- geometric model,
stars in spiral galaxies -- the Moon, Sun, and planets.

On May 13, 9:23 pm, WIlly wrote:
On May 13, 10:17 pm, Albertito wrote:



On May 13, 8:57 pm, WIlly wrote:

On May 12, 3:01 pm, Albertito wrote:

On May 12, 1:13 pm, Curt Jurgens wrote:

On May 12, 1:56 pm, Albertito wrote:

On May 11, 10:07 pm, PD wrote:

On May 9, 6:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

f' = Exp(-r) f ,
where r = arctanh(v/c),

can't still be experimentally distinguished from this one

f' = Exp(-v/c) f

There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.

PD

Interpretation under my model and that provided
by SR can't still be experimentally distinguished.

Consider this sum of betas,

v/c = v_p/c + v_mu'/c ,
with
v_p, pion speed at the moment of its decay wrt muon
detector.
v_mu', the relative speed of a muon with respect
to the point where the pion has decayed

Consider this sum of rapidities,

arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)

I spot a big mistake here, is atanh distributive?

tanh seems to be not that

tanh(x+y) = {tanh x + tanh y} / {1 + tanh x tanh y}

arctanh(v_p/c) = x - v_p/c = tanh(x),
arctanh(v_mu'/c) = y - v_mu'/c = tanh(y),
arctanh(w/c) = x + y,
w/c = tanh(x + y) = (tanh x + tanh y)/(1 + tanh x tanh y)

where is the great mistake?

You said this ****

v/c = v_p/c + v_mu'/c

Then you take arctanh of both sides

arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c)

Which you may not, but

arctanh(w/c) = arctanh(v_p/c + v_mu'/c)

Which gives something completely else

No, that's wrong. v and w are different variables.
Of course, we get
arctanh(v/c) = arctanh(v_p/c + v_mu'/c), and
arctanh(w/c) = arctanh(v_p/c) + arctanh(v_mu'/c),
and both expressions are correct, because w and v
are not the same variable.

Yes, sorry

The difference seems to be exactly the error I just made.

BTW, what is the difference?

The difference is that w is the relativistic sum of velocities
and v is the euclidean vector sum of velocities. Thus, w
cannot exceed c, but v can. And that's why to
express arctanh(v/c) = arctanh(v_p/c + v_mu'/c)
is meaningless for v c, because it yields indeterminate.
Therefore, the correct expression must be the simple sum,
v/c = v_p/c + v_mu'/c

On May 13, 12:35Â*pm, Albertito wrote:
On May 13, 9:07 pm, Eric Gisse wrote:



On May 13, 11:34 am, Albertito wrote:

On May 13, 8:23 pm, Eric Gisse wrote:

On May 13, 6:19 am, rbwinn wrote:

On May 13, 3:01�am, Eric Gisse wrote:

On May 13, 1:53�am, Albertito wrote:

On May 13, 10:04 am, PD wrote:

On May 13, 3:31 am, Albertito wrote:

On May 12, 11:21 pm, PD wrote:

On May 12, 6:56 am, Albertito wrote:

On May 11, 10:07 pm, PD wrote:

On May 9, 6:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c �below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

� � � � � r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't �still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

� � � � � � � �w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

� � � � � � �arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

� � � � � � � �f' = Exp(-r) f ,
� � � � � � � �where r = arctanh(v/c),

can't still be experimentally distinguished from this one

� � � � � � � �f' = Exp(-v/c) f

There is actually an abundance of these tests. I'll mention one.
Muon beamlines are created by allowing charged pions to decay in
flight. The pions have momenta such that their speed is close to that
of light. Since it is an exoenergetic decay, the muon has extra
kinetic energy from the decay. Since the pion's decay mechanism
doesn't give a whit whether the pion is in flight when it decays (and
in fact, the principle of relativity says the physics of the pion
decay has to be the same for pions at rest vs. pions in free flight),
we can guess what that kinetic energy profile is for the moving pions
by using the distribution for decaying pions at rest.. Or, put another
way, we can take the velocity distribution of muons in the pion rest
frame and boost them to the frame where the pions are moving close to
the speed of light. Using rapidities to perform that boost results in
a velocity distribution that agrees with direct measurement. The
Galilean transformations, on the other hand, predict that the muon
velocity would have a peak in the forward direction that exceeds c.
Muon time of flight can be measured directly via a triplet of
scintillator paddles, and in fact this is routinely done in muon
beamlines. The muon speed distribution never exceeds c.

PD

Interpretation under my model and that provided
by SR can't still be experimentally distinguished.

Consider this sum of betas,

� � v/c = v_p/c + v_mu'/c ,
� � with
� � v_p, pion speed at the moment of its decay wrt muon
� � � � � � detector.
� � v_mu', �the relative speed of a muon with respect
� � � � � � � � �to the point where the pion has decayed

Consider this sum of rapidities,

�arctanh(w/c) = �arctanh(v_p/c) �+ �arctanh(v_mu'/c)

In both sums, neither v_p nor v_m_u' exceed c.

SR assumes the final speed of the muon in the detector
must be

� � � � w = c tanh(arctanh(v_p/c) �+ �arctanh(v_mu'/c)),

� � � � but it needs also the assumption there is time dilation
� � � � of the muon lifetime, as t = t_0 / sqrt(1 - w^2/c^2),
� � � � in order to fit the prediction to the observation.

In my model, the final speed of the muon in the detector
is
� � � � � � �v �= v_p + v_mu',
� � � � � � without the assumption of any time dilation at all.

If it still is not clear to you, I can repeat,

� � � � � � �v/c =(v_p + v_mu')/c

And, if you do the math, you find out the left-hand-side of this
formula should be greater than 1, given the amount of kinetic energy
released in pions decaying to muons. Go ahead, plug in the numbers.

What's the problem with finding out the left-hand-side
of this formula is greater than 1?

Well, for one thing, it doesn't happen. As I mentioned to you, the
time of flight of muons is *measured*. We know what the speed is. It's
not greater than 1.

That's only a problem
if you use the relativistic kinetic energy.

No, sir. It's a problem with measurement.

Of course,
v/c = (v_p + v_mu')/c �is greater than 1, because v_p/c
is very close to 1, and v_mu'/c can be greater than
(1 - v_p/c).

You can't keep track of single muons to measure their
'time of flight'. All you can do is to perform statistics and
extrapolate to single muons. You can't say a muon called
A was emitted by a pion called P at time t=0, and then A
was detected at distance R at time t'. The extrapolation for
a single muon you perfom is pure nonsense. That's the
reason why we find out in particle physics there are three
odd generations of charged leptons, and nobody can guess
why there are actually those three generations and why they
exhibit the rest masses they exhibit. SR is unable to account for
the mass of muons and tau leptons with respect to the mass of
electrons. Should we call that impossibility a fraud?. If you avoid
SR, and try to explain why muon and tau lepton have the rest
masses they have wrt that of an electron, you will be able to find
out �those three generations of charged leptons, are just exact
ratios derived from multiples of c.

I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist.- Hide quoted text -

- Show quoted text -

Ptolemy was an actual published astronomer when Galileo decided that
the earth was rotating instead of the sun orbiting the earth.
Special relativity has just as many epicycles as Ptolemaic astronomy.
Robert B. Winn

Where are the epicycles, Robert?

Dark matter and dark energy are epicycles.

Bullet cluster? Casimir effect?

Just because _you do not understand_ does not make them "epicycles".

Under GR, dark matter is a hypothetical form of matter to
explain the variations in speed and direction of the apparent
motion of stars in spiral galaxies, which yield apparent flat
rotation curves.

Dark matter also is invoked to explain the big bang power spectrum and
the dynamic behavior of cluster collisions. Dark matter is assumed to
be a pressureless non-baryonic substance that is electromagnetically
transparent or otherwise inert. Somewhat different from "epicycle".


Under Ptolemaic system of astronomy, the epicycle is a
geometric model to explain the variations in speed and
direction of the apparent motion of the Moon, Sun, and
planets.

So you agree that dark matter is nothing like an epicycle? Dark matter
is a model with specific assumptions and specific predictions that
have been verified by observation. The point of the analogy really is
lost on me.


Thus,

Â* Â* Â*GR -- Ptolemaic system of astronomy,
Â* Â* Â*dark matter -- epicycle,
Â* Â* Â*hypothetical form of matter -- geometric model,
Â* Â* Â*stars in spiral galaxies -- the Moon, Sun, and planets.

On May 13, 12:34Â*pm, rbwinn wrote:
On May 13, 1:07Â*pm, Eric Gisse wrote:



On May 13, 11:34Â*am, Albertito wrote:

out �those three generations of charged leptons, are just exact
ratios derived from multiples of c.

I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist.- Hide quoted text -

- Show quoted text -

Ptolemy was an actual published astronomer when Galileo decided that
the earth was rotating instead of the sun orbiting the earth.
Special relativity has just as many epicycles as Ptolemaic astronomy.
Robert B. Winn

Where are the epicycles, Robert?

Dark matter and dark energy are epicycles.

Bullet cluster? Casimir effect?

Just because _you do not understand_ does not make them "epicycles".- Hide quoted text -

Well, I for sure do not understand the distance contraction. Â*I never
did believe in that.

Yes Robert, we know. You have repeatedly told us what you believe.

Â*What Galileo believed became "trendy" after his
death among younger scientists, so the Ptolemaic system of astronomy
was eventually taken out of use as old scientists faithful to
epicycles died off, and younger scientists did not remember the
complicated mathematics of epicycles. Â*Today there is no one left who
is even interested in what Ptolemy said.
Â* Â* I think eventually Einstein will fade away the same way.

Einstein published his theories four generations of physicists ago.
Einstein has been dead for two of them. What you think is obviously
irrelevant. Unless you want to argue that 120 year old physicists are
holding things back, you might want to think a little harder about
what you are saying.

Robert B. Winn

On May 13, 11:02*am, Albertito wrote:
On May 13, 6:59 pm, Igor wrote:



On May 13, 1:52 pm, Albertito wrote:

On May 13, 6:26 pm, Igor wrote:

On May 11, 2:26 pm, Albertito wrote:

On May 11, 7:10 pm, Igor wrote:

On May 9, 7:44 am, Albertito wrote:

Let us consider the simple case of addition of velocities
along a straight line. The incompleteness of SR resides
in the fact that a speed v can't currently be experimentally
distinguished from its rapidity r = arctanh(v/c), for values
of that beta = v/c *below the third-order term approximation.
The power series expansion of r = arctanh(v/c) is

* * * * * r = v/c + v*3/3c^3+ v^5/5c^5 + v^7/7c^7 + ...

Provide references of any experimental test of SR, showing
that the rapidity r can be distinguished from its beta v/c, beyond
its second-order approximation. Prove at least that the third-order
term v*3/3c^3 lives outside the error bars. Since, we can't *still
perform such accurate experimental tests, we must conclude
the addition of velocities still remains within the euclidean sum
of vectors

* * * * * * * *w = u + v.

That sum can't still be experimentally distinguished from the sum
of rapidities

* * * * * * *arctanh(w/c) = arctanh(u/c) + arctanh(v/c).

In addition, we must also conclude that the relativistic Doppler

* * * * * * * *f' = Exp(-r) f ,
* * * * * * * *where r = arctanh(v/c),

can't still be experimentally distinguished from this one

* * * * * * * *f' = Exp(-v/c) f

Uhh, you do understand that v/c = tanh r don't you? *I didn't think
so.

Really?, didn't you?
Didn't *you realize that both v/c and its rapidity,
r = arctanh(v/c), currently are experimentally
indistinguishable?

You must understand that if I know one of them, I know both of them.
So what is your point?

The point is that you can't know both of them.
How do you know that v/c is the beta you
have measured for a moving body, but it is
not its rapidity r = arctanh(v/c). Error bars
on (v/c ± d) do not allow you to discriminate
v/c from its rapidity beyond its second-order
term.

Now you're just being either silly, stupid, or both. *I don't need to
measure both of them. *Or maybe you haven't noticed that they are
functions of each other?

It's unnecessary to blurt out curses. I have not
said you need to measure both of them,
I've said that if you measure v/c, its error bars
make its rapidity useless for relativistic corrections,
for example relativistic addition of velocities.

Why don't you make a quantitative prediction for scattering angles and
energies of scattered particles in the Compton effect using your model?

On May 13, 2:13Â*pm, Eric Gisse wrote:
On May 13, 12:34Â*pm, rbwinn wrote:





On May 13, 1:07Â*pm, Eric Gisse wrote:

On May 13, 11:34Â*am, Albertito wrote:

out �those three generations of charged leptons, are just exact
ratios derived from multiples of c.

I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist.- Hide quoted text -

- Show quoted text -

Ptolemy was an actual published astronomer when Galileo decided that
the earth was rotating instead of the sun orbiting the earth.
Special relativity has just as many epicycles as Ptolemaic astronomy.
Robert B. Winn

Where are the epicycles, Robert?

Dark matter and dark energy are epicycles.

Bullet cluster? Casimir effect?

Just because _you do not understand_ does not make them "epicycles".- Hide quoted text -

Well, I for sure do not understand the distance contraction. Â*I never
did believe in that.

Yes Robert, we know. You have repeatedly told us what you believe.

Â*What Galileo believed became "trendy" after his
death among younger scientists, so the Ptolemaic system of astronomy
was eventually taken out of use as old scientists faithful to
epicycles died off, and younger scientists did not remember the
complicated mathematics of epicycles. Â*Today there is no one left who
is even interested in what Ptolemy said.
Â* Â* I think eventually Einstein will fade away the same way.

Einstein published his theories four generations of physicists ago.
Einstein has been dead for two of them. What you think is obviously
irrelevant. Unless you want to argue that 120 year old physicists are
holding things back, you might want to think a little harder about
what you are saying.

Ptolemy's ideas lasted a lot longer than that. Ptolemy only had
epicycles for the sun and the planets. Einstein has epicycles for
everything. Einstein has been dead less than 50 years.
Robert B. Winn

On May 13, 2:09Â*pm, rbwinn wrote:
On May 13, 2:13Â*pm, Eric Gisse wrote:

On May 13, 12:34Â*pm, rbwinn wrote:

On May 13, 1:07Â*pm, Eric Gisse wrote:

On May 13, 11:34Â*am, Albertito wrote:

out �those three generations of charged leptons, are just exact
ratios derived from multiples of c.

I'd like to take a moment to point out that you are arguing about
particle physics with an actual published particle physicist..- Hide quoted text -

- Show quoted text -

Ptolemy was an actual published astronomer when Galileo decided that
the earth was rotating instead of the sun orbiting the earth.
Special relativity has just as many epicycles as Ptolemaic astronomy.
Robert B. Winn

Where are the epicycles, Robert?

Dark matter and dark energy are epicycles.

Bullet cluster? Casimir effect?

Just because _you do not understand_ does not make them "epicycles".- Hide quoted text -

Well, I for sure do not understand the distance contraction. Â*I never
did believe in that.

Yes Robert, we know. You have repeatedly told us what you believe.

Â*What Galileo believed became "trendy" after his
death among younger scientists, so the Ptolemaic system of astronomy
was eventually taken out of use as old scientists faithful to
epicycles died off, and younger scientists did not remember the
complicated mathematics of epicycles. Â*Today there is no one left who
is even interested in what Ptolemy said.
Â* Â* I think eventually Einstein will fade away the same way.

Einstein published his theories four generations of physicists ago.
Einstein has been dead for two of them. What you think is obviously
irrelevant. Unless you want to argue that 120 year old physicists are
holding things back, you might want to think a little harder about
what you are saying.

Ptolemy's ideas lasted a lot longer than that. Â*Ptolemy only had
epicycles for the sun and the planets. Â*Einstein has epicycles for
everything. Â*Einstein has been dead less than 50 years.

So what's the epicycle that explains Compton scattering?

Robert B. Winn